Linear Algebraic Groups Contents

Home , Identity component, Linear algebraic group, Nilpotent group, Permutation group, Quotient group, Solvable group

Linear Algebraic Groups

These are the exercises (with solutions) and the exam for the course Algebra II, winter term 2014/2015 at Bonn. Students were not assumed to have heard algebraic geometry before, just algebra. In particular, basic aﬃne and projective geometry were introduced from scratch. The course covered the standard material up to the structure results about maximal tori and Borel subgroups. Root systems and data were covered, as well as how to associate a root system to a semisimple group.

Contents

1 Some pointers to the literature2

2 List of results proved in the course3

3 Exercises7

4 Solutions to Exercises 12

5 Exam and solutions 26 1 Some pointers to the literature

A. Borel: Linear Algebraic Groups Springer (1969, 1997) J. Humphreys: Linear Algebraic Groups Springer (1975, 1981) T. Springer: Linear Algebraic Groups Birkh¨auser(1981, 1998) P. Tauvel, R. Yu: Lie Algebras and Algebraic Groups Springer (2005) A. Onishchik, E. Vinberg: Lie Groups and Algebraic Groups Springer (1990) R. Goodman, N.Wallach: Symmetry, Representations and Invariants Springer (2009)

The first three books (Borel, Humphreys, Springer) are the classical textbooks on the subject. They share the following features: the ground field is algebraically closed and of arbitrary characteristic; rationality questions (i.e. non-algebraically closed fields of definition) are treated at the end; they cover the structure theory of linear algebraic groups including the classification of reductive/semisimple groups; they avoid schemes but use ringed spaces to define quotients. Of these books, I like Humphrey’s the best: it does a good job of explaining why things happen. The other three books all work over the complex numbers, and have different aims. Tauvel/Yu is really concerned with Lie algebras. It covers basically all of the foun- dational results otherwise cited (including commutative algebra, sheaves, group theory, projective geometry, root systems) — at the price that Lie algebras are introduced as late as in §19 and linear algebraic groups in §21. Moreover, the book is extremely light on examples. It treats a lot of the finer structure theory, including reductive, Borel, parabolic, Cartan groups. In my opinion, this volume is more a reference than a textbook to learn the subject from. The sections relevant to the course are §10,18,21–23,25–28. Goodman/Wallach starts with very explicit descriptions of the classical groups

(SLn, SOn, Spn) and then develops the theories of Lie groups and of algebraic groups in parallel. Thus it is very example-driven: for example, maximal tori and roots are all developed first for the classical groups. Later on, Chapter 11 is devoted to a rapid development of linear algebraic groups, up to Borel subgroups and maximal tori (but without the classification). I took the construction of homogenous spaces G/H and the proof that morphisms of linear algebraic groups have closed image from Appendix A. Onishchik/Vinberg is written in a unique style: the book grew out of a 1967 Moscow seminar; almost all of the theory is presented in a series of problems (generally with concise solutions). The book develops both (real and complex) Lie groups and complex linear algebraic groups, in a very quick and efficient fashion. They classify semisimple Lie algebras. To a large extent, I have been following the lecture notes of TamásSzamuely from a 2006 course at Budapast: http://www.renyi.hu/~szamuely/lag.pdf. It has the advantage of starting out with low technical demands (no schemes or sheaves are used; only quasi-projective varieties). The proof of conjugacy of maximal tori using group cohomology is from these notes. 2 List of results proved in the course

Affine algebraic geometry Proposition 1: Connected components of affine algebraic groups coincide with irreducible components. The identity component is a normal subgroup. Proposition 2: Mor(X,Y ) = Hom(A(Y ),A(X)) for affine varieties X,Y . For any finitely generated, reduced K-algebra there is a unique affine variety X with A(X) ∼= A. ∼ Proposition 3: A(X × Y ) = A(X) ⊗K A(Y ) for affine varieties X,Y .

Embedding theorem

Theorem: Any affine linear group G is a closed subgroup of some GLn. Chevalley Lemma: If H ⊂ G is a closed subgroup of a linear algebraic group, then there exists a finite-dimensional representation G → GL(V ) and a 1-dimensional linear subspace L ⊆ V such that H = Stab(L). Proposition 4: For a normal subgroup H ⊆ G of a linear algebraic group G, there is a finite-dimensional representation G → GL(V ) with kernel H.

Jordan decomposition Theorem: Jordan decomposition for a linear algebraic group G.

Theorem (Kolchin): If G ⊂ GLn is a unipotent group, then there exists a non-zero vector ﬁxed by G. Burnside Lemma: If V is a ﬁnite-dimensional K-vector space and A ⊆ End(V ) a subalgebra without A-stable subspaces except 0 and V , then A = End(V ).

Theorem (structure of commutative linear algebraic groups G): Gs,Gu ∼ are closed subgroups of G, and Gs ×Gu −→ G is an isomorphism of aﬃne algebraic groups.

Actions and representations Proposition 5: G diagonalisable ⇐⇒ G is isomorphic to a closed subgroup of Dn ⇐⇒ G = Gs is commutative ∗ n ∗ Lemma: X (Dn) = Z and X (Dn) is a basis of A(Dn). Theorem (structure of diagonalisable groups G): G diagonalisable ⇐⇒ X∗(G) is a finitely generated abelian group (without p-torsion if char(K) = p) and ∗ ∼ X (G) is a basis of A(G) ⇐⇒ G = µd1 × · · · × µdr × Dm with p 6 | di. Proposition 6: G diagonalisable ⇐⇒ all finite-dimensional G-representations decompose into characters Theorem (Weyl groups are finite): If G is a linear algebraic group and T ⊆ G a torus, then the quotient W (G, T ) := NG(T )/CG(T ) is finite. Connected solvable groups

Theorem (Lie-Kolchin): If G ⊂ GLn is a connected solvable subgroup, then there exists a complete ﬂag of G-invariant subspaces in V , i.e. G can be conjugated into Tn. Theorem (structure of connected nilpotent groups): If G is a connected nilpotent linear algebraic group, then Gu,Gs are closed, normal subgroups of G ∼ and Gs × Gu −→ G is an isomorphism of aﬃne algebraic groups, and Gs is a torus.

Tangent spaced and Lie algebras Lemma: If X is an aﬃne homogeneous space for a linear algebraic group G, then X is smooth. Theorem: If G is a connected, 1-dimensional linear algebraic group, then either G = Ga or G = Gm. [Only proved for char(K) = 0.]

Lemma: If G ⊆ GLn is a linear algebraic group, then the action of G on its Lie −1 algebra g = {A ∈ gln | XAf ∈ I(G) ∀f ∈ I(G)} is given by g · A = gAg . Proposition 7: For a linear algebraic group G, tangent space at the unit and Lie ∼ algebra coincide: L(G) = TeG.

