Lie Groups and Lie Algebras June 28, 2006

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Notes for Math 261A Lie groups and Lie algebras June 28, 2006

Contents

Contents ...... 1 How these notes came to be ...... 3 Lecture 1 ...... 4 Lecture 2 ...... 6 Tangent Lie algebras to Lie groups, 6 Lecture 3 ...... 9 Lecture 4 ...... 12 Lecture 5 ...... 16 Simply Connected Lie Groups, 16 Lecture 6 - Hopf Algebras ...... 21 The universal enveloping algebra, 24 Lecture 7 ...... 26 Universality of Ug, 26 Gradation in Ug, 27 Filtered spaces and algebras, 28 Lecture 8 - PBW theorem, Deformations of Associative Algebras . . 31 Deformations of associative algebras, 32 Formal deformations of associative algebras, 34 Formal deformations of Lie algebras, 35 Lecture 9 ...... 36 Lie algebra cohomology, 37 Lecture 10 ...... 41 Lie algebra cohomology, 41 H2(g, g) and Deformations of Lie algebras, 42 Lecture 11 - Engel’s Theorem and Lie’s Theorem ...... 46 The radical, 51 Lecture 12 - Cartan Criterion, Whitehead and Weyl Theorems . . . 52 Invariant forms and the Killing form, 52 Lecture 13 - The root system of a semi-simple Lie algebra ..... 59 Irreducible ﬁnite dimensional representations of sl(2), 63 Lecture 14 - More on Root Systems ...... 65 Abstract Root systems, 66 The Weyl group, 68 Lecture 15 - Dynkin diagrams, Classiﬁcation of root systems . . . . 71 Isomorphisms of small dimension, 76

1 2

Lecture 16 - Serre’s Theorem...... 78 Lecture 17 - Constructions of Exceptional simple Lie Algebras . . . 82 Lecture 18 - Representations ...... 87 Representations, 87 Highest weights, 89 Lecture 19 - The Weyl character formula ...... 93 The character formula, 94 Lecture 20 ...... 100 Compact Groups, 102 Lecture 21 ...... 105 What does a general Lie group look like?, 105 Comparison of Lie groups and Lie algebras, 107 Finite groups and Lie groups, 108 Algebraic groups (over R) and Lie groups, 109 Lecture 22 - Clifford algebras and Spin groups ...... 111 Clifford Algebras and Spin Groups, 112 Lecture 23 ...... 117 Clifford Groups, 118 Lecture 24 ...... 123 Spin representations of Spin and Pin groups, 123 Triality, 126 More about Orthogonal groups, 127 Lecture 25 - E8 ...... 129 Lecture 26 ...... 134 Construction of E8, 136 Lecture 27 ...... 140 Construction of the Lie group of E8, 143 Real forms, 144 Lecture 28 ...... 146 Working with simple Lie groups, 148 Every possible simple Lie group, 150 Lecture 29 ...... 152 Lecture 30 - Irreducible unitary representations of SL2(R) . . . . . 159 Finite dimensional representations, 159 The Casimir operator, 162 Lecture 31 - Unitary representations of SL2(R) ...... 165 Background about infinite-dimensional representations, 165 The group SL2(R), 166 Solutions to (some) Exercises...... 171 References ...... 176 Index ...... 178 3

How these notes came to be

Among the Berkeley professors, there was once Alan Knutson, who would teach Math 261. But it happened that professor Knutson was on sabbati- cal elsewhere (I don’t know where for sure), and eventually went there (or somewhere else) for good. During this turbulent time, Maths 261AB were cancelled two years in a row. The last of these four semesters (Spring 2006), some graduate students gathered together and asked Nicolai Reshetikhin to teach them Lie theory in a giant reading course. When the dust settled, there were two other professors willing to help in the instruction of Math 261A, Vera Serganova and Richard Borcherds. Thus Tag Team 261A was born. After a few lectures, professor Reshetikhin suggested that the students write up the lecture notes for the beneﬁt of future generations. The ﬁrst four lectures were produced entirely by the “editors”. The remaining lectures were LATEXed by Anton Geraschenko in class and then edited by the people in the following table. The columns are sorted by lecturer.

Nicolai Reshetikhin Vera Serganova Richard Borcherds 1 Anton Geraschenko 11 Sevak Mkrtchyan 21 Hanh Duc Do 2 Anton Geraschenko 12 Jonah Blasiak 22 An Huang 3 Nathan George 13 Hannes Thiel 23 Santiago Canez 4 Hans Christianson 14 Anton Geraschenko 24 Lilit Martirosyan 5 Emily Peters 15 Lilit Martirosyan 25 Emily Peters 6 Sevak Mkrtchyan 16 Santiago Canez 26 Santiago Canez 7 Lilit Martirosyan 17 Katie Liesinger 27 Martin Vito-Cruz 8 David Cimasoni 18 Aaron McMillan 28 Martin Vito-Cruz 9 Emily Peters 19 Anton Geraschenko 29 Anton Geraschenko 10 Qingtau Chen 20 Hanh Duc Do 30 Lilit Martirosyan 31 Sevak Mkrtchyan Richard Borcherds then edited the last third of the notes. The notes were further edited by Crystal Hoyt, Sevak Mkrtchyan, and Anton Geraschenko.

Send corrections and comments to [email protected]. Lecture 1 4

Lecture 1

Definition 1.1. A Lie group is a smooth manifold G with a group structure such that the multiplication µ : G G G and inverse map ι : G G are × → → smooth maps. ◮ Exercise 1.1. If we assume only that µ is smooth, does it follow that ι is smooth? n Example 1.2. The group of invertible endomorphisms of C , GLn(C), is a Lie group. The automorphisms of determinant 1, SLn(C), is also a Lie group. Example 1.3. If B is a bilinear form on Cn, then we can consider the Lie group A GL (C) B(Av, Aw)= B(v, w) for all v, w Cn . { ∈ n | ∈ } If we take B to be the usual dot product, then we get the group On(C). T 0 1m If we let n = 2m be even and set B(v, w) = v 1m 0 w, then we get Sp (C). − 2m Example 1.4. SU SL is a real form (look in lectures 27,28, and 29 for n ⊆ n more on real forms). Example 1.5. We’d like to consider infinite matrices, but the multiplication wouldn’t make sense, so we can think of GL GL via A A 0 , then n ⊆ n+1 7→ 0 1 define GL as n GLn. That is, invertible infinite matrices which look like ∞ the identity almost everywhere. S Lie groups are hard objects to work with because they have global charac- teristics, but we’d like to know about representations of them. Fortunately, there are things called Lie algebras, which are easier to work with, and representations of Lie algebras tell us about representations of Lie groups. Definition 1.6. A Lie algebra is a vector space V equipped with a Lie bracket [ , ] : V V V , which satisfies × → 1. Skew symmetry: [a, a] = 0 for all a V , and ∈ 2. Jacobi identity: [a, [b, c]] + [b, [c, a]] + [c, [a, b]] = 0 for all a, b, c V . ∈ A Lie subalgebra of a lie algebra V is a subspace W V which is closed ⊆ under the bracket: [W, W ] W . ⊆ Example 1.7. If A is a finite dimensional associative algebra, you can set [a, b] = ab ba. If you start with A = M , the algebra of n n matri- − n × ces, then you get the Lie algebra gl . If you let A M be the algebra n ⊆ n of matrices preserving a fixed flag V V V Cn, then you get 0 ⊂ 1 ⊂ ··· k ⊆ parabolicindexparabolic subalgebras Lie sub-algebras of gln. Lecture 1 5

Example 1.8. Consider the set of vector fields on Rn, Vect(Rn) = ℓ = i ∂ { Σe (x) [ℓ1,ℓ2]= ℓ1 ℓ2 ℓ2 ℓ1 . ∂xi | ◦ − ◦ } ◮ Exercise 1.2. Check that [ℓ1,ℓ2] is a first order differential operator. Example 1.9. If A is an associative algebra, we say that ∂ : A A is a → derivation if ∂(ab) = (∂a)b + a∂b. Inner derivations are those of the form [d, ] for some d A; the others are called outer derivations. We denote the · ∈ set of derivations of A by (A), and you can verify that it is a Lie algebra. D Note that Vect(Rn) above is just (C (Rn)). D ∞ The first Hochschild cohomology, denoted H1(A, A), is the quotient (A)/ inner derivations . D { } Definition 1.10. A Lie algebra homomorphism is a linear map φ : L L → ′ that takes the bracket in L to the bracket in L′, i.e. φ([a, b]L) = [φ(a), φ(b)]L′ . A Lie algebra isomorphism is a morphism of Lie algebras that is a linear isomorphism.1

A very interesting question is to classify Lie algebras (up to isomorphism) of dimension n for a given n. For n = 2, there are only two: the trivial bracket [ , ] = 0, and [e1, e2] = e2. For n = 3, it can be done without too much trouble. Maybe n = 4 has been done, but in general, it is a very hard problem. If e is a basis for V , with [e , e ]= ck e (the ck are called the structure { i} i j ij k ij constants of V ), then the Jacobi identity is some quadratic relation on the k 3n cij, so the variety of Lie algebras is some quadratic surface in C . Given a smooth real manifold M n of dimension n, we can construct Vect(M n), the set of smooth vector ﬁelds on M n. For X Vect(M n), we ∈ can deﬁne the Lie derivative LX by (LX f)(m) = Xm f, so LX acts on n · · C∞(M ) as a derivation. ◮ Exercise 1.3. Verify that [L ,L ]= L L L L is of the form X Y X ◦ Y − Y ◦ X L for a unique Z Vect(M n). Then we put a Lie algebra structure on Z ∈ Vect(M n)= (C (M n)) by [X,Y ]= Z. D ∞ There is a theorem (Ado’s Theorem2) that any Lie algebra g is isomorphic to a Lie subalgebra of gln, so if you understand everything about gln, you’re in pretty good shape.

1The reader may verify that this implies that the inverse is also a morphism of Lie algebras. 2Notice that if g has no center, then the adjoint representation ad : g → gl(g) is already faithful. See Example 7.4 for more on the adjoint representation. For a proof of Ado’s Theorem, see Appendix E of [FH91] Lecture 2 6

Lecture 2

Last time we talked about Lie groups, Lie algebras, and gave examples. Recall that M L is a Lie subalgebra if [M, M] M. We say that M is a ⊆ ⊆ Lie ideal if [M,L] M. ⊆ Claim. If M is an ideal, then L/M has the structure of a Lie algebra such that the canonical projection is a morphism of Lie algebras.

Proof. Take l , l L, check that [l + M, l + M] [l , l ]+ M. 1 2 ∈ 1 2 ⊆ 1 2 Claim. For φ : L L a Lie algebra homomorphism, 1 → 2 1. ker φ L is an ideal, ⊆ 1 2. im φ L is a Lie subalgebra, ⊆ 2 3. L1/ ker φ ∼= im φ as Lie algebras. ◮ Exercise 2.1. Prove this claim.

Tangent Lie algebras to Lie groups Let’s recall some differential geometry. You can look at [Lee03] as a reference. If f : M N is a differentiable map, then df : T M T N is → → the derivative. If G is a group, then we have the maps l : x gx and g 7→ r : x xg. Recall that a smooth vector field is a smooth section of the g 7→ tangent bundle T M M. → Definition 2.1. A vector field X is left invariant if (dl ) X = X l for g ◦ ◦ g all g G. The set of left invariant vector fields is called Vect (G). ∈ L dl g / T GO T GO X X lg G / G

Proposition 2.2. Vect (G) Vect(G) is a Lie subalgebra. L ⊆ Proof. We get an induced map l : C (G) C (G), and X is left invariant g∗ ∞ → ∞ if and only if LX commutes with lG∗ . Then X,Y left invariant [L ,L ] invariant [X,Y ] left invariant. ⇐⇒ X Y ⇐⇒ All the same stuﬀ works for right invariant vector ﬁelds VectR(G).

Deﬁnition 2.3. g = VectL(G) is the tangent Lie algebra of G. Lecture 2 7

Proposition 2.4. There are vector space isomorphisms Vect (G) T G L ≃ e and Vect (G) T G. Moreover, the Lie algebra structures on T G induced R ≃ e e by these isomorphisms agree.

Note that it follows that dim g = dim G.

Proof. Recall fibre bundles. dl : T G ∼ T G, so T G T G1. X is a g e −→ g ≃ e × section of T G, so it can be thought of as X : G T G, in which case the → e left invariant fields are exactly those which are constant maps, but the set of constants maps to TeG is isomorphic to TeG. If G is an n-dimensional Cω Lie group, then g is an n-dimensional Lie algebra. If we take local coordinates near e G to be x1,...,xn : U Rn ∈ e → with m : Rn Rn Rn the multiplication (defined near 0). We have a × → power series for m near 0,

m(x,y)= Ax + By + α (x,y)+ α (x,y)+ 2 3 · · · where A, B : Rn Rn are linear, α is degree i. Then we can consider → i the condition that m be associative (only to degree 3): m(x,m(y, z)) = m(m(x,y), z).

m(x,m(y, z)) = Ax + Bm(y, z)+ α (x,m(y, z)) + α (x,m(y, z)) + 2 3 · · · = Ax + B(Ay + Bz + α2(y, z)+ α3(y, z)))

+ α2(x,Ay + Bz + α2(y, z)) + α3(x,Ay + Bz) m(m(x,y), z) = [[compute this stuﬀ]]

Comparing ﬁrst order terms (remember that A, B must be non-singular), we can get that A = B = In. From the second order term, we can get that α2 is bilinear! Changing coordinates (φ(x) = x + φ2(x)+ φ3(x)+ , 1 · · · with φ− (x) = x φ2(x)+ φ˜3(x)+ ), we use the fact that mφ(x,y) = 1 − · · · φ− m(φx, φy) is the new multiplication, we have

m (x,y)= x + y + (φ (x)+ φ (y)+ φ (x + y))+ α (x,y)+ φ 2 2 2 2 · · · can be any symm form so we can tweak the coordinates| to make{z α2 skew-symmetric.} Looking at order 3, we have

α (x, α (y, z)) + α (x,y + z)= α (α (x,y), z)+ α (x + y, z) ( ) 2 2 3 2 2 3 ∗

◮ Exercise 2.2. Prove that this implies the Jacobi identity for α2. (hint: skew-symmetrize ( )) ∗ 1is this true? Lecture 2 8

Remarkably, the Jacobi identity is the only obstruction to associativity; all other coeﬃcients can be eliminated by coordinate changes.

a b Example 2.5. Let G be the set of matrices of the form 0 a−1 for a, b real, a> 0. Use coordinates x,y where ex = a, y = b, then x x′ m((x,y), (x′,y′)) = (x + x′, e y′ + ye− )

= (x + x′,y + y′ + (xy′ x′y)+ ). − · · · skew The second order term is skew symmetric, so these| {z are} good coordinates. There are H, E T G corresponding to x and y respectively so that [H, E]= ∈ e E2.

◮ Exercise 2.3. Think about this. If a, b commute, then eaeb = ea+b. If they do not commute, then eaeb = ef(a,b). Compute f(a, b) to order 3.

2what does this part mean? Lecture 3 9

Lecture 3

Last time we saw how to get a Lie algebra Lie(G) from a Lie group G.

Lie(G) = Vect (G) Vect (G). L ≃ R Let x1, ..., xn be local coordinates near e G, and let m(x,y)i be the ∈ ith coordinate of (x,y) m(x,y). In this local coordinate system, 7→ m(x,y)i = xi + yi + 1 ci xjyk + . If e , ..., e T G is the basis 2 jk · · · 1 n ∈ e induced by x1, ..., xn, (e ∂ ), then Pi ∼ i k [ei, ej]= cijek. Xk Example 3.1. Let G be GL , and let (g ) be coordinates. Let X : GL n ij n → T GLn be a vector ﬁeld. ∂f(g) LX (f)(g)= Xij(g) ∂gij Xi,j lh : g hg , where L (l∗(f))(g)= 7→ X h l (f)(g)= f(h 1g) h∗ − 1 1 ∂f(h− g) ∂(h− g)kl ∂f(x) = Xij(g) = Xij(g) x=h−1g ∂gij ∂gij ∂xkl | Xi,j Xi,j 1 ∂(h− g)kl 1 ∂gml 1 = = (h− ) = (h− ) δ h ∂g km ∂g ki jli ij m ij X =δimδlj 1 ∂f | {z } = Xij(g)(h− )ki x=h−1g ∂xkj | Xi,j,k 1 ∂f = (h− )kiXij(g) x=h−1g ! ∂xkj | Xj,k Xi 1 1 If we want X to be left invariant, i(h− )kiXij(g) = Xkj(h− g), then L (l (f)) = l (L (f)), (left invariance of X). X h∗ h∗ X P Example 3.2. All solutions are X (g) = (g M) , M-constant n n matrix. ij · ij × gives that left invariant vector ﬁelds on GL n n matrices = gl . The n ≈ × n “Natural Basis” is e = (E ), L = (g) ∂ . ij ij ij m mj ∂gmi P Example 3.3. Commutation relations between Lij are the same as commutation relations between eij. Lecture 3 10

Take τ T G. Define the vector field: v : G T G by v (g) = dl (τ), ∈ e τ → τ g where l : G G is left multiplication. v is a left invariant vector field by g → τ construction. Consider φ : I G, dφ(t) = v (φ(t)), φ(0) = e. → dt τ Proposition 3.4.

1. φ(t + s)= φ(t)φ(s)

2. φ extends to a smooth map φ : R G. → Proof. 1. Fix s and α(t)= φ(s)φ(t), β(t)= φ(s + t).

– α(0) = φ(s)= β(0) dβ(a) dφ(s+t) – dt = dt = vτ (β(t)) – dα(t) = d (φ(s)φ(t)) = dl v (φ(t)) = v (φ(s)φ(t)) = v (α(t)), dt dt φ(s) · τ τ τ where the second equality is because vτ is linear. = α satisﬁes same equation as β and same initial conditions, so by ⇒ uniqueness, they coincide for t <ǫ. | | 2. Now we have (1) for t + s < ǫ, t < ǫ, s < ǫ. Then extend φ to | | | | | | t < 2ǫ. Continue this to cover all of R. | |

This shows that for all τ T G, we have a mapping R G and it’s ∈ e → image is a 1-parameter (1-dimensional) Lie subgroup in G.

exp : g = T G G e → τ φ (1) = exp(τ) 7→ τ Notice that λτ exp(λτ)= φ (1) = φ (λ) 7→ λτ τ Example 3.5. GL , τ gl = T GL , dφ(t) = v (φ(t)) T GL gl . n ∈ n e n dt τ ∈ φ(t) n ≃ n dφ(t) v (φ(t)) = φ(t) τ, = φ(t) τ, φ(0) = I, τ · dt ·

∞ tnτ n φ(t) = exp(tI)= n! n=0 X exp : gln GLn → d L (f)(g)= f(γ(t)) γ(0)=g dt |t=0 Lecture 3 11

Baker-Campbell-Hausdorﬀ formula:

eX eY = eH(X,Y ) · 1 1 H(X,Y )= X + Y + [X,Y ] + ([X[X,Y ]] + [Y [Y, X]]) + 2 12 · · · sym skew symmetric | {z } Proposition 3.6. 1. Let| f{z: G} | H be a Lie{z group homomorphism,} → f / then the diagram GO HO is commutative. exp exp df Lie(G) e / Lie(H)

2. If G is connected, then (df)e deﬁnes the Lie group homomorphism f uniquely.

Proof. Next time.

Proposition 3.7. G, H Lie groups, G simply connected, then α : Lie(G) → Lie(H) is a Lie algebra homomorphism if and only if there is a Lie group homomorphism A : G H lifting α. → Proof. Next time. exp Lie algebras Lie groups(connected, simply connected) is an { } −−→{ } equivalence of categories. Lecture 4 12

Lecture 4

Theorem 4.1. Suppose G is a topological group. Let U G be an open ⊂ neighbourhood of e G. If G is connected, then ∈ G = U n. n 1 [≥ 1 Proof. Choose a non-empty open set V U such that V = V − , for example 1 n ⊂ V = U U − . Deﬁne H = n 1V , and observe H is an abstract subgroup, ∩ ∪ ≥ since V nV m V n+m. H is open since it is the union of open sets. If ⊆ σ / H, then σH H, since otherwise if h1, h2 H satisfy σh1 = h2, then ∈ 1 6⊂ ∈ σ = h h− H. Thus H is a complement of the union of all cosets not 2 1 ∈ containing H. Hence H is closed. Since G is connected, H = G.

Theorem 4.2. Let f : G H be a Lie group homomorphism. Then the → following diagram commutes:

(df)e Te / TeH exp exp f G / H

Further, if G is connected, (df)e determines f uniquely. Proof. 1) Commutative diagram. Fix τ T G and set η = df τ T H. ∈ e e ∈ e Recall we deﬁned the vector ﬁeld Vτ (g) = (dlg)(τ), then if φ(t) solves dφ = V (φ(t)) T G, dt τ ∈ φ(t) we have exp(τ)= φ(1). Let psi solve

dψ = V (ψ(t)), dt η so that exp(η)= ψ(1). Observe ψ˜(t)= f(φ(t)) satisﬁes

dψ˜ dφ = (df) = V (ψ˜), dt dt η so by uniqueness of solutions to ordinary diﬀerential equations, ψ = ψ˜. 2) Uniqueness of f. The exponential map is an isomorphism of a neighborhood of 0 g and a neighborhood of e G. But if G is connected, ∈ n ∈ G = n 1(nbd e) . ∪ ≥ Lecture 4 13

Theorem 4.3. Suppose G is a topological group, with G0 G the connected ⊂ component of e. Then 1) G0 is normal and 2) G/G0 is discrete.

Proof. 2) G0 G is open implies pr 1([e]) = eG0 is open in G, which in ⊂ − turn implies pr 1([g]) G/G0 is open for every g G. Thus each coset is − ∈ ∈ both open and closed, hence G/G0 is discrete. 1) Fix g G and consider the map G G defined by x gxg 1. This ∈ → 7→ − map fixes e and is continuous, which implies it maps G0 into G0. In other words, gG0g 1 G0, or G0 is normal. − ⊂ We recall some basic notions of algebraic topology. Suppose M is a connected topological space. Let x,y M, and suppose γ(t) : [0, 1] M ∈ → is a path from x to y in M. We sayγ ˜(t) is homotopic to γ if there is a continuous map h(s,t) : [0, 1]2 M satisfying → h(s, 0) = x, h(s, 1) = y • h(0,t)= γ(t), h(1,t) =γ ˜(t). • We call h the homotopy. On a smooth manifold, we may replace h with a smooth homotopy. Now fix x M. We define the first fundamental group 0 ∈ of M

π (M,x )= homotopy classes of loops based at x . 1 0 { 0} It is clear that this is a group with group multiplication composition of paths. It is also a fact that the deﬁnition does not depend on the base point x0:

π (M,x ) π (M,x′ ). 1 0 ≃ 1 0 By π (M) we denote the isomorphism class of π (M, ). Lastly, we say M is 1 1 · simply connected if π (M)= e , that is if all closed paths can be deformed 1 { } to the trivial one.

Theorem 4.4. Suppose G and H are Lie groups with Lie algebras g, h respectively. If G is simply connected, then any Lie algebra homomorphism ρ : g h lifts to a Lie group homomorphism R : G H. → → In order to prove this theorem, we will need the following lemma.

Lemma 4.5. Let ξ : R g be a smooth mapping. Then → dg = (dl )(ξ(t)) dt g has a unique solution on all of R with g(t0)= g0. Lecture 4 14

For convenience, we will write gξ := (dlg)(ξ). Proof. Since g is a vector space, we identify it with Rn and for sufficiently small r> 0, we identify B (0) g with a small neighbourhood of e, U (r) r ⊂ e ⊂ G, under the exponential map. Here Br(0) is measured with the usual Euclidean norm . Note for any g U (r) and t t sufficiently small, k · k ∈ e | − 0| we have gξ(t) C. Now according to Exercise 4.1, the solution with k k ≤ g(t )= e exists for sufficiently small t t and 0 | − 0| r g(t) Ue(r) t t0 < . ∈ ∀| − | C′ Now define h(t)= g(t)g so that h(t) U (r) for t t < r/C . That is, r 0 ∈ g0 | − 0| ′ and C′ do not depend on the choice of initial conditions, and we can cover R by intervals of length, say r/C′. ◮ Exercise 4.1. Verify that there is a constant C such that if t t is ′ | − 0| sufficiently small, we have

g(t) C′ t t . k k≤ | − 0| Proof of Theorem 4.4. We will construct R : G H. Beginning with g(t) : → [0, 1] G satisfying g(0) = e, g(1) = g, define ξ(t) g for each t by → ∈ d g(t)ξ(t)= g(t). dt Let η(t)= ρ(ξ(t)), and let h(t) : [0, 1] H satisfy → d h(t)= h(t)η(t), h(0) = e. dt Define R(g)= h(1). Claim: h(1) does not depend on the path g(t), only on g. Proof of Claim. Suppose g1(t) and g2(t) are two different paths connecting e to g. Then there is a smooth homotopy g(t,s) satisfying g(t, 0) = g1(t), g(t, 1) = g2(t). Define ξ(t,s) and η(t,s) by ∂g = g(t,s)ξ(t,s); ∂t ∂g = g(t,s)η(t,s). ∂s Observe ∂2g ∂ξ = gη ξ + g and (1) ∂s∂t ◦ ∂t ∂2g ∂η = gξ η + g , (2) ∂t∂s ◦ ∂s Lecture 4 15 and (1) is equal to (2) since g is smooth. Consequently

∂η ∂ξ = [η,ξ]. ∂t − ∂s Now deﬁne an s dependent family of solutions h( ,s) to the equations · ∂h (t,s)= h(t,s)ρ(ξ(t,s)), h(0,s)= e. ∂t Deﬁne θ(t,s) by

∂θ ∂ρ(ξ) = [ρ(ξ),θ] , ∂t − ∂s (3)  θ(0,s) = 0. 

Observe θ˜(t,s) = ρ(η(t,s)) also satisﬁes equation (3), so that θ = θ˜ by uniqueness of solutions to ODEs. Finally, ∂g ∂h gη(1,s)= (1,s)=0 = θ(1,s)=0 = (1,s) = 0, ∂s ⇒ ⇒ ∂s justifying the claim. We need only show R : G H is a homomorphism. Let g , g G and → 1 2 ∈ set g = g1g2. Letg ˜i(t) be a path from e to gi in G for each i = 1, 2. Then the pathg ˜(t) deﬁned by

g˜ (2t), 0 t 1 , g˜(t)= 1 ≤ ≤ 2 g g˜ (2t 1), 1 t 1 1 2 − 2 ≤ ≤ goes from e to g. Let h˜i for i = 1, 2 and h˜ be the paths in H corresponding tog ˜1,g ˜2, andg ˜ respectively and calculate

R(g1g2)= R(g)= h˜(1) = h˜1(1)h˜2(1) = R(g1)R(g2). Lecture 5 16

Lecture 5

Last time we talked about connectedness, and proved the following things:

n - Any connected topological group G has the property that G = n V , where V is any neighborhood of e G. ∈ S - If G is a connected Lie group, with α : Lie(G) Lie(H) a Lie algebra → homomorphism, then if there exists f : G H with df = α, it is → e unique.

- If G is connected, simply connected, with α : Lie(G) Lie(H) a Lie → algebra homomorphism, then there is a unique f : G H such that → dfe = α.

Simply Connected Lie Groups p The map p in Z X Y is a covering map if it is a locally trivial fiber → −→ bundle with discrete fiber Z. Locally trivial means that for any y Y there ∈ is a neighborhood U such that if f : U Z Z is the map defined by × → f(u, z)= u, then the following diagram commutes:

1 p− U ≃ U Z tt × tt tt tt f y tt Y U t ⊇ The exact sequence defined below is an important tool. Suppose we have a locally trivial fiber bundle with fiber Z (not necessarily discrete), with X,Y connected. Choose x X, z Z,y Y such that p(x )= y , 0 ∈ 0 ∈ 0 ∈ 0 0 and i : Z p 1(y ) is an isomorphism such that i(z )= x : → − 0 0 0

1 i x π− (y )o Z 0 ∈ 0

We can define p : π1(X,x0) π1(Y,y0) in the obvious way (π1 is a functor). ∗ → Also define i : π1(Z, z0) π1(X,x0). Then we can define ∂ : π1(Y,y0) ∗ → → π (Z)= connected components of Z by taking a loop γ based at y and 0 { } 0 lifting it to some pathγ ˜. This path is not unique, but up to fiber-preserving homotopy it is. The new pathγ ˜ starts at x0 and ends at x0′ . Then we define ∂ to be the map associating the connected component of x0′ to the homotopy class of γ. Lecture 5 17

Claim. The following sequence is exact:

i∗ p∗ π (Z, z ) / π (X,x ) / π (Y,y ) ∂ / π (Z) / 0 1 0 1 0 1 0 0 { } 1. im i = ker p ∗ ∗

2. ﬁbers of ∂ π1(Y,y0)/ im p { }≃ ∗ 3. ∂ is surjective. Proof.

1. ker p is the set of all loops which map to contractible loops in Y , ∗ 1 which are loops which are homotopic to a loop in π− (y0) based at x0. These are exactly the loops of im i . ∗ 2. The fiber of ∂ over the connected component Z Z is the set of z ⊆ all (homotopy classes of) loops in Y based at y0 which lift to a path 1 connecting x0 to a point in the connected component of π− (y0) containing i(Zz). If two loops β, γ based at y0 are in the same fiber, 1 homotope them so that they have the same endpoint. Thenγ ˜β˜− is a loop based at x0. So fibers of ∂ are in one to one correspondence with loops in Y based at y0, modulo images of loops in X based at x0, which is just π1(Y,y0)/ im p . ∗ 3. This is obvious, since X is connected.

Now assume we have a covering space with discrete ﬁber, i.e. maps

X o Z p Y such that π (Z, z )= e and π (Z)= Z. Then we get the sequence 1 0 { } 0 i∗ p∗ ∂ e / π1(X,x0) / π1(Y,y0) / Z / 0 { } { } and since p is injective, Z = π1(Y )/π1(X). ∗ Classifying all covering spaces of Y is therefore the same as describing all subgroups of π1(Y ). The universal cover of Y is the space Y˜ such that π1(Y˜ )= e , and for any other covering X, we get a factorization of covering {f } p maps Y˜ X Y . −→ −→ We construct X˜, the universal cover, in the following way: fix x0 X, ˜ ∈ and define Xx0 to be the set of basepoint-fixing homotopy classes of paths Lecture 5 18 connecting x to some x X. We have a natural projection [γ ] x, 0 ∈ x0,x 7→ and the fiber of this projection (over x0) can be identified with π1(X,x0). ˜ ˜ ′ ′ It is clear for any two basepoints x0 and x′ , Xx0 Xx via any path γx ,x 0 ≃ 0 0 0 . So we have ˜ o Xx0 π1(X) p X ˜ Claim. Xx0 is simply connected. ˜ Proof. We need to prove that π1(Xx0 ) is trivial, but we know that the fibers ˜ of p can be identified with both π1(X) and π1(X)/π1(Xx0 ), so we’re done.

Let G be a connected Lie group. We would like to produce a simply connected Lie group which also has the Lie algebra Lie(G). It turns out that the obvious candidate, G˜e, is just what we are looking for. It is not hard to see that G˜e is a smooth manifold (typist’s note: it is not that easy either. See [Hat02], pp. 64-65, for a description of the topology on G˜e. Once we have a topology and a covering space map, the smooth manifold structure of G lifts to G˜e. – Emily). We show it is a group as follows. Write γ for γ : [0, 1] G with endpoints e and g. Deﬁne multiplication g → by [γg][γh′ ] := [ γg(t)γh′ (t) t [0,1]]. The unit element is the homotopy class { } ∈ 1 of a contractible loop, and the inverse is given by [ γ(t)− t [0,1]]. { } ∈ Claim.

1. G˜ = G˜e is a group. 2. p : G˜ G is a group homomorphism. → 3. π (G) G˜ is a normal subgroup. 1 ⊆ 4. Lie(G˜)= Lie(G).

5. G˜ G is the universal cover (i.e. π (G) is discrete). → 1 Proof. 1. Associativity is inherited from associativity in G, composition with the identity does not change the homotopy class of a path, and the product of an element and its inverse is the identity.

2. This is clear, since p([γg][˜γh]) = gh.

3. We know π1(G) = ker p, and kernels of homomorphisms are normal. Lecture 5 19

4. The topology on G˜ is induced by the topology of G in the following way: If is a basis for the topology on G then fix a path γe,g for all U ˜ ˜ ˜ ˜ g G. Then = Uγe,g is a basis for the topology on G with Uγe,g ∈ U { } 1 defined to be the set of paths of the form γe,g− βγe,g with β a loop based at g contained entirely in U. Now take U a connected, simply connected neighborhood of e G. ∈ Since all paths in U from e to a fixed g G are homotopic, we have ∈ that U and U˜ are diffeomorphic and isomorphic, hence Lie isomorphic. Thus Lie(G˜)= Lie(G).

5. As established in (4), G and G˜ are diﬀeomorphic in a neighborhood 1 of the identity. Thus all points x p− (e) have a neighborhood which ∈ 1 does not contain any other inverse images of e, so p− (e) is discrete; 1 and p− (e) and π1(G) are isomorphic.

We have that for any Lie group G with a given Lie algebra Lie(G)= g, there exists a simply connected Lie group G˜ with the same Lie algebra, and G˜ is the universal cover of G.

Lemma 5.1. A discrete normal subgroup H G of a connected topological ⊆ group G is always central.

Proof. Consider the map φ : G H, g ghg 1h 1, which is continuous. h → 7→ − − Since G is connected, the image is also connected, but H is discrete, so the image must be a point. In fact, it must be e because φh(h) = e. So H is central.

Corollary 5.2. π1(G) is central, because it is normal and discrete. In particular, π1(G) is commutative. Corollary 5.3. G G/π˜ (G), with π (G) discrete central. ≃ 1 1 The following corollary describes all (connected) Lie groups with a given Lie algebra.

Corollary 5.4. Given a Lie algebra g, take G˜ with Lie algebra g. Then any other connected G with Lie algebra g is a quotient of G˜ by a discrete central subgroup of G˜.

Suppose G is a topological group and G0 is a connected component of e.

Claim. G0 G is a normal subgroup, and G/G0 is a discrete group. ⊆ Lecture 5 20

If we look at Lie groups Lie algebras , we have an “inverse” given { } → { } by exponential: exp(g) G. Then G0 = (exp g)n. So for a given Lie ⊆ n algebra, we can construct a well-deﬁned isomorphism class of connected, S simply connected Lie groups. When we say “take a Lie group with this Lie algebra”, we mean to take the connected, simply connected one. Coming Attractions: We will talk about Ug, the universal enveloping algebra, C(G), the Hopf algebra, and then we’ll do classiﬁcation of Lie algebras. Lecture 6 - Hopf Algebras 21

Lecture 6 - Hopf Algebras

Last time: We showed that a finite dimensional Lie algebra g uniquely determines a connected simply connected Lie group. We also have a “map” in the other direction (taking tangent spaces). So we have a nice correspondence between Lie algebras and connected simply connected Lie groups. There is another nice kind of structure: Associative algebras. How do these relate to Lie algebras and groups? Let Γ be a finite group and let C[Γ] := c g g Γ, c C be the { g g | ∈ g ∈ } C vector space with basis Γ. We can make C[Γ] into an associative algebra P by taking multiplication to be the multiplication in Γ for basis elements and linearly extending this to the rest of C[Γ].1 Remark 6.1. Recall that the tensor product V and W is the linear span of elements of the form v w, modulo some linearity relations. If V and W ⊗ are infinite dimensional, we will look at the algebraic tensor product of V and W , i.e. we only allow finite sums of the form a b . i ⊗ i We have the following maps P Comultiplication: ∆ : C[Γ] C[Γ] C[Γ], given by ∆( x g) = → ⊗ g x g g g ⊗ P PCounit: ε : C[Γ] C, given by ε( x g)= x . → g g Antipode: S : C[Γ] C[Γ] given byP S( x Pg)= x g 1. → g g − You can check that P P

– ∆(xy)=∆(x)∆(y) (i.e. ∆ is an algebra homomorphism),

– (∆ Id) ∆ = (Id ∆) ∆. (follows from the associativity of ), ⊗ ◦ ⊗ ◦ ⊗ – ε(xy)= ε(x)ε(y) (i.e. ε is an algebra homomorphism),

– S(xy)= S(y)S(x) (i.e. S is an algebra antihomomorphism).

Consider

∆ S Id,Id S m C[Γ] C[Γ] C[Γ] ⊗ ⊗ C[Γ] C[Γ] C[Γ]. −→ ⊗ −−−−−−−→ ⊗ −→ You get 1 m(S Id)∆(g)= m(g− g)= e ⊗ ⊗ so the composition sends xgg to ( g xg)e = ε(x)1A. So we have P P 1“If somebody speaks Danish, I would be happy to take lessons.” Lecture 6 - Hopf Algebras 22

1. A = C[Γ] an associative algebra with 1A 2. ∆ : A A A which is coassociative and is a homomorphism of → ⊗ algebras

3. ε : A C an algebra homomorphism, with (ε Id)∆ = (Id ε)∆ = Id. → ⊗ ⊗ Deﬁnition 6.2. Such an A is called a bialgebra, with comultiplication ∆ and counit ε.

We also have S, the antipode, which is an algebra anti-automorphism, so it is a linear isomorphism with S(ab)= S(b)S(a), such that

S Id A A ⊗ / A A O Id S ⊗ ⊗ ⊗ ∆ m

ε 1A A / C / A

Deﬁnition 6.3. A bialgebra with an antipode is a Hopf algebra.

If A is ﬁnite dimensional, let A∗ be the dual vector space. Deﬁne the multiplication, ∆ , S , ε , 1A∗ on A∗ in the following way: ∗ ∗ ∗ – lm(a) := (l m)(∆a) for all l,m A ⊗ ∈ ∗ – ∆ (l)(a b) := l(ab) ∗ ⊗ – S (l)(a) := l(S(a)) ∗

– ε (l) := l(1A) ∗

– 1A∗ (a) := ε(a)

Theorem 6.4. A∗ is a Hopf algebra with this structure, and we say it is dual to A. If A is ﬁnite-dimensional, then A∗∗ = A. ◮ Exercise 6.1. Prove it.

We have an example of a Hopf algebra (C[Γ]), what is the dual Hopf 2 algebra? Let’s compute A∗ = C[Γ]∗. Well, C[Γ] has a basis g Γ . Let δ be the dual basis, so δ (h) = 0 { ∈ } { g} g if g = h and 1 if g = h. Let’s look at how we multiply such things 6 – δ δ (h) = (δ δ )(h h)= δ (h)δ (h). g1 g2 g1 ⊗ g2 ⊗ g1 g2 – ∆ (δg)(h1 h2)= δg(h1h2) ∗ ⊗ 2 If you want to read more, look at S. Montgomery’s Hopf algebras, AMS, early 1990s. [Mon93] Lecture 6 - Hopf Algebras 23

1 – S (δg)(h)= δg(h− ) ∗

– ε (δg)= δg(e)= δg,e ∗

– 1A∗ (h) = 1. It is natural to think of A as the set of functions Γ C, where ∗ → ( xgδg)(h)= xgδg(h). Then we can think about functions

P– (f1f2)(h)=P f1(h)f2(h)

– ∆ (f)(h1 h2)= f(h1h2) ∗ × 1 – S (f)(h)= f(h− ) ∗ – ε (f)= f(e) ∗

– 1A∗ = 1 constant. So this is the Hopf algebra C(Γ), the space of functions on Γ. If Γ is any affine algebraic group, then C(Γ) is the space of polynomial functions on Γ, and all this works. The only concern is that we need C(Γ Γ) = C(Γ) C(Γ), which × ∼ ⊗ we only have in the finite dimensional case; you have to take completions of tensor products otherwise. So we have the notion of a bialgebra (and duals), and the notion of a Hopf algebra (and duals). We have two examples: A = C[Γ] and A∗ = C(Γ). A natural question is, “what if Γ is an infinite group or a Lie group?” and “what are some other examples of Hopf algebras?” Let’s look at some infinite dimensional examples. If A is an infinite dimensional Hopf algebra, and A A is the algebraic tensor product (finite ⊗ linear combinations of formal a b s). Then the comultiplication should ⊗ be ∆ : A A A. You can consider cases where you have to take some → ⊗ completion of the tensor product with respect to some topology, but we won’t deal with this kind of stuff. In this case, A∗ is too big, so instead of the notion of the dual Hopf algebra, we have dual pairs.

Deﬁnition 6.5. A dual pairing of Hopf algebras A and H is a pair with a bilinear map , : A H C which is nondegenerate such that h i ⊗ → (1) ∆a, l m = a,lm h ⊗ i h i (2) ab, l = a b, ∆ l h i h ⊗ ∗ i (3) Sa, l = a, S l h i h ∗ i (4) ε(a)= a, 1 , ε (l)= 1 , l h H i H h A i Lecture 6 - Hopf Algebras 24

Exmaple: A = C[x], then what is A∗? You can evaluate a polynomial at 0, or you can diﬀerentiate some number of times before you evaluate at 0. A∗ = span of linear functionals on polynomial functions of C of the form d n l (f)= f(x) . n dx x=0

A basis for C[x] is 1,xn with n 1, and we have ≥ m! ,n = m l (xm)= n 0 ,n = m 6 What is the Hopf algebra structure on A? We already have an algebra with identity. Define ∆(x) = x 1 + 1 x and extend it to an algebra ⊗ ⊗ homomorphism, then it is clearly coassociative. Define ε(1) = 1 and ε(xn)= 0 for all n 1. Define S(x)= x, and extend to an algebra homomorphism. ≥ − It is easy to check that this is a Hopf algebra. Let’s compute the Hopf algebra structure on A∗. We have l l (xN ) = (l l )(∆(xN )) n m n ⊗ m N N k k = (l l )( x − x ) n ⊗ m k ⊗ X ◮ d Exercise 6.2. Compute this out. The answer is that A∗ = C[y = dx ], and the Hopf algebra structure is the same as A. This is an example of a dual pair: A = C[x],H = C[y], with xn,ym = h i δn,mm!. Summary: If A is finite dimensional, you get a dual, but in the infinite dimensional case, you have to use dual pairs.