Quotients by normal subgroups Theorem: If H ⊆ G is a closed normal subgroup of a linear algebraic group, then G/H is a linear algebraic group. Proposition 8: If A ⊂ B are K-algebras, with B finitely generated over A, then the extension property holds: ∀ ∃ ∀ ∃ :Φ|A = ϕ . b∈B a∈A ϕ: A→K Φ: B→K b6=0 a6=0 ϕ(a)6=0 Φ(b)6=0 Proposition 9: If f : X → Y is a morphism of affine varieties, then f(X) contains an open subset of its closure f(X). Corollary: If f : G → H is a morphism of affine algebraic groups, then the image f(G) is a closed subgroup of H.

Quasi-projective varieties

d Proposition 10: The Plücker map pd : Gr(d, V ) → P(Λ V ) is a closed embedding. Lemma: X proper, Z ⊆ X closed ⇒ Z proper. X1,X2 proper ⇒ X1 × X2 proper. X proper, ϕ: X → Y morphism ⇒ ϕ(X) closed, proper subvariety of Y . X proper and affine ⇒ X is a finite set. Theorem: Projective varieties are proper. Nakamaya Lemma: If A is a commutative ring, M ⊂ A a maximal ideal and N a finitely generated A-module with N = MN, then there is an f ∈ A \ M with fN = 0. Homogeneous spaces and quotients Theorem: If H ⊂ G is a closed subgroup of a linear algebraic group, then G/H is a quasi-projective with a morphism π : G → G/H and if ϕ: G → X is any G-equivariant morphism of homogeneous G-spaces with ϕ(H) = ϕ(1), then there exists a unique morphism ψ : G/H → X with ϕ = ψπ. Moreover, if G acts on a quasi-projective variety Y such that H ⊆ Gy for some y ∈ Y , then the natural map G/H → G · y is a morphism. Proposition 11: Let B be a finitely generated K-algebra without zero divisors and A ⊂ B is a subalgebra. If there is b ∈ B, b 6= 0 such that all Φ: B → K with Φ(b) 6= 0 are uniquely determined by Φ|A, then B ⊆ Quot(A). Proposition 12: Let f : M → N and h: M → P be regular, dominant morphisms of affine varieties such that there exists a non-empty open set U ⊆ M with f(m1) = f(m2) ⇒ h(m1) = h(m2) ∀m1, m2 ∈ U, then there is a rational map g : N 99K P with h = gf.

Borel and parabolic subgroups Orbit lemma: If a linear algebraic group G acts on a quasi-projective variety X, then (i) every orbit is open in its closure, (i) orbits of minimal dimension are closed and, in particular, (iii) closed orbits exist. Borel Fixed Point Theorem: Any action of a connected solvable group G on a projective variety X has a ﬁxed point. Theorem: Any two Borel subgroups of a linear algebraic group are conjugate. Proposition 13: Borel subgroups are parabolic. 0 Proposition 14: If ϕ: G G is a surjective morphism of aﬃne algebraic groups and H ⊆ G is a parabolic (or Borel) subgroup, then so is ϕ(H) ⊆ G0.

Maximal tori Proposition 15: For a connected solvable group G, there exists a torus T ⊂ G such that T,→ G G/Gu is an isomorphism.

Corollary: A connected solvable group G is a semi-direct product G = Gu o T , ∼ where T = G/Gu is a maximal torus. Theorem: Any maximal tori in a connected linear algebraic group are conjugate.

Structure results Theorem: Let G be a connected linear algebraic group, B ⊆ G a Borel subgroup and T ⊆ G a maximal torus. Then G is covered by all Borel subgroups, and Gs is S −1 S −1 covered by all maximal tori: G = g∈G gBg and Gs = g∈G gT g . Proposition 16: Let G be a connected linear algebraic group and T ⊆ G a torus. Then CG(T ) is connected. Theorem: Given B ⊆ G, a Borel subgroup of a connected linear algebraic group, then NG(B) = B. Proposition 17: There is a bijection between the generalised ﬂag variety G/B and the set B of all Borel subgroups of G. 3 Exercises (ﬁve problems each week)

1. Let N be the set of all matrices in GLn(K) with exactly one non-zero entry in every row and every column. Show that N is a closed subgroup of GLn(K), that ◦ its identity component N = Dn is the subgroup of diagonal matrices, that N has n! connected components and that N is the normaliser of Dn.

2. Give examples of non-closed subgroups of GL2(C) and compute their closures.

3. Describe the Hopf algebra structures on the coordinate rings of Ga and GLn.

4. Prove that a T0 topological group is already T2. Show that an inﬁnite linear algebraic group is always T0 but never T2. Explain the discrepancy! 5. Show that the product of irreducible aﬃne K-varieties is again irreducible.

This fails for non-algebraically closed ﬁelds K: exhibit zero divisors in C ⊗R C.

6. Prove that the group Un of unipotent upper triangular matrices is nilpotent. 7. Prove that a group G is solvable if and only if it has a composition series with abelian factors, i.e. there is a chain of subgroups G = G0 ) G1 ) ··· ) Gm = {1} such that each Gi+1 is normal in Gi and all Gi/Gi+1 are abelian.

8. What is the Jordan decomposition for a ﬁnite group? For Ga?

9. Find a closed subgroup G of GL2 such that Gs is not a closed subset.

10. Compute the centre C of SL2(K), assuming char(K) 6= 2. Show that the quotient group PSL2(K) := SL2(K)/C is an aﬃne algebraic group. 4 (Hint: embed SL2 ⊂ A as a Zariski-closed subset, then check that the action of C 4 4 on SL2 extends to an action of C on A . Now map A /C to some aﬃne space as a Zariski-closed subset.)

11. Prove that the group Tn of upper triangular matrices is solvable.

12. Show that Ga and Gm are not isomorphic as aﬃne algebraic groups. 13. Let ϕ: X → Y be a morphism of aﬃne varieties. Show that ϕ is dominant (i.e. the image of X is dense in Y ) if and only if ϕ∗ : A(Y ) → A(X) is injective.

14. Let H ⊂ GLn be an arbitrary subgroup. Show that the Zariski-closure H is a linear algebraic group. Moreover, prove that closure preserves the following properties: H commutative; H normal; H solvable; H unipotent. 15. Show that none of the following implications among properties of linear algebraic groups can be reversed: unipotent

torus +3 diagonalisable +3 abelian +3 nilpotent +3 solvable 3 16. The group SL3 naturally acts on K . Writing x1, x2, x3 for the standard basis 3 of K , this induces an action of SL3 on polynomials p(x1, x2, x3) ∈ K[x1, x2, x3] by g · p = pg. Compute the weight spaces for the torus T := D3 ∩ SL3 on the vector space K[x1, x2, x3]2 of polynomials of degree 2, and draw the weights.

17. Show that the normaliser of D2 in GL2 is solvable, but not conjugated to a subgroup of T2.

18. Compute the Weyl group of GL3 with respect to the torus D3. T 19. For any linear algebraic group G, let H := χ∈X∗(G) ker(χ). Show that H is a closed, normal subgroup of G and that G/H is diagonalisable. Also show X∗(G) ∼= X∗(G/H).

20. (Continuation of 19.) Compute H and G/H for G = GLn.

21. Assume char(K) 6= 2 and let Γ ∈ M(n, K) with associated bilinear form Kn × Kn, (x, y) 7→ xtΓy, and O(Γ) its isometry group. Show that the tangent t space of O(Γ) at I = In is TI O(Γ) = {A ∈ M(n, K) | A Γ + ΓA = 0}. (Hint: use the Cayley transform c(A) = (I + A)(I − A)−1 for A ∈ M(n, K) with det(I − A) 6= 0 and show that c(A) ∈ O(Γ) if and only if AtΓ + ΓA = 0.)

22. Compute the dimensions of SOn and Spn. 23. Assume char(K) = 0. Let G be a linear algebraic group, all of whose elements have finite order. Show that G is finite. 24. For affine varieties X and Y , show that dim(X × Y ) = dim(X) + dim(Y ).

25. If X is an irreducible, smooth aﬃne variety and Z ( X a closed subvariety, prove dim(Z) < dim(X).

26. For a linear algebraic group G, use its comultiplication to deﬁne an associative, ∗ unital K-algebra structure on A(G) = HomK (A(G),K) such that the induced Lie algebra structure coincides with the one from left invariant vector ﬁelds.

27. Exhibit Gm as a closed subgroup of SO2. Then find a two-dimensional torus T ⊂ SO4 and compute the weights for the action induced on the adjoint representation, T,→ SO4 → GL(so4). 28. Let A be a finite-dimensional, associative and unital K-algebra. Show that the group of units is a linear algebraic group. What is its Lie algebra? 29. Show that the differential of the adjoint representation of a linear algebraic group G is given by ad := d(Ad)e : g → End(g), ad(A)(B) = [A, B]. 30. Show that a morphism ϕ: G → H of linear algebraic groups induces a homomorphism of Lie algebras dϕe : g → h. 31. Find a symmetric, non-degenerate bilinear form Γ on Kn such that T := SO(Γ) ∩ Dn is a torus of dimension m if n = 2m or n = 2m + 1. Prove that T is a maximal torus in SO(Γ): if H ⊆ SO(Γ) is abelian with T ⊆ H, then T = H.

n 32. Compute the orbits of the natural actions of GLn, Tn, Un, Dn on A and draw them for n = 2. Describe orbit closures as unions of orbits.

33. Show that the adjoint representation of SL2 has disconnected isotropy groups. 34. Let G be a unipotent linear algebraic group and X an aﬃne G-variety. Show that all orbits of G in X are closed.

35. For a homogeneous ideal I ⊆ K[x0, . . . , xn], geometrically compare the varieties V (I) ⊆ An+1 and V (I) ⊆ Pn. Prove the homogeneous Nullstellensatz.

2 36. Compute the orbits of the actions of GL3 on P , of GL4 on Gr(2, 4), and of 1 1 GL2 on P × P (diagonal action). Also compute isotropy groups for all orbits. 37. Show that any non-degenerate conic in P2 is isomorphic to P1. Deduce that homogeneous coordinate rings are not isomorphism invariants of projective varieties.

38. Show that A1 and P1 are homeomorphic, but A2 and P2 are not. 39. For an aﬃne, irreducible variety X and a point p ∈ X, show that the local ring of X at p is given by the localisation of A(X) at the maximal ideal Mp.

1 40. Show that Aut(P ) = PGL2.

41. Show that varieties are compact in the Zariski topology: any open cover has a ﬁnite subcover. 42. Show that a locally compact Hausdorﬀ space X is compact if and only if the projection X × Y → Y is a closed map for all topological spaces Y .

43. Let X = V (y2 − x3) ⊂ A2 be the cuspidal cubic. Show that the map A1 → X, t 7→ (t2, t3) is regular, birational and a homeomorphism, but not an isomorphism of varieties. Extend this to an example of a morphism of projective varieties with the same properties. 44. Let A be a commutative ring with unit and N a ﬁnitely generated A-module. Show that a surjective module endomorphism N → N is an isomorphism.

45. Find subgroups of SL2 such that the quotient is respectively projective, aﬃne or strictly quasi-aﬃne.

46. Compute the dimensions of Grassmannians Gr(d, n) and flag manifolds Fl(n). 47. Show that two irreducible varieties X and Y are birational, i.e. their function fields are isomorphic: K(X) ∼= K(Y ), if and only if there exist affine open subsets ∼ UX ⊆ X and UY ⊆ Y which are isomorphic: UX = UY .

p 48. Assume char(K) = p > 0. Show that the map Gm → Gm, x 7→ x is a bijective morphism of aﬃne algebraic groups, but is not an isomorphism. 49. Let G be a linear algebraic group, acting on a quasi-projective variety X. Show that orbits of minimal dimension are closed; in particular, closed orbits exist. (Hint: you can use the statement of exercise 25.) 50. Let G be a connected projective algebraic group. Show that G is commutative.

51. Find Borel subgroups in SO4, Sp4,T4 and U4.

52. Find all parabolic subgroups P with T4 ⊆ P ⊆ GL4. 53. Let the connected linear algebraic group G act on a quasi-projective variety X with ﬁnitely many orbits. Show that every irreducible closed G-invariant subset in X is the closure of a G-orbit. Find a counterexample for an action with inﬁnitely many orbits. 54. Find a connected linear algebraic group G and a maximal solvable subgroup U ⊂ G such that U is disconnected.

55. Classify all root systems in the Euclidean plane E := R2. (Note: A root system in an Euclidean space (E, (−, −)) is a subset Φ ⊂ E such that (RS1) Φ is ﬁnite, spans E and 0 ∈/ Φ; (RS2) for any α ∈ Φ, Rα ∩ Φ = {α, −α}; 2(x,α) (RS3) for any α ∈ Φ, the reﬂection sα : E → E, x 7→ x − (α,α) x preserves Φ; 2(β,α) (RS4) for any α, β ∈ Φ: hβ, αi := (α,α) ∈ Z.)