The universal enveloping algebra The idea is to construct a map from Lie algebras to associative algebras so that the representation theory of the associative algebra is equivalent to the representation theory of the Lie algebra. 1) let V be a vector space, then we can form the free associative algebra n (or tensor algebra) of V : T (V ) = C ( n 1V ⊗ ). The multiplication is ⊕ ⊕ ≥ given by concatenation: (v v ) (w w )= v v 1 ⊗···⊗ n · 1 ⊗···⊗ m 1 ⊗···⊗ n ⊗ w w . It is graded: T (V )T (V ) T (V ). It is also a Hopf 1 ⊗ ··· m n m ⊆ n+m algebra, with ∆(x) = x 1 + 1 x, S(x) = x, ε(1) = 1 and ε(x)=0. If ⊗ ⊗ − you choose a basis e1,...,en of V , then T (V ) is the free associative algebra e1,...,en . This algebra is Z+-graded: T (V ) = n 0Tn(V ), where the h i ⊕ ≥ degree of 1 is zero and the degree of each ei is 1. It is also a Z-graded bialgebra: ∆(Tn(V )) (Ti Tn i),S(Tn(V )) Tn(V ), ε : T (V ) C is a ⊆⊕ ⊕ − ⊂ → mapping of graded spaces ((C) = 0 ). n { } Lecture 6 - Hopf Algebras 25

Deﬁnition 6.6. Let A be a Hopf algebra. Then a two-sided ideal I A is ⊆ a Hopf ideal if ∆(I) A I + I A, S(I)= I, and ε(I) = 0. ⊆ ⊗ ⊗ You can check that the quotient of a Hopf algebra by a Hopf ideal is a Hopf algebra (and that the kernel of a map of Hopf algebras is always a Hopf ideal).

◮ Exercise 6.3. Show that I = v w w v v, w V = T (V ) T (V ) 0 h ⊗ − ⊗ | ∈ 1 ⊆ i is a homogeneous Hopf ideal.

Corollary 6.7. S(V )= T (V )/I0 is a graded Hopf algebra. Choose a basis e ,...,e in V , so that T (V )= e ,...,e and S(V )= 1 n h 1 ni e ,...,e / e e e e h 1 ni h i j − j ii ◮ Exercise 6.4. Prove that the Hopf algebra S(V ) is isomorphic to C[e ] 1 ⊗ C[e ]. ···⊗ n Remark 6.8. From the discussion of C[x], we know that S(V ) and S(V ∗) are dual.

◮ Exercise 6.5. Describe the Hopf algebra structure on T (V ∗) that is determined by the pairing v v , l l = δ l (v ) l (v ). h 1 ⊗···⊗ n 1 ⊗···⊗ mi m,n 1 1 · · · n n (free coalgebra of V ∗) Now assume that g is a Lie algebra.

Deﬁnition 6.9. The universal enveloping algebra of g is U(g)= T (g)/ x h ⊗ y y x [x,y] . − ⊗ − i Exercise: prove that x y y x [x,y] is a Hopf ideal. h ⊗ − ⊗ − i Corollary 6.10. Ug is a Hopf algebra.

If e ,...,e is a basis for V . Ug = e ,...,e e e e e = ck e , 1 n h 1 n| i j − j i k ij ki where ck are the structure constants of [ , ]. ij P Remark 6.11. The ideal e e e e is homogeneous, but x y y x [x,y] h i j − j ii h ⊗ − ⊗ − i is not, so Ug isn’t graded, but it is ﬁltered. Lecture 7 26

Lecture 7

Last time we talked about Hopf algebras. Our basic examples were C[Γ] and C(Γ) = C[Γ]∗. Also, for a vector space V , T (V ) is a Hopf algebra. Then S(V ) = T (V )/ x y y x x,y V . And we also have Ug = h ⊗ − ⊗ | ∈ i T g/ x y y x [x,y] x,y g . h ⊗ − ⊗ − | ∈ i Today we’ll talk about the universal enveloping algebra. Later, we’ll talk about deformations of associative algebras because that is where recent progress in representation theory has been.

Universality of Ug We have that g ֒ T g Ug. And σ : g ֒ Ug canonical embedding → → → (of vector spaces and Lie algebras). Let A be an associative algebra with τ : g L(A)= A [a, b]= ab ba a Lie algebra homomorphism such that → { | − } τ([x,y]) = τ(x)τ(y) τ(y)τ(x). − Proposition 7.1. For any such τ, there is a unique τ : Ug A homo- ′ → morphism of associative algebras which extends τ:

′ Ug τ / O > A }} σ } }}τ }} g }

Proof. Because T (V ) is generated (freely) by 1 and V , Ug is generated by 1 and the elements of g. Choose a basis e1,...,en of g. Then we have that τ(e )τ(e ) τ(e )τ(e ) = ck τ(e ). The elements e e (this is i j − j i k ij k i1 · · · ik a product) span Ug for indices i . From the commutativity of the diagram, Pj τ ′(ei)= τ(ei). Since τ ′ is a homomorphism of associative algebras, we have that τ (e e ) = τ (e ) τ (e ), so τ is determined by τ uniquely: ′ i1 · · · ik ′ i1 · · · ′ ik ′ τ (e e )= τ(e ) τ(e ). We have to check that the ideal we mod out ′ i1 · · · ik i1 · · · ik by is in the kernel. But that ideal is in the kernel because τ is a mapping of Lie algebras.

Deﬁnition 7.2. A linear representation of g in V is a pair (V, φ : g → End(V )), where φ is a Lie algebra homomorphism. If A is an associative algebra, then (V, φ : A End(V )) a linear representation of A in V . → Corollary 7.3. There is a bijection between representations of g (as a Lie algebra) and representations of Ug (as an associative algebra).

Proof. ( ) By the universality, A = End(V ), τ = φ. ( )g L(Ug) is a ⇒ ⇐ ⊂ Lie subalgebra. Lecture 7 27

Example 7.4 (Adjoint representation). ad : g End g given by x : → y [x,y]. This is also a representation of Ug. Let e1,...,en be a basis in g. 7→ k Then we have that adei (ej) = [ei, ej ]= k cijek, so the matrix representing the adjoint action of the element e is the matrix (ad ) = (ck ) of struc- i P ei jk ij tural constants. You can check that ad = (ad )(ad ) (ad )(ad ) [ei,ej ] ei ej − ej ei is same as the Jacobi identity for the ck . We get ad : Ug End(g) by ij → deﬁning it on the monomials ei1 ei as ade e = (ade ) (ade ) (the · · · k i1 ··· ik i1 · · · ik product of matrices). Let’s look at some other properties of Ug.

Gradation in Ug Recall that V is a Z -graded vector space if V = V . A linear map + ⊕n∞=0 n f : V W between graded vector spaces is grading-preserving if f(V ) → n ⊆ Wn. If we have a tensor product V W of graded vector spaces, it has ⊗ n a natural grading given by (V W )n = i=0Vi Wn i. The “geometric ⊗ ⊕ ⊗ − meaning” of this is that there is a linear action of C on V such that Vn = x t(x)= tn x for all t C . A graded morphism is a linear map respecting { | · ∈ } this action, and the tensor product has the diagonal action of C, given by t(x y)= t(x) t(y). ⊗ ⊗ d d Example 7.5. If V = C[x], dx is not grading preserving, x dx is. We say that (V, [ , ]) is a Z -graded Lie algebra if [ , ] : V V V is + ⊗ → grading-preserving. Example 7.6. Let V be the space of polynomial vector ﬁelds on C = n d n d Span(z )n 0. Then Vn = Cz . dz ≥ dz An associative algebra (V,m : V V V ) is Z -graded if m is grading- ⊗ → + preserving. Example 7.7.

(1) V = C[x], where the action of C is given by x tx. 7→ (2) V = C[x1,...,xn] where the degree of each variable is 1 ... this is the n-th tensor power of the previous example.

(3) Lie algebra: Vect(C)= a xn+1 d with Vect (C)= Cxn+1 d , { n 0 n dx } n dx deg(x) = 1. You can embed≥ Vect(C) into polynomial vector ﬁelds on P S1 (Virasoro algebra).

(4) T(V) is a Z+-graded associative algebra, as is S(V ). However, Ug is not because we have modded out by a non-homogeneous ideal. But the ideal is not so bad. Ug is a Z+-ﬁltered algebra: Lecture 7 28

Filtered spaces and algebras

Definition 7.8. V is a filtered space if it has an increasing filtration

V V V V 0 ⊂ 1 ⊂ 2 ⊂···⊂ such that V = Vi, and Vn = is a subspace of dimension less than or equal to n. f : V W is a morphism of filtered vector spaces if f(V ) W . → S n ⊆ n We can define filtered lie algebras and associative algebras as such that the bracket/multiplication are filtered maps. There is a functor from filtered vector spaces to graded associative algebras Gr : V Gr(V ), where Gr(V )= V V /V V /V . If f : V W → 0 ⊕ 1 0 ⊕ 2 1 · · · → is filtration preserving, it induces a map Gr(f) : Gr(V ) Gr(W ) functori- → ally such that this diagram commutes:

f V / W

Gr Gr Gr(f) Gr(V ) / Gr(W )

Let A be an associative ﬁltered algebra (i.e. A A A ) such that for i j ⊆ i+j all a Ai, b Aj, ab ba Ai+j 1. ∈ ∈ − ∈ − Proposition 7.9. For such an A,

(1) Gr(A) has a natural structure of an associative, commutative algebra (that is, the multiplication in A deﬁnes an associative, commutative multiplication in Gr(A)).

(2) For a A , b A , the operation aA , bA = aA bA bA aA ∈ i+1 ∈ j+1 { i j} i j − j i mod Ai+j is a lie bracket on Gr(A). (3) x,yz = x,y z + y x, z . { } { } { } Proof. Exercise1. You need to show that the given bracket is well deﬁned, and then do a little dance, keeping track of which graded component you are in.

Deﬁnition 7.10. A commutative associative algebra B is called a Poisson algebra if B is also a lie algebra with lie bracket , (called a Poisson { } bracket) such that x,yz = x,y z + y x, z (the bracket is a derivation). { } { } { } Lecture 7 29

Example 7.11. Let (M,ω) be a symplectic manifold (i.e. ω is a closed non-degenerate 2-form on M), then functions on M form a Poisson algebra. 2n We could have M = R with coordinates p1,...,pn,q1,...,qn, and ω = dp dq . Then the multiplication and addition on C (M) is the usual i i ∧ i ∞ one, and we can deﬁne f, g = pij ∂f ∂g , where ω = ω dxi dxj P { } ij ∂xi ∂xj ij ∧ and (pij) is the inverse matrix to (ω ). You can check that this is a Poisson Pij P bracket. Let’s look at Ug = 1, e e e e e = ck e . Then Ug is ﬁltered, with h i| i j − j i k ij ki (Ug) = Span e e k n . We have the obvious inclusion (Ug) n { i1 · · · ik | ≤ } P n ⊆ (Ug) and (Ug) = C 1. n+1 0 · Proposition 7.12.

(1) Ug is a ﬁltered algebra (i.e. (Ug) (Ug) (Ug) ) r s ⊆ r+s (2) [(Ug)r, (Ug)s] (Ug)r+s 1. ⊆ − Proof. 1) obvious. 2) Exercise2 (almost obvious).

Now we can consider Gr(Ug)= C 1 ( r 1(Ug)r/(Ug)r 1) · ⊕ ≥ − r Claim. (Ug)r/(Ug)r 1 S (g)= symmetricL elements of (C[e1,...,en])r. − ≃ Proof. Exercise3. So Gr(Ug) S(g) as a commutative algebra. ≃ S(g) ∼= Polynomial functions on g∗ = HomC(g, C). Consider C (M). How can we construct a bracket , which satisfies ∞ { } Liebniz (i.e. f, g g = f, g g + f, g g ). We expect that f, g (x) = { 1 2} { 1} 2 { 2} 1 { } pij(x) ∂f ∂g = p(x), df(x) dg(x) . Such a p is called a bivector field (it is ∂xi ∂xj h ∧ i a section of the bundle T M T M M). So a Poisson structure on C (M) ∧ → ∞ is the same as a bivector field p on M satisfying the Jacobi identity. You can check that the Jacobi identity is some bilinear identity on pij which follows from the Jacobi identity on , . This is equivalent to the Schouten identity, { } which says that the Schouten bracket of some things vanishes [There should be a reference here]. This is more general than the symplectic case because pij can be degenerate. Let g have the basis e1,...,en and corresponding coordinate functions 1 n 1 n x ,...,x . On g∗, we have that dual basis e ,...,e (you can identify these 1 n with the coordinates x ,...,x ), and coordinates x1,...,xn (which you can identify with the ei). The bracket on polynomial functions on g∗ is given by

k ∂p ∂q p,q = cijxk . { } ∂xi ∂xj X This is a Lie bracket and clearly acts by derivations. Lecture 7 30

Next we will study the following. If you have polynomials p,q on g∗, you can try to construct an associative product p q = pq + tm (p,q)+ . We ∗t 1 · · · will discuss deformations of commutative algebras. The main example will be the universal enveloping algebra as a deformation of polynomial functions on g∗. Lecture 8 - PBW theorem, Deformations of Associative Algebras 31

Lecture 8 - PBW theorem, Deformations of Asso- ciative Algebras

Last time, we introduced the universal enveloping algebra Ug of a Lie algebra g, with its universality property. We discussed graded and filtered spaces and algebras. We showed that under some condition on a filtered algebra A, the graded algebra Gr(A) is a Poisson algebra. We also checked that Ug satisfies this condition, and that Gr(Ug) S(g) as graded commutative algebras. ≃ The latter space can be understood as the space P ol(g∗) of polynomial functions on g∗. It turns out that the Poisson bracket on Gr(Ug), expressed in P ol(g∗), is given by

f, g (x)= x([df , dg ]) { } x x for f, g P ol(g ) and x g . Note that f is a function on g and x an ∈ ∗ ∈ ∗ ∗ element of g , so df is a linear form on T g = g , that is, df g. ∗ x x ∗ ∗ x ∈ Suppose that V admits a filtration V V V . Then, the asso- 0 ⊂ 1 ⊂ 2 ⊂··· ciated graded space Gr(V )= V (V /V ) is also filtered. (Indeed, 0 ⊕ n 1 n n+1 every graded space W = W admits≥ the filtration W W W n 0 nL 0 ⊂ 0 ⊕ 1 ⊂ W W W ) A natural≥ question is: When do we have V Gr(V ) 0 ⊕ 1 ⊕ 2 ⊂··· L ≃ as filtered spaces ? For the filtered space Ug, the answer is a consequence of the following theorem.

Theorem 8.1 (Poincaré-Birkhoff-Witt). Let e1,...,en be any linear basis for g. Let us also denote by e1,...,en the image of this basis in the universal enveloping algebra Ug. Then the monomials em1 emn form a basis for Ug. 1 · · · n Corollary 8.2. There is an isomorphism of filtered spaces Ug Gr(Ug). ≃ Proof of the corollary. In S(g), em1 emn also forms a basis, so we get an 1 · · · n isomorphism Ug S(g) of filtered vector spaces by simple identification ≃ of the bases. Since Gr(Ug) S(g) as graded algebras, the corollary is ≃ proved.

Remark 8.3. The point is that these spaces are isomorphic as ﬁltered vector spaces. Saying that two inﬁnite-dimensional vector spaces are isomorphic is totally useless.

Proof of the theorem. By deﬁnition, the unordered monomials e e for i1 · · · ik k p span the subspace T T of T (g), where T = g i. Hence, they ≤ 0 ⊕···⊕ p i ⊗ also span the quotient (Ug) := T T / x y y x [x,y] . The goal p 0 ⊕···⊕ p h ⊗ − ⊗ − i is now to show that the ordered monomials em1 emn for m + +m p 1 · · · n 1 · · · n ≤ still span (Ug) . Let’s prove this by induction on p 0. The case p = 0 p ≥ being trivial, consider e e e , with k p, and assume that i has i1 · · · ia · · · ik ≤ a Lecture 8 - PBW theorem, Deformations of Associative Algebras 32

the smallest value among the indices i1,...,ik. We can move eia to the left as follows

a 1 − e e e = e e eˆ e + e e [e , e ] eˆ e . i1 · · · ia · · · ik ia i1 · · · ia · · · ik i1 · · · ib−1 ib ia · · · ia · · · ik Xb=1 Using the commutation relations [e , e ]= cℓ e , we see that the term ib ia ℓ ibia ℓ to the right belongs to (Ug)k 1. Iterating this procedure leads to an equation − P of the form

m1 mn ei1 eia eik = e1 en + terms in (Ug)k 1, · · · · · · · · · − with m + + m = k p. We are done by induction. The proof of 1 · · · n ≤ the theorem is completed by the following homework.[This should really be done here]

◮ Exercise 8.1. Prove that these ordered monomials are linearly indepen- dant.

Let’s “generalize” the situation. We have Ug and S(g), both of which are quotients of T (g), with kernels x y y x [x,y] and x y y x . h ⊗ − ⊗ − i h ⊗ − ⊗ i For any ε C, consider the associative algebra S (g) = T (g)/ x y y ∈ ε h ⊗ − ⊗ x ε[x,y] . By construction, S (g) = S(g) and S (g) = Ug. Recall that − i 0 1 they are isomorphic as ﬁltered vector spaces. Remark 8.4. If ε = 0, the linear map φ : S (g) Ug given by φ (x) = εx 6 ε ε → ε for all x g is an isomorphism of ﬁltered algebras. So, we have nothing new ∈ here.

We can think of Sε(g) as a non-commutative deformation of the associative commutative algebra S(g). (Note that commutative deformations of the algebra of functions on a variety correspond to deformations of the variety.)

Deformations of associative algebras Let (A, m : A A A) be an associative algebra, that is, the linear map ⊗ → m satisﬁes the quadratic equation

m(m(a, b), c)= m(a, m(b, c)). ( ) †

Note that if ϕ : A A is a linear automorphism, the multiplication mϕ → 1 given by mϕ(a, b)= ϕ− (m(ϕ(a), ϕ(b))) is also associative. We like to think of m and mϕ as equivalent associative algebra structures on A. The “moduli space” of associative algebras on the vector space A is the set of solutions to ( ) modulo this equivalence relation. † Lecture 8 - PBW theorem, Deformations of Associative Algebras 33

One can come up with a notion of deformation for almost any kind of object. In these deformation theories, we are interested in some cohomology theories because they parameterize obstructions to deformations. The knowledge of the cohomology of a given Lie algebra g, enables us say a lot about the deformations of g. We’ll come back to this question in the next lecture. Let us turn to our original example: the family of associative algebras Sε(g). Recall that by the PBW theorem, we have an isomorphism of filtered ψ vector spaces S (g) S(g) = P ol(g ), but this is not an isomorphisms ε → ∗ of associative algebras. Therefore, the multiplication defined by f g := ∗ ψ(ψ 1(f) ψ 1(g)) is not the normal multiplication on S(g). We claim that − · − the result is of the form f g = fg + εnm (f, g), ∗ n n 1 X≥ where mn is a bidifferential operator of order n, that is, it is of the form I,J I J mn(f, g)= pn ∂ f∂ g, XI,J I,J where I and J are multi-indices of length n, and pn P ol(g ). The idea ∈ ∗ of the proof is to check this for f = ψ(er1 ern ) and g = ψ(el1 eln ) by 1 · · · n 1 · · · n writing er1 ern el1 eln = el1+r1 eln+rn + εkm (er1 ern , el1 eln ) 1 · · · n · 1 · · · n 1 · · · n k 1 · · · n 1 · · · n k 1 X≥ in Sε(g) using the commuting relations. I,J ◮ Exercise 8.2. Compute the pn for the Lie algebra g generated by X, Y , and H with bracket [X,Y ]= H, [H, X] = [H,Y ] = 0. This is called the Heisenberg Lie algebra.

So we have a family of products on P ol(g∗) which depend on ε in the following way: f g = fg + εnm (f, g) ∗ n n 1 X≥ Since f, g are polynomials and mn is a bidiﬀerential operator of order n, this series terminates, so it is a polynomial in ε. If we try to extend this product to C∞(g∗), then there are questions about the convergence of the product . There are two ways to deal with this problem. The ﬁrst one ∗ is to take these matters of convergence seriously, consider some topology on C∞(g∗) and demand that the series converges. The other solution is to forget about convergence and just think in terms of formal power series in ε. This is the so-called “formal deformation” approach. As we shall see, there are interesting things to say with this seemingly rudimentary point of view. Lecture 8 - PBW theorem, Deformations of Associative Algebras 34

Formal deformations of associative algebras

Let (A, m0) be an associative algebra over C. Then, a formal deformation of (A, m ) is a C[[h]]-linear map m : A[[h]] C A[[h]] A[[h]] such that 0 ⊗ [[h]] → n m(a, b)= m0(a, b)+ h mn(a, b) n 1 X≥ for all a, b A, and such that (A[[h]],m) is an associative algebra. We ∈ say that two formal deformations m andm ˜ are equivalent if there is a ϕ C[[h]]-automorphism A[[h]] A[[h]] such thatm ˜ = mϕ, with ϕ(x) = n −→ x + n 1 h ϕn(x) for all x A, where ϕn is an endomorphism of A. ≥ ∈ QuestionP : Describe the equivalence classes of formal deformations of a given associative algebra. When (A, m0) is a commutative algebra, the answer is known. Philo- sophically and historically, this case is relevant to quantum mechanics. In classical mechanics, observables are smooth functions on a phase space M, i.e they form a commutative associative algebra C∞(M). But when you quantize this system (which is needed to describe something on the order of the Planck scale), you cannot think of observables as functions on phase space anymore. You need to deform the commutative algebra C∞(M) to a noncommutative algebra. And it works... From now on, let (A, m0) be a commutative associative algebra. Let’s write m (a, b)= ab, and m(a, b)= a b. (This is called a star product, and 0 ∗ the terminology goes back to the sixties and the work of J. Vey). Then we have a b = ab + hnm (a, b). ∗ n n 1 X≥ Demanding the associativity of imposes an inﬁnite number of equations ∗ for the mn’s, one for each order: h0: a(bc) = (ab)c

1 h : am1(b, c)+ m1(a, bc)= m1(a, b)c + m1(ab, c) h2: . . . . .

◮ Exercise 8.3. Show that the bracket a, b = m (a, b) m (b, a) deﬁnes a { } 1 − 1 Poisson structure on A. This means that we can think of a Poisson structure on an algebra as the remnants of a deformed product where a b b a = ∗ − ∗ h a, b + O(h). { } Lecture 8 - PBW theorem, Deformations of Associative Algebras 35

One easily checks that if two formal deformations m andm ˜ are equivalent via ϕ (i.e:m ˜ = mϕ), then the associated m1, m˜ 1 are related by m (a, b) =m ˜ (a, b)+ ϕ (ab) ϕ (a)b aϕ (b). In particular, two equiv- 1 1 1 − 1 − 1 alent formal deformations induce the same Poisson structure. Also, it is possible to choose a representative in an equivalence class such that m1 is skew-symmetric (and then, m (a, b)= 1 a, b ). This leads to the following 1 2 { } program for the classification problem: 1. Classify all Poisson structures on A. 2. Given a Poisson algebra (A, , ), classify all equivalence classes of { } star products on A such that m (a, b)= 1 a, b . 1 2 { } Under some mild assumption, it can be assumed that a star product is symmetric, i.e. that it satisfies the equation m (a, b) = ( 1)nm (b, a) for all n − n n. The program given above was completed by Maxim Kontsevitch for the algebra of smooth functions on a manifold M. Recall that Poisson structures on C∞(M) are given by bivector fields on M that satisfy the Jacobi identity. Theorem 8.5 (Kontsevich, 1994). Let A be the commutative associative algebra C (M), and let us fix a Poisson bracket , on A. Equivalence ∞ { } classes of symmetric star products on A with m (a, b) = 1 a, b are in bi- 1 2 { } jection with formal deformations of , modulo formal diffeomorphisms of { } M. A formal deformation of , is a Poisson bracket , on A[[h]] such { } { }h that a, b = a, b + hnµ (a, b) { }h { } n n 1 X≥ for all a, b in A. A formal diffeomorphism of M is an automorphism ϕ of n A[[h]] such that ϕ(f) = f + n 1 h ϕn(f) and ϕ(fg) = ϕ(f)ϕ(g) for all f, g in A. ≥ P We won’t prove the theorem (it would take about a month) . As Poisson algebras are Lie algebras, it relates deformations of associative algebras to deformations of Lie algebras.

Formal deformations of Lie algebras Given a Lie algebra (g, [ , ]), you want to know how many formal deformations of g there are. Sometimes, there are none (like in the case of sln, as we will see later). Sometimes, there are plenty (as for triangular matrices). The goal is now to construct some invariants of Lie algebras which will tell you whether there are deformations, and how many of them there are. In order to do this, we should consider cohomology theories for Lie algebras. We n will focus ﬁrst on the standard complex C (g, g) = n 0 C (g, g), where Cn(g, g) = Hom(Λng, g). · ≥ L Lecture 9 36

Lecture 9

Let’s summarize what has happened in the last couple of lectures.

1. We talked about T (g), and then constructed three algebras:

– Ug = T (g)/ x y y x [x,y] , with Ug = S (g) S (g) as h ⊗ − ⊗ − i 1 ≃ ε filtered associative algebras, for all non-zero ε C. ∈ – S (g) = T (g)/ x y y x ε[x,y] is a family of associative ε h ⊗ − ⊗ − i algebras, with S (g) S (g) as filtered vector spaces. ε ≃ 0 – S (g) = P ol(g )= T (g)/ x y y x = S (g) is an associative, 0 ∼ ∗ h ⊗ − ⊗ i 0 commutative algebra with a Poisson structure defined by the Lie bracket.

2. We have two “pictures” of deformations of an algebra

(a) There is a simple “big” algebra B (such as B = T (g)) and a family of ideals Iε. Then we get a family B/Iε = Aε. This becomes a deformation family of the associative algebra A0 if we identify A A as vector spaces (these are called torsion ε ≃ 0 free deformations). Fixing this isomorphism gives a family of associative products on A0. We can think of this geometrically as a family of (embedded) varieties. (b) Alternatively, we can talk about deformations intrinsically (i.e., without referring to some bigger B). Suppose we have A0 and a family of associative products a b on A . ∗ε 0 φ Example 9.1. Let P ol(g∗) Sε(g) be the isomorphism of the −→ 1 PBW theorem. Then deﬁne f g = φ− (φ(f) φ(g)) = fg + n ∗ · n 1 ε mn(f, g). ≥ UnderstandingP deformations makes a connection between representation theory and Poisson geometry. A second course on Lie theory should discuss symplectic leaves of P ol(g∗), which happen to be coadjoint orbits and correspond to representations. This is why deformations are relevant to representation theory.

Let A be a Poisson algebra with bracket , , so it is a commutative { } algebra, and a Lie algebra, with the bracket acting by derivations. Typ- ically, A = C∞(M). Equivalence classes of formal (i.e., formal power series) symmetric (i.e.,m (f, g) = ( 1)nm (g,f) ) star products on C (M) n − n ∞ are in bijection with equivalence classes of formal deformations of , on { } C∞(M)[[h]]. Lecture 9 37

Apply this to the case A = C∞(g∗). The associative product on Sε(g) comes from the product on T (g). The question is, “how many equivalence classes of star products are there on A?” Any formal deformation of the Poisson structure on (A, , g) is a PBW deformation of some formal defor- { } mation of the Lie algebra C (g ) (with Lie bracket f, g (x)= x(df dg)). ∞ ∗ { } ∧ Such a deformation is equivalent to a formal deformation of the Lie algebra structure on g. This is one of the reasons that deformations of Lie algebras are important — they describe deformations of certain associative algebras. When one asks such questions, some cohomology theory always shows up.

Lie algebra cohomology Recall that (M, φ) is a g-module if φ : g End(M) is a Lie algebra homo- → q morphism. We will write xm for φ(x)m. Deﬁne C (g, M)= q 0 C (g, M) where Cq(g, M) = Hom(Λqg, M) (linear maps).· We deﬁne d : ≥Cq Cq+1 L → by dc(x x )= 1 ∧···∧ q+1 s+t 1 = ( 1) − c([x ,x ] x xˆ xˆ x ) − s t ∧ 1 ∧···∧ s ∧···∧ t ∧···∧ q+1 1 s

Motivation: If g = Vect( ), M = C ( ), then Cq(g, M)=Ωq( ), M ∞ M M with the Cartan formula

(dω)(ξ ξ )= 1 ∧···∧ q+1 s+t 1 = ( 1) − ω([ξ ,ξ ] ξ ξˆ ξˆ ξ ) − s t ∧ 1 ∧···∧ s ∧···∧ t ∧···∧ q+1 1 s

Proposition 9.2. C (g, C) Ω (G) Ω (G) where C is the 1-dimensional ≃ R ⊆ trivial module over g·(so xm =· 0). ·

◮ Exercise 9.2. Prove it. Lecture 9 38

Remark 9.3. This was Cartan’s original motivation for Lie algebra cohomology. It turns out that the inclusion Ω (G) ֒ Ω (G) is a homotopy R → equivalence of complexes (i.e. the two complexes· have the· same homology), and the proposition above tells us that C (g, C) is homotopy equivalent to ΩR(G). Thus, by computing the Lie algebra· cohomology of g (the homology of the complex C·(g, C)), one obtains the De Rham cohomology of G (the homology of the complex Ω·(G)). Deﬁne Hq(g, M) = ker(d : Cq Cq+1)/ im(d : Cq 1 Cq) as always. → − → Let’s focus on the case M = g, the adjoint representation: x m = [x,m]. · 0 0 H (g, g) We have that C = Hom(C, g) ∼= g, and dc(y)= y c = [y, c]. · so ker(d : C0 C1) is the set of c g such that [y, c] = 0 for all y g. → ∈ ∈ That is, the kernel is the center of g, Z(g). So H0(g, g)= Z(g).

H1(g, g) The kernel of d : C1(g, g) C2(g, g) is → µ : g g dµ(x,y)= µ([x,y]) [x,µ(y)] [µ(x),y] = 0 for all x,y g , { → | − − ∈ } which is exactly the set of derivations of g. The image of d : C0(g, g) → C1(g, g) is the set of inner derivations, dc : g g dc(y) = [y, c] . The { → | } Liebniz rule is satisﬁed because of the Jacobi identity. So

H1(g, g)= derivations / inner derivations =: outer derivations. { } { } H2(g, g) Let’s compute H2(g, g). Suppose µ C2, so µ : g g g is a linear ∈ ∧ → map. What does dµ = 0 mean?

dµ(x ,x ,x )= µ([x ,x ],x ) µ([x ,x ],x )+ µ([x ,x ],x ) 1 2 3 1 2 3 − 1 3 2 2 3 1 [x ,µ(x ,x )] + [x ,µ(x ,x )] [x ,µ(x ,x )] − 1 2 3 2 1 3 − 3 1 2 = µ(x , [x ,x ]) [x ,µ(x ,x )] + cyclic permutations − 1 2 3 − 1 2 3 Where does this kind of thing show up naturally? Consider deformations of Lie algebras:

n [x,y]h = [x,y]+ h mn(x,y) n 1 X≥ where the m : g g g are bilinear. The deformed bracket [ , ] n × → h must satisfy the Jacobi identity,

[a, [b, c]h]h + [b, [c, a]h]h + [c, [a, b]h]h = 0 Lecture 9 39

N which gives us relations on the mn. In degree h , we get

N 1 − [a, mN (b, c)] + mN (a, [b, c]) + mk(a, mN k(b, c))+ − Xk=1 N 1 − [b, mN (c, a)] + mN (b, [c, a]) + mk(b, mN k(c, a))+ − Xk=1 N 1 − [c, mN (a, b)] + mN (c, [a, b]) + mk(c, mN k(a, b)) = 0 ( ) − † Xk=1 ◮ Exercise 9.3. Derive ( ). †

Deﬁne [mK ,mN K](a, b, c) as −

mK a, mN K (b, c) + mK b, mN K(c, a) + mK c, mN K (a, b) . − − − Then ( ) can be written as † N 1 − dmN = [mk,mN k] ( ) − ‡ Xk=1 Theorem 9.4. Assume that for all n N 1, we have the relation n 1 N ≤1 − dmn = k=1− [mk,mn k]. Then d( k=1− [mk,mN k]) = 0. − − ◮ ExerciseP 9.4. Prove it. P

The theorem tells us that if we have a “partial deformation” (i.e. we N 1 have found m1,...,mN 1), then the expression k=1− [mk,mN k] is − a 3-cocycle. Furthermore, ( ) tells us that if we are to extend− our ‡ N 1 P deformation to one higher order, k=1− [mk,mN k] must represent zero in H3(g, g). − P Taking N = 1, we get dm = 0, so ker(d : C2 C3) = space of ﬁrst 1 → coeﬃcients of formal deformations of [ , ]. It will turn out that H2 is the space of equivalence classes of m1. It is worth noting that the following “pictorial calculus” may make some of the above computations easier. In the following pictures, arrows are considered to be oriented downwards, and trivalent vertices with two lines coming in and one going out represent the Lie bracket. So, for example, the antisymmetry of the Lie bracket is expressed as / = // − / Lecture 9 40 and the Jacobi identity is

// // $ t / / $ tt // + + /9ttt$$ = // 99$ / 9$ / / / / $ / / / / $ tt // / // // t$t / // + 9tt$ = 0 // − / 99$ / / 9$

We can also use pictures to represent cocycles. Take µ Hn(g, g). Then we ∈ draw µ as ? . . . ?? µ (/).*-+, with n lines going in. Then, the Cartan formula for the diﬀerential says that

i j i ? . . . / ? ?? / Ù ss ?? i+j+1 Ùs/Ù/ss d µ = ( 1) Ùs??/ + µ (/).*-+, − ?/ (/).*-+, 1 i j n+1 µ 1 i n+1 ≤ ≤X≤ (/).*-+, ≤X≤ and the bracket of two cocycles µ Hm and ν Hn is ∈ ∈ i i 9 0 5 9 0 5 9 0 55 Ù Õ 9 0 55 Ù Õ 990 ÙÙ ÕÕ 990 ÙÙ ÕÕ 900 µ Õ 900 ν Õ [µ,ν]= 990(/).*-+,ÕÕ 990(/).*-+,ÕÕ 90 Õ − 90 Õ 1 i n ν Õ 1 i m µÕ ≤X≤ (/).*-+, ≤X≤ (/).*-+,

◮ Exercise 9.5. Use pictures to show that d[µ,ν]= [dµ,ν] [µ,dν]. ± ± Also, these pictures can be used to do the calculations in Exercises 9.3 and 9.4. Lecture 10 41

Lecture 10

Here is the take-home exam, it’s due on Tuesday:

(1) B SL (C) are upper triangular matrices, then ⊂ 2

– Describe X = SL2(C)/B

– SL2(C) acts on itself via left multiplication implies that it acts on X. Describe the action.

0 x1 0 .. .. (2) Find exp  . .  0 xn 1  −   0 0      (3) Prove that if V,W are filtered vector spaces (with increasing filtration) and φ : V W satisfies φ(V ) W , and Gr(φ) : Gr(V ) ∼ Gr(W ) → i ⊆ i −→ an isomorphism, then φ is a linear isomorphism of filtered spaces.

Lie algebra cohomology Recall C (g, M) from the previous lecture, for M a finite dimensional representation· of g (and g finite dimensional). There is a book by D. Fuchs, Cohomology of -dimensional Lie algebras [Fuc50]. ∞ We computed that H0(g, g) = Z(g) g/[g, g] and that H1(g, g) is the ≃ space of exterior derivations of g. Say c Z1(g, g),1 so [c] H1(g, g). Define ∈ ∈ g˜c = g C∂c with the bracket [(x,t), (y,s)] = ([x,y] tc(y)+ sc(x), 0). So if ⊕ − k e1,...,en is a basis in g with the usual relations [ei, ej]= cijek, then we get 1 one more generator ∂c such that [∂c,x] = c(x). Then H (g, g) is the space of equivalence classes of extensions

0 g g˜ C 0 → → → → up to the equivalences f such that the diagram commutes:

0 / g / g˜ / C / 0

Id f Id 0 / g / g˜′ / C / 0

This is the same as the space of exterior derivations.

1Zn(g, M) is the space of n-cocycles, i.e. the kernel of d : Cn(g, M) → Cn+1(g, M). Lecture 10 42

H2(g, g) and Deformations of Lie algebras

A deformation of g is the vector space g[[h]] with a bracket [a, b]h = [a, b]+ n n 1 h mn(a, b) such that mn(a, b)= mn(b, a) and ≥ − P [a, [b, c]h]h + [b, [c, a]h]h + [c, [a, b]h]h = 0.

The hN order term of the Jacobi identity is the condition

N 1 − [a, mN (b, c)] + mN (a, [b, c]) + mk(a, mN k(b, c)) + cycle = 0 ( ) − †† Xk=1 For µ C2(g, g), we compute ∈ dµ(a, b, c)= [a, µ(b, c)] µ(a, [b, c]) + cycle. − − Deﬁne def mk,mN k (a, b, c) = mk(a, mN k(b, c)) + cycle { − } − This is called the Gerstenhaber bracket ... do a Google search for it if you like ... it is a tiny deﬁnition from a great big theory. Then we can rewrite ( ) as †† N 1 − dmN = mk,mN k . ( ) { − } ‡ Xk=1 2 In partiular, dm1 = 0, so m1 is in Z (g, g). 1 Equivalences: [a, b]h′ [a, b]h if [a, b]h′ = φ− ([φ(a), φ(b)]h) for some n ≃ φ(a)= a + n 1 h φn(a). then ≥

mP′ (a, b)= m (a, b) φ ([a, b]) + [a, φ (b)] + [φ (a), b]. 1 1 − 1 1 1 which we can write as m1′ = m1 + dφ1. From this we can conclude 2 Claim. The space of equivalence classes of possible m1 is exactly H (g, g).

Claim (was HW). If m1 is a 2-cocycle, and mN 1,...,m2 satisfy the equa- − tions we want, then N 1 − d mk,mN k = 0. { − }! Xk=1 N+1 3 This is not enough; we know that k=1 mk,mN k is in Z (g, g), but {3 − } to ﬁnd mN , we need it to be trivial in H (g, g) because of ( ). If the N+1 P ‡ 3 cohomology class of k=1 mk,mN k is non-zero, it’s class in H (g, g) is { − } called an obstruction to n-th order deformation. If H3(g, g) is zero, then any P Lecture 10 43

ﬁrst order deformation (element of H2(g, g)) extends to a deformation, but if H3(g, g) is non-zero, then we don’t know that we can always extend. Thus, H3(g, g) is the space of all possible obstructions to extending a deformation. Let’s keep looking at cohomology spaces. Consider C (g, C), where C is a one dimensional trivial representation of g given by x · 0 for any x g. 7→ ∈ First question: take Ug, with the corresponding 1-dimensional representation ε : Ug C given by ε(x)=0 for x g. → ∈ ◮ Exercise 10.1. Show that (Ug, ε, ∆,S) is a Hopf algebra with the ε above, ∆(x) = 1 x + x 1, and S(x)= x for x g. Remember that ∆ ⊗ ⊗ − ∈ and ε are algebra homomorphisms, and that S is an anti-homomorphism.

Let’s compute H1(g, C) (H0 is boring, just a point). This is ker(C1 d −→ C2). Well, C1(g, C) = Hom(g, C), C2(g, C) = Hom(Λ2g, C), and

dc(x,y)= c([x,y]).

So the kernel is the set of c such that c([x,y]) = 0 for all x,y g. Thus, ∈ ker(d) C1(g, C) is the space of g-invariant linear functionals. Recall that ⊆ 1 g acts on g by the adjoint action, and on g∗ = C (g, g) by the coadjoint action (x : l lx where lx(y)= l([x,y])). Under the coadjoint action, l g∗ 7→ 0 ∈ is g-invariant if lx = 0. Note that C is just one point, so its image doesn’t have anything in it. Now let’s compute H2(g, C) = ker(d : C2 C3)/ im(d : C1 C2). Let → → c Z2, then ∈ dc(x,y,z)= c([x,y], z) c([x, z],y)+ c([y, z],x) = 0 − for all x,y,z g. Now let’s ﬁnd the image of d : C1 C2: it is the set of ∈ → functions of the form dl(x,y)= l([x,y]) where l g . It is clear that l([x,y]) ∈ ∗ are (trivial) 2-cocycles because of the Jacobi identity. Let’s see what can we cook with this H2.

Deﬁnition 10.1. A central extension of g is a short exact sequence

0 C g˜ g 0. → → → → Two such extensions are equivalent if there is a Lie algebra isomorphism f : g˜ g˜ such that the diagram commutes: → ′ 0 / C / g˜ / g / 0

Id f Id 0 / C / g˜′ / g / 0 Lecture 10 44

Theorem 10.2. H2(g, C) is isomorphic to the space of equivalence classes of central extensions of g. Proof. Let’s describe the map in one direction. If c Z2, then consider ∈ g˜ = g C with the bracket [(x,t), (y,s)] = ([x,y], c(x,y)). Equivalences of ⊕ extensions boil down to c(x,y) c(x,y)+ l([x,y]). 7→ ◮ Exercise 10.2. Finish this proof.

Let’s do some (infinite dimensional) examples of central extensions. Example 10.3 (Affine Kac-Moody algebras). If g gl , then we define ⊆ n the loop space or loop algebra g to be the set of maps S1 g. To make L → the space more manageable, we only consider Laurent polynomials, z m 7→ m Z amz for am g with all but finitely many of the am equal to zero. ∈ ∈ The bracket is given by [f, g] g(z) = [f(z), g(z)]g. P L Since g gl , there is an induced trace tr : g C. This gives a ⊆ n → non-degenerate inner product on g: L 1 dz (f, g) := tr f(z− )g(z) . z I z =1 | | There is a natural 2-cocylce on g, given by L 1 dz c(f, g)= tr f(z)g′(z) = Res tr f(z)g′(z) , 2πi z z=0 I z =1 | | and a natural outer derivation ∂ : g g given by ∂x(z)= ∂x(z) . L →L ∂z The Kac-Moody algebra is g C∂ Cc. A second course on Lie theory L ⊕ ⊕ should have some discussion of the representation theory of this algebra. Example 10.4. Let gl be the algebra of matrices with finitely many non- zero entries. It is not very∞ interesting. Let gl1 be the algebra of matrices with finitely many non-zero diagonals. gl1 is∞ “more infinite dimensional” than gl , and it is more interesting. ∞ ∞ ◮ Exercise 10.3. Define I 0 J = . 0 I − For x,y gl , show that ∈ ∞ c(x,y)= tr(x[J, y]) is well defined (i.e. is a finite sum). Lecture 10 45

This c is a non-trivial 1-cocycle, i.e. [c] H2(gl1 , C) is non-zero. By the ∈ ∞ way, instead of just using linear maps, we require that the maps Λ2gl1 C 2 ∞ → are graded linear maps. This is Hgraded. Notice that in gln, tr(x[J, y]) = tr(J[x,y]) is a trivial cocycle (it is d of l(x)= tr(Jx). So we have that H2(gl , C)= 0 . n { } We can deﬁne a = gl Cc. This is some non-trivial central extension. ∞ ∞ ⊕

To summarize the last lectures:

1. We related Lie algebras and Lie Groups. If you’re interested in representations of Lie Groups, looking at Lie algebras is easier.