56. Compute the radicals R(GLn),R(SLn),R(Un). ∼ r 57. Let char(K) = 0 and U a commutative, unipotent group. Show that U = Ga for some r ∈ N. 58. Let G be a ﬁnite group and A a trivial G-module, i.e. an abelian group which has the trivial left G-action. Show that H1(G, A) = Hom(G, A). 59. Let ϕ: G → H be a surjective morphism of linear algebraic groups, and T ⊆ G a maximal torus (or a maximal connected normal unipotent subgroup, respectively). Show that ϕ(T ) ⊆ H has the same property. 60. Let (E, Φ) be a root system. A base of (E, Φ) is a subset S ⊆ Φ such that P S is a basis for E and every root β ∈ Φ can be written as β = α∈S mαα with either all mα ≥ 0 or all mα ≤ 0. The elements of S are also called simple roots, and their non-negative linear combinations positive roots and denoted Φ+. Draw bases and positive roots for the root systems in E = R2, of exercise 55. 61. Let G be a connected algebraic group, B ⊆ G a Borel subgroup and V a rational G-module. Show that the invariant subspaces coincide: V G = V B.

62. Let G be a linear algebraic group. Show that R(G)u = Ru(G), i.e. the unipotent part of the radical is the unipotent radical. 63. For a linear algebraic group G, show that G/R(G) is semisimple and that G/Ru(G) is reductive. 64. Let G be a group with subgroups H,N ⊆ G. Show that the following notions are equivalent — then G = N o H is called a semi-direct product of H by N: (1) There is a short exact sequence 1 → N −→ι G −→π H → 1 admitting a section σ : H → G, i.e. πσ = idH . (2) N is normal in G, and NH = G and N ∩ H = 1. (3) There is a homomorphism α: H → Aut(N) and G is isomorphic to the group 0 0 0 0 N oα H deﬁned by (n, h) · (n , h ) := (nα(h)(n ), hh ) on the set N × H. Also show that the existance of a retraction %: G → N, i.e. %ι = idN , is equivalent to a splitting G ∼= N × H of G as a direct product. 65. Show that the following subsets deﬁne root systems of rank n: n+1 {ej − ei | i, j ∈ {1, . . . , n + 1}, i 6= j} ⊂ Q (called type An) n {±ej ± ei | i, j ∈ {1, . . . , n}, i < j} ⊂ Q (called type Dn)

i 0 66. Show that centraliser of 0 −i ∈ PGL2(C) is disconnected. 67. For B ⊆ G a Borel subgroup of a connected linear algebraic group G, show that Z(B) = Z(G).

68. Prove directly for G = SLn and for G = SOn (with the bilinear form from Problem 31) that G is covered by Borel subgroups and that maximal tori coincide with their centralisers. 69. Let G be a connected linear algebraic group with a maximal torus and B be the set of Borel subgroups of G with its natural T -action. Show that there is a bijection between the ﬁxed point set BT and the Weyl group W (G). 70. Show that the following subsets deﬁne root systems of rank n: n {±ei | i ∈ {1, . . . , n}} ∪ {±ej ± ei | i, j ∈ {1, . . . , n}, i < j} ⊂ Q (type Bn) n {±2ei | i ∈ {1, . . . , n}} ∪ {±ej ± ei | i, j ∈ {1, . . . , n}, i < j} ⊂ Q (type Cn) 4 Algebra II — solutions to exercise sheet 1

Solution: It is clear that Dn ⊂ N and that σDn ⊂ N for any permutation σ ∈ Sn (letting permutations act on S rows, say). It is also clear that N = σ σDn and that this union is disjoint. We observe that Dn is connected: it n n is an open subset of A , hence irreducible (because A is irreducible — its aﬃne coordinate ring is K[x1, . . . , xn] and an integral domain), hence connected. This shows all statements except that N is the normaliser of Dn. This is an easy result purely about groups.

2. Give examples of non-closed subgroups of GL2(C) and compute their closures.

Solution: All of the following examples are closed in the norm topology of GL2(C) but not Zariski-closed. GL2(R). It is Zariski-dense in GL2(C): denoting G = GL2(R) and X = GL2(C), then G = {x ∈ X | x = c(x)} where c: X → X is complex conjugation (a bijection, but not a morphism of complex varieties). For any f ∈ A(X) with f(G) = 0, we also have c(f) ∈ A(X), and hence the two polynomials with real coefficients f + c(f) and (f − c(f))/i. However, G is the real affine variety in affine (n2 + 1)-sace with just one equation. Hence there can be no further polynomials vanishing on all of G, hence I(G) = 0 and G = V (I(G)) = V (0) = X. 1 n ∼ Z, embedded via n 7→ 0 1 . This subset is not closed, because it is discrete infinite. Its closure is U2 = Ga. 1 1 x 0 1 ∗ S , the 1-dimensional real sphere, embedded via S ,→ GL2(C), x 7→ 0 1/x . Its closure is Gm via S ⊂ C = Gm and the same embedding.

3. Describe the Hopf algebra structures on the coordinate rings of Ga and GLn. Solution: For G = Ga is A(G) = K[x] with comultiplication ∆(x) = x ⊗ 1 + 1 ⊗ x from m(g, h) = g + h. Moreover, coinverse ι(x) = −x and counit ε(x) = 0. P For G = GLn is A(G) = K[x11, . . . , xnn, y]/(1 − y · det(x)). Comultiplication ∆(xij ) = l xil ⊗ xlj ; coinverse i+j ι(xij ) = (−1) y det(ˆxij , wherex ˆij is the matrix obtained from x = (xij ) by cutting row i and column j — this is precisely the i, j-entry of the inverse matrix to x, computed using the adjoint matrix via X ·adj(X) = det(X)In, note that y = 1/ det(x) ∈ A(G); counit ε(xij ) = δij .

4. Prove that a T0 topological group is already T2. Show that an inﬁnite linear algebraic group is always T0 but never T2. Explain the discrepancy!

Solution: Let G be a T0 topological group. We first show that the singleton {e} is a closed subset, where e ∈ G is the neutral element. Given any element x ∈ G, there is either an open neighbourhood Ux of x with e∈ / U or an open neighbourhood V of e with x∈ / e. (This is the definition of T0, applied to the two points x, e.) In the latter case, we may assume that V = V −1 (if not, replace V with V ∩ V −1, this is still an open neighbourhood of e) and use the homeomorphism f : G → G, y 7→ xy, which maps e 7→ x. Let Ux := f(V ), this is an open −1 neighbourhood of x = f(e) with e∈ / Ux (otherwise e ∈ f(V ) = xV , hence x ∈ V , contradicting x∈ / V and −1 S V = V ). We now take the union U := x6=e Ux over all these neighbourhoods. By construction, this is an open set with U = G \{e}, so {e} is indeed closed. The map G × G → G, (g, h) 7→ gh−1 is continuous, since G is a topological group. Its preimage of the closed subset {e} is the diagonal ∆G = {(g, g) | g ∈ G}, which is therefore closed. It is a general (and easy) fact that a topological space X is T2 if and only if the diagonal ∆X is a closed subset of X × X. The statements about linear algebraic groups follow from general properties of the Zariski topology: points of affine varieties are closed because they correspond on maximal ideals; hence varieties are T0 (and even T1 but note that schemes have more points and are only T0 in general). Since any two non-empty open subsets of an irreducible variety meet, varieties are never T2. Explanation: this exercise shows that while linear algebraic groups are groups with a topology, they are not topological groups. The reason is that the topology on G × G for a variety is not the product topology.