2. From a Lie algebra g, we constructed Ug, the universal enveloping algebra. This got us thinking about associative algebras and Hopf algebras.

3. We learned about dual pairings of Hopf algebras. For example, C[Γ] and C(Γ) are dual, and Ug and C(G) are dual (if G is aﬃne algebraic and we are looking at polynomial functions). This pairing is a starting point for many geometric realizations of representations of G. Conceptually, the notion of the universal enveloping algebra is closely related to the notion of the group algebra C[Γ].

4. Finally, we talked about deformations. Lecture 11 - Engel’s Theorem and Lie’s Theorem 46

Lecture 11 - Engel’s Theorem and Lie’s Theorem

The new lecturer is Vera Serganova. We will cover 1. Classification of semisimple Lie algebras. This will include root systems and Dynkin diagrams. 2. (1) will lead to a classification of compact connected Lie Groups. 3. Finally, we will talk about representation theory of semisimple Lie algebras and the Weyl character formula. A reference for this material is Fulton and Harris [FH91]. The first part is purely algebraic: we will study Lie algebras. g will be a lie algebra, usually finite dimensional, over a field k (usually of characteristic 0). Any Lie algebra g contains the ideal (g) = [g, g], the vector subspace D generated by elements of the form [X,Y ] for X,Y g. ∈ ◮ Exercise 11.1. Show that g is an ideal in g. D ◮ Exercise 11.2. Let G be a simply connected Lie group with Lie algebra g. Then [G, G] is the subgroup of G generated by elements of the form ghg 1h 1 for g, h G. Show that [G, G] is a connected closed normal Lie − − ∈ subgroup of G, with Lie algebra g. D Warning 11.1. Exercise 11.2 is a tricky problem. Here are some poten- tial pitfalls: 1. For G connected, we do not necessarily know that the exponential map is surjective, because G may not be complete. For example, exp : sl (C) SL (C) is not surjective.1 2 → 2 2. If H G is a subgroup with Lie algebra h, then h g closed is not ⊆ ⊆ enough to know that H is closed in G. For example, take G to be a torus, and H to be a line with irrational slope. 3. This exercise requires G to be simply connected. Let 1 x y H :== 0 1 z S1   ×

 0 0 1 

 1 0 n 

n K :=  0 1 0 , c n Z H ∈ ⊆

( 0 0 1 ! ) ¡

1 −1 1 n iπ Assume 0 −1 is in the image, then its pre-image must have eigenvalues (2 + 1)

and −(2n + 1)iπ for some integer n. So the pre-image has distinct eigenvalues, so it is ¡ −1 1 diagonalizable. But that implies that 0 −1 is diagonalizable, contradiction. Lecture 11 - Engel’s Theorem and Lie’s Theorem 47

where c is an element of S1 of inﬁnite order. Then K is normal in H and G = H/K is a counterexample.

Definition 11.2. Define 0g = g, and ng = [ n 1g, n 1g]. This is D D D − D − called the derived series of g. We say g is solvable if ng = 0 for some n D sufficiently large.

Definition 11.3. We can also define 0g = g, and ng = [g, n 1g]. This D D D − is called the lower central series of g. We say that g is nilpotent if g = 0 Dn for some n sufficiently large.

We will denote g = 1g by g. Solvable and nilpotent lie algebras D1 D D are hard to classify. Instead, we will do the classiﬁcation of semisimple Lie algebras (see Deﬁnition 11.15). The following example is in some sense universal (see corollaries 11.7 and 11.12):

Example 11.4. Let gl(n) be the Lie algebra of all n n matrices, and let × b be the subalgebra of upper triangular matrices. I claim that b is solvable. To see this, note that b is the algebra of strictly upper triangular matrices, D and in general, kb has zeros on the main diagonal and the 2k 2 diagonals D − above the main diagonal (for k 2). Let n = b. You can check that n is ≥ D in fact nilpotent.

Useful facts about solvable/nilpotent Lie algebras:

1. If you have an exact sequence of Lie algebras

0 a g g/a 0 → → → → then g is solvable if and only if a and g/a are solvable.

2. If you have an exact sequence of Lie algebras

0 a g g/a 0 → → → → then if g is nilpotent, so are a and g/a.

Warning 11.5. The converse is not true. Diagonal matrices d is nilpotent, and we have 0 n b d 0. → → → → Note that b is not nilpotent, because b = b = 0 . D Dn 0 0∗ Lecture 11 - Engel’s Theorem and Lie’s Theorem 48

3. If a, b g are solvable ideals, then the sum a + b is solvable. To see ⊂ this, note that we have

0 a a + b (a + b)/a 0 → → → → b/(a b) ≃ ∩ a is solvable by assumption, and b/(a| {zb) is} a quotient of a solvable ∩ algebra, so it is solvable by (1). Applying (1) again, a + b is solvable.

4. If k F is a ﬁeld extension, with a Lie algebra g over k, we can make ⊆ a lie algebra g F over F . Note that g F is solvable (nilpotent) ⊗k ⊗k if and only if g is.

We will now prove Engel’s theorem and Lie’s theorem. For any Lie algebra g, we have the adjoint representation: X ad 7→ X ∈ gl(g) given by adX (Y ) = [X,Y ]. If g is nilpotent, then adX is a nilpotent operator for any X g. The converse is also true as we will see shortly ∈ (Cor. 11.9).

Theorem 11.6 (Engel’s Theorem). Let g gl(V ), and assume that X is ⊆ nilpotent for any X g. Then there is a vector v V such that g v = 0. ∈ ∈ · Note that the theorem holds for any representation ρ of g in which every element acts nilpotently; just replace g in the statement of the theorem by ρ(g).

Corollary 11.7. If V is a representation of g in which every element acts nilpotently, then one can ﬁnd 0 = V ( V ( ( V = V a complete ﬂag { } 0 1 · · · n such that g(Vi) Vi 1. That is, there is a basis in which all of the elements ⊆ − of g are strictly upper triangular.

Warning 11.8. Note that the theorem isn’t true if you say “suppose g is nilpotent” instead of the right thing. For example, the set of diagonal matrices d gl(V ) is nilpotent. ⊂ Proof. Let’s prove the theorem by induction on dim g.We ﬁrst show that g has an ideal a of codimension 1. To see this, take a maximal proper subalgebra a g. Look at the representation of a on the quotient space g/a. ⊂ This representation, a gl(g/a), satisﬁes the condition of the theorem,2 so → by induction, there is some X g such that ada(X) = 0 modulo a. So ∈ [a, X] a, so b = kX a is a new subalgebra of g which is larger, so it ⊆ ⊕ must be all of g. Thus, a must have had codimension 1. Therefore, a g is ⊆ actually an ideal (because [X, a] = 0).

2 For any X ∈ a, since X is nilpotent, adX is also nilpotent. Lecture 11 - Engel’s Theorem and Lie’s Theorem 49

Next, we prove the theorem. Let V = v V av = 0 , which is non- 0 { ∈ | } zero by the inductive hypothesis. We claim that gV V . To see this, take 0 ⊆ 0 x g, v V , and y a. We have to check that y(xv) = 0. But ∈ ∈ 0 ∈ yxv = x yv + [y,x] v = 0. 0 a ∈ Now, we have that g = kX a,|{z} and a|kills{z } V , and that X : V V is ⊕ 0 0 → 0 nilpotent, so it has a kernel. Thus, there is some v V which is killed by ∈ 0 X, and so v is killed by all of g.

Corollary 11.9. If ad is nilpotent for every X g, then g is nilpotent as X ∈ a Lie algebra.

Proof. Let V = g, so we have ad : g gl(g), which has kernel Z(g). By → Engel’s theorem, we know that there is an x g such that (ad g)(x) = 0. ∈ This implies that Z(g) = 0. By induction we can assume g/Z(g) is nilpotent. 6 But then g itself must be nilpotent as well because (g/Z(g)) = 0 implies Dn (g) = 0. Dn+1 Warning 11.10. If g gl(V ) is a nilpotent subalgebra, it does not imply ⊆ that every X g is nilpotent (take diagonal matrices for example). ∈ Theorem 11.11 (Lie’s Theorem). Let k be algebraically closed and of characteristic 0. If g gl(V ) is a solvable subalgebra, then all elements of g have ⊆ a common eigenvector in V .

This is a generalization of the statement that two commuting operators have a common eigenvector.

Corollary 11.12. If g is solvable, then there is a complete ﬂag 0 = V { } 0 V V = V such that g(V ) V . That is, there is a basis in which 1 · · · n i ⊆ i all elements of g are upper triangular.

Proof. If g is solvable, take any subspace a g of codimension 1 containing ⊂ g, then a is an ideal. We’re going to try to do the same kind of induction D as in Engel’s theorem. For a linear functional λ : a k, let → V = v V Xv = λ(X)v for all X a . λ { ∈ | ∈ } V = 0 for some λ by induction hypothesis. λ 6 Claim. g(V ) V . λ ⊆ λ Lecture 11 - Engel’s Theorem and Lie’s Theorem 50

Proof of Claim. Choose v V and X a, Y g. Then ∈ λ ∈ ∈ X(Y v)= Y ( Xv ) + [X,Y ]v

λ(X)v λ([X,Y ])v |{z} We want to show that λ[X,Y ] = 0. There is| a{z trick.} Let r be the largest integer such that v,Yv,Y 2v,...,Y rv is a linearly independent set. We know that Xv = λ(X)v for any X a. We claim that XY jv λ(X)Y jv ∈ ≡ mod (span v,Yv,...,Y j 1v ). This is clear for j = 0, and by induction, { − } we have

j j 1 j 1 XY v = YXY − v + [X,Y ]Y − v j−1 λ(X)Y v λ([X,Y ])Y j−1v ≡ j−2 ≡ − mod span v,...,Y v mod span v,...,Y j 2v {| {z } } { } j | {z } j 1 λ(X)Y v mod span v,...,Y − v ≡ { } So the matrix for X can be written as λ(X) on the diagonal and stuﬀ above the diagonal (in this basis). So the trace of X is (r + 1)λ(X). Then we have that tr([X,Y ]) = (r + 1)λ([X,Y ]), since the above statement was proved for any X a and [X,Y ] a. But the trace of a commutator is ∈ ∈ always 0. Since the characteristic of k is 0, we can conclude that λ[X,Y ] = 0. Claim To ﬁnish the proof, write g = kT a, with T : V V (we can do ⊕ λ → λ this because of the claim). Since k is algebraically closed, T has a non-zero eigenvector w in Vλ. This w is the desired common eigenvector. Remark 11.13. If k is not algebraically closed, the theorem doesn’t hold. For example, consider the (one-dimensional) Lie algebra generated by a rotation of R2. The theorem also fails if k is not characteristic 0. Say k is characteristic p, then let x be the permutation matrix of the p-cycle (p p 1 2 1) (i.e. − − · · · the matrix 0 Ip 1 ), and let y be the diagonal matrix diag(0, 1, 2,...,p 1). 1 0 − Then [x,y]= x, so the Lie algebra generated by x and y is solvable. However, y is diagonal, so we know all of its eigenvectors, and none of them is an eigenvector of x.

Corollary 11.14. Let k be of characteristic 0. Then g is solvable if and only if g is nilpotent. D If g is nilpotent, then g is solvable from the deﬁnitions. If g is solvable, D then look at everything over the algebraic closure of k, where g looks like upper triangular matrices, so g is nilpotent. All this is independent of D coeﬃcients (by useful fact (4)). Lecture 11 - Engel’s Theorem and Lie’s Theorem 51

The radical There is a unique maximal solvable ideal in g (by useful fact (3): sum of solvable ideals is solvable), which is called the radical of g.

Deﬁnition 11.15. We call g semisimple if rad g = 0.

◮ Exercise 11.3. Show that g/rad g is always semisimple.

If g is one dimensional, generated by X, then we have that [g, g] = 0, so g cannot be semisimple. If g is two dimensional, generated by X and Y , then we have that [g, g] is one dimensional, spanned by [X,Y ]. Thus, g cannot be semisimple because g is a solvable ideal. D There is a semisimple Lie algebra of dimension 3, namely sl2. Semisimple algebras have really nice properties. Cartan’s criterion (The- orem 12.7) says that g is semisimple if and only if the Killing form (see next lecture) is non-degenerate. Whitehead’s theorem (Theorem 12.10) says that if V is a non-trivial irreducible representation of a semisimple Lie algebra g, then Hi(g, V ) = 0 for all i. Weyl’s theorem (Theorem 12.14) says that every ﬁnite-dimensional representation of a semisimple Lie algebra is the direct sum of irreducible representations. If G is simply connected and compact, the g is semisimple . Lecture 12 - Cartan Criterion, Whitehead and Weyl Theorems 52

Lecture 12 - Cartan Criterion, Whitehead and Weyl Theorems

Invariant forms and the Killing form Let ρ : g gl(V ) be a representation. To make the notation cleaner, we → will write X¯ for ρ(X). We can deﬁne a bilinear form on g by BV (X,Y ) := tr(X¯Y¯ ). This form is symmetric because tr(AB) = tr(BA) for any linear operators A and B. We also have that

B ([X,Y ], Z)= tr(X¯ Y¯ Z¯ Y¯ X¯ Z¯)= tr(X¯Y¯ Z¯) tr(X¯Z¯Y¯ ) V − − = tr(X¯ Y¯ Z¯ X¯ Z¯Y¯ )= B (X, [Y, Z]), − V so B satisfies B([X,Y ], Z)= B(X, [Y, Z]). Such a form is called an invariant form. It is called invariant because it is implied by B being Ad-invariant. Assume that for any g G and X, Z g, ∈ ∈ we B(AdgX, AdgZ) = B(X, Z). Let γ be a path in G with γ′(0) = Y . We get that d B([Y, X], Z)+ B(X, [Y, Z]) = B Ad (X), Ad (Z) = 0. dt γ(t) γ(t) t=0 Definition 12.1. The Killing form, denoted by B, is the special case where ρ is the adjoint representation. That is, B(X,Y ) := tr(ad ad ). X ◦ Y ◮ Exercise 12.1. Let g be a simple Lie algebra over an algebraically closed field. Check that two invariant forms on g are proportional. ◮ Exercise 12.2 (In class). If g is solvable, then B(g, g) = 0. D Solution. First note that if Z = [X,Y ] g, then ad = [ad , ad ] ∈ D Z X Y ∈ (ad g) since the adjoint representation is a Lie algebra homomorphism. D Moreover, g solvable implies the image of the adjoint representation, ad(g) ≃ g/Z(g), is solvable. Therefore, in some basis of V of a representation of ad(g), all matrices of ad(g) are upper triangular (by Lie’s Theorem), and those of (adg) are all strictly upper triangular. The product of an upper D triangular matrix and a strictly upper triangular matrix will be strictly upper triangular and therefore have trace 0.

The converse of this exercise is also true. It will follow as a corollary of our next theorem (Corollary 12.6 below). Theorem 12.2. Suppose g gl(V ), char k = 0, and B (g, g) = 0. Then g ⊆ V is solvable. Lecture 12 - Cartan Criterion, Whitehead and Weyl Theorems 53

For the proof, we will need the following facts from linear algebra.

Lemma 12.3.1 Let X be a diagonalizable linear operator in V , with k algebraically closed. If X = A diag(λ ,...,λ ) A 1 and f : k k is a function, · 1 n · − → we deﬁne f(X) as A diag(f(λ ),...,f(λ )) A 1. Suppose tr(X f(X)) = 0 · 1 n · − · for any Q-linear map f : k k such that f is the identity on Q, then X = 0. → Proof. Consider only f such that the image of f is Q. Let λ1,...,λm be the eigenvalues of X with multiplicities n ,...,n . We obtain tr(X f(X)) = 1 m · n1λ1f(λ1) + . . . + nmλmf(λn) = 0. Apply f to this identity to obtain 2 2 n1f(λ1) + . . . + nmf(λm) = 0 which implies f(λi) = 0 for all i. If some λi is not zero, we can choose f so that f(λ ) = 0, so λ = 0 for all i. Since X i 6 i is diagonalizable, X = 0.

Lemma 12.4 (Jordan Decomposition). Given X gl(V ), there are unique ∈ X , X gl(V ) such that X is diagonalizable, X is nilpotent, [X , X ] = 0, s n ∈ s n s n and X = Xn + Xs. Furthermore, Xs and Xn are polynomials in X. Proof. All but the last statement is standard; see, for example, Corollay 2.5 of Chapter XIV of [Lan02]. To see the last statement, let the characteristic polynomial of X be (x λ )ni . By the chinese remainder theorem, we can i − i ﬁnd a polynomial f such that f(x) λ mod (x λ )ni . Choose a basis Q ≡ i − i so that X is in Jordan form and compute f(X) block by block. On a block with λ along the diagonal (X λ I)ni is 0, so f(X) is λ I on this block. i − i i Then f(X)= X is diagonalizable and X = X f(X) is nilpotent. s n − Lemma 12.5. Let g gl(V ). The adjoint representation ad : g gl(g) ⊆ → preserves Jordan decomposition: adXs = (adX )s and adXn = (adX )n. In particular, adXs is a polynomial in adX .

Proof. Suppose the eigenvalues of Xs are λ1,...,λm, and we are in a basis where X is diagonal. Check that ad (E ) = [X , E ] = (λ λ )E . So s Xs ij s ij i − j ij Xs diagonalizable implies adXs is diagonalizable (because it has a basis of eigenvectors). We have that adXn is nilpotent because the monomials in the k expansion of (adXn ) (Y ) have Xn to at least the k/2 power on one side of Y .

So we have that adX = adXs + adXn , with adXs diagonalizable, adXn nilpotent, and the two commute, so by uniqueness of the Jordan decomposition, adXs = (adX )s and adXn = (adX )n. Proof of Theorem 12.2. It is enough to show that g is nilpotent. Let X D ∈ g, so X = [Y , Z ]. It suffices to show that X = 0. To do this, let D i i s 1This is differentP from what we did in class. There is an easier way to do this if you are willing to assume k = C and use complex conjugation. See Fulton and Harris for this method. Lecture 12 - Cartan Criterion, Whitehead and Weyl Theorems 54 f : k k be any Q-linear map fixing Q. →

BV (f(Xs), Xs)= BV (f(Xs), X) (Xn doesn’t contribute)

= BV f(Xs), [Yi, Zi] Xi = BV ([f(Xs),Yi], Zi) (BV invariant) i g? X ∈ = 0 (assuming [f(X ),Y ] g) | {z } s i ∈

Then by Lemma 12.3, Xs = 0. To see that [f(X ),Y ] g, suppose the eigenvalues of X are λ ,...,λ . s i ∈ s 1 m Then the eigenvalues of f(Xs) are f(λi), the eigenvalues of adXs are of the form µ := λ λ , and eigenvalues of ad are ν := f(λ ) f(λ ) = ij i − j f(Xs) ij i − j f(µij). If we deﬁne g to be a polynomial such that g(µij)= νij, then adf(Xs) and g(adXs ) are diagonal (in some basis) with the same eigenvalues in the same places, so they are equal. So we have

[f(Xs),Yi]= g(adXs )(Yi) = h(ad )(Y ) g (using Lemma 12.5) X i ∈ for some polynomial h. The above arguments assume k is algebraically closed, so if it’s not apply the above to g k¯. Then g k¯ solvable implies g solvable as mentioned ⊗k ⊗k in the previous lecture.

Corollary 12.6. g is solvable if and only if B( g, g) = 0. D Proof. ( ) We have that B( g, g) = 0 implies B( g, g) = 0 which implies ⇐ D D D that ad( g) is solvable. The adjoint representation of g gives the exact D D sequence 0 Z( g) g ad( g) 0. → D → D → D → Since Z( g) and ad( g) are solvable, g is solvable by useful fact (1) of D D D Lecture 11, so g is solvable. ( ) This is exercise 12.2. ⇒ Theorem 12.7 (Cartan’s Criterion). The Killing form is non-degenerate if and only if g is semi-simple.

Proof. Say g is semisimple. Let a = ker B. Because B is invariant, we get that a is an ideal, and B a = 0. By the previous theorem (12.2), we have | that a is solvable, so a = 0 (by deﬁnition of semisimple). Suppose that g is not semisimple, so g has a non-trivial solvable ideal. Then the last non-zero term in its upper central series is some abelian ideal Lecture 12 - Cartan Criterion, Whitehead and Weyl Theorems 55 a g2. For any X a, the matrix of ad is of the form 0 with respect ⊆ ∈ X 0 0∗ to the decomposition g = a g/a, and for Y g, adY is of the form 0∗ ∗ . ⊕ ∈ ∗ Thus, we have that tr(ad ad )=0so X ker B, so B is degenerate. X ◦ Y ∈ Theorem 12.8. Any semisimple Lie algebra is a direct sum of simple algebras.

Proof. If g is simple, then we are done. Otherwise, let a g be an ideal, ⊆ then the restriction B a is nondegenerate. Note that a is also an ideal by | ⊥ invariance of B. Look at a a . On their intersection, B is zero, so the ∩ ⊥ intersection is a solvable ideal, so it is zero. Thus, we have that g = a a . ⊕ ⊥ By induction on dimension, we’re done.

Remark 12.9. In particular, if g = gi is semisimple, with each gi simple, we have that g = g . But g is either 0 or g , and it cannot be 0 D D i DLi i (lest g be a solvable ideal). Thus g = g. i L D Theorem 12.10 (Whitehead). If g is semisimple and V is an irreducible non-trivial representation of g, then Hi(g, V ) = 0 for all i 0. ≥ Proof. The proof uses the Casimir operator, C gl(V ). Assume for the V ∈ moment that g gl(V ). Choose a basis e ,...,e in g, with dual basis ⊆ 1 n f1,...,fn in g (dual with respect to BV , so BV (ei,fj)= δij). It is necessary that BV be non-degenerate for such a dual basis to exist, and this is where 3 we use that g is semisimple. The Casimir operator is deﬁned to be CV = e f gl(V ) (where is composition of linear operators on V ). The i ◦ i ∈ ◦ main claim is that [C , X] = 0 for any X g. This can be checked directly: P V ∈ put [X,fi] = aijfj, [X, ei] = bijej, then apply BV to obtain aji = B (e , [X,f ]) = B ([e , X],f ) = b , where the middle equality is by V i j P V i j P − ij invariance of BV .

[X,C ]= Xe f e Xf + e Xf e f X V i i − i i i i − i i Xi = [X, ei]fi + ei[X,fi] Xi = bijejfi + aijeifj Xi Xj = (aij + bji)eifj = 0. Xi Xj 2Quick exercise:why is a an ideal? 3 We will soon see that CV is independent of the basis e1,...,en, so the article “the” is apropriate. Lecture 12 - Cartan Criterion, Whitehead and Weyl Theorems 56

Suppose V is irreducible, and k is algebraically closed. Then the condition [CV , X] = 0 means precisely that CV is an intertwiner so by Schur’s lemma, CV = λId. We can compute

dim g

trV CV = tr(eifi) Xi=1 = BV (ei,fi) = dim g.

dim g X Thus, we have that λ = dim V , in particular, it is non-zero. For any representation ρ : g gl(V ), we can still talk about CV , but we → dim ρ(g) deﬁne it for the image ρ(g), so CV = dim V Id. We get that [CV , ρ(X)] = 0. The point is that if V is non-trivial irreducible, we have that CV is non-zero. Now consider the complex calculating the cohomology:

Hom(Λkg, V ) d Hom(Λk+1g, V ) −→ We will construct a chain homotopy4 γ : Ck+1 Ck between the zero map dim ρ(g) → on the complex and the map CV = dim V Id:

γc(x1,...,xk)= eic(fi,x1,...,xk) Xi

◮ Exercise 12.3. Check directly that (γd + dγ)c = CV c. dim ρ(g) Thus γd + dγ = CV = λId (where λ = dim V ). Now suppose dc = 0. d(γ(c)) Then we have that dγ(c) = λc, so c = λ . Thus, ker d/ im d = 0, as desired.

Remark 12.11. What is H1(g, k), where k is the trivial representation of g? Recall that the cochain complex is

k Hom(g, k) d Hom(Λ2g, k) . → −→ →··· If c Hom(g, k) and c ker d, then dc(x,y) = c([x,y]) = 0, so c is 0 on ∈ ∈ g = g. So we get that H1(g, k) = (g/ g) = 0. D D ∗ However, it is not true that Hi(g, k)=0 for i 2. Recall from Lecture ≥ 10 that H2(g, k) parameterizes central extensions of g (Theorem 10.2).

j ◮ Exercise 12.4. Compute H (sl2, k) for all j.

4Don’t worry about the term “chain homotopy” for now. It just means that γ satisﬁes the equation in Exercise 12.3. See Proposition 2.12 of [Hat02]. Lecture 12 - Cartan Criterion, Whitehead and Weyl Theorems 57

Remark 12.12. Note that for g semisimple, we have H1(g, M) =0 for any ﬁnite dimensional representation M (not just irreducibles). We have already seen that this holds when M is trivial and Whitehead’s Theorem shows this when M is non-trivial irreducible. If M is not irreducible, use short exact sequences to long exact sequences in cohomology: if

0 W M V 0 → → → → is an exact sequence of representations of g, then

H1(g, V ) H1(g, M) H1(g,W ) → → → → is exact. The outer guys are 0 by induction on dimension, so the middle guy is zero. We need a lemma before we do Weyl’s Theorem. Lemma 12.13. Say we have a short exact sequence

0 W M V 0. → → → → 1 If H (g, Homk(V,W )) = 0, then the sequence splits. V ∗ W ⊗ Proof. Let| X {z g.} Let XW represent the induced linear operator on W . ∈ XW c(X) Then we can write XM = . What is c(X)? It is an element of 0 XV Hom (V,W ). So c is a linear function from g to Hom (V,W ). It will be a k k 1-cocycle: we have [XM ,YM ] = [X,Y ]M because these are representations, which gives us

X c(Y ) c(Y )X Y c(X) c(X)Y = c([X,Y ]). W − V − W − V In general, dc(X,Y ) = c([X,Y ]) Xc(Y )+ Y c(X), where Xc(Y ) is given − by the action of X g on V W , which is not necessarily composition. ∈ ∗ ⊗ In our case this action is by commutation, where c(Y ) is extended to an endomorphism of V W by writing it as 0 c(Y ) . The line above says ⊕ 0 0 exactly that dc = 0. Put Γ = 1W K . Conjugating by Γ gives an equivalent representation. 0 1V We have 1 XW c(X)+ KXV XW K ΓX Γ− = − M 0 X V We’d like to kill the upper right part (to show that X acts on V and W separately). We have c Hom(g, V W ), K V W . Since the ﬁrst ∈ ∗ ⊗ ∈ ∗ ⊗ cohomology is zero, dc = 0, so we can ﬁnd a K such that c = dK. Since c(X) = dK(X) = X(K) = X K KX , the upper right part is indeed W − V 0. Lecture 12 - Cartan Criterion, Whitehead and Weyl Theorems 58

Theorem 12.14 (Weyl). If g is semi-simple and V is a ﬁnite-dimensional representation of g, then V is semi-simple5 (i.e. completely reducible).

Proof. The theorem follows immediately from Lemma 12.13 and Remark 12.12.

Weyl proved this using the unitary trick, which involves knowing about compact real forms. Remark 12.15. We know from Lecture 10 that deformations of g are enu- merated by H2(g, g). This means that semisimple Lie algebras do not have any deformations! This suggests that the variety of semisimple Lie algebras is discrete. Perhaps we can classify them. Whenever you have a Lie algebra g, you can ask about Aut g. Aut g will be a Lie group. What is the connected component of the identity? We can also ask about the Lie algebra of the group of automorphisms. Let X(t) be g d a path in Aut such that X(0) = 1. Then we compute dt X(t) t=0 = φ. We have that

[X(t)Y, X(t)Z]= X(t)([Y, Z]) We can diﬀerentiate this and get φ End g with φ[Y, Z] = [φY, Z] + [Y, φZ] ∈ – derivations. As an immediate corollary of the Jacobi identity, we get that adX is a derivation on g. So ad(g) er(g). ⊆ D ◮ Exercise 12.5. Check that ad(g) is an ideal.

We have seen in lecture 9 (page 38) that er(g)/ad(g) H1(g, g). The D ≃ conclusion is that er(g)= ad(g)— that is, all derivations on a semisimple D Lie algebra are inner. What does this say about the connected component of the identity of Aut g? It has the same Lie algebra as g, so it is just Ad G. Next time we’ll talk about root decomposition of semisimple Lie algebras.

′ ′ 5For any invariant subspace W ⊆ V , there is an invariant W ⊆ V so that V = W ⊕W . Lecture 13 - The root system of a semi-simple Lie algebra 59

Lecture 13 - The root system of a semi-simple Lie algebra

The goal for today is to start with a semisimple Lie algebra over a ﬁeld k (assumed algebraically closed and characteristic zero), and get a root system. Recall Jordan decomposition. For g gl(V ), any x g can be written ⊆ ∈ (uniquely) as x = xs + xn, where xs is semisimple and xn is nilpotent, both of which are polynomials in x. In general, xs and xn are in gl(V ), but not necessarily in g. Proposition 13.1. If g gl(V ) is semisimple, then x ,x g. ⊆ s n ∈ Proof. Notice that g acts on gl(V ) via commutator, and g is an invariant subspace. By complete reducibility (Theorem 12.14), we can write gl(V )= g m where m is g-invariant, so ⊕ [g, g] g and [g, m] m. ⊆ ⊆

We have that adxs and adxn are polynomials in adx (by Lemma 12.5), so [x , g] g , [x , g] g and [x , m] m , [x , m] m. ( ) n ⊆ s ⊆ n ⊆ s ⊆ † Take x = a + b g m, where a g and b m. We would like to show n ∈ ⊕ ∈ ∈ that b = 0, for then we would have that x g, from which it would follow n ∈ that x g. s ∈ Decompose V = V V with the V irreducible. Since x is a 1 ⊕···⊕ n i n polynomial in x, we have that x (V ) V , and a(V ) V since a g, so n i ⊆ i i ⊆ i ∈ b(V ) V . Moreover, we have that i ⊆ i [x , g] = [a, g] + [b, g] g, n ⊆ g m ∈ ∈ so [b, g] = 0 (i.e. b is an intertwiner).|{z} By| Schur’s{z} lemma, b must be a scalar operator on V (i.e. b = λ Id). We have tr (x ) = 0 because x is i |Vi i Vi n n nilpotent. Also tr (a) = 0 because g is semisimple implies g = g, so Vi D a = [xk,yk], and the traces of commutators are 0. Thus, trVi (b) = 0, so λ = 0 and b = 0. Now x = a g, and so x g. i P n ∈ s ∈ Since the image of a semisimple Lie algebra is semisimple, the proposition tells us that for any representation ρ : g gl(V ), the semisimple and → nilpotent parts of ρ(x) are in the image of g. In fact, the following corollary shows that there is an absolute Jordan decomposition x = xs + xn within g. Corollary 13.2. If g is semisimple, and x g, then there are x ,x g ∈ s n ∈ such that for any representation ρ : g gl(V ), we have ρ(x ) = ρ(x) and → s s ρ(xn)= ρ(x)n. Lecture 13 - The root system of a semi-simple Lie algebra 60

Proof. Consider the (faithful) representation ad : g gl(g). By the propo- → sition, there are some x ,x g such that (ad ) = ad and (ad ) = ad . s n ∈ x s xs x n xn Since ad is faithful, adx = adxn +adxs and ad[xn,xs] = [adxn , adxs ] = 0 tell us that x = xn + xs and [xs,xn] = 0. These are our candidates for the absolute Jordan decomposition. Given any surjective Lie algebra homomorphism σ : g g , we have that → ′ adσ(y)(σ(z)) = σ(ady(z)), from which it follows that adσ(xs) is diagonalizable and adσ(xn) is nilpotent (note that we’ve used surjectivity of σ). Thus, σ(x)n = σ(xn) and σ(x)s = σ(xs). That is, our candidates are preserved by surjective homomorphisms. Now given any representation ρ : g gl(V ), the previous paragraph → allows us to replace g by its image, so we may assume ρ is faithful. By the proposition, there are some y, z g such that ρ(x) = ρ(y), ρ(x) = ρ(z). ∈ s n Then [ρ(y), ] is a diagonalizable operator on gl ρ(g) = gl(g), and − gl(ρ(g)) ∼ [ρ(z), ] is nilpotent. Uniqueness of the Jordan decomposition implies − gl(ρ(g)) that ρ(y)= ρ(xs) and ρ(z)= ρ(xn). Since ρ is faithful, it follows that y = xs and z = xn. Deﬁnition 13.3. We say x g is semisimple if ad is diagonalizable. We ∈ x say x is nilpotent if adx is nilpotent. Given any representation ρ : g gl(V ) with g semisimple, the corol- → lary tells us that if x is semisimple, then ρ(x) is diagonalizable, and if x is nilpotent, then ρ(x) is nilpotent. If ρ is faithful, then x is semisimple (resp. nilpotent) if and only if ρ(x) is semisimple (resp. nilpotent).

Deﬁnition 13.4. We denote the set of all semisimple elements in g by gss. We call an x g regular if dim(ker ad ) is minimal (i.e. the dimension of ∈ ss x the centralizer is minimal).

Example 13.5. Let g = sln. Semisimple elements of sln are exactly the diagonalizable matrices, and nilpotent elements are exactly the nilpotent matrices. If x g is diagonalizable, then the centralizer is minimal ex- ∈ actly when all the eigenvalues are distinct. So the regular elements are the diagonalizable matrices with distinct eigenvalues.

Let h g be regular. We have that ad is diagonalizable, so we can ∈ ss h write g = g , where g = x g [h, x] = µx are eigenspaces of ad . µ k µ µ { ∈ | } h We know that∈g = 0 because it contains h. There are some other properties: L 0 6 1. [g , g ] g . µ ν ⊆ µ+ν 2. g g is a subalgebra. 0 ⊆ 3. B(g , g )=0if µ = ν (here, B is the Killing form, as usual). µ ν 6 − Lecture 13 - The root system of a semi-simple Lie algebra 61

4. B gµ g−µ is non-degenerate, and B g0 is non-degenerate. | ⊕ | Proof. Property 1 follows from the Jacobi identity: if x g and y g , ∈ µ ∈ ν then [h, [x,y]] = [[h, x],y] + [x, [h, y]] = µ[x,y]+ ν[x,y], so [x,y] g . Property 2 follows immediately from 1. Property 3 follows ∈ µ+ν from 1 because ad ad : g g , so B(x,y) = tr(ad ad ) = 0 x ◦ y γ → γ+µ+ν x ◦ y whenever µ + ν = 0. Finally, Cartan’s criterion says that B must be non- 6 degenerate, so property 4 follows from 3.

Proposition 13.6. In the situation above (g is semisimple and h g is ∈ ss regular), g0 is abelian. Proof. Take x g , and write x = x + x . Since ad is a polynomial of ∈ 0 s n xn ad , we have [x , h] = 0, so x g , from which we get x g . Since x n n ∈ 0 s ∈ 0 [xs, h] = 0, we know that adxs and adh are simultaneously diagonalizable (recall that ad is diagonalizable). Thus, for generic t k, we have that xs ∈ ker ad ker ad . Since h is regular, ker ad = ker ad = g . So h+txs ⊆ h xs h 0 [xs, g0] = 0, which implies that g0 is nilpotent by Corollary 11.9 to Engel’s Theorem. Now we have that ad : g g is nilpotent, and we want ad to x 0 → 0 x be the zero map. Notice that B(g , g ) = 0 since g is nilpotent, but B g 0 D 0 0 | 0 is non-degenerate by property 4 above, so g = 0, so g is abelian. D 0 0 Definition 13.7. We call h := g0 the Cartan subalgebra of g (associated to h). In Theorem 14.1, we will show that any two Cartan subalgebras of a semisimple Lie algebra g are related by an automorphism of g, but for now we just fix one. See [Hum78, 15] for a more general definition of Cartan § subalgebras. ◮ Exercise 13.1. Show that if g is semisimple, h consists of semisimple elements. All elements of h are simultaneously diagonalizable because they are all diagonalizable (by the above exercise) and they all commute (by the above proposition). For α h r 0 consider ∈ ∗ { } g = x g [h, x]= α(h)x for all h h α { ∈ | ∈ }

If this gα is non-trivial, it is called a root space and the α is called a root. The root decomposition (or Cartan decomposition) of g is g = h α h∗r 0 gα. ⊕ ∈ { } 1 0 Example 13.8. g = sl(2). Take H = 0 1 , a regularL element. The Cartan subalgebra is h = k H, a one-dimensional− subspace. We have 0 t 0 0· g2 = 0 0 and g 2 = t 0 , and g = h g2 g 2. − ⊕ ⊕ − Lecture 13 - The root system of a semi-simple Lie algebra 62

Example 13.9. g = sl(3). Take

x1 0 h = x x + x + x = 0 .  2  1 2 3  0 x  3    Let Eij be the elementary matrices. We have that [diag(x1,x2,x3), Eij ] = (x x )E . If we take the basis ε (x ,x ,x ) = x for h , then we have i − j ij i 1 2 3 i ∗ roots ε ε . They can be arranged in a diagram: i − j ε ε ε ε 2 3 X F 1 3 − 11 − 11 ε 11 ε 2 f 1 8 1 11 ε ε o 1 1 / ε ε 2 − 1 1 1 − 2 11 11 11 ε3 1 Ö 1 ε ε ε ε 3 − 1 3 − 2 We will see that this generalizes to sl(n) The rank of g is deﬁned to be dim h. In particular, the rank of sl(n) is going to be n 1. − Basic properties of the root decomposition are: 1. [g , g ] g . α β ⊆ α+β 2. B(g , g )=0if α + β = 0. α β 6

3. B gα g−α is non-degenerate. | ⊕ 4. B h is non-degenerate | Note that 3 implies that α is a root if and only if α is a root. − ◮ Exercise 13.2. Check these properties. Now let’s try to say as much as we can about this root decomposition. Deﬁne hα h as [gα, g α]. Take x gα and y g α and h h. Then ⊆ − ∈ ∈ − ∈ compute

hα gα ∈ ∈ B([x,y], h)= B(x, [y, h]) (B is invariant)

z }| { = α(h)Bz(}|x,y{ ) (since y gα) ∈

It follows that hα⊥ = ker(α), which is of codimension one. Thus, hα is one- dimensional. Lecture 13 - The root system of a semi-simple Lie algebra 63

Proposition 13.10. If g is semisimple and α is a root, then α(h ) = 0. α 6 Proof. Assume that α(hα) = 0. Then pick x gα, y g α such that ∈ ∈ − [x,y]= h =0. If α(h) = 0, then we have that [h, x]= α(h)x = 0, [h, y] = 0. 6 Thus x,y,h is a copy of the Heisenberg algebra, which is solvable (in h i fact, nilpotent). By Lie’s Theorem, adg(x) and adg(y) are simultaneously upper triangularizable, so adg(h) = [adg(x), adg(y)] is nilpotent. This is a contradiction because h is an element of the Cartan subalgebra, so it is semisimple.

For each root α, we will take H h such that α(H ) = 2 (we can α ∈ α α always scale Hα to get this). We can choose Xα gα and Yα g α such that ∈ ∈ − [Xα,Yα] = Hα. We have that [Hα, Xα] = α(Hα)Xα = 2Xα and [Hα,Yα] = 2Yα. That means we have a little copy of sl(2) ∼= Hα, Xα,Yα . Note that − ad h i .(this makes g a representation of sl via sl ֒ g ֒ gl(g 2 2 → → We normalize α(hα) to 2 so that we get the standard basis of sl2. This way, the representations behave well (namely, that various coeﬃcients are integers). Next we study these representations.

Irreducible ﬁnite dimensional representations of sl(2) Let H,X,Y be the standard basis of sl(2), and let V be an irreducible representation. By Corollary 13.2, the action of H on V is diagonalizable and the actions of X and Y on V are nilpotent. By Lie’s Theorem (applied to the solvable subalgebra generated by H and X), X and H have a common eigenvector v: Hv = λv and Xv = 0 (since X is nilpotent, its only eigenvalues are zero). Verify by induction that

r r 1 r 1 r r HY v = YHY − v + [H,Y ]Y − v = λ 2(r 1) Y v + 2Y v − − = (λ 2r)Y rv ( ) − ∗ r r 1 r 1 XY v = YXY − v + [X,Y ]Y − v r 1 r 1 = (r 1) λ (r 1) + 1 Y − v + λ 2(r 1) Y − v − − − − − r 1 = r(λ r + 1)Y − v ( ) − † Thus, the span of v,Yv,Y 2v,... is a subrepresentation, so it must be all of V (since V is irreducible). Since Y is nilpotent, there is a minimal n such that Y nv = 0. From ( ), we get that λ = n 1 is a non-negative n †1 − integer. Since v,Yv,...,Y − v have distinct eigenvalues (under H), they are linearly independent. Conclusion: For every non-negative integer n, there is exactly one irreducible representation of sl2 of dimension n + 1, and the H-eigenvalues on that representation are n,n 2,n 4,..., 2 n, n. − − − − Lecture 13 - The root system of a semi-simple Lie algebra 64

Remark 13.11. As a consequence, we have that in a general root decomposition, g = h α ∆ gα, each root space is one dimensional. Assume that ⊕ ∈ dim g α > 1. Consider an sl(2) in g, generated by Xα,Yα,Hα = [Xα,Yα] − L h i where Yα g α and Xα gα. Then there is some Z g α such that ∈ − ∈ ∈ − [Xα, Z] = 0 (since hα is one-dimensional). Hence, Z is a highest vector with respect to the adjoint action of this sl(2). But we have that ad (Z)= 2Z, Hα − and the eigenvalue of a highest vector must be positive! This shows that the choice of Xα and Yα is really unique. Deﬁnition 13.12. Thinking of g as a representation of sl = X ,Y ,H , 2 h α α αi the irreducible subrepresentation containing gβ is called the α-string through β.

Let ∆ denote the set of roots. Then ∆ is a ﬁnite subset of h∗ with the following properties:

1. ∆ spans h∗. 2. If α, β ∆, then β(H ) Z, and β β(H ) α ∆. ∈ α ∈ − α ∈ 3. If α, cα ∆, then c = 1. ∈ ± ◮ Exercise 13.3. Prove these properties. Lecture 14 - More on Root Systems 65

Lecture 14 - More on Root Systems

Assume g is semisimple. Last time, we started with a regular element h g ∈ ss and constructed the decomposition g = h g , where ∆ h is the set ⊕ α ∆ α ⊆ ∗ of roots. We proved that each g is one dimensional∈ (we do not call 0 a root). α L For each root, we associated an sl(2) subalgebra. Given Xα gα,Yα g α, ∈ ∈ − we set Hα = [Xα,Yα], and normalized so that α(Hα) = 2. Furthermore, we showed that

1. ∆ spans h∗, 2. α(H ) Z, with α α(H )β ∆ for all α, β ∆, and β ∈ − β ∈ ∈ 3. if α, kα ∆, then k = 1. ∈ ± How unique is this decomposition? We started with some choice of a regular semisimple element. Maybe a diﬀerent one would have produced a diﬀerent Cartan subalgebra.

Theorem 14.1. Let h and h′ be two Cartan subalgebras of a semisimple Lie algebra g (over an algebraically closed ﬁeld of characteristic zero). Then there is some φ Ad G such that φ(h) = h . Here, G is the Lie group ∈ ′ associated to g.

Proof. Consider the map

Φ : hreg g g g × α1 ×···× αN → (h, x ,...,x ) exp(ad ) exp(ad )h. 1 N 7→ x1 · · · xN Note that ad h is linear in both x and h, and each ad is nilpotent, so xi i gαi the power series for exp is ﬁnite. It follows that Φ is a polynomial function. Since d exp(ad )h = α (h)x g , the diﬀerential if Φ at (h, 0,..., 0) dt txi t=0 i i ∈ αi is

Idh 0 α (h)  1 0  DΦ (h,0,...,0) = . | 0 ..    0 α (h)   N    with respect to the decomposition g = h g g . DΦ is ⊕ α1 ⊕···⊕ αN |(h,0,...,0) non-degenerate because h hreg implies that α (h) = 0. So im Φ contains ∈ i 6 a Zariski open set. Let Φ′ be the analogous map for h′. Since Zariski open sets are dense, we have that im Φ im Φ = 0 . So there is some ∩ ′ 6 { } ψ Ad G, ψ′ Ad G, and h h, h′ h′ such that ψ(h) = ψ′(h′). Thus, we ∈ ∈ 1 ∈ ∈ have that h = ψ− ψ′(h). Lecture 14 - More on Root Systems 66

Abstract Root systems We’d like to forget that any of this came from a Lie algebra. Let’s just study an abstract set of vectors in h∗ satisfying some properties. We know that B is non-degenerate on h, so there is an induced isomorphism s : h h . By → ∗ deﬁnition, s(h), h = B(h, h ). h ′i ′ Let’s calculate

sH ,H = B(H ,H )= B(H ,H ) (B symmetric) h β αi β α α β = B(Hα, [Xβ,Yβ]) = B([Hα, Xβ],Yβ) (B invariant)

= B(Xβ,Yβ)β(Hα) 1 = B([H , X ],Y )β(H ) (2X = [H , X ]) 2 β β β α β β β 1 = B(H ,H )β(H ) (B invariant) 2 β β α

B(Hβ ,Hβ) Thus, we have that s(Hβ)= 2 β. Also, compute

1 (β, β) := β,s− β h i 2H = β β B(H ,H ) β β 4 = (β(Hβ) = 2) B(Hβ,Hβ)

2β ˇ So s(Hβ)= (β,β) . This is sometimes called the coroot of β, and denoted β. Thus, we may rewrite fact 2 from last time as:

2(α, β) 2(α, β) For α, β ∆, Z, and α β ∆. (2 ) ∈ (β, β) ∈ − (β, β) ∈ ′

Now you can define r : h h given by r (x) = x 2(x,β) β. This is β ∗ → ∗ β − (β,β) the reflection through the hyperplane orthogonal to β in h∗. The group generated by the r for β ∆ is a Coxeter group. If we want to study these β ∈ things, we’d better classify root systems. We want to be working in Euclidean space, but we are now in h∗. Let hr be the real span of the Hα’s. We claim that B is positive definite on hr. To see this, note that Xα,Yα,Hα make a little sl(2) in g, and that g is therefore a representation of sl(2) via the adjoint actions adXα , adYα , adHα . But we know that in any representation of sl(2), the eigenvalues of Hα must be integers. Thus, ad ad has only positive eigenvalues, so B(H ,H )= Hα ◦ Hα α α tr(ad ad ) > 0. Hα ◦ Hα Thus, we may think of our root systems in Euclidean space, where the def inner product on h is given by (µ,ν) = B(s 1(µ),s 1(ν)) = µ,s 1ν . ∗ − − h − i Lecture 14 - More on Root Systems 67

Definition 14.2. An abstract reduced root system is a finite set ∆ Rn r ⊆ 0 which satisfies { } (RS1) ∆ spans Rn,

(RS2) for α, β ∆, 2(α,β) Z, with r (∆) = ∆ ∈ (β,β) ∈ β (i.e. α, β ∆ r (α) ∆, with α r (α) Zβ ), and ∈ ⇒ β ∈ − β ∈ (RS3) α, kα ∆ k = 1 (this is the “reduced” part). ∈ ⇒ ± The number n is called the rank of ∆.

Notice that given root systems ∆ Rn, ∆ Rm, we have that 1 ⊂ 2 ⊂ ∆ ∆ Rn Rm is a root system. 1 2 ⊂ ⊕ Deﬁnition` 14.3. A root system is irreducible if it cannot be decomposed into the union of two systems of smaller rank.

◮ Exercise 14.1. Let g be a semisimple Lie algebra and let ∆ be its root system. Show that ∆ is irreducible if and only if g is simple.

2(α,β) 2(α,β) Now we will classify all systems of rank 2. Observe that (α,α) (β,β) = 4 cos2 θ, where θ is the angle between α and β. This thing must be an integer. Thus, there are not many choices for θ:

cos θ 0 1 1 √3 2 √2 2 π π±2π ±π 3π ±π 5π θ 2 3 , 3 4 , 4 6 , 6 Choose two vectors with minimal angle between them. If the minimal angle is π/2, then the system is reducible.

β O o / α Notice that α and β can be scaled independently. If the minimal angle is smaller than π/2, then r (α) = α, so the diﬀerence β 6 α r (α) is a non-zero integer multiple of β (in fact, a positive multiple of − β β since θ < π/2). If we assume α β (we can always switch them), k k ≤ k k we get that α r (α) < 2 α 2 β , so α r (α) = β. Thus, once we k − β k k k ≤ k k − β set the direction of β, its length is determined (relative to α). Then we can obtain the remaining elements of the root system from the condition that ∆ is invariant under rα and rβ, observing that no additional vectors can be Lecture 14 - More on Root Systems 68 added without violating RS2, RS3, or the prescribed minimal angle. Thus, all the irreducible rank two root systems are

A2,θ = π/3 B2,θ = π/4 G2,θO = π/6 1X F β ?_ O ? β 11 ?? f X F 8 1 ?? MMM 11 qq β 11 ?? MM 11 qq 1 ?? MM1 qq o 1 / α o ? / α o MqMq1 Mq / α 1 ?? qq 11 MM 11 ?? qq 1 MM 1 ?? qx qq Ö 1 MM& 11 ?? Ö rβ(α) rβ(α) rβ(α)

The Weyl group Given a root system ∆ = α ,...,α , we call the group generated by the { 1 N } rαi s the Weyl group, denoted W . Remark 14.4. Each r W can be constructed in terms of the group G of α ∈ g. If you have Xα,Yα,Hα generating an sl(2), then you know that there is 0 1 a subgroup SL(2) of G . There is a special element corresponding to 1 0 in there. It is given by s = exp(X ) exp( Y ) exp(X ). We have− that α α − α α β, Ad (h) = r β, h . So this reﬂection is coming from the embedding h sα i h α i of SL(2) in G. ◮ Exercise 14.2. Let g be a semisimple Lie algebra, and let h be a Cartan subalgebra. For each root α deﬁne

s = exp(ad ) exp( ad ) exp(ad ). α Xα − Yα Xα

Prove that sα(h)= h and

λ, s (h) = r (λ), h h α i h α i for any h h and λ h , where r is the reflection in α . ∈ ∈ ∗ α ⊥ Example 14.5. The root system An is the root system of sln+1. We pick n+1 an orthonormal basis ε1,...,εn+1 of R , the the root system is the set of all the differences: ∆ = ε ε i = j . We have that { i − j| 6 } ε k = i, j k 6 rε ε (εk)= εj k = i i− j   εi k = j is a transposition, so we have that W Sn+1 ≃ Now back to classification of abstract root systems. Draw a hyperplane in general position (so that it doesn’t contain any + + roots). This divides ∆ into two parts, ∆ = ∆ ∆−. The roots in ∆ ` Lecture 14 - More on Root Systems 69

are called positive roots, and the elements of ∆− are called negative roots. We say that α ∆+ is simple if it cannot be written as the sum of other ∈ positive roots. Let Π be the set of simple roots, sometimes called a base. It has the properties

1. Any α ∆+ is a sum of simple roots (perhaps with repitition): α = ∈ β Π mββ where mβ Z 0. ∈ ∈ ≥ 2.P If α, β Π, then (α, β) 0. ∈ ≤ This follows from the fact that if (α, β) > 0, then α β and β α are − − again roots (as we showed when we classiﬁed rank 2 root systems), and one of them is positive, say α β. Then α = β +(α β), contradicting − − simplicity of α.

3. Π is a linearly independent set. If they were linearly dependent, the relation a α = 0 must αi Π i i have some negative coeﬃcients (because all of Π is∈ in one half space), P so we can always write

0 = a α + + a α = a α + + a α 6 1 1 · · · r r r+1 r+1 · · · n n with all the a 0. Taking the inner product with the left hand side, i ≥ we get

a α + + a α 2 k 1 1 · · · r rk = (a α + + a α , a α + + a α ) 0 1 1 · · · r r r+1 r+1 · · · n n ≤ by 2, which is absurd.

Remark 14.6. Notice that the hyperplane is t⊥ for some t, and the positive roots are the α ∆ for which (t, α) > 0. This gives an order on the roots. ∈ You can inductively prove 1 using this order. Remark 14.7. Notice that when you talk about two roots, they always generate one of the rank 2 systems, and we know what all the rank 2 systems are.

Claim. Suppose we have chosen ∆+ Π. Then for α Π, we have that ⊃ ∈ r (∆+)=∆+ α r α . α ∪ {− } { } Proof. For β = α simple, we have r (β)= β + kα for some non-negative k; 6 α this must be on the positive side of the hyperplane, so it is a positive root. Now assume you have a positive root of the form γ = mα + m α , αi=α i i with m,m 0. Then we have that r (γ) = mα + m α 6 ∆. If i ≥ α − Pi′ i ∈ any of the m were non-zero, we know that some m is positive (by the i i′ P Lecture 14 - More on Root Systems 70 base case above), but every root can be (uniquely) written as either a non- negative or a non-positive combination of the simple roots. Since rα(γ) has some positive coeﬃcient, it must be a non-negative combination, so it is a positive root.

+ + Suppose we have two different sets of positive roots ∆ and ∆1 , then + + one can get ∆1 from ∆ by a chain of simple reflections. You have to be a little careful because the set of simple roots changes, so you’re picking your reflections from different sets. Fix a base Π = α ,...,α , and call { 1 n} ri := rαi simple reflections. Proposition 14.8. Simple reflections act transitively on the set of sets of positive roots.

+ + Proof. Start with ∆ , and apply ri to get a new ∆1 , then from the new + + + base apply some rα to get ∆2 , so you have ∆2 = rαri∆ . The important 1 thing is that for any element of the Weyl group, we have wrαw− = rwα. So if α = ri(αj), we have that rαri = rirj. Induct. Corollary 14.9. W is generated by simple reﬂections.

Proof. Any root is a simple root for some choice of positive roots. To show this, draw your hyperplane really close to your vector. Then we know that α is obtained from our initial set Π by simple reﬂections. We get that if α = r r (α ), then r = (r r )r (r r ) 1. i1 · · · ik j α i1 · · · ik j i1 · · · ik − Let’s see how it works for A . Choose a vector t = t ε + + t ε n 1 1 · · · n+1 n+1 in such a way that t > >t . Then we have that (ε ε ,t)= t t , 1 · · · n+1 i − j i − j so the positive roots are the ε ε for which i < j. Then the simple roots i − j are Π = ε ε ,...,ε ε { 1 − 2 n − n+1} Lecture 15 - Dynkin diagrams, Classiﬁcation of root systems 71

Lecture 15 - Dynkin diagrams, Classiﬁcation of root systems

Last time, we talked about root systems ∆ Rn. We constructed the Weyl ⊂ group W , the ﬁnite group generated by reﬂections. We considered Π ∆, ⊂ the simple roots. Π forms a basis for Rn, and showed that every root is a non-negative (or non-positive) linear combination of simple roots. 2(α,β) We showed that if you have α, β Π, then nαβ = Z 0. There ∈ (β,β) ∈ ≤ are not many options for this value. It can only be 0, 1, 2, or 3. Also, 2 − − − we have that nαβnβα = 4 cos θαβ. If nαβ = 0, the two are orthogonal. If n = 1, then the angle must be 2π/3 and the two are the same length. αβ − If n = 2, the angle must be 3π/4 and β = √2 α . If n = 3, the αβ − || || || || αβ − angle is 5π/6 and β = √3 α . Thus we get: || || || ||

nβα nαβ relationship Dynkin picture β O / α α 0 0 ØÙ ØÙ β β X 11 1 1 1 / α α β − − ØÙ ØÙ β ?_ ? ?? 2 1 ? / α α o β − − ØÙ ØÙ f β MMM 3 1 MM / α α o β − − ØÙ ØÙ We will encode this information in a Dynkin diagram. The nodes of the diagram are simple roots. We join two nodes by nαβnβα lines. The arrows go from the longer root to the shorter root in the case of two or three lines. (As always, the alligator eats the big one.) There is also something called a Coxeter diagram, which is not directed. You look at e = α , i = 1, ..., n. Then the number of lines between two i (α,α)1/2 2 vertices is n n = 4 (α,β) = 4(e , e )2. αβ βα · (α,α)(β,β) i j Example 15.1. A has n simple roots, given by ε ε . So the graphs n i − i+1 are

... Dynkin ØÙ ØÙ ØÙ ØÙ Coxeter ... • • • • Properties of the Coxeter graph:

(CX1) A subgraph of a Coxeter graph is a Coxeter graph. This is trivial. Lecture 15 - Dynkin diagrams, Classiﬁcation of root systems 72

(CX2) A Coxeter graph is acyclic.

Proof. Let e1,...,ek be a cycle in the Coxeter graph. Then look at

e , e = k + 2(e , e ) 0 i i i j ≤ i

(CX3) The degree of each vertex is less than or equal to 3, where we count double and triple edges as adding two and three to the degree, respectively.

Proof. Let e0 have e1,...,ek adjacent to it. Since there are no cycles, e1,...,ek are orthogonal to each other. So we can compute

e (e , e )e , e (e , e )e > 0 0 − 0 i i 0 − 0 i i X 1 X (e , e )2 > 0 − 0 i 2 X but (e0, ei) is one fourth of the number of edges connecting e0 and ei. So k cannot be bigger than 3.

(CX4) Suppose a Coxeter graph has a subgraph of type An, and only the endpoints of this subgraph have additional edges (say Γ1 at one end and Γ2 at the other end). Then we can forget about the stuﬀ in the middle and just fuse Γ1 with Γ2, and we will still get a Coxeter graph. WW g WW g Γ WWgW ... gWWggg Γ Γ WWgWgWWggg Γ 8?9>:=;<1gggg • • WWW8?9>:=;<2 −→ 8?9>:=;<1gggg • WWW8?9>:=;<2 To see this, let e ,...,e be the vertices in the A . Let e = e + +e . 1 k k 0 1 · · · k Then we can compute that (e0, e0) = 1, with (e0, es) = (e1, es) and (e0, et) = (ek, et). Thus, a Coxeter graph must be one of the following forms. There can only be one fork (two could be glued to give valence 4), and only one double edge, and if there is a triple edge, nothing else can be connected to it. So the only possible Dynkin diagraphs so far are of the form

\\\\\ \\ . . ØÙ ØÙ \. . .b\b\b\ ... bbbbb bbb ØÙ ØÙ ØÙ ØÙ ØÙ ...... ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ

ØÙ ØÙ Lecture 15 - Dynkin diagrams, Classiﬁcation of root systems 73

Now we use the property that a subgraph of a Dynkin graph is a Dynkin graph. Now we will calculate that some things are forbidden. We label the vertex corresponding to αi with a number mi, and check that

miαi, miαi = 0. X X 1 2 3 / 4 2 1 2 3 o 2 1 ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ 1 2 3 2 1 1 2 3 4 3 2 1 ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ

ØÙ 2 ØÙ 2

ØÙ 1 2 4 6 5 4 3 2 1 ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ

ØÙ 3 Thus we have narrowed our list of possible Dynkin diagrams to a short list. Connected Dynkin diagrams: n is the total number of vertices.

... ØÙ ØÙ ØÙ ØÙ (An) ... / ØÙ ØÙ ØÙ ØÙ (Bn) ... o ØÙ ØÙ ØÙ ØÙ (Cn) n nnn ØÙ ... PnP (Dn) ØÙ ØÙ ØÙ PPP ØÙ Those are the “classical diagrams”, and the others are

/ ØÙ ØÙ (G2)

/ ØÙ ØÙ ØÙ ØÙ (F4)

ØÙ ØÙ ØÙ ØÙ ØÙ (E6)

ØÙ

ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ (E7)

ØÙ Lecture 15 - Dynkin diagrams, Classiﬁcation of root systems 74

ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ (E8)

ØÙ Each of these is indeed the diagram of some root system, which is what we will show now. We’ve already constructed An. So let’s jump to the next ones. Root systems: we start with Dn. Let ε1,...,εn be an orthonormal basis for Rn. Then let the roots be

∆= (ε ε ) i < j n . {± i ± j | ≤ } We choose the simple roots to be

ε1 ε2 ... εn 2 εn 1 εn 1 εn − − − − − −

εn 1 + εn − To get the root system for B , take D and add ε i n , in which n n {± i| ≤ } case the simple roots are

ε1 ε2 ... εn 1 εn / εn − − − To get C , take D and add 2ε i n , then the simple roots are n n {± i| ≤ }

ε1 ε2 ... εn 1 εn o 2εn − − − 2α Recall that we can deﬁne corootsα ˇ = (α,α) . This will reverse the arrows in the Dynkin diagram. The dual root system is usually the same as the original, but is sometimes diﬀerent. For example, Cn and Bn are dual. Those are all the classical diagrams. Now let’s do the others. We’ve already done G2. Now we will do F4 and E8, then we know that the rest are subsets. F4 comes from some special properties of a cube in 4-space. Let ε1, ε2, 4 ε3, ε4 be an orthonormal basis for R . Then let the roots be (ε ε ε ε ) (ε ε ), ε , ± 1 ± 2 ± 3 ± 4 ± i ± j ± i 2 These are orthogonal to some faces of the 4-cube. The simple roots are

ε1 ε2 ε3 ε4 ε ε ε ε / ε4 − − − 2 − 3 3 − 4 2 Lecture 15 - Dynkin diagrams, Classiﬁcation of root systems 75

There are 48 roots total. Remember that the dimension of the lie algebra is the number of roots plus the dimension of the cartan (the rank of g, which is 4 here), so the dimension is 52 in this case. 9 Construction of E8. Look at R with our usual orthonormal basis. The trick is that we are going to project on to the plane orthogonal to ε + +ε 1 · · · 9 (so just throw in the relation that that sum is 0, so that we work in the hyperplane). The roots are

ε ε i = j (ε + ε + ε ) i = j = k { i − j| 6 }∪{± i j k | 6 6 } The total number of roots is ∆ = 9 8 + 2 9 = 240. So the dimension of | | · 3 the algebra is 248! The simple roots are ε ε ε ε ε ε ε ε ε ε ε ε ε ε 1 − 2 2 − 3 3 − 4 4 − 5 5 − 6 6 − 7 7 − 8

ε6 + ε7 + ε8

These are all the indecomposable root systems, the rest are just direct sums.

Theorem 15.2 (Serre). Say k algebraically closed, characteristic 0. For each indecomposable root system, there exists (unique up to isomorphism) a simple Lie algebra with the given root system.

There are several ways to do this. You could construct all of them explicitly. The other way is to use Serre’s relations, which is what we will do later. Let’s construct the Lie algebras for An,Bn,Cn, Dn. We know that An is sl(n + 1). Let’s consider so(2n), the linear maps of k2n preserving the form. 2n 0 1n We can choose a basis for k so that the matrix of the form is I = 1n 0 . Let X so(2n), then we have that XtI + IX = 0. So we can write ∈ A B X = C At − Where Bt = B,Ct = C. What should the Cartan subalgebra be? It − − should be a maximal abelian subalgebra. Notice that if we take A = diag(t1,...,tn) to be diagonal and B = C = 0, then we get an abelian subalgebra h. What are the root spaces? We have that E E has ij − j+n,i+n eigenvalue t t . We have that E E has eigenvalue t + t , and i − j i,j+n − j,i+n i j the symmetric term with eigenvalue t t . So we get the root system D . − i − j n Now let’s look at sp(2n), the linear operators which preserve a non- 0 1n degenerate skew-symmetric form. In some basis, the form is J = 1n 0 , with XtJ + JX = 0. We get the same conditions, except that B = −Bt,C = Lecture 15 - Dynkin diagrams, Classiﬁcation of root systems 76

Ct. The root system is the same as for so(2n) plus roots corresponding to Ei,i+n, with eigenvalue 2ti. We still have to look at so(2n + 1). We’ll choose a basis so that our form is a little diﬀerent. We have a bilinear symmetric form, and we will write it as 1 0 B = , 0 I 0 1n where I = 1n 0 . So we have that so(2n) so(2n + 1) and ⊆ 0 x x y y 1 · · · n 1 · · · n y1  −.  . A B   X =  y   − n   x   1   −. t   . C A     y   − n −    Isomorphisms of small dimension

Let’s say that we believe the theorem. Then you can see that B2 and C2 are the same. Also, D2 is just A1 A1, and D3 is A3:

B2 = C2 D2 `= A1 A1 D3 = A3 ØÙ ØÙ MMM / ` MM ØÙ ØÙ qqØÙ qqq ØÙ ØÙ Thus, we have that sp(4) so(5), so(4) sl(2) sl(2), so(6) sl(4) as ≃ ≃ ⊕ ≃ Lie algebras. You can see some of these directly: let’s construct a map of groups SL(2) SL(2) SO(4), whose kernel is discrete. Let W be the space of × → 2 2 matrices, then the left group acts on W : (X,Y )w = XwY 1, preserving × − the determinant of w = a b . The quadratic form ad bc is preserved. The c d − Lie group preserving such a form is SO(4), so we have a map. You can check that the kernel is the matrices of the form (I,I) and ( I, I), and it − − is surjective (count dimensions). Thus, you get an isomorphism on the level of Lie algebras. Let’s also do so(6) sl(4). The approach is the same. Let V be the ≃ standard 4-dimensional representation of sl(4). Let W = Λ2V , which is a 6-dimensional representation of sl(4). Note that you have a pairing

Λ2V Λ2V Λ4V k × → ≃ Lecture 15 - Dynkin diagrams, Classiﬁcation of root systems 77 where the last isomorphism is as representations of SL(4) (because determinant is 1 in SL(4)). Thus, W = Λ2V has some SL(4)-invariant symmetric bilinear form which is preserved, so you get a map SL(4) SO(6). It is → not hard to check that the kernel is I, and the dimensions match, so it is ± surjective. Thus, you get an isomorphism of Lie algebras. Next time, we’ll do this theorem and we’ll talk about the Exceptional algebras. Lecture 16 - Serre’s Theorem 78

Lecture 16 - Serre’s Theorem

Today we’re going to prove Serre’s theorem, and then work with some constructions of the exceptional algebras. Start with a semi-simple Lie algebra g (over k = k¯), with Cartan subalgebra h g. Then we have the root system ∆, with a ﬁxed set of simple ⊂ roots Π = α ,...,α . We have X := X ,Y ,H associated to each simple { 1 n} i αi i i root. 2(αi,αj ) The Cartan matrix is aij = αj(Hi)= . We know that αij Z 0 (αi,αi) ∈ ≤ for i = j and a = 2. 6 ii Then we claim that the following relations are satisﬁed:

[H ,H ]=0 [H , X ]= a X [H ,Y ]= a Y [X ,Y ]= δ H (1) i j i j ij j i j − ij j i j ij i

+ 1 aij 1 aij (θij ) (adXi ) − Xj = 0 (adYi ) − Yj = 0 (θij−) (2) the second group are called Serre’s relations (though Serre called them Weyl relations). We verify that these relations hold in our Lie algebra. The tricky one from (1) is that [X ,Y ] = 0 when i = j. You know that i j 6 [gα , g α ] gα α , but αi αj ∆! So gα α = 0. i − j ⊆ i− j − 6∈ i− j Example 16.1. g = sl(n + 1). We have Xi = Ei,i+1,Yi = Ei+1,i,Hi = [X ,Y ]= E E . i i ii − i+1,i+1 Let’s check that the second group of relations hold. We want to show 1 aij 1 ai,i+1 that (adXi ) − Xj = 0. What is aij? We know that (adXi ) − Xi+1 = 0 by a calculation ... just do it.

To see why these relations hold in general, consider θij−. We have that adXi (Yj) = 0, so Yj is a highest vector with respect to this sl(2) action. We also have that adHi (Yj) = aijYj, so the highest weight is aij. We know − − 1 aij what representations of sl(2) look like, so we know that (adYi ) − Yj = 0. + Thus, θij− holds. To show θij, you can switch Xi and Yi, and negate Hi. We don’t know that these are all the relations yet. We also do not yet know that for every root system there is an algebra. But we know that if such an algebra exists, then it satisﬁes these relations.

Theorem 16.2. Let g be the Lie algebra with generators H , X ,Y for 1 i i i ≤ i n, with a Cartan matrix a = (α , αˇ ) from some root system and the ≤ ij i j relations given, then g is a ﬁnite dimensional semi-simple Lie algebra with the Cartan subalgebra spanned by H1,...,Hn, and with the given root system ∆.

Note that this shows existence. But since any such Lie algebra satisﬁes these relations, we have uniqueness (you cannot add more relations without making the root system smaller). Lecture 16 - Serre’s Theorem 79

Remark 16.3. When we talk about a Lie algebra with some generators and relations, how do you think about it? What is a free Lie algebra? It is tricky to think about the Jacobi identity, not like in an associative algebra. But a Lie algebra always has an associative algebra Ug. What is Ug going to be? The free associative algebra with the same generators . If you have generators g1,...,gn, look at the tensor algebra T (g1,...,gn), and realize g inside as a Lie subalgebra generated by g1,...,gn. Proof. Step 1: Take the free Lie algebra and quotient by the relations in group (1). Call the result g˜. Then the claim is that

+ g˜ = n˜− h n˜ ⊕ ⊕ as vector spaces, where h is the abelian Lie algebra generated by H1,...,Hn, and n˜± are free Lie algebras generated by the Xi (resp. Yi). They should be free because there are no relations among the Xi’s, but it is not so easy. There is a standard trick for doing such things. + Ug˜ = U(n˜−)U(h)U(n˜ ) (easy to show from relations (1)). In fact, we + want U(g˜)= U(n˜−) U(h) U(n˜ ). We construct a representation of g˜ in ⊗ ⊗ T (n˜−) S(h) T (n˜+) T (n˜ ) S(h). − ⊗ | {z } | {z } | {z } We say Y (1 1) = Y 1, h(1 1) = 1 h, and X (1 1) = 0. Then we i ⊗ i ⊗ ⊗ ⊗ i ⊗ deﬁne things inductively:

Y (Y a b)= Y Y a b i j ⊗ def i j ⊗ H (Y a b)= Y (H (a b)) a Y (a b) i j ⊗ j i ⊗ − ij j ⊗ X (Y a b)= Y (X (a b)) + δ H (a b) i j ⊗ j i ⊗ ij i ⊗ Your job is to check that this gives a representation (note that 1 1 generates ⊗ the whole representation). So we get an isomorphism of U(n˜−) and T (n˜−). We deﬁne U(n˜− + h) և T (n˜ ) S(h) given by x x(1 1). We do the same thing for n˜+. − ⊗ 7→ ⊗ + 0 U(n˜ ) U(g˜) T (n˜−) S(h) 0 → → → ⊗ → + 0 U(n˜−) U(g˜) T (n˜ ) S(h) 0 → → → ⊗ → Hence U(g˜) T (n˜ ) S(h) T (n˜+). So g˜ = n˜ h n˜+. ≃ − ⊗ ⊗ − ⊕ ⊕ Step 2: We now throw in relations from group (2). We have that θ± ij ∈ n˜±. We have to verify that + [Yk,θij] = 0 which sort of follows from relations from group (1) (check for yourself). We have that + + [Hk,θij]= ckijθij Lecture 16 - Serre’s Theorem 80

So the ideal j+ n˜+ generated by θ+ is an ideal in the whole algebra g˜. ⊂ { ij} Similarly with j n˜ generated by θ− . − ⊂ − { ij} Now we want g = n h n+, where n = n˜ /j . The main point is − ⊕ ⊕ ± ± ± that throwing in Serre’s relations doesn’t disturb the Cartan subalgebra. From relations in (1), h acts semi-simply on g, so we can write g = h α h∗ gα, where the gα are eigenspaces ⊕ ∈ L g = x g [h, x]= α(h)x . ( ) α { ∈ | } ∗ We can look at R = α h∗ g = 0 h∗ { ∈ | α 6 }⊆ We know that α ,..., α R. We have yet to show that we get the ± 1 ± n ∈ right root system and the g is semisimple and finite dimensional. If α R, then it must be either a non-negative or non-positive com- ∈ bination of the α (because of the decomposition ( ). We still have that i ∗ [g , g ] g . So we know that R is R+ R (property (a)). We also α β ⊆ α+β − have that R Z = α , because [X , X ] = 0 (property (b)). ∩ αi {± i} i i ` Step 3: Now we’ll construct a Weyl group. The main idea is that the relations (1) and (2) imply that adXi , adYi are locally nilpotent operators on g (A is locally nilpotent on V is for any vector v V , there is some n(v) ∈ such that An(v)v = 0). You can check it on generators, and then by the Jacobi identity, you get it for any vector. Let s = exp(ad ) exp( ad ) exp(ad ). Then s Aut g i Xi − Yi Xi i ∈ ◮ Exercise 16.1. Check that s (h) h i ⊆ We have that λ, s (h) = r (λ), h ( ) h i i h i i ∗∗ for any h h and any λ h , where r (λ)= λ λ(H )α . ∈ ∈ ∗ i − i i Let W be the group generated by r1,...,rn. We have that w(R) = R because of the identity ( ) ∗∗ On the other hand, we know that W is the Weyl group of the root system ∆. Hence we have an embedding ∆ R. Moreover, we have that dim g = 1 ⊆ α for any α ∆ because if we write w = r r and s = s s , and if ∈ i1 · · · ik i1 · · · ik α = w(αi), then gα = s(gαi ). Step 4: Prove that ∆ = R. Let λ R r ∆. Then λ is not proportional ∈ to any α ∆. One can find some h in the real span of the H such that ∈ i λ, h = 0 and α, h = 0 for all α ∆. This divides ∆ = ∆+ ∆ . Choose h i h i 6 ∈ − Π ∆+ a new basis of simple roots β ,...,β . We showed that W acts ⊆ { 1 n} ` transitively on the sets of simple roots, so we can find some w W such 1 ∈ that w(αi)= βi. Then look at w− (λ). 1 Since λ = miβi for some mi > 0, mj < 0, and w− (λ) = miαi for m > 0,m < 0, w 1(λ) is not a linear combination of only positive roots i j P − P Lecture 16 - Serre’s Theorem 81

+ or of only negative roots. But R = R R−, so this is a contradiction and hence ∆ = R. ` Step 5: Check that g is semi-simple. We have to show that there are no nontrivial abelian ideals in g. We already know that g = h g ⊕ α ∆ α and that g is 1-dimensional. We also know that the Serre relations∈ hold. α L Notice that for any ideal a, a adh-invariant implies that a = h′ α S gα. If ⊕ ∈ g a, then X a, so [Y , X ]= H a, so [H , X ] = 2X a. Thus, α ⊆ α ∈ α α α ∈ α α α ∈ we have the whole sl(2) in a, so it cannot be abelian. Thus we have that a = h . Take h h non-zero. But α ,...,α span h , so α are such that ′ ∈ ′ 1 n ∗ i α (h) = 0, then [h, X ]= α (h)X a. Contradiction. i 6 i i i ∈ So we know in each case what the Lie algebra is and how to work with them. In the non-exceptional cases, we have nice geometric descriptions of them. Next time, we will give some nice explanations of the exceptional Lie algebras. Next time will be the last lecture on the structure theory, and then we will do representation theory. Lecture 17 - Constructions of Exceptional simple Lie Algebras 82

Lecture 17 - Constructions of Exceptional simple Lie Algebras

We’ll begin with the construction of G2. Consider the picture

v2 1 • 11 11 11 w3 1 w1 M u 11 MMM qq 1 • MMqqq • 1 qqq• MMM 11 qqq MMM 1 qq MM11 v1 q q M v3 • w•2 • This is the projective plane over F2. Consider the standard 7-dimensional representation of gl(7), call it E. Take a basis given by points in the projective plane above. ◮ Exercise 17.1. Prove that Λ3E has an open gl(7)-orbit, by looking at the particular vector ω = v v v + w w w + u v w + u v w + u v w 1 ∧ 2 ∧ 3 1 ∧ 2 ∧ 3 ∧ 1 ∧ 1 ∧ 2 ∧ 2 ∧ 3 ∧ 3 and checking that gl(7)ω = Λ3E. (Remember x(v v )= x(v ) v + v 1 ∧ 2 1 ∧ 2 1 ∧ x(v2). ) Claim. g = x gl(7) xω = 0 is a simple Lie algebra with root system G . { ∈ | } 2 Proof. We have to choose a Cartan subalgebra of this g. Consider linear operators which are diagonal in the given basis u, v1, v2, v3, w1, w2, w3, take h = diag(c, a1, a2, a3, b1, b2, b3). We want that hω = 0, which gives some linear equations. It is easy to check that hω = 0 implies that c = 0, a1 + a + a = 0, and b = a . This is a 2-dimensional subalgebra, it is our 2 3 i − i Cartan subalgebra. If you consider the root diagram for G2 and look at only the long roots, you get a copy of A . This should mean that you have an embedding A 2 2 ⊆ G2, so we should look for a copy of sl(3) in our g. We can write E = ku V W , where V = v , v , v and W = w , w , w . Let’s consider the ⊕ ⊕ h 1 2 3i h 1 2 3i subalgebra which not only kills ω, but also kills u. Let g = x g xu = 0 . 0 { ∈ | } For x g we have x(v v v )=0and x(w w w ) = 0, so x preserves ∈ 0 1 ∧ 2 ∧ 3 1 ∧ 2 ∧ 3 V and W . It also must kill the 2-form α = v w + v w + v w , since 1 ∧ 1 2 ∧ 2 3 ∧ 3 0= x(u α)= xu α + u xα = u xα forces xα = 0. This 2-form gives ∧ ∧ ∧ ∧ a pairing, so that V W . We can compute exactly what the pairing is, ≃ ∗ v , w = δ . Therefore the operator x must be of the form h i ji ij 0 0 0 x = 0 A 0   0 0 At −   Lecture 17 - Constructions of Exceptional simple Lie Algebras 83

[[where A has trace zero?]] The total dimension of G2 is 14, which we also know by the exercise is the dimension of g. We saw the Cartan subalgebra has dimension 2, and this g0 piece has dimension 6. So we still need another 6 dimensional piece. For each v V , deﬁne a linear operator X which acts by X (u) = v. ∈ v v Λ2V W is dual to V (since Λ3V = C). Then X acts by X (v )= γ(v v ) ≃γ v v ′ ∧ ′ and X (w) = 2 v, w u. Check that this kills α, and hence is in g. v h i Similarly, you can deﬁne X for each w W by X (u) = w, X (w )= w ∈ w w ′ γ(w w ), X (v) = 2 w, v u. ∧ ′ w h i If you think about a linear operator which takes u v , it must be in 7→ i some root space, this tells you about how it should act on V and W . This is how we constructed Xv and Xw, so we know that Xvi and Xwi are in some root spaces. We can check that their roots are the short roots in the diagram, X F 11 1 Xw3 11 O 11 Xv Mf 1 8 Xv 2 MMM 1 qqq 1 o MM1 qq / q M q11M qqq 1MMM Xw qx q 1 M& Xw 1 11 2 11 X 1 Ö v3 1 and so they span the remaining 6 dimensions of G2. To properly complete this construction, we should check that this is semisimple, but we’re not going to.

We saw here that G2 is isomorphic to the Lie algebra of automorphisms of a generic 3-form in 7-dimensional space. The picture of the projective plane is related to Cayley numbers, an important nonassociative division algebra, of which G2 is the algebra of automorphisms. Let’s analyze what we did with G2, so that we can do a similar thing to construct E8. We discovered certain phenomena, we can write g = g g g . This gives us a Z/3-grading: [g , g ] g . As an 0 ⊕ 1 ⊕ 2 i j ⊆ i+j(mod 3) sl(3) V W sl(3) representation, it has three components: ad, standard, and the dual to |{z} |{z} |{z} the standard. We get that W = V Λ2V . Similarly, V Λ2W . This is ∼ ∗ ≃ ≃ called Triality. More generally, say we have g0 a semi-simple Lie algebra, and V,W Lecture 17 - Constructions of Exceptional simple Lie Algebras 84

representations of g0, with intertwining maps

α : Λ2V W → β : Λ2W V → V W ∗. ≃

We also have γ : V W V V g (representations are semi-simple, so ⊗ ≃ ⊗ ∗ → 0 the map g gl(V ) V V splits). We normalize γ in the following way. 0 → ≃ ⊗ ∗ Let B be the Killing form, and normalize γ so that B(γ(v w), X)= w,Xv . ⊗ h i Make a Lie algebra g = g V W by defining [X, v]= Xv and [X, w]= Xw 0 ⊕ ⊕ for X g , v V, w W . We also need to define [ , ] on V ,W and between ∈ 0 ∈ ∈ V and W . These are actually forced, up to coefficients:

[v , v ]= aα(v v ) 1 2 1 ∧ 2 [w , w ]= bβ(w w ) 1 2 1 ∧ 2 [v, w]= cγ(v w). ⊗ There are some conditions on the coeﬃcients a, b, c imposed by the Jacobi identity; [x, [y, z]] = [[x,y], z] + [y, [x, z]]. Suppose x g , with y, z g ∈ 0 ∈ i for i = 0, 1, 2, then there is nothing to check, these identities come for free because α, β, γ are g0-invariant maps. There are only a few more cases to check, and only one of them gives you a condition. Look at

[v , [v , v ]] = caγ(v α(v v )) (RHS) 0 1 2 0 ⊗ 1 ∧ 2 and it must be equal to

[[v0, v1], v2]+[v1, [v0, v2]] = acγ(v α(v v )) + acγ(v α(v v )) (LHS) − 2 ⊗ 0 ∧ 1 1 ⊗ 0 ∧ 1 This doesn’t give a condition on ac, but we need to check that it is satisﬁed. It suﬃces to check that B(RHS,X) = B(LHS,X) for any X g . This ∈ 0 gives us the following condition:

α(v v ), Xv = α(v v ), Xv α(v v ), Xv h 1 ∧ 2 0i h 0 ∧ 2 1i − h 0 ∧ 1 2i

The fact that α is an intertwining map for g0 gives us the identity:

α(v v ), Xv = α(Xv v ), v α(v Xv ), v h 1 ∧ 2 0i h 1 ∧ 2 0i − h 1 ∧ 2 0i and we also have that

α(v v ), v = α(v v ), v = α(v v ), v h 1 ∧ 2 0i h 0 ∧ 2 1i h 0 ∧ 1 2i Lecture 17 - Constructions of Exceptional simple Lie Algebras 85

With these two identities it is easy to show that the equation (and hence this Jacobi identity) is satisﬁed. We also get the Jacobi identity on [w, [v1, v2]], which is equivalent to: abβ(w α(v v )) = c(γ(v w)v γ(v w)v ) ∧ 1 ∧ 2 1 ⊗ 2 − 2 ⊗ 1 It suﬃces to show for any w W that the pairings of each side with w are ′ ∈ ′ equal,

ab w′, β(w α(v v ) = cB(γ(v w), γ(v w′)) h ∧ 1 ∧ 2 i 1 ⊗ 2 ⊗ cB(γ(v w), γ(v w′)) − 2 ⊗ 1 ⊗ This time we will get a condition on a, b, and c. You can check that any of the other cases of the Jacobi identity give you the same conditions. Now we will use this to construct E . Write g = g V W , where we 8 0 ⊕ ⊕ take g0 = sl(9). Let E be the 9-dimensional representation of g0. Then take 3 3 6 3 6 V = Λ E and W = Λ E∗ Λ E. We have a pairing Λ E Λ E k, so ≃ 2 ⊗ → we have V W ∗. We would like to construct α : Λ W , but this is just ≃ 6 → 2 given by v1 v2 including into Λ E W . Similarly, we get β : Λ W V . ∧ ≃ 9 → You get that the rank of g is 8 (= rank of g0). Notice that dim V = 3 = 84, which is the same as dim W , and dim g = dim sl(9) = 80. Thus, we have 0 that dim g = 84 + 84 + 80 = 248, which is the dimension of E8, and this is indeed E8. Remember that we previously got E7 and E6 from E8. Look at the diagram for E8: ε ε ε ε ε ε ε ε 1 − 2 3 − 4 5 − 6 7 − 8 ØÙ ε ØÙ ε ØÙ ε ØÙ ε ØÙ ε ØÙ ε ØÙ 2 − 3 4 − 5 6 − 7

ε6 + εØÙ 7 + ε8

The extra guy, ε6 + ε7 + ε8, corresponds to the 3-form. When you cut out ε ε , you can ﬁgure out what is left and you get E . Then you can 1 − 2 7 additionally cut out ε2 ε3 and get E6. − / Fianally, we construct F4: ØÙ ØÙ ØÙ ØÙ We know that any simple algebra can be determined by generators and relations, with a Xi,Yi,Hi for each node i. But sometimes our diagram has a symmetry, like switching the horns on a Dn, which induces an automorphism of the Lie algebra given by γ(Xi)= Xi for i

Theorem 17.1. (Aut g)/(Aut0 g) = AutΓ. So the connected component of the identity gives some automorphisms, and the connected components are parameterized by automorphisms of the diagram. Lecture 17 - Constructions of Exceptional simple Lie Algebras 86

D is the diagram for SO(2n). We have that SO(2n) O(2n), and the n ⊂ group of automorphisms of SO(2n) is O(2n). This isn’t true of SO(2n + 1), because the automorphisms given by O(2n + 1) are the same as those from SO(2n + 1). This corresponds the the fact that Dn has a nontrivial automorphism, but Bn doesn’t. Notice that E6 has a symmetry; the involution:

z } ! $ ØÙ ØÙ ØÙ ØÙ ØÙ

ØÙ

Define X1′ = X1 + X5, X2′ = X2 + X4, X3′ = X3, X6′ = X6, and the same with Y ’s, the fixed elements of this automorphism. We have that H1′ = H1 + H5 (you have to check that this works), and similarly for the other H’s. As the set of fixed elements, you get an algebra of rank 4 (which must be our F ). You can check that α (H )= 1, α (H )= 1, α (H )= 4 1′ 2′ − 2′ 1′ − 3′ 1′ 0, α (H ) = 2, α (H ) = 1, so this is indeed F as desired. In fact, any 3′ 2′ − 2′ 3′ − 4 diagram with multiple edges can be obtained as the fixed algebra of some automorphism:

◮ Exercise 17.2. Check that G2 is the ﬁxed algebra of the automorphism of D4: ØÙ Z

× ØÙ ØÙ 11 11 0 1 ØÙ Check that Bn,Cn can be obtained from A2n, A2n+1

z } . . . ! $ ØÙ ØÙ ØÙ ØÙ next time, we’ll do representation theory ... highest weight, etc. Lecture 18 - Representations 87

Lecture 18 - Representations

Representations Let g be a semi-simple lie algebra over an algebraically closed field k of characteristic 0. Consider the root decomposition g = h g , and let ⊕ α α V be a finite-dimensional representation of g. Because all elements of h are L semisimple, and because Jordan decomposition is preserved, the elements of h can be simultaneously diagonalized. That is, we may write V = µ h∗ Vµ, where V = v V hv = µ(h)v h h . We call V a weight space,∈ and µ µ { ∈ | ∀ ∈ } µ L the weights. Define P (V ) = µ h V = 0 . The multiplicity of a weight { ∈ ∗| µ 6 } µ P (V ) is dim V , and is denoted m . ∈ µ µ Example 18.1. You can take V = k (the trivial representation). Then P (V )=0 and m0 = 1. Example 18.2. If V = g and we take the adjoint representation, then we have that P (V )=∆ 0 , with m = 1 for α ∆, and m = the rank of ∪ { } α ∈ 0 g.