5. Show that the product of irreducible aﬃne K-varieties is again irreducible.

This fails for non-algebraically closed ﬁelds K: exhibit zero divisors in C ⊗R C.

Solution: Let A := A(X) and B := A(Y ) be the aﬃne coordinate rings. We know that A(X × Y ) =∼ A ⊗K B. Also, X irreducible corresponds to I(X) being a prime ideal or, equivalently, A(X) being an integral domain. Therefore, we have to show that if A and B are integral domains, then so is A ⊗K B. P Let f = i ai ⊗ bi ∈ A ⊗ B be a zero divisor. We can assume that the bi are linearly independent. If x ∈ X, then P f(x, −) = i ai(x)bj ∈ B is a zero divisor in B, hence zero. With the bj linearly independent, we ﬁnd ai(x) = 0 for all i, thus ai = 0 (Nullstellensatz!) and so f = 0.

In C ⊗R C, one computes (1 ⊗ i + i ⊗ 1)(1 ⊗ 1 + i ⊗ i) = 0. Algebra II — solutions to exercise sheet 2

6. Prove that the group Un of unipotent upper triangular matrices is nilpotent. 2 Solution: This a straightforward computation with commutators of matrices. For example, C (Un) = [Un, Un] n consists of matrices with zeros on the secondary diagonal. In the central series, C (Un) is trivial. 7. Prove that a group G is solvable if and only if it has a composition series with abelian factors, i.e. there is a chain of subgroups G = G0 ) G1 ) ··· ) Gm = {1} such that each Gi+1 is normal in Gi and all Gi/Gi+1 are abelian. Solution: We use the fact that for a subgroup H ⊆ G holds: H ⊇ D(G) = [G, G] if and only if H is normal in G and G/H is abelian. n i If G is solvable, then by deﬁnition the derived series D (G) trivialises. Put Gi := D (G). Then by the fact, each Gi+1 = D(Gi) is normal in Gi with abelian quotient. The series terminates after ﬁnitely many steps because G is solvable. If we are given the chain of subgroups, then invoking the fact in the reverse direction, we get D(Gi) ⊆ Gi+1. i Thus inductively, D (G) ⊆ Gi and therefore the derived series trivialises.

8. What is the Jordan decomposition for a ﬁnite group? For Ga?

Solution: If char(K) = 0, then all elements of a finite group G (considered embedded in some GLn) are semisimple. To see this, note that a unipotent matrix U has all eigenvalues 1. If U 6= In, then U has non-trivial Jordan blocks k and U 6= In for k ≥ 1. Therefore U can never have finite order, i.e. belong to G. Note that this argument fails 1 1 in finite characteristic: for example, the matrix U = 0 1 over a field K with char(K) = 2 is unipotent with 2 U 6= I2 but U = I2. ∼ Ga = U2, so all elements of Ga are unipotent.

9. Find a closed subgroup G of GL2 such that Gs is not a closed subset. ∗ ∗ Solution: The subgroup T2 of upper triangular matrices 0 ∗ . The subset of semisimple elements is a b ∗ (T2)s = 0 d | a, d ∈ K , b ∈ K with a 6= d or a = d, b = 0 . a b It is the complement of matrices 0 d with a = d, b 6= 0, and this subset is not Zariski-open: for example, its a b closure are the matrices 0 d with a = d (and arbitrary b), which is not all of T2.

1 0 −1 0 Solution: By direct computation, C = { 0 1 , 0 −1 } is the group with two elements. With SL2 = 4 4 {(x1, x2, x3, x4) ∈ A | x1x4 − x2x3 = 1}, the group is a closed subset of A . The group C acts by ±id, 4 4 which obviously extends to A . An element of A /C is a quadruple up to sign, and we can map

4 6 A /C → A , ±(x1, x2, x3, x4) 7→ (x1x2, x1x3, x1x4, x2x3, x2x4, x3x4).

It is straightforward (and necessary) to check that this map is (a) well-deﬁned and (b) cut out by the three 4 6 polynomial equations y1y6 = y2y5 = y3y4, with coordinates A(A ) = K[x1, . . . , x4] and A(A ) = K[y1, . . . , y6]. The map is not injective, but it is injective on the subset PSL2. 6 This exhibits PSL2 = SL2/C ⊂ A as a Zariski-closed subset. (The equation det = 1, i.e. x1x4 − x2x3 = 1, for SL2 is invariant under C.) It remains to observe that multiplication PSL2 × PSL2 → PSL2 and inversion PSL2 → PSL2 are given by polynomial maps.

4 10 2 2 2 2 Alternative: the map A /C → A , ±(x1, x2, x3, x4) 7→ (x1, x2, x3, x4, x1x2, x1x3, x1x4, x2x3, x2x4, x3x4) is injective and can therefore also be used to embed PSL2 in an aﬃne space. Note: This exercise shows that some linear algebraic groups do not come with a natural embedding into a linear group. For such groups, it is easier to check that they are aﬃne algebraic groups. Algebra II — solutions to exercise sheet 3

11. Prove that the group Tn of upper triangular matrices is solvable.

Solution: Start by checking D(Tn) = Un and then use the method of exercise 6. See Tauvel/Yu §10.8.

12. Show that Ga and Gm are not isomorphic as affine algebraic groups. ∗ Solution: In fact, the underlying varieties are not isomorphic: X := Ga = K and Y := Gm = K have affine coordinate rings A(X) = K[x] and A(Y ) = K[y, y−1]. From the equivalence between affine varieties and reduced, finitely generated algebras we know that X =∼ Y if and only if K[x] =∼ K[y, y−1]. One way to see that these rings ∗ ∗ ∗ ∗ −1 ∗ ∗ are not isomorphic is by their groups of units: A(X) = K[x] = K whereas A(Y ) = K[y, y ] = K × Z, mapping (λ, n) 7→ λyn. Note that the field K is fixed from the outset: in A(X), every automorphism is a multiple of the identity, whereas this is not the case for A(Y ). (Abstractly, it may be hard to see that the groups K∗ ∗ 1 ∗ and K × Z are non-isomorphic. There are many bizarre isomorphisms, for example S =∼ C as groups.) This solution is equivalent to comparing the automorphism groups as affine varieties: Aut(X) = A(X)∗.