We can plot the set of weights in h∗: Example 18.3. Let g = sl(3): ε ε ε ε 2 3 X F 1 3 − 11 − 1 ε3 1 −KS ε 11 ε 2 f 1 8 1 11 ε ε o 1 1M / ε ε 2 1 qqqq M1MMM 1 2 − qqqq 1 MMMM − t| q 11 "* ε1 1 ε2 − 11 − ε3 1 Ö 1 ε ε ε ε 3 − 1 3 − 2 For the adjoint representation, the weights are shown in single lines. If E is the standard 3-dimensional representation, then we get weights ε , ε , ε , shown in dotted lines. { 1 2 3} We can also take the dual representation E∗. In general, the weights of the dual representation are the negatives of the original representation because hφ, v = φ, hv , so we get the double lines. h i −h i Note that the sum of these is a G2. (We noted earlier that G2 has some sl(3)s sitting around). Properties: For V ﬁnite-dimensional (1) µ(H ) Z for any root α and µ P (V ). α ∈ ∈ This is because all weights of any representation of sl(2) are integers, so restricting to our little sl(2), we get this result. Lecture 18 - Representations 88

(2) g V V for α ∆,µ P (V ). α µ ⊆ µ+α ∈ ∈ To check this:

h(xαv)= xαhv + [h, xα]v

= xαµ(h)v + α(h)xαv

= (µ + α)(h)xαv

(3) If µ P (V ) and w W , then w(µ) P (V ) and m = m . ∈ ∈ ∈ µ w(µ) To see this, it is sufficient to check for simple reflections w = ri. Recall that for each reflection, there is a special element of the group G itself (coming from the small SL(2) G). It is exp X exp( Y )exp X s ⊆ i − i i i ∈ G. Then si(h)= h and si(Vµ)= Vriµ.

Example 18.4. If g = sl(3), then we get W = S3 = D3. The orbit of a point can have a couple of diﬀerent forms. If the point is on a mirror, then you get a triangle. For a generic point, you get a hexagon (which is not regular, but still symmetric). It is pretty clear that knowing the weights and multiplicities gives us a lot of information about the representation. However, it is hard to draw pictures in more than three dimensions, so instead you write the following expression: µ ch V = mµe µ P (V ) ∈X You can formally add and multiply these expressions: ch is additive with respect to direct sum and multiplicative with respect to tensor products:

ch(V W )= ch V + ch W ⊕ ch(V W ) = (ch V )(ch W ) ⊗ This is because Vµ Wν (V W )µ+ν. You can also check that the ch of ⊗µ ⊆ ⊗ the dual is mµe− . If you don’t care about multiplication, then you can do inﬁnite-dimensional stuﬀ. P Relations to group characters: if you have G GL(V ), then we have → the characters χV (g) = trV (g). How are these related? Well, you can exponentiate h to get a subgroup H G. To compute χ (exp h) for h h, ⊆ V ∈ you just need to add the exp’s of the eigenvalues of h:

µ(h) χV (exp h)= tr(exp h)= mµe .

Recall that group characters are constantX on conjugacy classes. If you look at diﬀerent Cartan subalgebras, then something ... the character is determined by its values on the Cartan. . .This is a good way to keep track of weights. Lecture 18 - Representations 89

Highest weights Fix a set of simple roots Π = α . A highest weight is a λ P (V ) such { i} ∈ that α + λ P (V ), α Π. i 6∈ ∀ ∈ Say V is irreducible and λ is some highest weight. Let v V . We have ∈ λ that g = n h n+, and that n+v = 0. Since v generates V (because V is − ⊕ ⊕ irreducible), where V = (Ug)v. But by PBW, U(g)= Un Uh Un+, so − ⊕ ⊕ V = Un−v. Thus, all weights of the irreducible representation are given by applying negative roots. There are some relations among the diﬀerent ways of applying roots: [Yα1 ,Yα2 ]= Yα1+α2 . Note that in particular, we get that mλ = 1. o o α1 λ 1 11 11 1 α2 1 α3 11 1 Ö 1 o 1o 1 11 1 11 11 Ö 1 1 The highest weight is unique in an irreducible representation: suppose µ is another one. But all weights lie “below” λ. So we get λ µ and µ λ, ≤ ≤ where a b means that b = a + + n α with n 0. (Note that we ≤ α ∆ α α ≥ are using the irreducible hypothesis∈ to say that all weights are below the P highest weight). If λ is a highest weight of an irreducible representation with respect to some choice of simple roots α ,...,α , then for any w W , w(λ) will be the 1 n ∈ highest weight with respect to w(α1),...,w(αn). Thus, we get that P (V ) is a subset in the convex hull of the orbit of the highest weight under the Weyl group, W λ. If λ is a highest weight of an irreducible V , then we know that λ(H ) i ∈ Z 0. Sometimes this is written (λ, αˇi) = λ(Hi) 0. To see this, note that ≥ ≥ we are restricting to our little sl(2) again. There are a few lattices in h∗. If you look at the lattice generated by the roots, then we call that the root lattice, and it is denoted Q:

Q = Zα Zα . 1 ⊕···⊕ n There is another one, called the weight lattice, and is called P :

P = µ h∗ (µ, αˇ ) Z . { ∈ | i ∈ } Let P + = λ P (λ, αˇ ) 0 . These are called the integral dominant { ∈ | i ≥ } elements. Both P and Q have the same rank, but P is usually bigger (and never smaller, obviously). Example 18.5. For sl(2), the root lattice is the even integers (because [H, X] = 2X) While the weight lattice is the integers (because you can get any integer as a weight). Lecture 18 - Representations 90

Thus, you can look at P/Q, which is the center of the simply connected group corresponding to g. We have said that the highest weight is in P +.

◮ Exercise 18.1. Show that P + is the fundamental domain of the W action on P . (This follows from the fact that W acts simply and transitively on the set of bases). In the case of sl(3), we get:

P + • • • r • • • r • • • r 1X r α2• 1 • r • 1 r • 1•1 r • • r • / • r α1 • r • • r • • • r r • • • • • • • • For every µ P , there is a unique element w W such that wµ P +. ∈ ∈ ∈ Theorem 18.6. There is a bijection between P + and the set of (isomorphism classes of) ﬁnite-dimensional irreducible representations of g.

If you have an irreducible representation V , then to it we associate its highest weight in P +. We already know that the map is well-defined (because the highest weights are unique). For injectivity we must show that two different representations cannot have the same highest weight. For surjectivity, we need to show that every element of P + is the highest weight of some irreducible representations. We are going to use Verma modules because they are nice. A lot of people avoid using them, but I don’t think that is a good approach. Let V be our representation with highest weight λ. That means that Vλ is a 1-dimensional representation of the subalgebra b = h n+ g. There is ⊕ ⊆ an embedding V V , which is a homomorphism of Ub-modules (that is, λ → of representations of b). Thus, there is an induced object U(g) b V , and ⊗U λ there is a homomorphism of Ug-modules from U(g) b V to V . However, ⊗U λ there is a problem: when we do the induction, we get something infinite dimensional.

Deﬁnition 18.7. A Verma module is M(λ)= Ug b V . ⊗U λ Verma Modules are universal in the sense that all representations with highest weight λ are quotients of M(λ). To understand what M(λ) is as a vector space, we use PBW to get that Ug = Un Uh Un+ = U(n ) − ⊗ ⊗ − ⊗ Ub (these tensor products are as vector spaces). Thus, we get M(λ) = Lecture 18 - Representations 91

Un V . The Verma module is something universal ... . We can choose − ⊗k λ a basis U(n−), say α1,...,αn, αn+1,...,αN (the positive roots, ordered). k1 kN Then Yαi i N is a basis for n−, and by PBW Yα Yα v is a basis in { } ≤ { 1 · · · N } U(n ) V M(λ). Even though the Verma module is inﬁnite-dimensional, − ⊗k λ it still has a weight decomposition:

h(Y k1 Y kN v) = (λ k α )(h)(Y k1 Y kN v) α1 · · · αN − 1 1 −··· α1 · · · αN

11 1 11 11 11 1 1 1 2 2 1 11 11 11 1 1 1 3 2 1 '!&"%#$ 1 1 1 11 11 11 11 11 11 All the multiplicities along the edges are 1. Then you get multiplicity 2 just below λ because there is a linear dependence. You can check that parallel to the edges, you keep getting multiplicity 2. And in general, the multiplicity grows by 1 with every step you take from the wall.

◮ Exercise 18.2. Check this out.

We have a nice formula:

m =# λ µ = k α µ { − α } α ∆+ X∈ This is called the Kostant partition function. Let’s calculate the multiplicity circled in the previous picture. We have that 2α3 = α3 + α2 + α1 = 2α1 + 2α2, so the multiplicity is 3. These calculations will prepare you for the Weyl formula.

Claim. A Verma module M(λ) has a unique proper maximal submodule N(λ).

Proof. N being proper is equivalent to N V = 0. This property is clearly ∩ λ preserved under taking sums, so you get a unique maximal submodule.

If V and W are irreducible representations that have the same highest weight, then they are both the unique irreducible quotient M(λ)/N(λ), so they are isomorphic. Thus, the map from irreducibles to P + is injective. Now we need to show that if λ P +, then the quotient M(λ)/N(λ) is ∈ ﬁnite-dimensional. The trick is to look at sl(2). The picture of the Verma module is that you have all the weights λ kα with multiplicity 1. If you − mod out by any submodule, you get something ﬁnite-dimensional. Lecture 18 - Representations 92

Sketch of Proof. If w is a weight vector in a Verma module such that Xiw = 0 for i = 1,...,n, then we claim that w N(λ). To see this, you note that ∈ you can never get back to λ. ni+1 Suppose λ, αˇi = ni Z 0. Then w = Yi v and Xiw = 0. We have ≥ ni+1 h ni+1i ∈ XjY v = Yi Xjv = 0 because [Xj,Yi]=0 for i = j. 6 ni+1 Consider the quotient M(λ)/N(λ)= V (λ). Then we have that Yi v = 0, so the Yi act locally nilpotently on V (λ). And Xi also act locally nilpotently (even on the Verma module, not just the quotient). Then si = exp X exp( Y )exp X , and we get that P (V (λ)) is W invariant because of i − i i something. But then you can start with λ and look at the orbit. The weights are in the convex hull, which is bounded, so V (λ) is ﬁnite-dimensional.

We introduced these Verma modules because we will use them later to write down the character formula. Next time:

(1) Fundamental representations, etc.

(2) Weyl character formula Lecture 19 - The Weyl character formula 93

Lecture 19 - The Weyl character formula

+ If λ P (i.e. (λ, αˇi) Z 0 for all i), then we can construct a representation ∈ ∈ ≥ V (λ) which is a representation with λ as the highest weight. We deﬁne the fundamental weights ω1,...,ωn of a representation to be those weights for which (ω , αˇ ) = δ . It is clear that any weight λ = λ ω + + λ ω for i j ij 1 1 · · · n n λ 0. So people often draw the Dynkin diagrams with the fundamental i ≥ weights labelled: 1 0 ... 0 0 ØÙ ØÙ ØÙ ØÙ This is V (ω1), the ﬁrst fundamental representation for sl(n). Here is the adjoint representation:

1 0 ... 0 1 ØÙ ØÙ ØÙ ØÙ Observe that if v V a highest vector, and w W another highest weight ∈ ∈ vector in another representation, then v w V W is a highest weight ⊗ ∈ ⊗ vector. It follows that every finite dimensional irreducible representation can be realized as a subrepresentation in a tensor product of fundamental representations. Fundamental Example 19.1. Let’s calculate the fundamental weights for sl(n+1). Recall weights of that we have simple roots ε ε ,...,ε ε , which are also the coroots. sl(n + 1) 1 − 2 n − n+1 It follows that ωi = ε1 + + εi for i = 1,...,n. k · · · Consider Λ E, where E is the standard representation. Let e1,...,en+1 be a weight basis for E (i.e. the vectors e1,...,en are a basis, with decreasing k weights ε1,...,εn). We’d like to write down the weights P (Λ E). Note that ΛkE is spanned by the vectors e e , which have weights ε + +ε . i1 ∧···∧ ik i1 · · · ik Thus, the highest weight is ε + +ε = ω ,soΛkE = V is a fundamental 1 · · · k k ωk representation. Notes also that the weights form the single orbit Wωk. k Λ E = Vωk Representations where the weights form a single orbit are called minuscule. Remark 19.2. If λ is the highest weight of the representation V , then λ is − Highest a weight of V ∗. To compute the highest weight of V ∗, we need to get back weights of into P +, so we apply the longest word in the Weyl group: w(λ) is the duals − highest weight of V ∗. Thus, there is a fixed involution of the Weyl chamber (namely, w) which takes the highest weight of V to the highest weight of − V . It is clear that w preserves the set of simple roots and preserves inner ∗ − products, so it is an involution of the Dynkin diagram. In the case of sl(n + 1), the involution is s s . . . + + . In ØÙ ØÙ ØÙ ØÙ particular, the dual of the standard representation Vω1 is Vωn . The bijection between P + and irreducible representations implies Corollary 19.3. If V,W are finite dimensional representations, with ch V = ch W , then V W . ≃ Lecture 19 - The Weyl character formula 94

µ How do you calculate ch V ? Recall that ch V = mµe , where mµ is the multiplicity of the weight µ. You can do it with Verma modules. P Remember that for Verma modules, you still get a weight decomposition.

Example 19.4. g = sl(3). Consider the case λ = 2ω1 + ω2:

2 1 ØÙ ØÙ 11 11 0 0 1 1 1 1 11 '!&"%#$ 11 1 1 11 11 1 0 111 1 2 2 1 1 11 11 1 1 1 11 0 1 1 11 2 1 0 1 11 1 '!&"%#$ 1 1 1 11 11 11 1 0 1 1 1 1 0 0 11 1 11 1 1 11 11 1 0 0 0 0 0 11 Explain this In the quotient, the shown weights disappear. What about the one below picture it? Its multiplicity in the Verma module is 2. The multiplicity in the Verma better module centered at 2ω1 is 1, so in the quotient, you get multiplicity 1. How about right below it? You had 3 from the big Verma module, and you remove 1 and 1, so you get 1 left. Furthur down and to the left, you get all zeros. Similarly, we can ﬁll in the other numbers. You get the leftover hexagon.

We have the Weyl character formula, which is nice if you have a com- puter:

The character formula First some notation:

1. W is the Weyl group. ( 1)w = det(w) = ( 1)sgn(w). − − 1 2. ρ = + α. Note that r (ρ)= ρ α = ρ (ρ, αˇ )α , so we know 2 α ∆ i − i − i i that (ρ, αˇ∈) = 1 for all i. Thus, ρ is the sum of all the fundamental P i weights.

Then the result is that

( 1)wew(λ+ρ) ch V (λ)= w W − (Weyl Character Formula1) ∈ w w(ρ) w W ( 1) e P ∈ − 1 This formula may lookP ugly, but it is sweet. It says that you can compute the character of Vλ in the following way. Translate the Weyl vector ρ around by the Weyl group; this Lecture 19 - The Weyl character formula 95 note that the numerator and denominator is skew with respect to the action of W , so ch is W -symmetric. The denominator is called the Weyl denominator. For those who know something about the representation theory of gl(n), I’d like to show the connection between these two things. If g = gl(n) ⊃ sl(n). Let z = eεi , so z z = 1. We know that W S , which acts on i 1 · · · n ≃ n the zi by permutation. We have that 1 ρ = ε ε 2 i − j Xi

a1+n 1 a1+n 1 z1 − ... zn − = det . Dλ  .  zan  1    will form some polytope. Make a piece of cardboard shaped like this polytope (ok, so maybe this is only practical for rank 2), and put (−1)w at the vertex w(ρ). This is your cardboard denominator. Now the formula tells you that when you center your cardboard denominator around any weight, and then add the multiplicities of the weights of Vλ at the vertices with the appropriate sign, you’ll get zero (unless you centered your cardboard at w(λ + ρ), in which case only one non-zero multiplicity shows up in the sum, so you’ll get ±1). Since we know that the highest weight has multiplicity 1, 1 1 1 01 we can use this to compute the rest of the character. 1 For sl(3), your cardboard denominator will be a hexagon, and one 1 ? ?1 1 0 1 step of computing the character of V2ω1+ω2 might look like: 1 ? 1 0 − − − ch V 0 = 0 1+ ? 1 + 0 0, so ? = 2. Since is symmetric with 1 1 respect to W , all three of the ?s must be 2. Lecture 19 - The Weyl character formula 96

So the character is the Schur polynomial.

Proof of character formula. Express ch V (λ) as a linear combination of characters of Verma modules. This is what we did with the picture for sl(3).

Consider sl(2). We know what the representations are: V (n). The Verma module is inﬁnite in the negative direction, but at n 2, there is another Verma module, which is also inﬁnite. − − α We have that ρ = ω = 2 picture Thus, we claim that ch V (n)= ch M(n) ch M( n 2). To see nω (n 2)−ω − nω− α this, we compute ch M(n)= e + e − + = e (1 + e− + − − 2α enω · · · e( n 2)ω e + ) = − , and similarly, ch M( n 2) = − . − · · · 1 e α − − 1 e α − e(n+1)ω − So we get that ch M(n) = − and ch M( n 2) = (eα/2 e α/2) − − − − e (n+1)ω (eα/2 e−α/2) − e(n+1)ω e(−n−1)ω So we have that ch V (λ)= − eα/2−e α/2 − Consider M(λ) for λ P . Then ∈ λ α 2α ch M(λ)= e (1 + e− + e− + ) · · · α ∆+ Y∈ where m =# λ µ = n α µ { − α } α ∆+ X∈ We claim that this character is

α γ (1 + e− + )= m e · · · γ α ∆+ Y∈ X where m =# γ = n α . Thus, we have that γ { α } P eλ eλ+ρ ch M(λ)= = α α/2 α/2 α ∆+ (1 e− ) α ∆+ (e e− ) ∈ − ∈ − 1 1 Q Q where ρ = 2 α ∆+ α. Now recall the∈ Casimir operator. If e is a basis for g, with dual basis P { i} f . Then the Casimir operator is Ω = e f Ug. We showed before { i} i i ∈ that Ω is in the center of Ug, so that Ωg = gΩ for all g g. P ∈ Claim. Ω M(λ) = (λ + 2ρ, λ)Id . This is 2 | 2 Lecture 19 - The Weyl character formula 97

We will use the identity (λ + 2ρ, λ) = (λ + ρ, λ + ρ) (ρ, ρ). It is enough − to show that something highest weight, then use stuﬀ Yα Let ui be an orthonormal basis for h. Xα α ∆, dual basis. { } ∈ (Xα,Yα) We can write n o n 2 XαYα YαXα Ω= ui + + (Xα,Yα) (Xα,Yα) i=1 α ∆+ X X∈ Let v be the highest vector. then we have that

n 2 λ(Hλ) Ωv = ( λ(ui) )v + (Xα,Yα)) 1 α ∆+ X X∈ since ui(v)= λ(ui)v, Xαv = 0,YαXαv = 0, XαYα = Hα + YαXα, XαYαv = λ(Hα)v. 2(λ,α) 1 2 We have that λ(Hα)= (α,α) and (Xα,Yα)= 2 (Hα,Hα)= (α,α) . We get that

Ωv = (λ, λ)v + (λ, α)v = what we want α ∆+ X∈ Remark 19.5. Let

Ωλ = µ P (λ + ρ, λ + ρ) = (µ + ρ, µ + ρ) { ∈ | } geometrically, this is a sphere centered somewhere. This is a finite set. This is 3. We have that M(λ) has a unique irreducible quotient, which we denote V (λ). If λ P +, then this is finite-dimensional, but in general, it is not. ∈ For example, sometimes the Verma module itself is irreducible (like in the sl(2) case). Point number 4 is that there exists a finite filtration of the Verma module

0= F M(λ) F M(λ) F M(λ)= M(λ) 0 ⊂ 1 ⊂···⊂ k such that the consecutive quotients are irreducible with Fi/Fi 1 V (µ) for − ≃ some µ Ωλ. ∈ We say that a vector w is singular of weight µ if hw = µ(h)w and n+w = 0. Suppose a module is generated by some highest vector, and all singular vectors are proportional to the highest vector, then the representation is Lecture 19 - The Weyl character formula 98 irreducible. Why is this so? The best way is to draw a picture:

11 Verma module X 11 11 0 1 / 1 1 11 11 1 11 11 singular 11 1 11 u 1 11 · 11 11 submodule 1

Suppose you have some u in a proper submodule. Apply the positive roots to it ... you will eventually get to a place where you would kill it by applying positive roots, but that will be a singular vector. Take a smallest singular vector w in M(λ). Because it is smallest, there are no more singular vectors below it. Take F1M(λ) to be the submodule generated by w. Then this submodule is irreducible and generated by the highest vector w (with some weight µ). Why does µ lie in this sphere Ωλ? Because the Casimir must act on µ in the same way as it acted on λ. Now take the quotient and do the same. The filtration is finite because of point 3, that the sphere is finite. You can use each weight only as much as the multiplicity in the Verma module, and all the multiplicities are finite. We get a nice conclusion:

ch M(λ)= bµch(V (µ)) µ λ µX≤Ωλ ∈

[[box this too]] where λ µ = nαα, nα Z 0. We have that bλ = 1. − ∈ ≥ Now you can invert this kind of equation to get P ch V (λ)= c ch(M(µ)) ( ) µ ∗ µ λ µX≤Ωλ ∈

[[more box maybe ... this is the fact we need]] with cλ = 1. If you like, you get the c’s by inverting the matrix of b’s. Now c eµ+ρ ch V (λ)= µ α/2 α/2 α ∆+ (e e− ) ∈ P − call the denominator Γ, so Q

µ+ρ Γch(V (λ)) = cµe X Lecture 19 - The Weyl character formula 99

You have an action of W on the LHS. You can check that w(Γ) = ( 1)wΓ, − so the RHS must also be skew-symmetric with respect to the action of W . Because of this, you can combine the terms by orbits of the Weyl group Try to ﬁnish this yourself. Calculate the cµ. Next time, we’ll ﬁnish, and talk about the relation to compact groups. Lecture 20 100

Lecture 20

This is Serganova’s last lecture. We have that ch V (λ)= c ch M(µ) ( ) µ ∗ µ λ ≤ cXλ=1 which we get from ch M(λ)= ch V (µ) µ λ ≤ bXλ=1 where µ P , but not in P +. The b are Kazhdan-Lusztig multiplicities, ∈ µ and are hard to compute. But we can compute the c when λ P +. µ ∈ There was once a conjecture that all the c were 1, but this is false, µ ± and leads to some interesting stuﬀ. Since we know the characters of all the Verma modules, we can rewrite ( ) as ∗ α/2 α/2 µ+λ (e e− )ch V (λ)= c e − µ α ∆+ µ Ωλ,µ λ Y∈ ∈ X≤ we know that w(LHS) = ( 1)wLHS, so the same must be true of the RHS: − w( c eµ+ρ) = ( 1)w c eµ+ρ µ − µ X X w This is equivalent to saying that cw(µ+ρ) ρ = ( 1) cµ. We also know − − that cλ = 1, so you get

( 1)wew(λ+ρ) + ( ( 1)wew(µ+ρ)) − − µ<λ w W X µ+Xρ P + X∈ ∈ W acts in such a way that every orbit contains an element of P +, so we can take µ + ρ P +. We want to show that c = 0 for µ < λ. µ Ωλ. We have ∈ µ ∈ that (µ + ρ, µ + ρ) = (λ + ρ, λ + ρ) and

λ µ = n α − i i X with ni Z 0. So we get that (λ µ, λ+µ+2ρ)=0,so( niαi, λ+µ+2ρ)= ∈ ≥ − 0, so for λ + µ + ρ P +, we have that ∈ P (λ + µ + ρ, α ) 0 i ≥

(ρ, αi) > 0, recall that (ρ, αˇi) = 1.

α/2 α/2 w w(λ+ρ) (e e− )ch V (λ)= ( 1) e − − α ∆+ w W Y∈ X∈ Lecture 20 101

λ = 0, ch V (0) = e0 = 1. gives the Weyl denominator identity: α/2 α/2 w w(ρ) (e e− )= ( 1) e = − − D α ∆+ w W Y∈ X∈ So we get the formula 1 ch V (λ)= ( 1)wew(λ+ρ) − w W D X∈ Dimension formula: we can get (λ + ρ, α) dim V (λ)= (ρ, α) α ∆+ Y∈ In the sl(n) case, you get ε + + ε = 0, and you get ρ = stuff, so 1 · · · n you can compute dimension as (ρ, α) = (n 1)!(n 2)! = (n 1)!! − − · · · − α ∆+ Y∈ If λ = aiεi, then we get P (a + a + j i) i j − Yi

Compact Groups So far we classiﬁed semi-simple Lie algebras over a characteristic 0 algebraically closed ﬁeld. How is this connected to compact groups. Let G be a compact connected group (over R), like SU(2), which is topologically S3.

SU(n)= X GL(n, C) XtX = Id, det X = 1 { ∈ | } ◮ Exercise 20.1. If G is abelian, then it is a product of circles, so it is Tn.

There exists the G-invariant volume form satisfying

(1) G ω = 1 R (2) G f(gh)ω = G f(hg)ω = G f(g)ω It is aR nice exerciseR to constructR this ... you take a volume near the identity and translate it around. So you pick ω Λtop(T G) and deﬁne ω = e ∈ e ∗ g Lg∗−1 ωe. Check for yourself that this form is right invariant. From this you get a nice theorem:

Theorem 20.1. If V is a ﬁnite dimensional representation of G (over R), then there exists a positive deﬁnite G-invariant inner product on V : (gv,gw) = (v, w).

Proof. Start with some positive deﬁnite inner product Q(v, w), and deﬁne

(v, w)= Q(gv,gw)ω ZG which is invariant.

It follows that any ﬁnite dimensional representation of a compact group G is completely reducible (i.e. splits into a direct sum of irreducibles). By the way, representations of Lie groups are always taken to be smooth. In particular, look at Ad : G GL(g), which gives ad : g gl(g), → → which must split into irreducible ideals: g = g g a, each of which is 1 ⊕··· k ⊕ one-dimensional or simple (dump all the one-dimensional guys into a, which is then the center). Thus, the Lie algebra of a compact group is the direct sum of its center and a semi-simple Lie algebra. Such a Lie algebra is called reductive. If G is simply connected, then I claim that a is trivial. This is because the simply connected group connected to a must be a torus, so a center gives you some fundamental group. Thus, if G is simply connected, then g is semi-simple. Criterion: If the Killing form B on g is negative deﬁnite, then the corresponding group is compact. Lecture 20 103

If you have g gl(g), and you know that g has an invariant positive → deﬁnite product, so it lies in so(g). Here you have At = A, so you have to − check that tr(A2) < 0. It is not hard to check that the eigenvalues of A are imaginary (as soon as At = A), so we have that the trace of the square is − negative. g is a Lie algebra over R How to classify compact Lie algebras? We know the classiﬁcation over C, so we can always take g gC = g R C, which remains semi-simple. ⊗ However, this process might not be injective. For example, take su(2) = a b a Ri, b C and sl(2, R), then they complexify to the same { ¯b a | ∈ ∈ } thing.− g in this case is called a real form of gC. So you can start with gC and classify all real forms.

Theorem 20.2 (Cartan?). Every semi-simple Lie algebra has exactly one (up to isomorphism) compact real form.

For example, for sl(2) it is su(2). Classical Lie groups: SL(n, C), SO(n, C) (SO has lots of real forms of this, because in the real case, you get a signiture of a form; in the complex case, all forms are isomorphic), Sp(2n, C). What are the corresponding compact simple Lie groups? Compact real forms: SU(n) = the group of linear operators on Cn preserving a positive definite Hermitian form. SL(n) = the group of linear operators on Rn preserving a positive definite symmetric bilinear form. Sp(2n) =the group of linear operators on Hn preserving a positive definite Hermitian form We’re not going to prove this theorem because we don’t have time, but let’s show existence.

Proof of existence. Suppose gC = g R C = g ig. Then you can construct ⊗ ⊕ σ : gC gC “complex conjugation”. Then σ preserves the commutator, → but it is an anti-linear involution. Classifying real forms amounts to classifying all anti-linear involutions. There should be one that corresponds to the compact algebra. Take X1,...,Xn,H1,...,Hn,Y1,...,Yn generators for the algebra. Then we just need to deﬁne σ on the generators: σ(X ) = Y ,σ(Y ) = X ,σ(H ) = H , and extend anti-linearly. This i − i i − i i − i particular σ is called the Cartan involution. Now we claim that g = (gC)σ = X σ(X)= X is a compact simple Lie { | } algebra. We just have to check that the Killing form is negative deﬁnite. If you take h h, written as h = a H , then σ(h)= h implies that all the a ∈ i i i are purely imaginary. This implies that the eigenvalues of h are imaginary, P which implies that B(h, h) < 0. You also have to check it on Xi,Yi. The Lecture 20 104

ﬁxed things will be of the form (aX aY¯ ) g. The Weyl group action i − i ∈ shows that B is negative on all of the root space. Look at exp hσ G (simply connected), which is called the maximal ⊂ torus T . I’m going to tell you several facts now. You can always think of T n n as R /L. The point is that R can be identiﬁed with hre, and hre∗ has two natural lattices: Q (the root lattice) and P (the weight lattice). So one can n identify T = R /L = hre/Pˇ, where Pˇ is the natural dual lattice to P , the set of h h such that ω, h Z for all ω P . G is simply connected, and ∈ h i∈ ∈ when you quotient by the center, you get Ad G, and all other groups with the same algebra are in between. Ad T = hre/Qˇ. We have the sequence 1 Z(G) T Ad T 1 . You can check that any element is semi- { }→ → → → { } simple in a compact group, so the center of G is the quotient P/Q Q/ˇ Pˇ. ≃ Observe that P/Q = the determinant of the Cartan matrix. For example, | | 2 1 if g = sl(3), then we have det 1− 2 = 3, and the center of SU(3) is the set of all elements of the form diag− (ω,ω,ω) where ω3 = 1. G2 has only one real form because the det is 1? Orthogonality relations for compact groups:

1 χ(g)ψ¯(g− )ω = δχ,ψ ZG where χ and ψ are characters of irreducible representations. You know that the character is constant on conjugacy classes, so you can integrate over the conjugacy classes. There is a nice picture for SU(2).

• •

• T •

The integral can be written as 1 χ(t)ψ¯(t)V ol(C(t))dt W | | ZT And V ol(C(t)) = (t) ¯(t). You divide by W because that is how many D D | | times each class hits T . You know that ch V (λ)= eλ + lowerterm Lecture 21 105

Lecture 21

The new lecturer is Borcherds. This is the last third of the course: Examples and calculations. The (unoﬃcial) goal is to prove no theorems. We’ll try to do

1. Cliﬀord algebras and spin groups.

2. Construction of all Lie groups and all representations. You might say this is impossible, so let’s just try to do all simple ones, and in particular E8, E7, E6.

3. Representations of SL2(R). Today we’ll give an overview of Lie algebras and Lie groups. We’ll compare with algebraic groups (matrix groups), and ﬁnite groups.

What does a general Lie group look like? Example 21.1. Let G be the group of all shape-preserving (translations, rotations, reﬂections, dilations) transformations of R4, sometimes called R4 GO (R).1 The is pretty much the smallest possible Lie group with · 4 all the properties of a general Lie group.

(1) G has a connected component Gconn, which is a normal subgroup, and G/Gconn is a discrete group. Any Lie group can be built from a connected Lie group and a discrete group (0-dimensional group). Classifying discrete groups is completely hopeless.

(2) We let Gsol be the maximal normal connected solvable subgroup of G . Recall that solvable means that there is a chain G G conn sol ⊇ sol′ ⊇ 1 where all consecutive quotients are abelian. So you build ··· ⊃ solvable groups out of abelian groups.

Gconn/Gsol is connected and has no solvable normal subgroups. It turns out that it is much easier to deal with the group if you kill off all the solvable subgroups like this. Groups like this are semi-simple. They are almost a product of simple groups. A semi-simple group G modulo its discrete center is a product of simple groups. The nice thing is that connected simple Lie groups are all completely classified. The list is quite long, and it has a lot of infinite series and a lot of exceptional guys. Some examples are P SU(n), PSLn(R), PSLn(C). You can go from simple Lie groups to simple Lie algebras

1R4 is translations, G is multiplication by scalers, O is reﬂections and rotations Lecture 21 106

over R, then complexify to get Lie algebras over C. This last step is finite-to-one, and they can be classified fairly easily. For example, the Lie algebras of sl2(R) and su2(R) are different, but they have the same complexification, sl2(C), which, by the way, is actually simple as a real Lie algebra. Its complexification is not simple, it is sl (C) sl (C). 2 ⊕ 2 Suppose Gsol is solvable. The only thing you can say is that its derived subgroup G is nilpotent. That is, G [G , G ] [G , [G , G ]] sol′ ′ ⊇ ′ ′ ⊇ ′ ′ ′ ⊇ 1, or rather there is a chain such that G /G the center of ··· ⊇ i i+1 ⊆ G1/Gi+1. Thus, nilpotent groups can be built up by taking central extensions. Nilpotent groups are in practice not classifiable. The simply connected ones are all isomorphic to subgroups of upper-triangular matrices 1 ..  . ∗   1     0 1      If you put arbitrary things on the diagonal, that is what the solvable groups look like. We have the picture

G discrete; classification hopeless Gconn  simples; classified  G  ∗ Q    abelian discrete  all connected Gsol classification trivial     abelian  G  nil   nilpotent; classification a mess    1     4 In our case, G is R GO4(R). The connected component Gconn is 4 + · 4 + R GO4 (R), which is the stuff that preserves orientation. Gsol is R R · 4 + · (rescaling). Gnil is R . Gconn/Gsol is SO4(R), Gsol/Gnil is R . Gconn/G is ∗ P SO4(R) SO3(R) SO3(R). G /Gsol is Z/2Z. ≃ × ∗ Note: What is O (R)? It almost is the product O (R) O (R). To 4 3 × 3 see this is to look at the quaternions H, which is an associative, but not commutative algebra. Notice that qq¯ = a2 + b2 + c2 + d2, which is > 0 if q = 0, so any non-zero quaternion has an inverse. Thus, H is a division 6 algebra. Put R4 to be the quaternions, and S3 is the unit sphere in this R4 Lecture 21 107 consisting of the quaternions such that q = qq¯ = 1. It is easy to check k k that pq = p q . So left or right multiplication by an element of S3 give k k k k k k rotations of R4. Thus, we have that S3 S3 O (R), and the image is × → 4 SO (R). This is not quite injective because the kernel is (1, 1), ( 1, 1) . 4 { − − } So you get S3 S3/( 1, 1) SO (R). × − − ≃ 4 Let R3 be the purely imaginary quaternions. Then q S3 acts by con- ∈ jugation on R3, so we get S3 O (R), and again it is not quite injective or → 3 surjective. The image is SO (R) and the kernel is 1, 1 , so S3/ 1, 1 3 { − } { − }≃ SO (R). So we almost have that SO isomorphic to SO SO . If you take 3 4 3 × 3 a double cover of SO4(R), it is the product of two copies of a double cover of SO3(R). Orthogonal groups in dimension 4 have a strong tendency to split up like this.

Comparison of Lie groups and Lie algebras

Let g be a Lie algebra. We can set gsol to be the maximal solvable ideal (normal subalgebra), and gnil = gsol′ and you get a chain g simples; classiﬁcation known gsol Q abelian; easy to classify g nil  nilpotent; classiﬁcation a mess 0   We have a map from Lie groups to Lie algebras. How much structure does this detect? This is an “isomorphism” (equivalence of categories) from simply connected Lie groups to Lie algebras. It cannot detect

(1) non-trivial components of G. SO4 and O4 have the same Lie algebra (2) the (discrete) center of G. If Z is any discrete subgroup in the center of G, then G and G/Z have the same Lie algebra. For example SU(2) and P SU(2) have the same Lie algebra.

If G is connected, with universal cover G˜, which is simply connected, then G = G/˜ discrete subgroup of the center. So if you know the universal cover, then you can read off all the other connected groups with the same Lie algebra. Let’s find all the groups with the algebra o4(R). First let’s find a simply connected group with this Lie algebra. You might guess SO4(R), but that isn’t simply connected. The simply connected one is S3 S3 as we saw before × (it is a product of two simply connected groups, so it is simply connected). Lecture 21 108

What is the center? The center of S3 is generated by 1, so the center of − S3 S3 is (Z/2Z)2, the Klein four group. There are three subgroups of order × 2 (Z/2Z)2 L qqq LL qq LLL qqq LL qqq LL ( 1, 1) (1, 1) ( 1, 1) h − − i h − i h − i OOO qq OOO qqq OOO qqq OOO qqq 1 q There are 5 groups with this Lie algebra therefore:

P SO4(R) Q ppp QQQ pp QQQ ppp QQQ ppp QQ 3 3 SO4(R) SO3(R) S S SO3(R) O × × OOO mmm OOO mmm OO mmm OO mmm S3 S3 × These orthogonal groups tend to have double covers, which are important if you want to study fermions.

Finite groups and Lie groups Analogies: (1) The classification of finite groups resembles the classification of simply Lie groups when n 2. ≥ For example, PSLn(R) is a simple Lie group, and PSLn(Fq) is a simple finite group except for n = 2,q 3. Simple finite groups form about ≤ 18 series similar to Lie groups, and 26 or 27 exceptions, called sporadic groups, which don’t seem to have any analogues for Lie groups. (2) Finite groups and Lie groups are both built up from simple and abelian groups. However, the way that finite groups are built is much more complicated than the way Lie groups are built. Finite groups can contain simple subgroups in very complicated ways; not just as direct factors. For example, there are wreath products. Let G be finite simple, and H be simple, acting on a set of n points. Then form Gn, acted on by H in the obvious way. Then you can form the semi- direct product Gn H. There is no analogue for (finite-dimensional) · Lie groups (There is an analogue for infinite-dimensional Lie groups, which is why there isn’t much theory there.). Lecture 21 109

(3) Solvable ﬁnite groups need not have nilpotent group as a derived subgroup. Example: S4, with derived group A4, which is not nilpotent.

Algebraic groups (over R) and Lie groups Any algebraic group gives you a Lie group. By algebraic group, we mean an algebraic variety which is also a group, such as GLn(R). Probably all the Lie groups you’ve come across have been algebraic groups. Since they are so similar, we’ll list diﬀerences. Diﬀerences between algebraic groups and Lie groups:

(1) Unipotent and semi-simple abelian algebraic groups are totally dif- 1 ferent, but for Lie groups they are nearly the same. R 0 1∗ is a 0 ≃ unipotent and R −1 is semi-simple. The homomorphism × ≃ 0 a exp : R R is not rational, so you need something transcenden- → × tal to see that they are almost the same as Lie groups.

(2) Abelian varieties are diﬀerent from aﬃne algebraic groups.

Example 21.2. Take the elliptic curve y2 = x3 + x, and look at the a b 2 2 1 circle group of matrices b a with a + b = 1. These are both S as Lie groups, but completely− different as algebraic groups because of the different algebraic structures (complete or affine varieties).

(3) Some Lie groups do not correspond to ANY algebraic groups.

(a) The Heisenberg group: the group of symmetries of L2(R) generated by translations (f(x) f(x + y)), multiplication by eixy 7→ (f(x) eixyf(x)), and multiplication by eiy (f(x) eiyf(x)). 7→ 7→ This can also be modeled as 1 x y 1 0 n 0 1 z 0 1 0 n Z     ∈  0 0 1 , 0 0 1        

It has the property that in any ﬁnite dimensional representa tion, the center acts trivially, so it is not isomorphic to any algebraic group.

(b) The metaplectic group. Look at SL2(R), with algebra sl2(R) = a b a + d = 0 . Find all groups with this Lie algebra. There { c d | } are two obvious ones: SL (R) and PSL (R). There aren’t any 2 2 other ones that can be represented as groups of ﬁnite-dimensional matrices. However, if you look at SL2(R), you’ll ﬁnd that it is not simply connected. Lecture 21 110

Iwasawa decomposition: SL (R) K A N as a topological 2 ≃ × × space, where K is the complex subgroup of things of the form cos θ sin θ 1 a 0 1 n sin θ cos θ , which is S , A is 0 a−1 , and N is 0 1 , so it has fundamental− group Z. ^ So we take the universal cover SL2(R), which has center Z. ^ Any ﬁnite-dimensional representation of SL2(R) factors through ^ SL2(R). So SL2(R) cannot be written as a group of ﬁnite- dimensional matrices. Representing this group is a real pain.