A different approach uses automorphisms of Ga and Gm as affine algebraic groups. The only automorphisms of −1 Gm as an affine algebraic group are λ 7→ λ and λ 7→ λ . By contrast, Ga has no such automorphisms at all. t Note that the Lie group analogs of Ga and Gm are isomorphic via R → R>0, t 7→ e . 13. Let ϕ: X → Y be a morphism of affine varieties. Show that ϕ is dominant (i.e. the image of X is dense in Y ) if and only if ϕ∗ : A(Y ) → A(X) is injective. Solution: Assume ϕ dominant and let f ∈ A(Y ) with 0 = ϕ∗(f) = fϕ ∈ A(X). As a continuous function Y → K, the vanishing locus of f is a closed subset of Y . However, f is zero on the image of ϕ, hence on a dense set. So the smallest possible vanishing locus is Y , hence f = 0. Now assume that ϕ∗ : A(Y ) → A(X) is injective. Put Z := im(ϕ), this is the Zariski-closure of the image of ϕ. If ϕ is not dominant, i.e. Z ( Y , then I(Z) ) I(Y ). Hence there is f ∈ I(Z) with f∈ / I(Y ); this is a function f : Y → K which is non-zero and vanishes on Z. Therefore, the composition fϕ = ϕ∗(f) = 0, contradicting the injectivity of ϕ∗.

14. Let H ⊂ GLn be an arbitrary subgroup. Show that the Zariski-closure H is a linear algebraic group. Moreover, prove that closure preserves the following properties: H commutative; H normal; H solvable; H unipotent. Solution: (We repeatedly use: if f : X → Y is a continuous map with f(A) ⊆ B for subsets A ⊆ X,B ⊆ Y , then f(A) ⊆ f(A). In fact, this property (for all A) is equivalent to continuity.) For any h ∈ H, left multiplication lh : GLn → GLn is a homeomorphism preserving the subset H. Therefore, it induces a homeomorphism of closures, lh : H → H. As this works for all h ∈ H, we get H · H ⊆ H. Now right multiplication by h ∈ H is a homeomorphism GLn → GLn restricting to H → H, hence mapping H → H. −1 We get H · H = H. Analogously, inversion GLn → GLn, g 7→ g is a homeomorphism preserving H, hence it provides a homeomorphism of H. Therefore, H is a subgroup of GLn, and thus a linear algebraic group. If H is commutative, then for any h ∈ H, the map [h, −]: H → H has trivial image {1}. As the set {1} is closed, the induced map on closures is [h, −]: H → {1} and so [h, g] = 1 for all h ∈ H, g ∈ H. Now for g ∈ H, the map [−, g]: H → {1} induces [−, g]: H → {1}. Hence H is commutative as well. For H normal, use the map H → H, h 7→ ghg−1 for g ∈ G. Similar reasoning works for H solvable. For H n unipotent, use that unipotency in GLn is given by the equations (In − g) = 0. 15. Show that none of the following implications among properties of linear algebraic groups can be reversed: unipotent

torus +3 diagonalisable +3 abelian +3 nilpotent +3 solvable

Solution: Any ﬁnite abelian subgroup of GL2 is diagonalisable (char(K) = 0) but not a torus. ∼ Ga = U2 is abelian but not diagonalisable (it is unipotent). U3 is nilpotent but not abelian. T2 is solvable but not nilpotent. Any torus is nilpotent but not unipotent. Wanted: an example of a nilpotent group which is neither abelian nor unipotent! Algebra II — solutions to exercise sheet 4

3 16. The group SL3 naturally acts on K . Writing x1, x2, x3 for the standard basis 3 of K , this induces an action of SL3 on polynomials p(x1, x2, x3) ∈ K[x1, x2, x3] by g · p = pg. Compute the weight spaces for the torus T := D3 ∩ SL3 on the vector space K[x1, x2, x3]2 of polynomials of degree 2, and draw the weights. 2 2 2 Solution: The vector space V := K[x1, x2, x3]2 has a basis x1, x1x2, x1x3, x2, x2x3, x3. The torus T = D3 ∩ SL3 is isomorphic to D2, and the basis of V also turns out to give all simultaneous eigenvectors for the T -module:

  t1 0 0 2 2 2 −1 t =  0 t2 0  , t · x1 = t1x1, t · x1x2 = t1t2x1x2, t · x1x3 = t2 x1x3 etc. 0 0 1/t1t2

∗ ∗ 2 2 a1 a2 We have X (T ) = X (D2) = Z , and the character associated to (a1, a2) ∈ Z is χ(a1,a2) : T → Gm, t 7→ t1 t2 . Therefore, we get the following six weight spaces (this is the full set due to dim(V ) = 6)

2 2 2 Vχ(2,0) = Kx1,Vχ(0,2) = Kx2,Vχ(−1,−1) = Kx3,Vχ(1,1) = Kx1x2,Vχ(0,−1) = Kx1x3,Vχ(−1,0) = Kx2x3.

2 The set of weights is {(2, 0), (0, 2), (1, 1), (−1, −1), (0, −1), (−1, 0)} ⊂ Z .

17. Show that the normaliser of D2 in GL2 is solvable, but not conjugated to a subgroup of T2. ` 0 ∗ Solution: The normaliser already appeared in Exercise 1: N2 = D2 A2 (disjoint union), where A2 = { ∗ 0 } is the subset of matrices with zeroes on the diagonal. The group N2 is solvable because 1 ⊂ D2 ⊂ N2 is a ∼ composition series with abelian factors (N2/D2 = Z/2Z). 2 2 We show that there are no invariant subspaces for the action of N2 on K : as a D2-module, K has the invariant ∗ 0 subspaces 0 and ∗ but neither of these is A2-invariant. Therefore, N2 cannot embed into T2. This exercise shows that both assumptions of connected, solvable are necessary in the Lie-Kolchin theorem.

18. Compute the Weyl group of GL3 with respect to the torus D3.

Solution: Put G := GL3 and T := D3. A direct computation shows that the centraliser CG(T ) = T . Let t ∈ T be a diagonal matrix and g ∈ G. We want to see when gtg−1 is diagonal. Replacing g by (det(g))−1/3g, if necessary, we can assume det(g) = 1. (This simplifies the formula for g−1 via the adjoint matrix a bit.) A lengthy computation by hand then shows that g has exactly one non-zero entry in each row, and each column. Therefore g is obtained from t via a permutation matrix, and W (D3, GL3) = S3, the symmetric group on 3 letters. T 19. For any linear algebraic group G, let H := χ∈X∗(G) ker(χ). Show that H is a closed, normal subgroup of G and that G/H is diagonalisable. Also show X∗(G) ∼= X∗(G/H). Solution: Each ker(χ) = χ−1(1) is a closed subgroup, hence their intersection H is closed. Also kernels are normal, so again their intersection H is normal. In order to show that G/H is diagonalisable, we will prove that it is commutative and consists of semisimple ∗ −1 −1 elements. For any g1, g2 ∈ G and χ ∈ X (G), we have χ([g1, g2]) = χ(g1)χ(g2)χ(g1) χ(g2) = 1, hence [G, G] ⊆ ker(χ), and so [G, G] ⊆ H. This implies G/H commutative. Denoting by π : G → G/H the projection, any element of G/H is of the form π(g). From the properties of Jordan decomposition we know that π(g) = π(g)sπ(g)u = π(gs)π(gu). We claim that χ(u) = 1 for any u ∈ Gu. This will imply u ∈ H, and thus π(g) = π(g)s. Now a character is an action on a 1-dimensional vector space, so it has to be linearisable — but then the eigenvalues of a unipotent element are all 1, hence χ(u) = 1. The homomorphism X∗(π) = π∗ : X∗(G/H) → X∗(G) is injective, because π is surjective. Moreover, X∗(π) is also surjective: if χ ∈ X∗(G), then χι = 1 by the definition of H, where ι: H → G denotes the embedding. Hence χ gives a well-defined character G/H → Gm.

20. (Continuation of 19.) Compute H and G/H for G = GLn.

Solution: Any character of G is trivial on D(G) = [G, G]. For GLn, the commutator subgroup is D(GLn) = SLn. ∼ (This requires a computation.) Hence H ⊇ SLn. Since SLn is normal in GLn with quotient GLn/SLn = Gm and there is the non-trivial character det: GLn → Gm, we ﬁnd H = SLn. Algebra II — solutions to exercise sheet 5

t Solution: [Goodman/Wallach, exercise 1.4.5.5] An element g ∈ GLn is a Γ-isometry if and only if (gx) Γ(gy) for all x, y ∈ Kn. This is equivalent to gtΓg = Γ. We compute (where the last step uses 2 6= 0 in K) c(A) ∈ O(Γ) ⇐⇒ c(A)tΓc(A) = Γ ⇐⇒ (I − A)−t(I + A)tΓ(I + A)(I − A)−1 = Γ ⇐⇒ (I + A)tΓ(I + A) = (I − A)tΓ(I − A) ⇐⇒ Γ + AtΓA + AtΓ + Γ = Γ + AtΓA − AtΓ − Γ ⇐⇒ 2(AtΓ + ΓA) = 0 ⇐⇒ AtΓ + ΓA = 0. t 1 1 Let now A ∈ M(n, K) with A Γ + ΓA = 0. The set of t ∈ A such that det(I − tA) = 0 is finite; let U ⊂ A be its 1 open complement. (Note that U is also an affine variety, similarly to how Gm ⊂ A is.) We consider the morphism of affine varieties, γ : U → GLn, t 7→ c(tA). Obviously γ(0) = I and by the above, γ(t) ∈ O(Γ) for all t ∈ U. Its differential at 0 is a map dγ0 : T0U = K → TI O(Γ). We will show that dγ0(1) = dγ(t)/dt(0) = 2A; this will imply t TI O(Γ) ⊇ {A ∈ M(n, K) | A Γ+ΓA = 0}. For the computation, apply the usual rules for differentiating products −1 −1 −2 and powers: d/dt|t=0((I+tA)·(I−tA) ) = d/dt|t=0(I+tA)·I+I·d/dt|t=0(I−tA) = A−(I−0·A) (−A) = 2A; −1 this is possible since A and I commute. (Over K = C, one can also use the geometric series for (I − tA) .) 1 t t For B ∈ M(n, K), consider the regular function ψB : GLn → A , g 7→ tr((g Γg − Γ)B). Note g Γg − Γ = 0 ⇐⇒ tr((gtΓg − Γ = 0)B) = 0 ∀B; this follows from the non-degeneracy of the trace bilinear form on M(n, K). Via ψB , we restrict to deriving scalar-valued functions. For A ∈ M(n, K), we have the directional derivative DA = P t ij Aij ∂/∂xij , and DAψB (I) = tr((A Γ + ΓA)B). This can be seen in matrix coordinates: writing Γ = (Γij ) t P t etc., we have tr((g Γg − Γ = 0)B) = ijkl(gkiΓklglj − Γkl)bji. Thus TI O(Γ) ⊆ {A ∈ M(n, K) | A Γ + ΓA = 0}.

22. Compute the dimensions of SOn and Spn.

Solution: We apply the previous exercise to compute the tangent spaces TI SOn and TI Spn. The dimensions of these tangent spaces coincide with the dimensions of the groups, since algebraic groups are smooth. t For the orthogonal group On, the bilinear form is given by Γ = In, hence TI SOn = {A ∈ M(n, K) | A + A = 0}. The condition At + A = 0 is equivalent to A skew-symmetric. Hence the diagonal entries are zero, and A is 1 determined by its upper triangular part. Therefore dim(TI On) = (n − 1) + (n − 2) + ··· + 1 = 2 n(n − 1). Moreover, On and SOn have the same tangent space at I, because SOn is the identity component of On. 2n 0 In For the symplectic group Sp , the bilinear form on K is given by Γ = J = , where In is the identity n −In 0 t BC matrix in M(n, K). We get TI Spn = {A ∈ M(2n, K) | A J + JA = 0}. (Here I = I2n.) Writing A = DE as a block matrix, we find the condition AtJ + JA = 0 turns out to be equivalent to D = Dt,C = Ct,B = −Et. Hence C and D are symmetric n × n-matrices, B is an arbitrary n × n-matrix and E is completely determined 1 2 by B. The dimension therefore is dim(TI Spn) = 2 · 2 (n + 1)n + n = n(2n + 1). 23. Assume char(K) = 0. Let G be a linear algebraic group, all of whose elements have finite order. Show that G is finite. Solution: We can assume that G is connected because for any linear algebraic group, G/G◦ is a finite group (G◦ t connected component of the neutral element). Put Gt := {g ∈ G | g = 1}. This is a closed subset of G and by S assumption G = t Gt. We deduce G = Gt for some fixed t from the following general fact: A union of countably many subvarieties Zi ( X does not equal X: we can assume that Zi = V (fi) are hypersurfaces and we do induction on dim(X). The case dim(X) = 1 is obvious (K algebraically closed and char(K) = 0 implies K uncountable). Now take two non-proportional functions h1, h2 ∈ A(X) and consider the pencil of 1 hypersurfaces Lt := V (h1 + th2 with t ∈ A . These are uncountably many pairwise different hypersurfaces, so not all Zi can be among the Lt. This gives us points lying in X but not on any Zi. The map f : G → G, g 7→ gt is a morphism of affine algebraic groups; in particular, it is continuous. We have −1 −1 f (1) = G, due to G = Gt. However, the fibre f (1) is finite, as can be seen from embedding G ⊆ GLn — here we use char(K) = 0. Hence G is finite (and with the assumption G connected, even G = {1}).