Most important case: the metaplectic group Mp2(R) is the connected double cover of SL2(R). It turns up in the theory of modular forms of half-integral weight and has a representation called the metaplectic representation.

next class we’ll do Cliﬀord algebras and spin groups. Lecture 22 - Cliﬀord algebras and Spin groups 111

Lecture 22 - Cliﬀord algebras and Spin groups

Today we’ll mainly talk about spin groups and Clifford algebras. First let’s look at the most “important” Lie groups. Dimension 1: There are just R and S1 = R/Z. Dimension 2: You get the abelian groups, which are all the quotients of R2 by some discrete subgroup (of the center, which is all of R2). You get three cases: R2, R2/Z = R S1, and R2/Z2 = S1 S1. There is one other × × group in 2 dimensions, which is the smallest non-abelian solvable Lie group: a b the set of all matrices of the form 0 a−1 , where a> 0. The Lie algebra is the subalgebra of 2 2 matrices of the form a b , which is generated by × 0 a two elements as X,Y , with [X,Y ]= X. − h i Dimension 3: You get the group SL2(R), which is the most important Lie group of all! There are some variations, like the quotient PSL2(R) = SL2(R)/center (P always means that you’ve quotiented out by the center); the center of SL (R) is just 1. As we saw before, SL (R) is not simply 2 ± 2 connected ... it has the double cover Mp2(R), which is also not simply connected. PSL2(R) is familiar because it acts on the upper half plane by fractional linear transformations. There is a compact version of this: SU2 (S means determinant 1, U stands for unitary, which means it preserves the unitary form z z¯ + z z¯ on C2). SU S3, and it surjects onto SO (R), 1 1 2 2 2 ≃ 3 with kernel 1 (O means it preserves an inner product). We also call SU ± 2 Spin3(R). The Lie algebras sl2(R) and su2(R) are different over the reals, but when you complexify, they become equal. There is another interesting 3-dimensional algebra: the Heisenberg algebra: generated by X,Y,Z, with [X,Y ] = Z, and Z central. You can think of this as strictly upper triangular matrices. There are some less interesting algebras and groups. There are abelian groups, which is boring, and there are lots of boring solvable ones, such as R1+some 2-dimensional solvable or R2 R1 with some action of R1 on R2. · As you go up in dimension, the number of boring solvable algebras goes up a lot, and nobody can do anything about them, so you forget about them. Dimension 6: (nothing interesting happens in dimensions 4,5) You get SL2(C) ∼= Spin1,3(R) Dimension 8: SU3(R) and SL3(R), which is the first one with a non- trivial root system. Dimension 14: G2, which we might discuss. Dimension 248: E8, which we will discuss because it is my favorite. This class is mostly about finite-dimensional algebras, but let’s mention some infinite-dimensional Lie groups or Lie algebras. 1. Aut(Hilbert space). 2. Diff(Manifold) ... physics stuff. Lecture 22 - Clifford algebras and Spin groups 112

3. Gauge groups: (continuous, smooth, analytic, or whatever) maps from a manifold M to a group G.

i3 i 4. Virasoro algebra: [L ,L ] = (i j)L + − c. If you forget about i j − i+j 12 the c, you get vector ﬁelds on S1, so this is some central extension of Vect(S1).

5. Aﬃne Kac-Moody algebras, which are more or less central extensions of certain gauge groups over the circle.

Cliﬀord Algebras and Spin Groups With Lie algebras of small dimensions, there are accidental isomorphisms. Almost all of these can be explained with Cliﬀord algebras and Spin groups. Motivational examples:

1 1 1. SO2(R)= S : you can double cover S by itself

3 2. SO3(R): has a simply connected double cover S 3. SO (R): has a simply connected double cover S3 S3 4 × 4 2 4 4. SO5(C): Look at Sp4(C), which acts on C and on Λ (C ), which is 6-dimensional, and decomposes as 5 1. Λ2(C4) has a symmetric 2 4 2 4 ⊕ 4 4 bilinear form given by Λ (C ) Λ (C ) Λ (C ) C, and Sp4(C) ⊗ → ≃ 5 preserves this form. You get that Sp4(C) acts on C , preserving a symmetric bilinear form, so it maps to SO5(C). You can check that the kernel is 1. So Sp (C) is a double cover of SO (C) ± 4 5 4 2 4 5. SL4(C): acts on C , and we still have our 6-dimensional Λ (C ), with a symmetric bilinear form. From this you get

1 SL (C) SO (C) ± → 4 → 6 So we have double covers S1, S3, S3 S3, Sp (C), SL (C) of the orthogonal × 4 4 groups in dimensions 2,3,4,5, and 6, respectively. What is the next one? All of these look completely unrelated. Examples of Cliﬀord algebras:

1. C: generated by R, together with i, with i2 = 1 − 2. H: generated by R, together with i, j, each squaring to 1, with ij + − ji = 0. Lecture 22 - Cliﬀord algebras and Spin groups 113

2 2 2 2 3. Dirac wanted a square root for the operator ∂ + ∂ + ∂ ∂ (the ∂x2 ∂y2 ∂z2 − ∂t2 wave operator in 4 dimensions). What he tried doing was supposing ∂ ∂ ∂ ∂ that the square root is of the form A = γ1 ∂x + γ2 ∂y + γ3 ∂z + γ4 ∂t . Then set A2 = and compare coeﬃcients, and you ﬁnd that γ2 = 1, ∇ 1 γ2 = 1, γ2 = 1, γ2 = 1, and γ γ + γ γ = 0 for i = j. 2 3 4 − i j j i 6 Dirac solved this by taking the γ to be 4 4 complex matrices. The i × operator A acts on vector-valued functions on space-time.

Definition 22.1. A general Clifford algebra over R should be generated by 2 elements γ1,...,γn such that γi is some given real, and γiγj + γjγi = 0 for i = j. 6 Definition 22.2 (better (invariant) definition). Suppose we have some vector space V over some field K (usually R), with some quadratic form N : V K. Then the Clifford algebra over K is generated by the vector → space V , with relations v2 = N(v). Note that N(λv) = λ2N(v), and the expression (a, b) := N(a + b) − N(a) N(b) is bilinear, so if the characteristic of K is not 2, we have N(a)= (a,a) − 2 . Thus, you can work with the bilinear form instead of the quadratic form. We’ll use quadratic forms so that everything works in characteristic 2.

Example 22.3. Take V = R2 with basis i, j, with N(xi + yj)= x2 y2. − − Then the relations are (xi + yj)2 = x2 y2, which gives you exactly the − − relations for the quaternions.

Warning 22.4. A few authors (mainly in index theory) use the relations v2 = N(v). Some people add a factor of 2, which usually doesn’t − matter, but is wrong in characteristic 2.

Let’s work out a few more small examples: C (R) is the Cliﬀord algebra over R of the quadratic form x2 + + m,n 1 · · · x2 x2 x2 (m pluses and n minuses). m − m+1 −···− m+n Problem: Find the structure of all the Cm,n(R). We know that C0,0(R) is R 2 C1,0(R) is R[ε]/(ε 1) = R R 2 − ⊕ C0,1(R) is R[i]/(i +1) = C 2 2 C2,0(R) is R[α, β]/(α 1, β 1, αβ + βα), which we can model by α = 1 0 0 1 − − 0 1 and β = 1 0 . −We have a homomorphism C (R) M (R), which is onto (because 2,0 → 2 the two matrices generate M2(R)); the dimension of the later is 4, and the dimension of C2,0(R) is at most 4 because it is spanned by 1, α, β, αβ. So this map is actually an isomorphism. Thus C (R) M (R). Note that the 2,0 ≃ 2 Lecture 22 - Cliﬀord algebras and Spin groups 114

dim V same argument shows that CV (K) has dimension at most 2 . The tough part of Cliﬀord algebras is to show that it cannot be smaller than that. 2 2 1 0 C1,1(R) is R[α, β]/(α 1, β +1, αβ+βα), which we model by α = 0 1 0 1 − − and β = 1 0 − 0 1 234 0 R C HH H M (H) ⊕ 2 1 R R M (R) M (C) ⊕ 2 2 2 M (R) M (R) M (R) 2 2 ⊕ 2 3 M2(C)

What is CU V in terms of CU and CV . You might guess CU V = CU CV . ⊕ ⊕ ∼ ⊗ Answer 1: No for obvious definition of : C = C C . ⊗ 1,1 6 1,0 ⊗ 0,1 Answer 2: Yes for the correct definition of : Clifford algebras are Z/2Z- ⊗ graded. C (K)= C0 (K) C1 (K) where C0 (K) is spanned by the products V V ⊕ V V of even numbers of elements of V . Superalgebra of Z/2Z-graded algebras ⊗ is (a b)(c d) = ( 1)deg b deg cac bd. ⊗ ⊗ − · ⊗ What is CU V (K)? Let’s take dim U to be even for now. Take α1, . . . , ⊕ αm to be an orthogonal basis for U and let β1,...,βn to be an orthogonal basis for V . Then set γ = α α α β . What are the relations between i 1 2 · · · m i the αi and the γj?

α γ = α α α α β i j i 1 2 · · · m i = α α α β α 1 2 · · · m i i

(since dim U even, αi anti-commutes with everything)

γ γ = γ α α β i j i 1 · · · m j = α α γ β 1 · · · m i j = α α α α β β 1 · · · m 1 · · · m i j = γ γ − j i Finally, you can compute

γ2 = α α α α β β i 1 · · · m 1 · · · m i i = α2 α2 β2 ± 1 · · · m i which is some real number. So the γi’s satisfy the relations of some Clifford algebra and commute with the αi. The reason for throwing in all those αi’s, you get some things that commute with the αi’s, but satisfy the relations of a different Clifford algebra. We see that CU V (K) = CU (K) CW (K), where W is V with the ⊕ ∼ ⊗ quadratic form multiplied by α2 α2 . If you work it out carefully, you ± 1 · · · m get that the α2 α2 is ( 1)dim U/2 discriminant(U). ± l · · · m − · Lecture 22 - Clifford algebras and Spin groups 115

Take dim U = 2 and see what we get. We find that C (R) = M (R) m+2,n ∼ 2 ⊗ C (R), where the indices have reversed because we have a factor of 1, n,m − so we must change the sign of the form, which means that we change the subscripts. Similarly, we find that C (R) = M (R) C (R), where m+1,n+1 ∼ 2 ⊗ m,n the indices are in the right order. Finally, we have that C (R) = H m,n+2 ∼ ⊗ Cn,m(R). Using these formulas, we can reduce any Clifford algebra to tensor products of things like R, C, H, and M2(R). Recall the rules for taking tensor products of matrix algebras:

1. R X = X ⊗ ∼ 2. C H = M (C) because all 2-dimensional quadratic forms look the ⊗ ∼ 2 same over C

3. C R C = C C ⊗ ∼ ⊕ 4. H C H = M (R), which you can see by thinking of the H’s by acting ⊗ ∼ 4 by left and right multiplication on R4.

5. Mm(Mn(X)) ∼= Mmn(X) 6. M (A) M (B) = M (A B) m ⊗ n ∼ mn ⊗ Now we can work out every possible real Clifford algebra: see the table1. You don’t need to fill in the middle because you just take Mn to M2n. Then you find that C = C M (R). You also get a mod 8 period- 8+m,n ∼ m,n ⊗ 16 icity going down.

1The table will look all wrong in the dvi; look at the pdf or ps instead. Lecture 22 - Cliﬀord algebras and Spin groups 116 ) R ( 16 M ) R ( 8 M ⊕ ) R ( 16 M ) R ( 8 M ) C ( 4 M ) H ( 2 M H ⊕ H ) C ( H 2 etc. % % M L L L L L L L ) L L L L R L L ( L L 2 L L ) L M R ( C ⊕ 2 ) % % M R L L ( L L 2 L L ) L L M L L H L L ( L L 2 L L ) ) ) ) L R M R C H H ( ( ( ( 012345678 R ⊕ ⊕ 2 2 2 4 ) R M M M M H ( 2 n M 1 2 3 4 5 6 7 8 0 m Lecture 23 117

Lecture 23

Last time we defined the Clifford algebra CV (K), where V a vector space 2 over K with a quadratic form N. CV (K) is generated by V with x = N(x). C (R) uses the form x2 + + x2 x2 x2 . We found that m,n 1 · · · m − m+1 −···− m+n the structure depends heavily on m n mod 8. − Remark 23.1. This mod 8 periodicity turns up in several other places: (1) Clifford algebras

(2) Bott periodicity, which says that stable homotopy groups of orthogonal groups are periodic mod 8.

(3) Real K-theory, which also has a mod 8 periodicity.

m,n (4) Even unimodular lattices (such as the E8 lattice) in R exist if and only if m n 0 mod 8. − ≡ (5) The Super Brauer group of R is Z/8Z. The Super Brauer group consists of super division algebras over R

C[ε+] R[ε+] H[ε ] • − • •

R H • •

R[ε ]• •H[ε+] − C[•ε ] − where ε are odd and ε2 = 1 and ε z =zε ¯ for z C. ± ± ± ± ± ∈ 0 1 1 Recall that CV (R)= CV (R) CV (R), where CV (R) is the odd part and 0 ⊕ CV (R) is the even part. It turns out that we will need to know the structure 0 of Cm,n(R). Fortunately, this is easily reduced to the previous case. Look at 0 CU V (R) with dim U = 1, with γ a basis for U and γ1,...,γn form a basis ⊕ 0 for V . Then CU V (R) is generated by γγ1,...,γγn. What are the relations? we have that ⊕ γγ γγ = γγ γγ i · j − j i for i = j, and 6 (γγ )2 = ( γ2)γ2 i − i 0 So the generators γγi of CU V (R) also generate a Cliﬀord algebra, CV ( disc), where we’ve multiplied the⊕ form on V by γ2. Over R, we ﬁnd− that − Lecture 23 118

0 1 Cm+1,n(R) ∼= Cn,m(R) and Cm,n+1(R) ∼= Cm,n(R). So we’ve worked out all the real algebras and all the even and odd parts of them. Remark 23.2. For Complex Clifford algebras, it is similar, but easier. You find that C (C) = M m (C) and C (C) = M m (C) M m (C). Notice 2m ∼ 2 2m+1 ∼ 2 ⊕ 2 that we have mod 2 periodicity now. Notice also that Bott periodicity for the unitary group is mod 2. You could figure these out by tensoring the real algebras with C if you wanted.

Cliﬀord Groups

Usually the Cliﬀord groups are denoted ΓV (K). We will show that ΓV (K) is a subgroup of CV (K)× and we’ll ﬁnd an exact sequence

central 1 K× Γ (K) O (K) 1 → −−−−→ V → V → Deﬁnition 23.3. Γ (K)= homogeneous (either degree 1 or degree 0, not V { really needed) invertible elements x of C (K) such that xV α(x) 1 V V − ⊆ } (recall that V C (K)), where α is the automorphism of C (K) induced ⊆ V V by 1 on V . − Many books leave out the α, which is a mistake (though not a serious 1 1 one). Why do we use xV α(x)− instead of xV x− , which is also possible? Reasons:

(1) It is the correct “superalgebra” sign. The superalgebra convention says that whenever you exchange two elements of odd degree, you put in a 1, and you think of V and x as odd. − (2) Putting α in makes the theory much cleaner in ODD dimensions. For example, we will get a map C (K) O (K) which is onto if we use V → V α, but not if we do not. (You get SOV (K) without the α, which isn’t too bad, but is still annoying.)

1 Lemma 23.4. (V non-degenerate) The elements of ΓV (K) acting trivially on V are the elements of K Γ (K) C (K). × ⊆ V ⊆ V Proof. Suppose a + a Γ0 (K)+Γ1 (K) acts trivially on V . Then (a + 0 1 ∈ V V 0 a )v = v(a a ) (the minus sign comes because we’re using α), so a v = va 1 0 − 1 0 0 and a v = va . Choose an orthogonal basis γ ,...,γ for V .2 Write a as 1 − 1 1 n 0 a linear combination of monomials in this basis, so that

a0 = m + γ1n

1I promised no Lemmas or Theorems, but I was lying to you. 2All these results are true in characteristic 2, but you have to work harder ... you can’t go around choosing orthogonal bases. Lecture 23 119 where m C0 (K) and n C1 (K) and neither m nor n contain a factor of ∈ V ∈ V γ1. Apply the relations

a v = va and a v = va 0 0 1 − 1 with v = γ1 and use orthogonality to conclude that a0 contains no monomial with a factor γ1. Repeat with v equal to the other basis elements. Similarly, write a1 as a linear combination of monomials in the basis, and apply the relations to v varying among the basis elements. ◮ Exercise 23.1. Check that you must have a K and a = 0. (If you 0 ∈ 1 don’t use α, a could be γ γ when n is odd.) 1 1 · · · n So a + a = a K Γ (K)= K . 0 1 0 ∈ ∩ V ∗ T Now we define the anti-automorphism on CV (K) to be 1 on V , and extend to C (K) (anti means that (ab)T = bT aT ). Do not confuse a V 7→ α(a) (automorphism), a aT (anti-automorphism), and a α(aT ) (anti- 7→ 7→ automorphism). Now we define the spinor norm N by N(a)= aaT , N α(a)= aα(a)T for a C (K). Main properties: (1)N is a homomorphism Γ (K) C (K) ∈ V V → V with image in K× (N is a mess on the rest of CV (K)) and (2) the image of ΓV (K) in Aut V lies in OV (K). Notice that on V , N coincides with the quadratic form N (many authors seem not to have noticed this, and use different letters ... sometimes they use a stupid sign convention which makes them different).

Proof. We show that if a Γ (K), then N α(a) acts trivially on V . Calcu- ∈ V late

α α 1 T T 1 1 α(N (a))v(N (a))− = α(a)a vα(a )− a− 1 1 = α(a)α(a− )vaa− (apply T ; use T = 1) |V = v

So by the previous lemma, N α(a) K . ∈ × α This implies that N is a homomorphism on ΓV (K) because

N α(a)N α(b)= aα(a)T N α(b) = aN α(b)α(a)T (N α(b) is central) = abα(b)T α(a)T = (ab)α(ab)T = N α(ab)

After all this work with N α, what we’re really interested is N. N = N α on Γ0 (K) and N = N α on Γ1 (K). Since Γ (K)=Γ0 (K) Γ1 (K), which V − V V V ∪ V Lecture 23 120 shows that N is also a homomorphism. Observe that the Cliﬀord action on V preserves N on V because N is a homomorphism, but N is just the |V quadratic form on V . So we get a homomorphism from C (K) O (K). V → V What is its kernel, and is it surjective? The kernel is easy to ﬁgure out, because we did it a few minutes ago. The kernel is the elements of ΓV (K) acting trivially on V , which is just the elements of K×. Is it onto? To see this, we should write some explicit elements. Say r V and N(r) = 0. ∈ 6 Then r Γ (K) (we’ll see this in a second). What is its action on V ? ∈ V 1 r rvα(r)− = rv − N(r) = r if v=r − = v if (v, r) = 0

So it acts by reflection through the hyperplane r⊥. This is another good reason for putting the α ... you would get something more annoying otherwise. You might think that we’ll conclude that the map is surjective because OV (K) is generated by reflections. That’s wrong! Orthogonal groups need not be generated by reflections. Here is a counterexample. Take the field F and V = F4 = H H, where 2 2 ⊕ H is the quadratic form x2 + y2 + xy (each H has three elements of norm 1). Then the reflections generate the group S S of order 36 ... there is 3 × 3 no way to flip the two H’s by reflections.

◮ Exercise 23.2. check this

But the orthogonal group has order 72. It turns out that this is the ONLY counterexample. Any other field, or any other non-degenerate form on any other space works just fine. In fact, the map C (K) O (K) is always onto even in the exam- V → V ple above. For groups over R (which is what we are interested in in this course), this follows easily from the fact that OV (R) is indeed generated by reflections. Summarizing: We have the exact sequence

1 K× Γ (K) O (K) 1 → → V → V → where K× lands in the center of ΓV (K).

/ / / / 1 K× ΓV (K) OV (K) 1

N N k 2 / x x / / 2 / 1 K× 7→ K× K×/(K×) 1 Lecture 23 121

Definition 23.5. Pin (K)= elements of Γ (K) of spinor norm 1 , and V { V } Spin (K)= even elements of Pin (K) = Pin0 (K) Pin1 (K) . V { V V ∪ V } 2 Note that the kernel of N on K× is the elements x with x = 1, so it is elements = 1. By taking elements of norm 1, we get sequences ± 1 1 Pin (K) O (K) →± → V → V 1 1 Spin (K) SO (K) →± → V → V These maps are NOT always onto, but there are many important cases when they are, like when V has a positive definite quadratic form. The image is the set of elements of OV (K) or SOV (K) which have spinor norm 1 in 2 K×/(K×) . What is N : O (K) K /(K )2? It is the UNIQUE homomorphism V → × × such that N(a) = N(r) if a is reflection in r⊥, and r is a vector of norm N(r). Example 23.6. Take V to be a positive definite vector space over R. Then N maps to 1 in R /(R )2 = 1 (because N is positive definite). So the × × ± spinor norm on OV (R) is TRIVIAL. So if V is positive definite, we get double covers

1 1 Pin (R) O (R) 1 →± → V → V → 1 1 Spin (R) SO (R) 1 →± → V → V → This will account for the weird double covers we saw before. What if V is negative definite. Every reflection now has image 1 2 − in R×/(R×) , so the spinor norm N is the same as the determinant map O (R) 1. V →± So in order to find interesting examples of the spinor norm, you have to look at cases that are neither positive definite nor negative definite. Let’s look at Losrentz space: R1,3.

norm>0 u norm=0 ?? ?? ?? ?? ?? norm<0 ?? ?? ?? ?

Reﬂection through a vector of norm < 0 (spacelike vector, P : parity reversal) has spinor norm 1, det 1 and reﬂection through a vector of − − norm > 0 (timelike vector, T : time reversal) has spinor norm +1, det 1. − So O1,3(R) has 4 components (it is not hard to check that these are all the components), usually called 1, P , T , and P T . Lecture 23 122

Remark 23.7. For those who know Galois cohomology. We get an exact sequence of algebraic groups

1 GL Γ O 1 → 1 → V → V → (algebraic group means you don’t put a ﬁeld). You do not necessarily get an exact sequence when you put in a ﬁeld. If 1 A B C 1 → → → → is exact, 1 A(K) B(K) C(K) → → → is exact. What you really get is

1 H0(Gal(K/K¯ ), A) H0(Gal(K/K¯ ),B) H0(Gal(K/K¯ ),C) → → → → H1(Gal(K/K¯ ), A) → →··· 1 1 It turns out that H (Gal(K/K¯ ), GL1) = 1. However, H (Gal(K/K¯ ), 1) = 2 ± K×/(K×) . So from 1 GL Γ O 1 → 1 → V → V → you get

1 1 K× Γ (K) O (K) 1= H (Gal(K/K¯ ), GL ) → → V → V → 1 However, taking 1 µ Spin SO 1 → 2 → V → V → you get

N 2 1 1 1 Spin (K) SO (K) K×/(K×) = H (K/K,µ¯ ) →± → V → V −→ 2 so the non-surjectivity of N is some kind of higher Galois cohomology.

Warning 23.8. Spin SO is onto as a map of ALGEBRAIC V → V GROUPS, but Spin (K) SO (K) need NOT be onto. V → V Example 23.9. Take O (R) = SO (R) 1 as 3 is odd (in general 3 ∼ 3 × {± } O (R) = SO (R) 1 ). So we have a sequence 2n+1 ∼ 2n+1 × {± } 1 1 Spin (R) SO (R) 1. →± → 3 → 3 → Notice that Spin (R) C0(R) = H, so Spin (R) H , and in fact we saw 3 ⊆ 3 ∼ 3 ⊆ × that it is S3. Lecture 24 123

Lecture 24

Last time we constructed the sequences

1 K× Γ (K) O (K) 1 → → V → V → N 2 1 1 Pin (K) O (K) K×/(K×) →± → V → V −→ N 2 1 1 Spin (K) SO (K) K×/(K×) →± → V → V −→ Spin representations of Spin and Pin groups Notice that Pin (K) C (K) , so any module over C (K) gives a repre- V ⊆ V × V sentation of PinV (K). We already figured out that CV (K) are direct sums of matrix algebras over R, C, and H. What are the representations (modules) of complex Clifford algebras? Recall that C2n(C) ∼= M2n (C), which has a representations of dimension n 2 , which is called the spin representation of PinV (K) and C2n+1(C) ∼= M n (C) M n (C), which has 2 representations, called the spin representa- 2 × 2 tions of Pin2n+1(K). What happens if we restrict these to SpinV (C) PinV (C)? To do that, 0 ⊆ 0 we have to recall that C (C) = M n−1 (C) M n−1 (C) and C (C) = 2n ∼ 2 × 2 2n+1 ∼ M2n (C). So in EVEN dimensions Pin2n(C) has 1 spin representation of n n 1 dimension 2 splitting into 2 HALF SPIN representations of dimension 2 − and in ODD dimensions, Pin2n+1(C) has 2 spin representations of dimension n 2 which become the same on restriction to SpinV (C). Now we’ll give a second description of spin representations. We’ll just do the even-dimensional case (odd is similar). Say dim V = 2n, and say 2 we’re over C. Choose an orthonormal basis γ1,...,γ2n for V , so that γi = 1 and γ γ = γ γ . Now look at the group G generated by γ ,...,γ , which i j − j i 1 2n is finite, with order 21+2n (you can write all its elements explicitly). You can see that representations of CV (C) correspond to representations of G, with 1 acting as 1 (as opposed to acting as 1). So another way to look − − at representations of the Clifford algebra, you can look at representations of G. Let’s look at the structure of G:

(1) The center is 1. This uses the fact that we are in even dimensions, ± lest γ γ also be central. 1 · · · 2n (2) The conjugacy classes: 2 of size 1 (1 and 1), 22n 1 of size 2 − − ( γ γ ), so we have a total of 22n + 1 conjugacy classes, so we ± i1 · · · in should have that many representations. G/center is abelian, isomorphic to (Z/2Z)2n, which gives us 22n representations of dimension 1, Lecture 24 124

so there is only one more left to ﬁnd! We can ﬁgure out its dimension by recalling that the sum of the squares of the dimensions of irreducible representations gives us the order of G, which is 22n+1. So 22n 11 + 1 d2 = 22n+1, where d is the dimension of the mystery × × representation. Thus, d = 2n, so d = 2n. Thus, G, and therefore ± n CV (C), has an irreducible representation of dimension 2 (as we found earlier in another way).

Example 24.1. Consider O (R). As before, O (R) = SO (R) ( 1), 2,1 2,1 ∼ 2,1 × ± and SO2,1(R) is not connected: it has two components, separated by the spinor norm N. We have maps

1 1 Spin (R) SO (R) N 1. →± → 2,1 → 2,1 −→± Spin (R) C (R) = M (R), so Spin (R) has one 2-dimensional spin 2,1 ⊆ 2∗,1 ∼ 2 2,1 representation. So there is a map Spin (R) SL (R); by counting dimen- 2,1 → 2 sions and such, you can show it is an isomorphism. So Spin2,1(R) ∼= SL2(R). Now let’s look at some 4-dimensional orthogonal groups

Example 24.2. Look at SO4(R), which is compact. It has a complex spin representation of dimension 24/2 = 4, which splits into two half spin representations of dimension 2. We have the sequence

1 1 Spin (R) SO (R) 1 (N = 1) →± → 4 → 4 →

Spin4(R) is also compact, so the image in any complex representation is contained in some unitary group. So we get two maps Spin (R) SU(2) 4 → × SU(2), and both sides have dimension 6 and centers of order 4. Thus, we ﬁnd that Spin (R) = SU(2) SU(2) = S3 S3, which give you the two half 4 ∼ × ∼ × spin representations.

So now we’ve done the positive deﬁnite case.

Example 24.3. Look at SO3,1(R). Notice that O3,1(R) has four components distinguished by the maps det, N 1. So we get →± 1 1 Spin (R) SO (R) N 1 1 →± → 3,1 → 3,1 −→± → We expect 2 half spin representations, which give us two homomorphisms Spin (R) SL (C). This time, each of these homomorphisms is an iso- 3,1 → 2 morphism (I can’t think of why right now). The SL2(C)s are double covers of simple groups. Here, we don’t get the splitting into a product as in the positive deﬁnite case. This isomorphism is heavily used in quantum ﬁeld theory because Spin3,1(R) is a double cover of the connected component of Lecture 24 125

the Lorentz group (and SL2(C) is easy to work with). Note also that the center of Spin3,1(R) has order 2, not 4, as for Spin4,0(R). Also note that the group PSL (C) acts on the compactiﬁed C by a b (τ)= aτ+b . 2 ∪ {∞} c d cτ+d Subgroups of this group are called KLEINIAN groups. On the other hand, + 3 the group SO3,1(R) (identity component) acts on H (three-dimensional hyperbolic space). To see this, look at

norm= 1 u − norm=0 ? u ?? ?? u norm=1 ?

?? ? o norm= 1 −

One sheet of norm 1 hyperboloid is isomorphic to H3 under the induced − metric. In fact, we’ll deﬁne hyperbolic space that way. If you’re a topolo- gist, you’re very interested in hyperbolic 3-manifolds, which are H3/(discrete subgroup of SO3,1(R)). If you use the fact that SO3,1(R) ∼= PSL2(R), then you see that these discrete subgroups are in fact Klienian groups.

There are lots of exceptional isomorphisms in small dimension, all of which are very interesting, and almost all of them can be explained by spin groups.

Example 24.4. 0 O2,2(R) has 4 components (given by det, N); C2,2(R) ∼= M (R) M (R), which induces an isomorphism Spin (R) SL (R) 2 × 2 2,2 → 2 × SL2(R), which give you the two half spin representations. Both sides have dimension 6 with centers of order 4. So this time we get two non-compact groups. Let’s look at the fundamental group of SL2(R), which is Z, so the fundamental group of Spin (R) is Z Z. As we recall, Spin (R) and 2,2 ⊕ 4,0 Spin3,1(R) were both simply connected. This shows that SPIN GROUPS NEED NOT BE SIMPLY CONNECTED. So we can take covers of it. What do the corresponding covers (e.g. the universal cover) of Spin2,2(R) look like? This is hard to describe because for FINITE dimensional complex representations, you get finite dimensional representations of the lie algebra L, which correspond to the finite-dimensional representations of L C, which ⊗ correspond to the finite-dimensional representations of L′ = Lie algebra of Spin4,0(R), which correspond to the finite-dimensional representations of Spin4,0(R), which has no covers because it is simply connected. This means that any finite-dimensional representation of a cover of Spin2,2(R) actually factors through Spin2,2(R). So there is no way you can talk about Lecture 24 126 these things with finite matrices, and infinite-dimensional representations are hard. To summarize, the ALGEBRAIC GROUP Spin2,2 is simply connected (as an algebraic group) (think of an algebraic group as a functor from rings to groups), which means that it has no algebraic central extensions. However, the LIE GROUP Spin2,2(R) is NOT simply connected; it has fundamental group Z Z. This problem does not happen for COMPACT Lie groups ⊕ (where every finite cover is algebraic).

We’ve done O4,0, O3,1, and O2,2, from which we can obviously get O1,3 and O0,4. Note that O4,0(R) ∼= O0,4(R), SO4,0(R) ∼= SO0,4(R), Spin4,0(R) ∼= Spin (R). However, Pin (R) = Pin (R). These two are hard to distin- 0,4 4,0 6∼ 0,4 guish. We have Pin4,0(R) Pin0,4(R)

O4,0(R) = O0,4(R)

Take a reflection (of order 2) in O4,0(R), and lift it to the Pin groups. What is the order of the lift? The reflection vector v, with v2 = 1 lifts to the ± element v Γ (R) C (R). Notice that v2 = 1 in the case of R4,0 and ∈ V ⊆ V∗ v2 = 1 in the case of R0,4, so in Pin (R), the reflection lifts to something − 4,0 of order 2, but in Pin0,4(R), you get an element of order 4!. So these two groups are different. Two groups are isoclinic if they are confusingly similar. A similar phe- nomenon is common for groups of the form 2 G 2, which means it has a · · center of order 2, then some group G, and the abelianization has order 2. Watch out. ◮ Exercise 24.1. Spin3,3(R) ∼= SL4(R).

Triality This is a special property of 8-dimensional orthogonal groups. Recall that O8(C) has the Dynkin diagram D4, which has a symmetry of order three:

ØÙ Z

× ØÙ ØÙ 11 11 0 1 ØÙ But O8(C) and SO8(C) do NOT have corresponding symmetries of order three. The thing that does have the symmetry of order three is the spin group! The group Spin8(R) DOES have “extra” order three symmetry. You can see it as follows. Look at the half spin representations of Lecture 24 127

Spin8(R). Since this is a spin group in even dimension, there are two. C (R) = M 8/2−1 (R) M 8/2−1 (R) = M (R) M (R). So Spin (R) has 8,0 ∼ 2 × 2 ∼ 8 × 8 8 two 8-dimensional real half spin representations. But the spin group is compact, so it preserves some quadratic form, so you get 2 homomorphisms Spin (R) SO (R). So Spin (R) has THREE 8-dimensional representa- 8 → 8 8 tions: the half spins, and the one from the map to SO8(R). These maps Spin (R) SO (R) lift to Triality automorphisms Spin (R) Spin (R). 8 → 8 8 → 8 The center of Spin8(R) is (Z/2) + (Z/2) because the center of the Cliﬀord group is 1, γ γ . There are 3 non-trivial elements of the center, and ± ± 1 · · · 8 quotienting by any of these gives you something isomorphic to SO8(R). This is special to 8 dimensions.

More about Orthogonal groups

Is OV (K) a simple group? NO, for the following reasons: (1) There is a determinant map O (K) 1, which is usually onto, so V → ± it can’t be simple.

(2) There is a spinor norm map O (K) K /(K )2 V → × × (3) 1 center of O (K). − ∈ V (4) SOV (K) tends to split if dim V = 4, abelian if dim V = 2, and trivial if dim V = 1.

It turns out that they are usually simple apart from these four reasons why they’re not. Let’s mod out by the determinant, to get to SO, then look at Spin (K), then quotient by the center, and assume that dim V 5. Then V ≥ this is usually simple. The center tends to have order 1,2, or 4. If K is a FINITE field, then this gives many finite simple groups. Note that SOV (K) is NOT a subgroup of OV (K), elements of determinant 1 in general, it is the image of Γ0 (K) Γ (K) O (K), which is the V ⊆ V → V correct definition. Let’s look at why this is right and the definition you know 0 is wrong. There is a homomorphism ΓV (K) Z/2Z, which takes ΓV (K) to 1 → 0 and ΓV (K) to 1 (called the DICKSON INVARIANT). It is easy to check that det(v) = ( 1)dickson invariant(v). So if the characteristic of K is not 2, − det = 1 is equivalent to dickson = 0, but in characteristic 2, determinant is the wrong invariant (because determinant is always 1). Special properties of O1,n(R) and O2,n(R). O1,n(R) acts on hyperbolic space Hn, which is a component of norm 1 vectors in Rn,1. O (R) acts on − 2,n the “Hermitian symmetric space” (Hermitian means it has a complex structure, and symmetric means really nice). There are three ways to construct this space: Lecture 24 128

(1) It is the set of positive deﬁnite 2-dimensional subspaces of R2,n

(2) It is the norm 0 vectors ω of PC2,n with (ω, ω¯) = 0.

(3) It is the vectors x + iy R1,n 1 with y C, where the cone C is the ∈ − ∈ interior of the norm 0 cone.

◮ Exercise 24.2. Show that these are the same.

Next week, we’ll mess around with E8. Lecture 25 - E8 129

Lecture 25 - E8

In this lecture we use a vector notation in which powers represent repetitions: 2 so (18) = (1, 1, 1, 1, 1, 1, 1, 1) and ( 1 , 06) = ( 1 , 1 , 0, 0, 0, 0, 0, 0). ± 2 ± 2 ± 2 Recall that E8 has the Dynkin diagram

5 3 ( 1 , 1 ) − 2 2 ØÙ e e e e e e 2 − 3 4 − 5 6 − 7 e ØÙ e ØÙ e ØÙ e ØÙ e ØÙ e ØÙ e ØÙ e 1 − 2 3 − 4 5 − 6 7 − 8 where each vertex is a root r with (r, r) = 2; (r, s) = 0 when r and s are not joined, and (r, s)= 1 when r and s are joined. We choose an orthonormal − basis e1,...,e8, in which the roots are as given. We want to figure out what the root lattice L of E8 is (this is the lattice generated by the roots). If you take e e ( 15, 13) (all { i − i+1} ∪ − the A7 vectors plus twice the strange vector), they generate the D8 lattice = (x ,...,x ) x Z, x even . So the E lattice consists of two cosets { 1 8 | i ∈ i } 8 of this lattice, where the other coset is (x ,...,x ) x Z+ 1 , x odd . P { 1 8 | i ∈ 2 i } Alternative version: If you reflect this lattice through the hyperplane e , P 1⊥ then you get the same thing except that xi is always even. We will freely use both characterizations, depending on which is more convenient for the P calculation at hand. We should also work out the weight lattice, which is the vectors s such that (r, r)/2 divides (r, s) for all roots r. Notice that the weight lattice of E8 is contained in the weight lattice of D8, which is the union of four cosets 7 1 8 1 1 7 of D8: D8, D8 + (1, 0 ), D8 + ( 2 ) and D8 + ( 2 , 2 ). Which of these have 1 5 1−3 integral inner product with the vector ( 2 , 2 )? They are the first and the − 7 last, so the weight lattice of E is D D + ( 1 , 1 ), which is equal to the 8 8 ∪ 8 − 2 2 root lattice of E8. In other words, the E8 lattice L is UNIMODULAR (equal to its dual L′), where the dual is the lattice of vectors having integral inner product with all lattice vectors. This is also true of G2 and F4, but is not in general true of Lie algebra lattices. The E8 lattice is EVEN, which means that the inner product of any vector with itself is always even. Even unimodular lattices in Rn only exist if 8 n (this 8 is the same 8 | that shows up in the periodicity of Clifford groups). The E8 lattice is the only example in dimension equal to 8 (up to isomorphism, of course). There are two in dimension 16 (one of which is L L, the other is D some ⊕ 16∪ coset). There are 24 in dimension 24, which are the Niemeier lattices. In 32 dimensions, there are more than a billion! Lecture 25 - E8 130

The Weyl group of E8 is generated by the reflections through s⊥ where s L and (s,s) = 2 (these are called roots). First, let’s find all the roots: ∈ (x ,...,x ) such that x2 = 2 with x Z or Z + 1 and x even. If 1 8 i i ∈ 2 i x Z, obviously the only solutions are permutations of ( 1, 1, 06), of i ∈ P ±P ± which there are 8 22 = 112 choices. In the Z + 1 case, you can choose 2 × 2 the first 7 places to be 1 , and the last coordinate is forced, so there are 27 ± 2 choices. Thus, you get 240 roots. Let’s find the orbits of the roots under the action of the Weyl group. We don’t yet know what the Weyl group looks like, but we can find a large subgroup that is easy to work with. Let’s use the Weyl group of D8, which consists of the following: we can apply all permutations of the coordinates, or we can change the sign of an even number of coordinates (e.g., reflection in (1, 1, 06) swaps the first two coordinates, and reflection − in (1, 1, 06) followed by reflection in (1, 1, 06) changes the sign of the first − two coordinates.) Notice that under the Weyl group of D8, the roots form two orbits: the 8 set which is all permutations of ( 12, 06), and the set ( 1 ). Do these ± ± 2 become the same orbit under the Weyl group of E8? Yes; to show this, we just need one element of the Weyl group of E8 taking some element of 1 8 the first orbit to the second orbit. Take reflection in ( 2 )⊥ and apply it to 2 6 (12, 06): you get ( 1 , 1 ), which is in the second orbit. So there is just one 2 − 2 orbit of roots under the Weyl group. What do orbits of W (E8) on other vectors look like? We’re interested in this because we might want to do representation theory. The character of a representation is a map from weights to integers, which is W (E8)-invariant. 2 Let’s look at vectors of norm 4 for example. So xi = 4, xi even, and 1 xi Z or xi Z + 2 . There are 8 2 possibilities which are permutations of ∈ 7 ∈ 8 4 × 4P 4 P 7 ( 2, 0 ). There are 4 2 permutations of ( 1 , 0 ), and there are 8 2 ± ×7 ± × permutations of ( 3 , 1 ). So there are a total of 240 9 of these vectors. ± 2 ± 2 × There are 3 orbits under W (D8), and as before, they are all one orbit under 8 the action of W (E ). Just reflect (2, 07) and (13, 1, 04) through ( 1 ). 8 − 2 ◮ Exercise 25.1. Show that the number of norm 6 vectors is 240 28, and × they form one orbit

(If you’ve seen a course on modular forms, you’ll know that the number of vectors of norm 2n is given by 240 d3. If you let call these c , × d n n then c qn is a modular form of level 1 (E| even, unimodular), weight 4 n P 8 (dim E /2).) P8 For norm 8 there are two orbits, because you have vectors that are twice a norm 2 vector, and vectors that aren’t. As the norm gets bigger, you’ll get a large number of orbits. Lecture 25 - E8 131

What is the order of the Weyl group of E8? We’ll do this by 4 diﬀerent methods, which illustrate the diﬀerent techniques for this kind of thing:

(1) This is a good one as a mnemonic. The order of E8 is given by

numbers on the Weight lattice of E8 W (E8) = 8! 1 | | × aﬃne E8 diagram × Root lattice of E8 Y 3ØÙ = 8! 1 × 1ØÙ 2ØÙ 3ØÙ 4ØÙ 5ØÙ 6ØÙ 4ØÙ 2ØÙ × = 214 35 52 7 × × ×

We can do the same thing for any other Lie algebra, for example,

1 2 3 4 2 W (F ) = 4! ( / ) 1 | 4 | × ØÙ ØÙ ØÙ ØÙ ØÙ × = 27 32 × (2) The order of a reﬂection group is equal to the products of degrees of the fundamental invariants. For E8, the fundamental invariants are of degrees 2,8,12,14,18,20,24,30 (primes +1).

(3) This one is actually an honest method (without quoting weird facts). The only fact we will use is the following: suppose G acts transitively on a set X with H = the group fixing some point; then G = H X . | | | |·| | This is a general purpose method for working out the orders of groups. First, we need a set acted on by the Weyl group of E8. Let’s take the root vectors (vectors of norm 2). This set has 240 elements, and the Weyl group of E acts transitively on it. So W (E ) = 240 subgroup 8 | 8 | ×| fixing (1, 1, 06) . But what is the order of this subgroup (call it G )? − | 1 Let’s find a set acted on by this group. It acts on the set of norm 2 vectors, but the action is NOT transitive. What are the orbits? G1 fixes s = (1, 1, 06). For other roots r, G obviously fixes (r, s). So − 1 how many roots are there with a given inner product with s?

(s, r) number choices 2 1 s 6 1 56 (1, 0, 16), (0, 1, 16), ( 1 , 1 , 1 ) ± − ± 2 − 2 2 0 126 1 56 − 2 1 s − − 1These are the numbers giving highest root. Lecture 25 - E8 132

So there are at least 5 orbits under G1. In fact, each of these sets is a single orbit under G1. How can we see this? Find a large subgroup of G1. Take W (D6), which is all permutations of the last 6 coordinates and all even sign changes of the last 6 coordinates. It is generated by reflections associated to the roots orthogonal to e1 and e2 (those that start with two 0s). The three cases with inner product 1 are three orbits under W (D6). To see that there is a single orbit under G1, we just need some reflections that mess up these orbits. If you take a 6 vector ( 1 , 1 , 1 ) and reflect norm 2 vectors through it, you will get 2 2 ± 2 exactly 5 orbits. So G1 acts transitively on these orbits. We’ll use the orbit of vectors r with (r, s)= 1. Let G be the vectors − 2 fixing s and r: s r We have that G1 = G2 56. ØÙ ØÙ | | | | · Keep going ... it gets tedious, but here are the answers up to the last step:

Our plan is to chose vectors acted on by Gi, fixed by Gi+1 which give us the Dynkin diagram of E8. So the next step is to try to find vectors t that give us the picture s r t , i.e, they have inner product ØÙ ØÙ ØÙ 1 with r and 0 with s. The possibilities for t are ( 1, 1, 0, 05) (one − − − of these), (0, 0, 1, 1, 04) and permutations of its last five coordinates ± 5 (10 of these), and ( 1 , 1 , 1 , 1 ) (there are 16 of these), so we get 27 − 2 − 2 2 ± 2 total. Then we could check that they form one orbit, which is boring. Next find vectors which go next to t in our picture: s r t , i.e., whose inner product is 1 with t and zero ØÙ ØÙ ØÙ ØÙ − with r, s. The possibilities are permutations of the last four coords of 4 (0, 0, 0, 1, 1, 03 ) (8 of these) and ( 1 , 1 , 1 , 1 , 1 ) (8 of these), so ± − 2 − 2 − 2 2 ± 2 there are 16 total. Again check transitivity. Find a fifth vector; the possibilities are (04, 1, 1, 02) and perms of the 4 ±3 last three coords (6 of these), and ( 1 , 1 , 1 ) (4 of these) for a total − 2 2 ± 2 of 10. For the sixth vector, we can have (05, 1, 1, 0) or (05, 1, 0, 1) (4 pos- 5 2 ± ± sibilites) or ( 1 , 1 , 1 ) (2 possibilities), so we get 6 total. − 2 2 ± 2 NEXT CASE IS TRICKY: finding the seventh one, the possibilities are (06, 1, 1) (2 of these) and (( 1 )6, 1 , 1 ) (just 1). The proof of tran- ± − 2 2 2 sitivity fails at this point. The group we’re using by now doesn’t even act transitively on the pair (you can’t get between them by changing an even number of signs). What elements of W (E8) fix all of s r t these first 6 points ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ? We want to find roots perpendicular to all of these vectors, and the only possibility is 1 8 (( 2 ) ). How does reflection in this root act on the three vectors above? Lecture 25 - E8 133

2 (06, 12) (( 1 )6, 1 ) and (06, 1, 1) maps to itself. Is this last vector 7→ − 2 2 − in the same orbit? In fact they are in diﬀerent orbits. To see this, look for vectors

? ØÙ s r t (06, 1, 1) ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ± 6 completing the E8 diagram. In the (0 , 1, 1) case, you can take the vector (( 1 )5, 1 , 1 , 1 ). But in the other case, you can show that − 2 2 2 − 2 there are no possibilities. So these really are diﬀerent orbits. Use the orbit with 2 elements, and you get

order of W (E6) W (E ) = 240 56 27 16 10 6 2 1 | 8 | × × × × × × × z order of W}|(E7) { because the group ﬁxing all| 8 vectors must{z be trivial. You} also get that W (A2 A1) | × | W (“E ”) = 16 10 6 2 1 | 5 | × × × × W (A4) | z }|| { | {z ØÙ } where “E5” is the algebra with diagram ØÙ ØÙ ØÙ ØÙ (that is, D5). Similarly, E is A and E is A A . 4 4 3 2 × 1 We got some other information. We found that the Weyl group of E8 acts transitively on all the conﬁgurations

ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ but not on ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ (4) We’ll slip this in to next lecture

Also, next time we’ll construct the Lie algebra E8. Lecture 26 134

Lecture 26

Today we’ll finish looking at W (E8), then we’ll construct E8. Remember that we still have a fourth method of finding the order of W (E8). Let L be the E8 lattice. Look at L/2L, which has 256 elements. Look at this as a set acted on by W (E8). There is an orbit of size 1 (represented by 0). There is an orbit of size 240/2 = 120, which are the roots (a root is congruent mod 2L to it’s negative). Left over are 135 elements. Let’s look at norm 4 vectors. Each norm 4 vector, r, satisfies r r mod 2, and ≡− there are 240 9 of them, which is a lot, so norm 4 vectors must be congruent · to a bunch of stuff. Let’s look at r = (2, 0, 0, 0, 0, 0, 0, 0). Notice that it is congruent to vectors of the form (0 2 . . . 0), of which there are 16. It ···± is easy to check that these are the only norm 4 vectors congruent to r mod 2. So we can partition the norm 4 vectors into 240 9/16 = 135 subsets of · 16 elements. So L/2L has 1+120+135 elements, where 1 is the zero, 120 is represented by 2 elements of norm 2, and 135 is represented by 16 elements of norm 4. A set of 16 elements of norm 4 which are all congruent is called a FRAME. It consists of elements e ,..., e , where e2 = 4 and (e , e ) = 1 ± 1 ± 8 i i j for i = j, so up to sign it is an orthogonal basis. 6 Then we have

W (E ) = (# frames) subgroup fixing a frame | 8 | × | | because we know that W (E8) acts transitively on frames. So we need to know what the automorphisms of an orthogonal base are. A frame is 8 subsets of the form (r, r), and isometries of a frame form the group (Z/2Z)8 S , − · 8 but these are not all in the Weyl group. In the Weyl group, we found a (Z/2Z)7 S , where the first part is the group of sign changes of an EVEN · 8 number of coordinates. So the subgroup fixing a frame must be in between these two groups, and since these groups differ by a factor of 2, it must be one of them. Observe that changing an odd number of signs doesn’t preserve the E lattice, so it must be the group (Z/2Z)7 S , which has order 27 8!. 8 · 8 · So the order of the Weyl group is # norm 4 elements 135 27 8! = 27 S · · | · 8|× 2 dim L × Remark 26.1. Similarly, if Λ is the Leech lattice, you actually get the order of Conway’s group to be # norm 8 elements 212 M | · 24| · 2 dim Λ × where M24 is the Mathieu group (one of the sporadic simple groups). The Leech lattice seems very much to be trying to be the root lattice of the Lecture 26 135 monster group, or something like that. There are a lot of analogies, but nobody can make sense of it. 8 W (E8) acts on (Z/2Z) , which is a vector space over F2, with quadratic (a,a) form N(a)= 2 mod 2, so you get a map 1 W (E ) O+(F ) ± → 8 → 8 2 which has kernel 1 and is surjective. O+ is one of the 8-dimensional orthog- ± 8 onal groups over F2. So the Weyl group is very close to being an orthogonal group of a vector space over F2. What is inside the root lattice/Lie algebra/Lie group E8? One obvious way to find things inside is to cover nodes of the E8 diagram:

ØÙ

ØÙ ØÙ ×ØÙ ØÙ ØÙ ØÙ ØÙ If we remove the shown node, you see that E contains A D . We can do 8 2 × 5 better by showing that we can embed the aﬃne E˜8 in the E8 lattice.

ØÙ

highest rootØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ − simple roots

Now you can remove nodes| here and get some{z bigger sub-diagram} s. For example, if we cover

ØÙ

ØÙ ×ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ you get that an A E in E . The E consisted of 126 roots orthogonal to 1 × 7 8 7 a given root. This gives an easy construction of E7 root system, as all the elements of the E lattice perpendicular to (1, 1, 0 . . . ) 8 − We can cover ØÙ

ØÙ ØÙ ×ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ Then we get an A E , where the E are all the vectors with the ﬁrst 3 2 × 6 6 coordinates equal. So we get the E6 lattice for free too. If you cover

ØÙ

ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ×ØÙ Lecture 26 136

you see that there is a D8 in E8, which is all vectors of the E8 lattice with integer coordinates. We sort of constructed the E8 lattice this way in the ﬁrst place. We can ask questions like: What is the E8 Lie algebra as a representation of D8? To answer this, we look at the weights of the E8 algebra, considered as a module over D , which are the 112 roots of the form ( 1 1 . . . ) 8 ···± ···± and the 128 roots of the form ( 1/2,... ) and 1 vector 0, with multiplicity ± 8. These give you the Lie algebra of D8. Recall that D8 is the Lie algebra of SO16. The double cover has a half spin representation of dimension 16/2 1 2 − = 128. So E8 decomposes as a representation of D8 as the adjoint representation (of dimension 120) plus a half spin representation of dimension 128. This is often used to construct the Lie algebra E8. We’ll do a better construction in a little while. We’ve found that the Lie algebra of D8, which is the Lie algebra of SO16, is contained in the Lie algebra of E8. Which group is contained in the the compact form of the E8? We found that there were groups

Spin16(R) h XXXX hhhh XXXXX hhhhh XX SO16(R) Spin16(R)/(Z/2Z) ∼= Spin16(R)/(Z/2Z) VVV 89?> :;=< VVVV fffff VV ffffff P SO16(R) corresponding to subgroups of the center (Z/2Z)2:

1 S kkk SSS kkk SSS kkk SS Z/2Z Z/2Z Z/2Z R RRR lll RR lll (Z/2Z)2

We have a homomorphism Spin (R) E (compact). What is the kernel? 16 → 8 The kernel are elements which act trivially on the Lie algebra of E8, which is equal to the Lie algebra D8 plus the half spin representation. On the Lie algebra of D8, everything in the center is trivial, and on the half spin representation, one of the elements of order 2 is trivial. So the subgroup that you get is the circled one. ◮ Exercise 26.1. Show SU(2) E (compact)/( 1, 1) is a subgroup of × 7 − − E8 (compact). Similarly, show that SU(9)/(Z/3Z) is also. These are similar to the example above.

Construction of E8 Earlier in the course, we had some constructions: Lecture 26 137

1. using the Serre relations, but you don’t really have an idea of what it looks like

2. Take D8 plus a half spin representation Today, we’ll try to find a natural map from root lattices to Lie algebras. The idea is as follows: Take a basis element eα (as a formal symbol) for each root α; then take the Lie algebra to be the direct sum of 1-dimensional α spaces generated by each e and L (L root lattice ∼= Cartan subalgebra) . Then we have to define the Lie bracket by setting [eα, eβ]= eα+β, but then we have a sign problem because [eα, eβ] = [eβ, eα]. Is there some way to 6 − resolve the sign problem? The answer is that there is no good way to solve this problem (not true, but whatever). Suppose we had a nice functor from root lattices to Lie algebras. Then we would get that the automorphism group of the lattice has to be contained in the automorphism group of the Lie algebra (which is contained in the Lie group), and the automorphism group of the Lattice contains the Weyl group of the lattice. But the Weyl group is NOT usually a subgroup of the Lie group. We can see this going wrong even in the case of sl2(R). Remember that a 0 0 b the Weyl group is N(T )/T where T = −1 and N(T ) = T −1 , 0 a ∪ b 0 and this second part is stuff having order 4, so you cannot possibly− write this as a semi-direct product of T and the Weyl group. So the Weyl group is not usually a subgroup of N(T ). The best we can do is to find a group of the form 2n W N(T ) where n is the rank. · ⊆ For example, let’s do it for SL(n + 1, R) Then T = diag(a1,...,an) with a a = 1. Then we take the normalizer of the torus to be N(T ) =all 1 · · · n permutation matrices with 1’s with determinant 1, so this is 2n S , and ± · n it does not split. The problem we had with signs can be traced back to the fact that this group doesn’t split. We can construct the Lie algebra from something acted on by 2n W (but · not from something acted on by W ). We take a CENTRAL EXTENSION of the lattice by a group of order 2. Notation is a pain because the lattice is written additively and the extension is nonabelian, so you want it to be written multiplicatively. Write elements of the lattice in the form eα formally, so we have converted the lattice operation to multiplication. We will use the central extension

1 1 eˆL eL 1 →± → → → ∼=L We wante ˆL to have the property thate ˆα|{z}eˆβ = ( 1)(α,β)eˆβeˆα, wheree ˆα is − something mapping to eα. What do the automorphisms ofe ˆL look like? We Lecture 26 138 get 1 (L/2L) Aut(êL) Aut(eL) → → → (Z/2)rank(L) for α L/2L, we get the| {z map} e ˆβ ( 1)(α,β)eˆβ. The map turns out to ∈ → − be onto, and the group Aut(eL) contains the reflection group of the lattice. This extension is usually non-split. Now the Lie algebra is L 1-dimensional spaces spanned by (êα, eˆα) ⊕ { − } for α2 = 2 with the convention that eˆα ( 1 in the vector space) is eˆα (-1 − − − in the groupe ˆL). Now define a Lie bracket by the “obvious rules” [α, β] = 0 for α, β L (the Cartan is abelian), [α, eˆβ ] = (α, β)êβ (êβ is in the root space ∈ of β), and [êα, eˆβ] = 0 if (α, β) 0 (since (α + β)2 > 2), [êα, eˆβ] =e ˆαeˆβ if ≥ L α α 1 (α, β) < 0 (product in the groupe ˆ ), and [ê , (ê )− ]= α. Theorem 26.2. Assume L is positive definite. Then this Lie bracket forms a Lie algebra (so it is skew and satisfies Jacobi).

Proof. Easy but tiresome, because there are a lot of cases; let’s do them (or most of them). We check the Jacobi identity: We want [[a, b], c]+ [[b, c], a]+ [[c, a], b] = 0

1. all of a, b, c in L. Trivial because all brackets are zero.

2. two of a, b, c in L. Say α, β, eγ

[[α, β], eγ ] + [[β, eγ ], α] +[[eγ , α], β]

0 (β,α)( α,β)eγ − and similar for the| third{z term,} | giving{z a} sum of 0.

3. one of a, b, c in L. α, eβ, eγ . eβ has weight β and eγ has weight γ and eβeγ has weight β + γ. So check the cases, and you get Jacobi:

[[α, eβ ], eγ ] = (α, β)[eβ , eγ ] [[eβ, eγ ], α]= [α, [eβ, eγ ]] = (α, β + γ)[eβ , eγ ] − − [[eγ , α], eβ ]= [[α, eγ ], eβ ] = (α, γ)[eβ , eγ ], − so the sum is zero.

4. none of a, b, c in L. This is the really tiresome one, eα, eβ, eγ . The main point of going through this is to show that it isn’t as tiresome as you might think. You can reduce it to two or three cases. Let’s make our cases depending on (α, β), (α, γ), (β, γ).

(a) if 2 of these are 0, then all the [[ , ], ] are zero. ∗ ∗ ∗ Lecture 26 139

(b) α = β. By case a, γ cannot be orthogonal to them, so say − (α, γ)=1(γ, β)= 1; adjust so that eαeβ = 1, then calculate − [[eγ , eβ], eα] [[eα, eβ], eγ ] + [[eα, eγ ], eβ] − = eαeβeγ (α, γ)eγ + 0 − = eγ eγ = 0. − (c) α = β = γ, easy because [eα, eγ ] = 0 and [[eα, eβ], eγ ] = − [[eγ , eβ], eα] − (d) We have that each of the inner products is 1, 0 or 1. If some − (α, β) = 1, all brackets are 0.

This leaves two cases, which we’ll do next time Lecture 27 140

Lecture 27

Last week we talked aboute ˆL, which was a double cover of eL. L is the root lattice of E8. We had the sequence 1 1 eˆL eL 1. →± → → → The Lie algebra structure one ˆL was given by [α, β] = 0 [α, eβ] = (α, β)eβ 0 if (α, β) 0 ≥ [eα, eβ]= eαeβ if (α, β)= 1  − α if (α, β)= 2 − α The Lie algebra is L 2 eˆ. ⊕ α =2 Let’s ﬁnish checking the Jacobi identity. We had two cases left: L [[eα, eβ], eγ ] + [[eβ, eγ ], eα] + [[eγ , eα], eβ] = 0 (α, β) = (β, γ) = (γ, α) = 1, in which case α + β + γ = 0. then − − [[eα, eβ], eγ ] = [eαeβ, eγ ] = α + β. By symmetry, the other two terms are β + γ and γ + α;the sum of all three terms is 2(α + β + γ) = 0. (α, β) = (β, γ) = 1, (α, γ) = 0, in which case [eα, eγ ] = 0. We − − check that [[eα, eβ], eα] = [eαeβ, eγ ]= eαeβeγ (since (α + β, γ)= 1). − Similarly, we have [[eβ, eγ ], eα] = [eβeγ , eα] = eβeγ eα. We notice that eαeβ = eβeα and eγ eα = eαeγ so eαeβeγ = eβeγeα; again, the sum − − of all three terms in the Jacobi identity is 0. This concludes the veriﬁcation of the Jacobi identity, so we have a Lie algebra. Is there a proof avoiding case-by-case check? Good news: yes! Bad news: it’s actually more work. We really have functors as follows:

elementary, Dynkin but tedious / Double / Lie algebras diagrams cover Lˆ only for positive O KK deﬁnite lattices KK KK KK i KK KK KK W KK K% , Root lattice L Vertex algebras

these work for any even lattice Lecture 27 141

where Lˆ is generated bye ˆαi (the i’s are the dots in your Dynkin diagram), withe ˆαi eˆαj = ( 1)(αi,αj )eˆαj eˆαi , and 1 is central of order 2. − − Unfortunately, you have to spend several weeks learning vertex algebras. In fact, the construction we did was the vertex algebra approach, with all the vertex algebras removed. So there is a more general construction which gives a much larger class of inﬁnite-dimensional Lie algebras. Now we should study the double cover Lˆ, and in particular prove its existence. Given a Dynkin diagram, we can construct Lˆ as generated by the α elements e i for αi simple roots with the given relations. It is easy to check that we get a surjective homomorphism Lˆ L with kernel generated by z → with z2 = 1. What’s a little harder to show is that z = 1 (i.e., show that 6 Lˆ = L). The easiest way to do it is to use cohomology of groups, but since 6 we have such an explicit case, we’ll do it bare hands: Problem: Given Z, H groups with Z abelian, construct central extensions

1 Z G H 1 → → → → (where Z lands in the center of G). Let G be the set of pairs (z, h), and set the product (z , h )(z , h ) = (z z c(h , h ), h h ), where c(h , h ) Z 1 1 2 2 1 2 1 2 1 2 1 2 ∈ (c(h1, h2) will be a cocycle in group cohomology). We obviously get a homomorphism by mapping (z, h) h. If c(1, h) = c(h, 1) = 1 (normalization), 7→ then z (z, 1) is a homomorphism mapping Z to the center of G. In par- 7→ ticular, (1, 1) is the identity. We’ll leave it as an exercise to ﬁgure out what the inverses are. When is this thing associative? Let’s just write everything out:

(z1, h1)(z2, h2) (z3, h3) = (z1z2z3c(h1, h2)c(h1h2, h3), h1h2h3) (z1, h1) (z2, h2)(z3, h3) = (z1z2z3c(h1, h2h3)c(h2, h3), h1h2h3) so we must have

c(h1, h2)c(h1h2, h3)= c(h1h2, h3)c(h2, h3).

This identity is actually very easy to satisfy in one particular case: when c is bimultiplicative: c(h1, h2h3) = c(h1, h2)c(h1, h3) and c(h1h2, h3) = c(h , h )c(h , h ). That is, we have a map H H Z. Not all cocycles 1 3 2 3 × → come from such maps, but this is the case we care about. To construct the double cover, let Z = 1 and H = L (free abelian). ± If we write H additively, we want c to be a bilinear map L L 1. × → ± It is really easy to construct bilinear maps on free abelian groups. Just take any basis α1,...,αn of L, choose c(α1, αj) arbitrarily for each i, j and extend c via bilinearity to L L. In our case, we want to find a double × cover Lˆ satisfyinge ˆαeˆβ = ( 1)(α,β)eˆβeˆα wheree ˆα is a lift of eα. This just − Lecture 27 142 means that c(α, β) = ( 1)(α,β)c(β, α). To satisfy this, just choose c(α , α ) − i j on the basis α so that c(α , α ) = ( 1)(αi,αj )c(α , α ). This is trivial to { i} i j − j i do as ( 1)(αi,αi) = 1. Notice that this uses the fact that the lattice is even. − There is no canonical way to choose this 2-cocycle (otherwise, the central extension would split as a product), but all the different double covers are isomorphic because we can specify Lˆ by generators and relations. Thus, we have constructed Lˆ (or rather, verified that the kernel of Lˆ L has order → 2, not 1). Let’s now look at lifts of automorphisms of L to Lˆ.

◮ Exercise 27.1. Any automorphism of L preserving ( , ) lifts to an automorphism of Lˆ

There are two special cases:

1. 1 is an automorphism of L, and we want to lift it to Lˆ explicitly. − α α α 1 First attempt: try sendinge ˆ toe ˆ− := (ˆe )− , which doesn’t work 1 because a a− is not an automorphism on non-abelian groups. 7→ 2 Better: ω :e ˆα ( 1)α /2(ˆeα) 1 is an automorphism of Lˆ. To see 7→ − − this, check

α β (α2+β2)/2 α 1 β 1 ω(ê )ω(ê ) = ( 1) (ê )− (ê )− − α β (α+β)2 /2 β 1 α 1 ω(ê eˆ ) = ( 1) (ê )− (ê )− − which work out just right

2. If r2 = 2, then α α (α, r)r is an automorphism of L (reﬂection 7→ − (α,r) through r ). You can lift this bye ˆα eˆα(ˆer) (α,r) ( 1)( 2 ). This ⊥ 7→ − × − is a homomorphism (check it!) of order (usually) 4! Remark 27.1. Although automorphisms of L lift to automorphisms of Lˆ, the lift might have larger order.

This construction works for the root lattices of An, Dn, E6, E7, and E8; these are the lattices which are even, positive definite, and generated by vectors of norm 2 (in fact, all such lattices are sums of the given ones). What about Bn, Cn, F4 and G2? The reason the construction doesn’t work for these cases is because there are roots of different lengths. These all occur as fixed points of diagram automorphisms of An, Dn and E6. In fact, we have a functor from Dynkin diagrams to Lie algebras, so and automorphism of the diagram gives an automorphism of the algebra Lecture 27 143

Involution Fixed points Involution Fixed Points ØÙ

ØÙ ØÙ Y . z | ... " $ . = Cn+1 × / ØÙ ØÙ ØÙ ØÙ D4 = ØÙ ØÙ 11 ØÙ ØÙ = A2n+1 11 = G ØÙ / 2 ØÙ ØÙ ØÙ E = ØÙ Z 6 z | " $ ØÙ Dn = ... ? ... o = F4 ØÙ ØÙ ?? ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ? Ô = Bn ØÙ ØÙ ØÙ ØÙ A2n doesn’t really give you a new algebra: it corresponds to some superalgebra stuﬀ.

Construction of the Lie group of E8 It is the group of automorphisms of the Lie algebra generated by the elements exp(λAd(êα)), where λ is some real number,e ˆα is one of the basis elements of the Lie algebra corresponding to the root α, and Ad(êα)(a)=[êα, a]. In other words,

λ2 exp(λAd(êα))(a)=1+ λ[êα, a]+ [êα, [êα, a]]. 2 and all the higher terms are zero. To see that Ad(êα)3 = 0, note that if β is a root, then β + 3α is not a root (or 0).

Warning 27.2. In general, the group generated by these automorphisms is NOT the whole automorphism group of the Lie algebra. There might be extra diagram automorphisms, for example.

We get some other things from this construction. We can get simple groups over finite fields: note that the construction of a Lie algebra above works over any commutative ring (e.g. over Z). The only place we used division is in exp(λAd(êα)) (where we divided by 2). The only time this α α term is non-zero is when we apply exp(λAd(ê )) toe ˆ− , in which case we find that [êα, [êα, eˆ α]] = [êα, α] = (α, α)êα, and the fact that (α, α) = 2 − − cancels the division by 2. So we can in fact construct the E8 group over any commutative ring. You can mumble something about group schemes over Z at this point. In particular, we have groups of type E8 over finite fields, which are actually finite simple groups (these are called Chevalley groups; it takes work to show that they are simple, there is a book by Carter called Finite Simple Groups which you can look at). Lecture 27 144

Real forms

So we’ve constructed the Lie group and Lie algebra of type E8. There are in fact several diﬀerent groups of type E8. There is one complex Lie algebra of type E8, which corresponds to several diﬀerent real Lie algebras of type E8. Let’s look at some smaller groups:

a b Example 27.3. sl2(R) = c d with a, b, c, d real a + d = 0; this is not compact. On the other hand, su (R) = a b with d = a imaginary 2 c d − b = c¯, is compact. These have the same Lie algebra over C. − Let’s look at what happens for E8. In general, suppose L is a Lie algebra with complexification L C. How can we find another Lie algebra M with ⊗ the same complexification? L C has an anti-linear involution ω : l z ⊗ L ⊗ 7→ l z¯. Similarly, it has an anti-linear involution ω . Notice that ω ω is ⊗ M L M a linear involution of L C. Conversely, if we know this involution, we can ⊗ reconstruct M from it. Given an involution ω of L C, we can get M as ⊗ the fixed points of the map a ω ω(a)“=” ω(a). Another way is to put 7→ L L = L+ L , which are the +1 and 1 eigenspaces, then M = L+ iL . ⊕ − − ⊕ − Thus, to find other real forms, we have to study the involutions of the complexification of L. The exact relation is kind of subtle, but this is a good way to go. Example 27.4. Let L = sl (R). It has an involution ω(m)= mT . su (R) 2 − 2 is the set of fixed points of the involution ω times complex conjugation on sl2(C), by definition.

So to construct real forms of E8, we want some involutions of the Lie algebra E8 which we constructed. What involutions do we know about? There are two obvious ways to construct involutions:

2 1. Lift 1 on L toe ˆα ( 1)α /2(ˆeα) 1, which induces an involution on − 7→ − − the Lie algebra.

2. Take β L/2L, and look at the involutione ˆα ( 1)(α,β)eˆα. ∈ 7→ − (2) gives nothing new ... you get the Lie algebra you started with. (1) only gives you one real form. To get all real forms, you multiply these two kinds of involutions together. Recall that L/2L has 3 orbits under the action of the Weyl group, of size 1, 120, and 135. These will correspond to the three real forms of E8. How do we distinguish diﬀerent real forms? The answer was found by Cartan: look at the signature of an invariant quadratic form on the Lie algebra! A bilinear form ( , ) on a Lie algebra is called invariant if ([a, b], c)+ (b[a, c]) = 0 for all a, b, c. This is called invariant because it corresponds to Lecture 27 145 the form being invariant under the corresponding group action. Now we can construct an invariant bilinear form on E8 as follows:

1. (α, β)in the Lie algebra = (α, β)in the lattice

α α 1 2. (ˆe , (ˆe )− ) = 1 3. (a, b)=0if a and b are in root spaces α and β with α + β = 0. 6 This gives an invariant inner product on E8, which you prove by case-by-case check

◮ Exercise 27.2. do these checks

Next time, we’ll use this to produce bilinear forms on all the real forms and then we’ll calculate the signatures. Lecture 28 146

Lecture 28

Last time, we constructed a Lie algebra of type E , which was L eˆα, 8 ⊕ where L is the root lattice and α2 = 2. This gives a double cover of the root L lattice: 1 1 eˆL eL 1. →± → → → 2 We had a lift for ω(α) = α, given by ω(êα) = ( 1)(α /2)(êα) 1. So ω − − − becomes an automorphism of order 2 on the Lie algebra. eα ( 1)(α,β)eα 7→ − is also an automorphism of the Lie algebra. Suppose σ is an automorphism of order 2 of the real Lie algebra L = + L + L− (eigenspaces of σ). We saw that you can construct another real + form given by L + iL−. Thus, we have a map from conjugacy classes of automorphisms with σ2 = 1 to real forms of L. This is not in general in isomorphism. Today we’ll construct some more real forms of E8. E8 has an invari- α α 1 ant symmetric bilinear form (e , (e )− ) = 1, (α, β) = (β, α). The form is unique up to multiplication by a constant since E8 is an irreducible representation of E8. So the absolute value of the signature is an invariant of the Lie algebra. For the split form of E8, what is the signature of the invariant bilinear form (the split form is the one we just constructed)? On the Cartan subalgebra L, ( , ) is positive definite, so we get +8 contribution to the signature. On eα, (eα) 1 , the form is 0 1 , so it has signature 0 120. Thus, the { − } 1 0 · signature is 8. So if we find any real form with a different signature, we’ll have found a new Lie algebra. Let’s first try involutions eα ( 1)(α,β)eα. But this doesn’t change the 7→ − 0 1 0 1 signature. L is still positive definite, and you still have 1 0 or 1− 0 on the other parts. These Lie algebras actually turn out to be isomorphic− to what we started with (though we haven’t shown that they are isomorphic). 2 Now try ω : eα ( 1)α /2(eα) 1, α α. What is the signature of the 7→ − − 7→ − form? Let’s write down the + and eigenspaces of ω. The + eigenspace − will be spanned by eα e α, and these vectors have norm 2 and are − − − orthogonal. The eigenspace will be spanned by eα + e α and L, which − − have norm 2 and are orthogonal, and L is positive definite. What is the Lie algebra corresponding to the involution ω? It will be spanned by eα e α − − where α2 = 2 (norm 2), and i(eα + e α) (norm 2), and iL (which is now − − − negative definite). So the bilinear form is negative definite, with signature 248(= 8). − 6 ± With some more work, you can actually show that this is the Lie algebra of the compact form of E8. This is because the automorphism group of E8 preserves the invariant bilinear form, so it is contained in O0,248(R), which is compact. Lecture 28 147

Now let’s look at involutions of the form eα ( 1)(α,β)ω(eα). Notice 7→ − that ω commutes with eα ( 1)(α,β)eα. The β’s in (α, β) correspond to 7→ − L/2L modulo the action of the Weyl group W (E8). Remember this has three orbits, with 1 norm 0 vector, 120 norm 2 vectors, and 135 norm 4 vectors. The norm 0 vector gives us the compact form. Let’s look at the other cases and see what we get. Suppose V has a negative definite symmetric inner product ( , ), and suppose σ is an involution of V = V+ V (eigenspaces of σ). What is ⊕ − the signature of the invariant inner product on V+ iV ? On V+, it is ⊕ − negative definite, and on iV it is positive definite. Thus, the signature is − dim V dim V+ = tr(σ). So we want to work out the traces of these − − − involutions. Given some β L/2L, what is tr(eα ( 1)(α,β)eα)? If β = 0, the ∈ 7→ − traces is obviously 248 because we just have the identity map. If β2 = 2, we need to figure how many roots have a given inner product with β. Recall that this was determined before: (α, β) # of roots α with given inner product eigenvalue 2 1 1 1 56 -1 0 126 1 -1 56 -1 -2 1 1

Thus, the trace is 1 56 + 126 56 + 1 + 8 = 24 (the 8 is from the Cartan − − subalgebra). So the signature of the corresponding form on the Lie algebra is 24. We’ve found a third Lie algebra. − If we also look at the case when β2 = 4, what happens? How many α with α2 = 2 and with given (α, β) are there? In this case, we have:

(α, β) # of roots α with given inner product eigenvalue 2 14 1 1 64 -1 0 84 1 -1 64 -1 -2 14 1

The trace will be 14 64 + 84 64+14+8 = 8. This is just the split form − − − again. Summary: We’ve found 3 forms of E8, corresponding to 3 classes in L/2L, with signatures 8, 24, 248, corresponding to involutions eα − − 7→ ( 1)(α,β)e α of the compact form. If L is the compact form of a simple − − Lie algebra, then Cartan showed that the other forms correspond exactly to the conjugacy classes of involutions in the automorphism group of L (this Lecture 28 148 doesn’t work if you don’t start with the compact form — so always start with the compact form). In fact, these three are the only forms of E8, but we won’t prove that.

Working with simple Lie groups As an example of how to work with simple Lie groups, we will look at the general question: Given a simple Lie group, what is its homotopy type? Answer: G has a unique conjugacy class of maximal compact subgroups K, and G is homotopy equivalent to K.

Proof for GLn(R). First pretend GLn(R) is simple, even though it isn’t; whatever. There is an obvious compact subgroup: On(R). Suppose K is any compact subgroup of GLn(R). Choose any positive deﬁnite form ( , ) on Rn. This will probably not be invariant under K, but since K is compact, we can average it over K get one that is: deﬁne a new form

(a, b)new = K (ka, kb) dk. This gives an invariant positive definite bilinear form (since integral of something positive definite is positive definite). Thus, R any compact subgroup preserves some positive definite form. But the subgroup fixing some positive definite bilinear form is conjugate to a subgroup of On(R) (to see this, diagonalize the form). So K is contained in a conjugate of On(R). Next we want to show that G = GLn(R) is homotopy equivalent to On(R) = K. We will show that G = KAN, where K is On, A is all diagonal matrices with positive coefficients, and N is matrices which are upper triangular with 1s on the diagonal. This is the Iwasawa decomposition. In general, we get K compact, A semi-simple abelian, and N is unipotent. The proof of this you saw before was called the Grahm-Schmidt process for n orthonormalizing a basis. Suppose v1,...,vn is any basis for R . 1. Make it orthogonal by subtracting some stuff, you’ll get v , v v , 1 2 − ∗ 1 v v v , . . . . 3 − ∗ 2 − ∗ 1 2. Normalize by multiplying each basis vector so that it has norm 1. Now we have an orthonormal basis.

This is just another way to say that GLn can be written as KAN. Making things orthogonal is just multiplying by something in N, and normalizing is just multiplication by some diagonal matrix with positive entries. An orthonormal basis is an element of On. Tada! This decomposition is just a topological one, not a decomposition as groups. Uniqueness is easy to check. n(n 1)/2 Now we can get at the homotopy type of GLn. N ∼= R − , and + n A ∼= (R ) , which are contractible. Thus, GLn(R) has the same homotopy type as On(R), its maximal compact subgroup. Lecture 28 149

If you wanted to know π1(GL3(R)), you could calculate π1(O3(R)) ∼= Z/2Z, so GL3(R) has a double cover. Nobody has shown you this double cover because it is not algebraic.

Example 28.1. Let’s go back to various forms of E8 and ﬁgure out (guess) the fundamental groups. We need to know the maximal compact subgroups.

1. One of them is easy: the compact form is its own maximal compact subgroup. What is the fundamental group? Remember or quote the weight lattice fact that for compact simple groups, π1 ∼= root lattice , which is 1. So this form is simply connected.

2. β2 = 2 case (signature 24). Recall that there were 1, 56, 126, 56, and − 1 roots α with (α, β) = 2, 1, 0, 1, and -2 respectively, and there are − another 8 dimensions for the Cartan subalgebra. On the 1, 126, 1, 8 parts, the form is negative definite. The sum of these root spaces gives a Lie algebra of type E7A1 with a negative definite bilinear form (the 126 gives you the roots of an E7, and the 1s are the roots of an A1). So it a reasonable guess that the maximal compact subgroup has something to do with E7A1. E7 and A1 are not simply connected: the compact form of E7 has π1 = Z/2 and the compact form of A1 also has 2 π1 = Z/2. So the universal cover of E7A1 has center (Z/2) . Which part of this acts trivially on E8? We look at the E8 Lie algebra as a representation of E A . You can read off how it splits form the 7 × 1 picture above: E = E A 56 2, where 56 and 2 are irreducible, 8 ∼ 7 ⊕ 1 ⊕ ⊗ and the centers of E and A both act as 1 on them. So the maximal 7 1 − compact subgroup of this form of E8 is the simply connected compact form of E A /( 1, 1). This means that π (E ) is the same as π 7 × 1 − − 1 8 1 of the compact subgroup, which is (Z/2)2/( 1, 1) = Z/2. So this − − ∼ simple group has a nontrivial double cover (which is non-algebraic).

3. For the other (split) form of E8 with signature 8, the maximal compact subgroup is Spin16(R)/(Z/2), and π1(E8) is Z/2.

You can compute any other homotopy invariants with this method.

Let’s look at the 56-dimensional representation of E7 in more detail. We had the picture (α, β) # of α’s 2 1 1 56 0 126 -1 56 -2 1 Lecture 28 150

The Lie algebra E7 ﬁxes these 5 spaces of E8 of dimensions 1, 56, 126 + 8, 56, 1. From this we can get some representations of E7. The 126+ 8 splits as 1 + (126 + 7). But we also get a 56-dimensional representation of E7. Let’s show that this is actually an irreducible representation. Recall that in calculating W (E8), we showed that W (E7) acts transitively on this set of 56 roots of E8, which can be considered as weights of E7. An irreducible representation is called minuscule if the Weyl group acts transitively on the weights. This kind of representation is particularly easy to work with. It is really easy to work out the character for example: just translate the 1 at the highest weight around, so every weight has multiplicity 1. So the 56-dimensional representation of E7 must actually be the irreducible representation with whatever highest weight corresponds to one of the vectors.

Every possible simple Lie group We will construct them as follows: Take an involution σ of the compact + + form L = L + L− of the Lie algebra, and form L + iL−. The way we constructed these was to first construct An, Dn, E6, and E7 as for E8. Then construct the involution ω : eα e α. We get B , C , F , and G as 7→ − − n n 4 2 fixed points of the involution ω. Kac classified all automorphisms of finite order of any compact simple Lie group. The method we’ll use to classify involutions is extracted from his method. We can construct lots of involutions as follows:

1. Take any Dynkin diagram, say E8, and select some of its vertices, corresponding to simple roots. Get an involution by taking eα eα 7→ ± where the sign depends on whether α is one of the simple roots we’ve selected. However, this is not a great method. For one thing, you get a lot of repeats (recall that there are only 3, and we’ve found 28 this way).

'!&"%#$ ØÙ 1 1 1 ØÙ '!&"%#$ ØÙ ØÙ '!&"%#$ ØÙ ØÙ ØÙ ØÙ 2. Take any diagram automorphism of order 2, such as

z } ! $ ØÙ ØÙ ØÙ ØÙ ØÙ

ØÙ This gives you more involutions. Lecture 28 151

Next time, we’ll see how to cut down this set of involutions. Lecture 29 152

Lecture 29

α Split form of Lie algebra (we did this for An, Dn, E6, E7, E8): A = eˆ L. 2 ⊕ Compact form A+ + iA , where A eigenspaces of ω :e ˆα ( 1)α /2eˆ α. − ± 7→ − L − We talked about other involutions of the compact form. You get all the other forms this way. The idea now is to ﬁnd ALL real simple Lie algebras by listing all involutions of the compact form. We will construct all of them, but we won’t prove that we have all of them. We’ll use Kac’s method for classifying all automorphisms of order N of a compact Lie algebra (and we’ll only use the case N = 2). First let’s look at inner automorphisms. Write down the AFFINE Dynkin diagram

ØÙ 3

highest weight = − 1ØÙ 2ØÙ 3ØÙ 4ØÙ 5ØÙ 6ØÙ 4ØÙ 2ØÙ

Choose ni with nimi = N where the mi are the numbers on the diagram. We have an automorphism eαj e2πinj /N eαj induces an automorphism of P 7→ order dividing N. This is obvious. The point of Kac’s theorem is that all inner automorphisms of order dividing N are obtained this way and are conjugate if and only if they are conjugate by an automorphism of the Dynkin diagram. We won’t actually prove Kac’s theorem because we just want to get a bunch of examples. See [Kac90] or [Hel01].

Example 29.1. Real forms of E8. We’ve already found three, and it took us a long time. We can now do it fast. We need to solve nimi = 2 where n 0; there are only a few possibilities: i ≥ P maximal compact nimi = 2 # of ways how to do it subgroup K P ØÙ 2 1 one way E (compact form) × ×ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ 8 ØÙ 1 2 two ways A E × ØÙ ×ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ 1 7 ØÙ D (split form) ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ×ØÙ 8 1 1 + 1 1 noways × × The points NOT crossed oﬀ form the Dynkin diagram of the maximal compact subgroup. Thus, by just looking at the diagram, we can see what all the real forms are!

Example 29.2. Let’s do E7. Write down the aﬃne diagram: 1 2 3 4 3 2 1 ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ

ØÙ 2 Lecture 29 153

We get the possibilities maximal compact nimi = 2 # of ways how to do it subgroup K P 2 1 one way* E7 (compact form) × ×ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ 1 2 two ways* A1D6 × ØÙ ×ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ A7 (split form)** ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ 1 1 + 1 1 one way × E6 R *** × × ×ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ×ØÙ ⊕ ØÙ (*) The number of ways is counted up to automorphisms of the diagram. (**) In the split real form, the maximal compact subgroup has dimension equal to half the number of roots. The roots of A look like ε ε for 7 i − j i, j 8 and i = j, so the dimension is 8 7+7=56= 112 . ≤ 6 · 2 (***) The maximal compact subgroup is E R because the ﬁxed subal- 6 ⊕ gebra contains the whole Cartan, and the E6 only accounts for 6 of the 7 dimensions. You can use this to construct some interesting representations of E6 (the minuscule ones). How does the algebra E7 decompose as a representation of the algebra E R? 6 ⊕ We can decompose it according to the eigenvalues of R. The E R is 6 ⊕ the zero eigenvalue of R [why?], and the rest is 54-dimensional. The easy way to see the decomposition is to look at the roots. Remember when we computed the Weyl group we looked for vectors like

ØÙ ØÙ ØÙ or ØÙ ØÙ ØÙ The 27 possibilities (for each) form the weights of a 27-dimensional representation of E6. The orthogonal complement of the two nodes is an E6 root system whose Weyl group acts transitively on these 27 vectors (we showed that these form a single orbit, remember?). Vectors of the E7 root system are the vectors of the E6 root system plus these 27 vectors plus the other 27 vectors. This splits up the E7 explicitly. The two 27s form single orbits, so they are irreducible. Thus, E = E R 27 27, and the 27s are minuscule. 7 ∼ 6 ⊕ ⊕ ⊕ Let K be a maximal compact subgroup, with Lie algebra R + E6. The factor of R means that K has an S1 in its center. Now look at the space G/K, where G is the Lie group of type E7, and K is the maximal compact subgroup. It is a Hermitian symmetric space. Symmetric space means that it is a (simply connected) Riemannian manifold M such that for each point p M, there is an automorphism fixing p and acting as 1 on the tangent ∈ − space. This looks weird, but it turns out that all kinds of nice objects you know about are symmetric spaces. Typical examples you may have seen: Lecture 29 154 spheres Sn, hyperbolic space Hn, and Euclidean space Rn. Roughly speaking, symmetric spaces have nice properties of these spaces. Cartan classified all symmetric spaces: they are non-compact simple Lie groups modulo the maximal compact subgroup (more or less ... depending on simply connectedness hypotheses ’n such). Historically, Cartan classified simple Lie groups, and then later classified symmetric spaces, and was surprised to find the same result. Hermitian symmetric spaces are just symmetric spaces with a complex structure. A standard example of this is the upper half plane x + iy y > 0 . It is acted on by SL (R), which acts by a b τ = aτ+b . { | } 2 c d cτ+d Let’s go back to this G/K and try to explain why we get a Hermitian symmetric space from it. We’ll be rather sketchy here. First of all, to make it a symmetric space, we have to find a nice invariant Riemannian metric on it. It is sufficient to find a positive definite bilinear form on the tangent space at p which is invariant under K ... then you can translate it around. We can do this as K is compact (so you have the averaging trick). Why is it Hermitian? We’ll show that there is an almost complex structure. We have S1 acting on the tangent space of each point because we have an S1 in the center of the stabilizer of any given point. Identify this S1 with complex numbers of absolute value 1. This gives an invariant almost complex structure on G/K. That is, each tangent space is a complex vector space. Almost complex structures don’t always come from complex structures, but this one does (it is integrable). Notice that it is a little unexpected that G/K has a complex structure (G and K are odd-dimensional in the case of G = E , K = E R, 7 6 ⊕ so they have no hope of having a complex structure).

Example 29.3. Let’s look at E6, with aﬃne Dynkin diagram

1 2 3 2 1 ØÙ ØÙ ØÙ ØÙ ØÙ

ØÙ 2

ØÙ 1 Lecture 29 155

We get the possibilities maximal compact nimi = 2 # of ways how to do it subgroup K P 2 1 one way E6 (compact form) × ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ ØÙ 1 2 one way × A1A5 × ØÙ ØÙ ØÙ ØÙ ØÙ

×ØÙ ØÙ 1 1 + 1 1 one way D5 R × × ×ØÙ ØÙ ØÙ ØÙ ØÙ ⊕ ØÙ

×ØÙ In the last one, the maximal compact subalgebra is D R. Just as before, 5 ⊕ we get a Hermitian symmetric space. Let’s compute its dimension (over C). The dimension will be the dimension of E6 minus the dimension of D R, all divided by 2 (because we want complex dimension), which is 5 ⊕ (78 46)/2 = 16. − So we have found two non-compact simply connected Hermitian symmetric spaces of dimensions 16 and 27. These are the only “exceptional” cases; all the others fall into inﬁnite families! There are also some OUTER automorphisms of E6 coming from the diagram automorphism ØÙ σ ØÙ Õ Ù ØÙ ØÙ ØÙ ØÙ ØÙ −→ ØÙ

ØÙ ØÙ The ﬁxed point subalgebra has Dynkin diagram obtained by folding the E6 on itself. This is the F4 Dynkin diagram. The ﬁxed points of E6 under the diagram automorphism is an F4 Lie algebra. So we get a real form of E6 with maximal compact subgroup F4. This is probably the easiest way to construct F4, by the way. Moreover, we can decompose E6 as a representation of F4. dim E = 78 and dim F = 52, so E = F 26, where 26 turns out to be 6 4 6 4 ⊕ irreducible (the smallest non-trivial representation of F4 ... the only one anybody actually works with). The roots of F look like (..., 1, 1 . . . ) 4 ± ± (24 of these) and ( 1 1 ) (16 of these), and (..., 1 . . . ) (8 of them) ... ± 2 ···± 2 ± the last two types are in the same orbit of the Weyl group. Lecture 29 156

The 26-dimensional representation has the following character: it has all norm 1 roots with multiplicity 1 and 0 with multiplicity 2 (note that this is not minuscule). There is one other real form of E6. To get at it, we have to talk about Kac’s description of non-inner automorphisms of order N. The non-inner automorphisms all turn out to be related to diagram automorphisms. Choose a diagram automorphism of order r, which divides N. Let’s take the standard thing on E6. Fold the diagram (take the fixed points), and form a TWISTED affine Dynkin diagram (note that the arrow goes the wrong way from the affine F4)

>ØÙ 1 : 2 1 2 3 2 1 ØÙ / Twisted Aﬃne F4 gg3 ØÙ ØÙ ØÙ ØÙ ØÙ r 2 1 ggg 3ØÙ ØÙ ØÙ $ 2 1 2 3 4 2 ØÙ / Aﬃne F ØÙ ØÙ ØÙ ØÙ ØÙ 4 ØÙ 1

There are also numbers on the twisted diagram, but nevermind them. Find ni so that r nimi = N. This is Kac’s general rule. We’ll only use the case N = 2. P If r> 1, the only possibility is r = 2 and one n1 is 1 and the corresponding mi is 1. So we just have to ﬁnd points of weight 1 in the twisted aﬃne Dynkin diagram. There are just two ways of doing this in the case of E6

/ and / ØÙ ØÙ ØÙ ØÙ ×ØÙ ×ØÙ ØÙ ØÙ ØÙ ØÙ one of these gives us F4, and the other has maximal compact subalgebra C4, which is the split form since dim C4 = #roots of F4/2 = 24.

Example 29.4. 1 2 3 / 4 2 F4. The aﬃne Dynkin is ØÙ ØÙ ØÙ ØÙ ØÙ We can cross out one node of weight 1, giving the compact form (split form), or a node of weight 2 (in two ways), giving maximal compacts A1C3 or B4. This gives us three real forms.

Example 29.5. G2. We can actually draw this root system ... UCB won’t supply me with a four-dimensional board. The construction is to take the D4 algebra and look at the ﬁxed points of:

ØÙ Z

× ρ ØÙ ØÙ 11 11 0 1 ØÙ Lecture 29 157

We want to find the fixed point subalgebra. Fixed points on Cartan subalgebra: ρ fixes a two-dimensional space, and has 1-dimensional eigenspaces corresponding to ω andω ¯, where ω3 = 1. The 2-dimensional space will be the Cartan of G2. Positive roots of D4 as linear combinations of simple roots (not fundamental weights):

0 1 0 0 gf ed gf ed

1 0 0 0 0 0 0 1 1 1 1 1 11 11 11 11 `a 0 0 bc1 `a bc0

0 1 0 1 gf ed gf ed

1 1 0 1 0 1 1 1 1 1 1 1 11 11 11 11 `a 0 0 bc1 `a bc1

1 1 0 1 gf ed gf ed

1 1 0 1 1 1 1 2 1 1 1 1 11 11 11 11 `a 0 1 bc1 `a bc1

projections of norm 2/3 projections of norm 2 There| are six orbits under{zρ, grouped above. It obviously} | acts{z on the} negative roots in exactly the same way. What we have is a root system with six roots of norm 2 and six roots of norm 2/3. Thus, the root system is G2: 1 1 •11 1 1 11 1 1 1 1 1 •1 • •11 • 11 1 1 2 11 1 1 1 1 1 • 1 • •11 11 1 1 1 1 11 1 • 1• 1 •1 • 11 1 •1 One of the only root systems to appear on a country’s national ﬂag. Now let’s work out the real forms. Look at the aﬃne: 1 2 / 3 . we can ØÙ ØÙ ØÙ Lecture 29 158 delete the node of weight 1, giving the compact form: / . We ×ØÙ ØÙ ØÙ can delete the node of weight 2, giving A1A1 as the compact subalgebra: / ... this must be the split form because there is nothing else the ØÙ ØÙ ØÙ split form× can be. Let’s say some more about the split form. What does the Lie algebra of G look like as a representation of the maximal compact subalgebra A A ? 2 1 × 1 In this case, it is small enough that we can just draw a picture:

?>1=< • 1 1 1 1 1 ?>1=< 1 1 ?> =< • • • • 89 • • • • :; 1 2 1 89?> • • :;=< −→ 1 1 • 1• •1 • •1 •1 •1 •1 89 :;

89•1:;

We have two orthogonal A1s, and we have leftover the stuﬀ on the right. This thing on the right is a tensor product of the 4-dimensional irreducible representation of the horizontal and the 2-dimensional of the vertical. Thus, (horizontal) G2 = 3 1 + 1 3 + 4 2 as irreducible representations of A1 (vertical)× ⊗ ⊗ ⊗ A1 . Let’s use this to determine exactly what the maximal compact subgroup is. It is a quotient of the simply connected compact group SU(2) SU(2), × with Lie algebra A A . Just as for E , we need to identify which elements 1 × 1 8 of the center act trivially on G . The center is Z/2 Z/2. Since we’ve 2 × decomposed G2, we can compute this easily. A non-trivial element of the center of SU(2) acts as 1 (on odd dimensional representations) or 1 (on − even dimensional representations). So the element z z SU(2) SU(2) × ∈ × acts trivially on 3 1 + 1 3 + 4 2. Thus the maximal compact subgroup ⊗ ⊗ × of the non-compact simple G is SU(2) SU(2)/(z z) = SO (R), where 2 × × ∼ 4 z is the non-trivial element of Z/2.

So we have constructed 3 + 4 + 5 + 3 + 2 (from E8, E7, E6, F4, G2) real forms of exceptional simple Lie groups. There are another 5 exceptional real Lie groups: Take COMPLEX groups E8(C), E7(C), E6(C), F4(C), and G2(C), and consider them as REAL. These give simple real Lie groups of dimensions 248 2, 133 2, 78 2, 52 2, × × × × and 14 2. × Lecture 30 - Irreducible unitary representations of SL2(R) 159

Lecture 30 - Irreducible unitary representations of SL2(R)

SL2(R) is non-compact. For compact Lie groups, all unitary representations are finite dimensional, and are all known well. For non-compact groups, the theory is much more complicated. Before doing the infinite-dimensional representations, we’ll review finite-dimensional (usually not unitary) representations of SL2(R).

Finite dimensional representations Finite-dimensional complex representations of the following are much the same: SL2(R), sl2R, sl2C [branch SL2(C) as a complex Lie group] (as a complex Lie algebra), su2R (as a real Lie algebra), and SU2 (as a real Lie group). This is because finite-dimensional representations of a simply connected Lie group are in bijection with representations of the Lie algebra. Complex representations of a REAL Lie algebra L correspond to complex representations of its complexification L C considered as a COMPLEX Lie ⊗ algebra. Note: Representations of a COMPLEX Lie algebra L C are not the ⊗ same as representations of the REAL Lie algebra L C = L + L. The rep- ⊗ ∼ resentations of the real Lie algebra correspond roughly to (reps of L) (reps ⊗ of L). Strictly speaking, SL2(R) is not simply connected, which is not important for finite-dimensional representations. Recall the main results for representations of SU2: 1. For each positive integer n, there is one irreducible representation of dimension n.

2. The representations are completely reducible (every representation is a sum of irreducible ones). This is perhaps the most important fact.

The ﬁnite-dimensional representation theory of SU2 is EASIER than the representation theory of the ABELIAN Lie group R2, and that is because representations of SU2 are completely reducible. For example, it is very diﬃcult to classify pairs of commuting nilpotent matrices.

Completely reducible representations:

1. Complex representations of ﬁnite groups.

2. Representations of compact groups (Weyl character formula) Lecture 30 - Irreducible unitary representations of SL2(R) 160

3. More generally, unitary representations of anything (you can take orthogonal complements of subrepresentations)

4. Finite-dimensional representations of semisimple Lie groups.

Representations which are not completely reducible:

1. Representations of a finite group G over fields of characteristic p G . | | | 2. Infinite-dimensional representations of non-compact Lie groups (even if they are semisimple).

1 0 We’ll work with the Lie algebra sl2R, which has basis H = 0 1 , E = 0 1 , and F = 0 0 . H is a basis for the cartan a 0 . E spans− 0 0 1 0 0 a the root space of the simple root. F spans the root space of the− negative of the simple root. We ﬁnd that [H, E] = 2E, [H, F ]= 2F (so E and F are − eigenvectors of H), and you can check that [E, F ]= H.

2 0 2 o weights = eigenvalues under H − • • • F [ H B E

Weyl group of order 2

0 1 2 1 0 The Weyl group is generated by ω = 1 0 and ω = −0 1 . Let V be a ﬁnite-dimensional irreducible− complex representatio− n of sl R. 2 First decompose V into eigenspaces of the Cartan subalgebra (weight spaces) (i.e. eigenspaces of the element H). Note that eigenspaces of H exist because V is FINITE-DIMENSIONAL (remember this is a complex representation). Look at the LARGEST eigenvalue of H (exists since V is ﬁnite-dimensional), with eigenvector v. We have that Hv = nv for some n. Compute

H(Ev) = [H, E]v + E(Hv) = 2Ev + Env = (n + 2)Ev

So Ev = 0 (lest it be an eigenvector of H with higher eigenvalue). [E, ] − increases weights by 2 and [F, ] decreases weights by 2, and [H, ] ﬁxes − − weights. We have that E kills v, and H multiplies it by n. What does F do to v?

nv (n 2)F v (n 4)F 2v (n 6)F 3v . . . O − O − O − O H H H H F F F + , , 0 h v j F v k F 2v l F 3v . . . E E E E n (2n 2) (3n 6) × × − × − Lecture 30 - Irreducible unitary representations of SL2(R) 161

What is E(F v)? Well,

EF v = F Ev + [E, F ]v =0+ Hv = nv

In general, we have

H(F iv) = (n 2i)F iv − i i 1 E(F v) = (ni i(i 1))F − v − − F (F iv)= F i+1v

So the vectors F iv span V because they span an invariant subspace. This gives us an inﬁnite number of vectors in distinct eigenspaces of H, and V is ﬁnite-dimensional. Thus, F kv = 0 for some k. Suppose k is the SMALLEST integer such that F kv = 0. Then

k k 1 0= E(F v) = (nk k(k 1)) EF − v − − =0 6 | {z } So nk k(k 1) = 0, and k = 0, so n (k 1) = 0, so k = n + 1 . So V − − 6 − − has a basis consisting of v,Fv,...,F nv. The formulas become a little better F 2v F 3v F nv if we use the basis wn = v, wn 2 = Fv,wn 4 = , ,..., . − − 2! 3! n! 1 2 3 4 5 6 E ! ! ! w 6 w 4 w 2 w0 w2 w4 w6 − a − a − a _ _ _ 6 5 4 3 2 1 F

This says that E(w2) = 5w4 for example. So we’ve found a complete description of all finite-dimensional irreducible complex representations of sl2R. This is as explicit as you could possibly want. These representations all lift to the group SL2(R): SL2(R) acts on ho- a b mogeneous polynomials of degree n by c d f(x,y) = f(ax + by, cx + dy). This is an n + 1-dimensional space, and you can check that the eigenspaces i n i are x y − . We have implicitly constructed VERMA MODULES. We have a basis wn, wn 2,...,wn 2i,... with relations H(wn 2i) = (n 2i)wn 2i, Ewn 2i = − − − − − − (n i + 1)wn 2i+2, and F wn 2i = (i + 1)wn 2i 2. These are obtained by − − − − − copying the formulas from the finite-dimensional case, but allow it to be infinite-dimensional. This is the universal representation generated by the highest weight vector wn with eigenvalue n under H (highest weight just means E(wn) = 0). Let’s look at some things that go wrong in infinite dimensions. Lecture 30 - Irreducible unitary representations of SL2(R) 162

Warning 30.1. Representations corresponding to the Verma modules do NOT lift to representations of SL2(R), or even to its universal cover. 0 1 The reason: look at the Weyl group (generated by 1 0 ) of SL2(R) acting on H ; it changes H to H. It maps eigenspaces− with eigenvalue m to h i − eigenvalue m. But if you look at the Verma module, it has eigenspaces − n,n 2,n 4,... , and this set is obviously not invariant under changing − − sign. The usual proof that representations of the Lie algebra lifts uses the exponential map of matrices, which doesn’t converge in infinite dimensions. ^ Remark 30.2. The universal cover SL2(R) of SL2(R), or even the double cover Mp2(R), has NO faithful finite-dimensional representations. Proof. Any finite dimensional representation comes from a finite dimensional representation of the Lie algebra sl2R. All such finite dimensional representations factor through SL2(R).

All finite-dimensional representations of SL2(R) are completely reducible. Weyl did this by Weyl’s unitarian trick: Notice that finite-dimensional representations of SL2(R) are isomorphic (sort of) to finite-dimensional representations of the COMPACT group SU2 (because they have the same complexified Lie algebras. Thus, we just have to show it for SU2. But representations of ANY compact group are completely reducible. Reason: 1. All unitary representations are completely reducible (if U V , then ⊆ V = U U ). ⊕ ⊥ 2. Any representation V of a COMPACT group G can be made unitary: take any unitary form on V (not necessarily invariant under G), and average it over G to get an invariant unitary form. We can average because G is compact, so we can integrate any continuous function over G. This form is positive definite since it is the average of positive definite forms (if you try this with non-(positive definite) forms, you might get zero as a result).

The Casimir operator Set Ω = 2EF + 2F E + H2 U(sl R). The main point is that Ω commutes ∈ 2 with sl2R. You can check this by brute force: [H, Ω] = 2 ([H, E]F + E[H, F ]) + · · · 0 [E, Ω] = 2[|E, E]F +{z 2E[F, E] +} 2[E, F ]E + 2F [E, E] + [E,H]H + H[E,H] = 0 [F, Ω] = Similar Lecture 30 - Irreducible unitary representations of SL2(R) 163

Thus, Ω is in the center of U(sl2R). In fact, it generates the center. This doesn’t really explain where Ω comes from. Remark 30.3. Why does Ω exist? The answer is that it comes from a symmetric invariant bilinear form on the Lie algebra sl2R given by (E, F ) = 1, (E, E) = (F, F ) = (F,H) = (E,H) = 0, (H,H) = 2. This bilinear form is an invariant map L L C, where L = sl R, which by duality gives an in- ⊗ → 2 variant element in L L, which turns out to be 2E F +2F E+H H. The ⊗ ⊗ ⊗ ⊗ invariance of this element corresponds to Ω being in the center of U(sl2R).

Since Ω is in the center of U(sl2R), it acts on each irreducible representation as multiplication by a constant. We can work out what this constant is for the ﬁnite-dimensional representations. Apply Ω to the highest vector wn:

2 (2EF + 2F E + HH)wn = (2n +0+ n )wn 2 = (2n + n )wn

So Ω has eigenvalue 2n + n2 on the irreducible representation of dimension n + 1. Thus, Ω has DISTINCT eigenvalues on different irreducible representations, so it can be used to separate different irreducible representations. The main use of Ω will be in the next lecture, where we’ll use it to deal with infinite-dimensional representation. To finish today’s lecture, let’s look at an application of Ω. We’ll sketch an algebraic argument that the representations of sl2R are completely reducible. Given an exact sequence of representations

0 U V W 0 → → → → we want to ﬁnd a splitting W V , so that V = U W . → ⊕ Step 1: Reduce to the case where W = C. The idea is to look at

0 HomC(W, U) HomC(W, V ) HomC(W, W ) 0 → → → → and HomC(W, W ) has an obvious one-dimensional subspace, so we can get a smaller exact sequence

0 HomC(W, U) subspace of HomC(W, V ) C 0 → → → → and if we can split this, the original sequence splits. Step 2: Reduce to the case where U is irreducible. This is an easy induction on the number of irreducible components of U.

◮ Exercise 30.1. Do this. Lecture 30 - Irreducible unitary representations of SL2(R) 164

Step 3: This is the key step. We have

0 U V C 0 → → → → with U irreducible. Now apply the Casimir operator Ω. V splits as eigenvalues of Ω, so is U C UNLESS U has the same eigenvalue as C (i.e. unless ⊕ U = C). Step 4: We have reduced to

0 C V C 0 → → → → 1 which splits because sl2(R) is perfect (no homomorphisms to the abelian 0 algebra 0 0∗ ). Next time, in the ﬁnal lecture, we’ll talk about inﬁnite-dimensional unitary representations.

1L is perfect if [L, L]= L Lecture 31 - Unitary representations of SL2(R) 165

Lecture 31 - Unitary representations of SL2(R)

Last lecture, we found the ﬁnite-dimensional (non-unitary) representations of SL2(R).

Background about inﬁnite-dimensional representations (of a Lie group G) What is an ﬁnite-dimensional representation?

1st guess Banach space acted on by G? This is no good for some reasons: Look at the action of G on the functions on G (by left translation). We could use L2 functions, or L1 or Lp. These are completely diﬀerent Banach spaces, but they are essentially the same representation.

2nd guess Hilbert space acted on by G? This is sort of okay.

The problem is that ﬁnite dimensional representations of SL2(R) are NOT Hilbert space representations, so we are throwing away some interesting representations.

Solution (Harish-Chandra) Take g to be the Lie algebra of G, and let K be the maximal compact subgroup. If V is an inﬁnite-dimensional representation of G, there is no reason why g should act on V . The simplest example fails. Let R act on L2(R) by left translation. d d 2 Then the Lie algebra is generated by dx (or i dx ) acting on L (R), but d 2 2 dx of an L function is not in L in general.

Let V be a Hilbert space. Set Vω to be the K-finite vectors of V , which are the vectors contained in a finite-dimensional representation of K. The point is that K is compact, so V splits into a Hilbert space direct sum finite-dimensional representations of K, at least if V is a Hilbert space. Then Vω is a representation of the Lie algebra g, not a representation of G. Vω is a representation of the group K. It is a (g, K)-module, which means that it is acted on by g and K in a “compatible” way, where compatible means that

1. they give the same representations of the Lie algebra of K. 2. k(u)v = k(u(k 1v)) for k K, u g, and v V . − ∈ ∈ ∈ The K-ﬁnite vectors of an irreducible unitary representation of G is ADMISSIBLE, which means that every representation of K only oc- curs a ﬁnite number of times. The GOOD category of representations is the representations of admissible (g, K)-modules. It turns out that this is a really well behaved category. Lecture 31 - Unitary representations of SL2(R) 166

We want to ﬁnd the unitary irreducible representations of G. We will do this in several steps:

1. Classify all irreducible admissible representations of G. This was solved by Langlands, Harish-Chandra et. al.

2. Find which have hermitian inner products ( , ). This is easy.

3. Find which ones are positive deﬁnite. This is VERY HARD. We’ll only do this for the simplest case: SL2(R).

The group SL2(R)

We found some generators (in Lie(SL2(R)) C last time: E, F , H, with ⊗ 0 1 [H, E] = 2E, [H, F ]= 2F , and [E, F ]= H. We have that H = i 1 0 , 1 1 i −1 1 i 1 0 −0 1 − E = 2 i 1 , and F = 2 i −1 . Why not use the old 0 1 , 0 0 , and 0 0 − − − − 1 0 ? a 0 Because SL (R) has two different classes of Cartan subgroup: −1 , 2 0 a spanned by 1 0 , and cos θ sin θ , spanned by 0 1 , and the second 0 1 sin θ cos θ 1 0 one is COMPACT.− The point− is that non-compact− (abelian) groups need not have eigenvectors on infinite-dimensional spaces. An eigenvector is the same as a weight space. The first thing you do is split it into weight spaces, and if your Cartan subgroup is not compact, you can’t get started. We work with the compact subalgebra so that the weight spaces exist. Given the representation V , we can write it as some direct sum of eigenspaces of H, as the Lie group H generates is compact (isomorphic to S1). In the finite-dimensional case, we found a HIGHEST weight, which gave us complete control over the representation. The trouble is that in infinite dimensions, there is no reason for the highest weight to exist, and in general they don’t. The highest weight requires a finite number of eigenvalues. A good substituted for the highest weight vector: Look at the Casimir operator Ω = 2EF + 2F E + H2 + 1. The key point is that Ω is in the center of the universal enveloping algebra. As V is assumed admissible, we can conclude that Ω has eigenvectors (because we can find a finite-dimensional space acted on by Ω). As V is irreducible and Ω commutes with G, all of V is an eigenspace of Ω. We’ll see that this gives us about as much information as a highest weight vector. Let the eigenvalue of Ω on V be λ2 (the square will make the interesting representations have integral λ; the +1 in Ω is for the same reason). Suppose v V , where V is the space of vectors where H has eigenvalue ∈ n n n. In the finite-dimensional case, we looked at Ev, and saw that HEv = (n+2)Ev. What is F Ev? If v was a highest weight vector, we could control this. Notice that Ω = 4F E +H2 +2H +1 (using [E, F ]= H), and Ωv = λ2v. Lecture 31 - Unitary representations of SL2(R) 167

This says that 4F Ev + n2v + 2nv + v = λ2v. This shows that F Ev is a multiple of v. Now we can draw a picture of what the representation looks like: E vn 4 vn 2 vn vn+2 vn+4 H

· · · ] −] −] ] ] ] · · · 2 2 F n +2n+1−λ 4

Thus, Vω is spanned by Vn+2k, where k is an integer. The non-zero elements among the Vn+2k are linearly independent as they have diﬀerent eigenvalues. The only question remaining is whether any of the Vn+2k vanish. There are four possible shapes for an irreducible representation

% % % % % E – inﬁnite in both directions: e e e e e H · · · · · · · · · · F – a lowest weight, and inﬁnite in the other direction:

% % % % % E e e e e e H · · · · · · · · F – a highest weight, and inﬁnite in the other direction:

% % % % % E e e e e e H · · · · · · · · F – we have a highest weight and a lowest weight, in which case it is ﬁnite- % % % % E dimensional e e e e H · · · · · · · F We’ll see that all these show up. We also see that an irreducible representation is completely determined once we know λ and some n for which V = 0. n 6 The remaining question is to construct representations with all possible values of λ C and n Z. n is an integer because it must be a representations ∈ ∈ of the circle. If n is even, we have

λ−7 λ−5 λ−3 λ−1 λ+1 λ+3 λ+5 λ+7 2 2 2 2 2 2 2 2 E 6 4 2 0 2 4 6 H · · · Z − Z − Z − Z Z Z Z Z · · · λ+7 λ+5 λ+3 λ+1 λ−1 λ−3 λ−5 λ−7 F 2 2 2 2 2 2 2 2

It is easy to check that these maps satisfy [E, F ]= H, [H, E] = 2E, and [H, F ]= 2F − Lecture 31 - Unitary representations of SL2(R) 168

◮ Exercise 31.1. Do the case of n odd.

Problem: These may not be irreducible, and we want to decompose them into irreducible representations. The only way they can fail to be irreducible if if Evn = 0 of F vn = 0 for some n (otherwise, from any vector, you can generate the whole space). The only ways that can happen is if

n even: λ an odd integer n odd: λ an even integer.

What happens in these cases? The easiest thing is probably just to write out an example.

Example 31.1. Take n even, and λ = 3, so we have

2 101 23 45 − − E 6 4 2 0 2 4 6 H · · · Z − Z − Z − Z Z Z Z Z · · · 5 4 3 2 1 0 1 2 F − −

You can just see what the irreducible subrepresentations are ... they are shown in the picture. So V has two irreducible subrepresentations V and − V+, and V/(V V+) is an irreducible 3-dimensional representation. − ⊕ Example 31.2. If n is even, but λ is negative, say λ = 3, we get −

5 4 3 2 1 0 1 2 − − − − − E 6 4 2 0 2 4 6 H · · · Z − Z − Z − Z Z Z Z Z · · · 2 1 0 1 2 3 4 5 F − − − − −

Here we have an irreducible ﬁnite-dimensional representation. If you quotient out by that subrepresentation, you get V+ V . ⊕ − ◮ Exercise 31.2. Show that for n odd, and λ = 0, V = V+ V . ⊕ − So we have a complete list of all irreducible admissible representations:

1. if λ Z, you get one representation (remember λ λ). This is the 6∈ ≡− bi-inﬁnite case.

2. Finite-dimensional representation for each n 1 (λ = n) ≥ ± 3. Discrete series for each λ Zr 0 , which is the half inﬁnite case: you ∈ { } get a lowest weight when λ< 0 and a highest weight when λ> 0. Lecture 31 - Unitary representations of SL2(R) 169

4. two “limits of discrete series” where n is odd and λ = 0. Which of these can be made into unitary representations? H = H, E = † − † F , and F † = E. If we have a hermitian inner product ( , ), we see that 2 (v , v )= (Ev , v ) j+2 j+2 λ + j + 1 j j+2 2 = (v , F v ) λ + j + 1 j − j+2 2 λ j 1 = − − (v , v ) > 0 −λ + j + 1 2 j j

λ 1 j where we ﬁx the sign errors. So we want − − to be real and positive − λ+j+1 whenever j, j + 2 are non-zero eigenvectors. So (λ 1 j)(λ +1+ j)= λ2 + (j + 1)2 − − − − should be positive for all j. Conversely, when you have this, blah. This condition is satisﬁed in the following cases: 1. λ2 0. These representations are called PRINCIPAL SERIES repre- ≤ sentations. These are all irreducible except when λ = 0and n is odd, in which case it is the sum of two limits of discrete series representations 2. 0 <λ< 1 and j even. These are called COMPLEMENTARY SERIES. They are annoying, and you spend a lot of time trying to show that they don’t occur. 3. λ2 = n2 for n 1 (for some of the irreducible pieces). ≥ If λ = 1, we get

3 2 10 1 2 3 4 − − − E 6 4 2 0 2 4 6 H · · · Z − Z − Z − Z Z Z Z Z · · · 4 3 2 1 0 1 2 3 F − − −

We see that we get two discrete series and a 1-dimensional representation, all of which are unitary For λ = 2 (this is the more generic one), we have

2 10 1 2 3 4 − − E 5 3 1 1 3 5 H · · · Z − Z − Z − Z Z Z Z · · · 4 3 2 1 0 1 2 F − − Lecture 31 - Unitary representations of SL2(R) 170

The middle representation (where (j + 1)2 < λ2 = 4 is NOT unitary, which we already knew. So the DISCRETE SERIES representations ARE unitary, and the FINITE dimensional representations of dimension greater than or equal to 2 are NOT.

Summary: the irreducible unitary representations of SL2(R) are given by

1. the 1-dimensional representation

2. Discrete series representations for any λ Zr 0 ∈ { } 3. Two limit of discrete series representations for λ = 0

4. Two series of principal series representations:

j even: λ iR, λ 0 ∈ ≥ j odd: λ iR, λ> 0 ∈ 5. Complementary series: parameterized by λ, with 0 <λ< 1.

The nice stuﬀ that happened for SL2(R) breaks down for more complicated Lie groups. Representations of ﬁnite covers of SL2(R) are similar, except j need not be integral. For example, for the double cover SL\(R)= Mp (R), 2j Z. 2 2 ∈ ◮ Exercise 31.3. Find the irreducible unitary representations of Mp2(R). Solutions to (some) Exercises 171

Solutions to (some) Exercises

Solution 1.1. Yes. Consider µ 1(e) G G. We would like to use − ⊆ × the implicit function theorem to show that there is a function f (which is as smooth as µ) such that (h, g) µ 1(e) if and only if g = f(h). This ∈ − function will be ι. You need to check that for every g, the derivative of left 1 1 multiplication by g at g− is non-singular (i.e. that dlg(g− ) is a non-singular matrix). This is obvious because we have an inverse, namely dlg−1 (e).

Solution 1.2. Just do it.

Solution 4.1. We calculate: d d g(t) 2 = 2 g, g dtk k hdt i d 2 g g ≤ dt k k

2 2 ξ g . ≤ k kk k That is, η(t) := g(t) 2 satisﬁes the diﬀerential inequality: k k d η(t) ξ η(t), dt ≤ k k

which in turn implies (Gronwall’s inequality) that 2 t ξ(s) ds η(t) e t0 k k ≤

so that t ξ(s) ds g e t0 k k k k ≤ C′ t t ≤ | − 0| since for t t suﬃciently small, exponentiation is Lipschitz. | − 0| Solution 8.2. We would like to compute the coeﬃcients of the product (XaHbY r)(XsHcY d) once it is rewritten in the PBW basis by repeatedly applying the relations XY Y X = εH, HX = XH, and HY = YH. Check − by induction that

r s ∞ n n r s r n l s n Y X = ε ( 1) n! X − H Y − . − n n n=0 X I,J It follows that pn is zero unless I = (Y,...,Y ) and J = (X,...,X), in I,J ( 1)n n which case pn = −n! H . Solutions to (some) Exercises 172

n Solution 9.3. We have [a, b]h := n∞=0 h mn(a, b), where m0(a, b) = [a, b]. Now we compute P l [a, [b, c]h]h = [a, h ml(b, c)]h l 0 X≥ l k = h h mk(a, ml(b, c)) l 0 k 0 X≥ X≥ N = h mk(a, mN k(b, c)) (N = k + l) − N 0 X≥ Adding the cyclic permutations and looking at the coeﬃcient of hN , we get the desired result. Solution 11.1. Obvious.

1 1 1 Solution 11.2. [G, G] is normal because r[g, h]r− = [rgr− , rhr− ]. To see that [G, G] is connected, let γ : [0, 1] G be a path from g to h and deﬁne gh → the path t gγ(t)g 1γ(t) 1 in [G, G] from the identity to [g, h]. Since all 7→ − − the generators of [G, G] are connected to e G by paths, all of [G, G] is ∈ connected to e. Now we show that the Lie algebra of [G, G] is g. Consider the Lie D algebra homomorphism π : g g/ g. Since G is simply connected, this → D lifts to some Lie group homomorphism p : G H: → g / g π / g/ g D D exp exp p [G, G] / / n G H ∼= R where H is the simply connected Lie group with Lie algebra g/ g. Note D that the Lie algebra of the kernel of p must be contained in ker π = g. D Also, g/ g is abelian, so H is abelian, so [G, G] is in the kernel of p. This D shows that Lie([G, G]) g. ⊆ D To see that g Lie([G, G]), assume that g gl(V ). Then for X,Y D ⊆ ⊆ ∈ g consider the path γ(t) = exp(X√t) exp(Y √t) exp( X√t) exp( Y √t) in − − [G, G]: 1 1 γ(t)= 1+ X√t + X2t + 1+ Y √t + Y 2t + 2 · · · 2 · · · × 1 1 1 X√t + X2t + 1 Y √t + Y 2t + − 2 · · · − 2 · · · =1+ √t(X + Y X Y )+ − − t(XY X2 XY Y X Y 2 + X2 + Y 2)+ − − − − · · · =1+ t[X,Y ]+ O(t3/2) Solutions to (some) Exercises 173

So γ′(0) = [X,Y ]. This shows that [G, G] is a connected component of the kernel of p. 1 Since [G, G] is a connected component of p− (0), it is closed in G.

Solution 11.3. Let π : g g/rad g be the canonical projection, and assume → a g/rad g is solvable. Then ka = 0 for some k, so kπ 1(a) rad g. ∈ D D − ⊆ Since rad g is solvable, we have that N π 1(a) = 0forsome N. By deﬁnition D − of rad g, we get that π 1(a) rad g, so a = 0 g/rad g. Thus, g/rad g is − ⊆ ⊆ semi-simple.

Solution 12.1. An invariant form B induces a homomorphism g g . → ∗ Invariance says that this homomorphism is an intertwiner of representations of g (with the adjoint action on g and the coadjoint action on g∗). Since g is simple, these are both irreducible representations. By Schur’s Lemma, any two such homomorphisms must be proportional, so any two invariant forms must be proportional.

Solution 12.2. Done in class.

Solution 12.3. yuck.

Solution 12.4. The complex for computing cohomology is

d d d 0 k 0 Hom(sl , k) 1 Hom(Λ2sl , k) 2 Hom(Λ3sl , k) 0 −→ −−→ 2 −−→ 2 −−→ 2 −→ c dc(x)= x c = 0 7−→ − · f df(x,y)= f([x,y]) 7−→ α dα(x,y,z)= α([x,y], z) α([x, z],y) 7→ − +α([y, z],x)

0 We have that ker d1 = k, so H (sl2, k) = k. The kernel of d1 is zero, as we 2 computed in Remark 12.11. Since Hom(sl2, k) and Hom(Λ sl2, k) are both three-dimensional, it follows that d1 is surjective, and since the kernel of d2 must contain the image of d1, we know that d2 is the zero map. This tells 1 2 3 3 us that H (sl2, k) = 0, H (sl2, k) = 0, and H (sl2, k) = Hom(Λ sl2, k) ∼= k. Solution 12.5. Let D er(g) and let X,Y g. Then ∈ D ∈

[D, adX ] er(g)(Y )= D([X,Y ]) [X, D(Y )] D − = [D(X),Y ] + [X, D(Y )] [X, D(Y )] − = adD(X)(Y ).

Solution 13.1. Let x h, so [x, h] = 0. Since ad is a polynomial in ad , ∈ xn x we get that [x , h] = 0, so x h. Thus, it is enough to show that any n n ∈ nilpotent element in h is zero. We do this using property 4, that the Killing Solutions to (some) Exercises 174 form is non-degenerate on h. If y h, then B(x ,y) = tr(ad ad ). h ∈ n xn ◦ y is abelian by the proposition, so [xn,y] = 0, so adxn commutes with ady.

Thus, we can simultaneously upper triangularize adxn and ady. Since adxn is nilpotent, it follows that tr(ad ad ) = 0. So x = 0 by non-degeneracy xn ◦ y n of B. Solution 13.2. Since ∆ is a finite set in h , we can find some h h so that ∗ ∈ α(h) = β(h) for distinct roots α and β. Then this h is a regular element 6 which gives the right Cartan subalgebra, and the desired properties follow from the properties on page 60. Solution 13.3. If ∆ does not span h , then there is some non-zero h h ∗ ∈ such that α(h) = 0 for all α ∆. This means that all of the eigenvalues of ∈ adh are zero. Since h is semisimple, adh = 0. And since ad is faithful, we get h = 0, proving property 1. To prove 2, consider the α-string through β. It must be of the form gβ+nα gβ+(n 1)α gβ+mα for some integers n 0 m. From the ⊕ − ⊕···⊕ ≥ ≥ characterization of irreducible finite dimensional representations of sl2, we know that each eigenvalue of H is an integer, so β(H ) = r Z (since α α ∈ [Hα, Xβ] = β(Hα)Xβ). We also know that the eigenvalues of Hα are symmetric around zero, so we must have r = (β + sα)(H ) for some s for − α which gβ+sα is in the α-string through β. Then we get β(Hα)+ sα(Hα)= β(α) + 2s = r = β(α), from which we know that s = β(H ). Thus, − − − α gβ β(Hα)α = 0, so β β(Hα) α is a root. − 6 − Finally, we prove 3. If α and β = cα are roots, then by property 2, we know that α(Hβ) = 2/c and β(Hα) = 2c are integers (note that Hβ = Hα/c). It follows that c = 1 , 1, or 2. Therefore, it is enough to show that α ± 2 ± ± and 2α cannot both be roots. To see this, consider the α-string through 2α. We have that [Hα, X2α] = 2α(Hα)X2α = 4X2α, so the α-string must have a non-zero element [Y , X ] g , which is spanned by X . But then we α 2α ∈ α α would have that X2α is a multiple of [Xα, Xα] = 0, which is a contradiction. Solution 14.1. If ∆ is reducible, with ∆ = ∆ ∆ , then set h to be the 1 ∪ 2 i span of ∆i, and set gi = hi gα (for i = 1, 2). Then we have that ⊕ α ∆i g = g g as a vector space. We∈ must check that g is an ideal (the by 1 ⊕ 2 L 1 symmetry, g will also be an ideal). From the relation [g , g ] g , we 2 α β ⊆ α+β know that it is enough to check that for α ∆ , ∈ 1 [gα, g α] h1, and (1) − ⊆ [gα, h2] = 0. (2) Letting β ∆ , we have that β([X ,Y ]) = β(H ) = 2(α,β) = 0 because ∈ 2 α α α (β,β) ∆1 and ∆2 are orthogonal; 1 follows because ∆2 spans the orthogonal complement of h1 in h. Similarly, we have [Xα,Hβ] = α(Hβ)Xα = 0; 2 follows because the Hβ span h2. Solutions to (some) Exercises 175

If, conversely, g = g g , then since g and g are ideals, we get that 1 ⊕ 2 1 2 the roots coming from g1 are orthogonal to the roots from g2, showing that the root system of g is reducible. References 176

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Index adjoint representation, 27 gln, 4, 47 Ado’s Theorem, 5 α-string, 64 Heisenberg algebra, 33, 63 antipode, 21 Hochschild cohomology, 5 Hopf Algebras, 21 b, 47 Hopf ideal, 25 Baker-Campbell-Hausdorff, 11 bialgebra, 22 invariant form, 52 Borcherds, Richard E., 3, 105–170 Jacobi identity, 4 Cartan decomposition, 61 joke, 178 Cartan subalgebra, 61 Jordan decomposition, 53 Cartan’s Criterion, 54 absolute, 59, 60 Casimir operator, 55 central extension, 43 Kac-Moody algebra, 44 cohomology Killing form, 52 of Lie algebras, 37 Knutson, Alan, 3 Complete reducibility, see Weyl’s Kontsevitch, Maxim, 35 Theorem LATEX, 3 comultiplication, 21 Lie algebra, 4 coroot, 66 of a Lie group, 6 counit, 21 Lie algebra cohomology, 37 covering map, 16 Lie derivative, 5 Coxeter diagram, 71 Lie group, 4 Coxeter group, 66 Lie ideal, 6 d, 47 Lie’s Theorem, 49 Danish, 21 loop algebra, 44 deformation loop space, 44 of a Lie algebra, 42 lower central series, 47 deformations nilpotent, 47 of associative algebras, 32 element, 60 derived series, 47 dual pairing, 23 one-parameter subgroup, 10 Dynkin diagram, 71 PBW, see Poincaré-Birkhoff-Witt Engel’s Theorem, 48 Poincaré-Birkhoff-Witt, 31 exponential map, 10 rank, 62, 67 filtered space, 28 real form, 4 gl , 44 regular element, 60 ∞ Index 179 representations, 26 Reshetikhin, Nicolai, 3–45 root, 61 positive, 69 simple, 69 root space, 61 root decomposition, 61 root system abstract, 67 irreducible, 67 semisimple, 51 element, 60 Serganova, Vera, 3, 46–104 sl2, 61, 63 sl3, 62 sln, 60 solvable, 47 star product, 34 unitary trick, 58 universal cover, 17 universal enveloping algebra, 24– 32 upper triangular, see b useful facts, 47–48 variety of Lie algebras, 5 Verma module, 90–101

Weyl group, 68 Weyl’s Theorem, 58 Whitehead’s Theorem, 55

Zariski open set, 65