24. For aﬃne varieties X and Y , show that dim(X × Y ) = dim(X) + dim(Y ). ∼ Solution: This boils down to showing T(x,y)(X × Y ) = TxX ⊕ TyY which is easy in either of the deﬁnitions of tangent space. Now let U ⊆ X and V ⊆ Y be the open, dense subsets where the dimensions of tangent spaces are minimal. It is then clear that dim(X) + dim(Y ) is the minimum attained on the open subset U × V ⊆ X × Y .

25. If X is an irreducible, smooth aﬃne variety and Z ( X a closed subvariety, prove dim(Z) < dim(X).

∗ Solution: Denote the inclusion morphism ι: Z,→ X; its induced morphism ι : A(X) A(Z) is then surjective. 2 2 We get induced surjections MX,P MZ,P of maximal ideals and also MX,P MZ,P . Hence, we obtain a 2 2 surjection MX,P /MX,P MZ,P /MZ,P (of cotangent spaces at P ). There exist regular functions f ∈ MX,P 2 which don’t vanish on Z (as X is irreducible). Choose such an f of minimal degree. Then f cannot be in MX,P , 2 2 so MX,P /MX,P MZ,P /MZ,P is not an isomorphism, so dim(Z) ≤ dim(TpZ) < dim(TpX) = dim(X) for p ∈ Z. Algebra II — solutions to exercise sheet 6

∗ Solution: For two functionals ψ1, ψ2 ∈ A(G), we deﬁne their tensor product to be ψ1 ⊗ ψ2 ∈ (A(G) ⊗ A(G)) ∗ with (ψ1 ⊗ψ2)(f1 ⊗f2) := ψ1(f1)·ψ2(f2) ∈ K for any f1, f2 ∈ A(G). Using the comultiplication ∆ = µ : A(G) → ∗ A(G) ⊗ A(G), we deﬁne ψ1 · ψ2 := (ψ1 ⊗ ψ2)∆ ∈ A(G) . This product is associative: coassociativity of ∆ means (id ⊗ ∆)∆ = (∆ ⊗ id)∆: A(G) → A(G)⊗3. Composing ⊗3 ψ1 ⊗ ψ2 ⊗ ψ3 : A(G) → K gives (ψ1 · ψ2) · ψ3 = ((ψ1 ⊗ ψ2)∆ ⊗ ψ3)∆ = (ψ1 ⊗ (ψ2 ⊗ ψ3)∆)∆ = ψ1 · (ψ2 · ψ3). / λ(G) Now we use the isomorphisms θ : DerK (A(G),K(e)) o DerK (A(G),A(G)) : ε where ε(d)(f) = d(f)(e) and ∗ θ(v)(f) = Dv(f) with Dv(f)(g) = v(λgf) for f ∈ A(G) and g ∈ G. Let ψ1, ψ2 ∈ DerK (A(G),K(e)) ⊂ A(G) , and P P f ∈ A(G) with ∆(f) = a1⊗b1+···+ar ⊗br. Then Dψ1 (f) = i ψ1(bi)·ai and Dψ2 Dψ1 (f)(e) = i ψ1(bi)·ψ2(ai): P P Dψ1 (f)(g) = ψ1(λgf) = ψ1 i ai(g) · bi = i ai(g) · ψ1(bi) and P P Dψ2 Dψ1 (f)(e) = ψ2(Dψ1 (f)) = ψ2 i ψ1(bi) · ai = i ψ1(bi) · ψ2(ai).

Hence ε[Dψ2 ,Dψ1 ] = ψ2 · ψ1 − ψ1 · ψ2 = [ψ2, ψ1].

27. Exhibit Gm as a closed subgroup of SO2. Then find a two-dimensional torus T ⊂ SO4 and compute the weights for the action induced on the adjoint representation, T,→ SO4 → GL(so4). a b Solution: First, c d ∈ O2 ⇐⇒ a = d, b = −c. Solving for a linear combination of Gm → Gm, x 7→ x and x 7→ x−1 with determinant 1, we find a = x/2 + 1/2x and b = −ix/2 + i/2x, where i ∈ K is a fixed root of −1; : x/2+1/2x −ix/2+i/2x hence that D(x) = ix/2−i/2x x/2+1/2x ∈ SO2. : 2 D(x) 0 Using block matrices, we obtain T = D2 = (Gm) ,→ SO4,(x, y) 7→ 0 D(y) . As T is diagonalisable, the −1 t representation T,→ SO4 → GL(so4), t·M = tMt for t = (x, y) ∈ T and M ∈ so4 = {M ∈ M4(K) | M = −M} ∗ 2 has exactly two non-trivial eigenspaces. The weights are (1, 0), (0, 1) ∈ X (T ) = Z and the weight space for χ(1,0) AB 0 a b1 b2 0 1 −1 0 1 is { | A = ∈ so2,B = }, by explicit computation (note D(x) D(x ) = ): −B 0 −a 0 ib1 ib2 −1 0 −1 0 −1 D(x) 0 AB D(x ) 0 ! (x, y) · M = = χ1,0(x, y)M = xM 0 D(y) −BC 0 D(y−1)

28. Let A be a ﬁnite-dimensional, associative and unital K-algebra. Show that the group of units is a linear algebraic group. What is its Lie algebra?

Solution: For any unit a ∈ A, the map La : A → A given by left multiplication with a is an isomorphism. Hence ∗ we get an injection A ,→ GL(A), a 7→ La. In order to see that the image is closed, we consider T ∈ End(A) and show that TRb = RbT for all b ∈ A is equivalent to T = La for some a ∈ A. The implication (⇐=) is trivial, so assume that T commutes with all Rb and put a := T (1); then T (b) = TRb(1) = RbT (1) = ab = La(b). Hence ∗ A ⊂ GL(A) is the zero set of ﬁnitely many commutator equations [−,Rb] = 0, which are polynomial. ∗ The Lie algebra of A is A with Lie bracket from the algebra commutator: [a1, a2] = a1a2 − a2a1, since for any −1 a ∈ A, because derivating the equation gRbg = Rb gives GRb + RbG = 0 for G ∈ End(A). 29. Show that the diﬀerential of the adjoint representation of a linear algebraic group G is given by ad := d(Ad)e : g → End(g), ad(A)(B) = [A, B]. Solution: Wallach/Goodman, Theorem 1.5.7

30. Show that a morphism ϕ: G → H of linear algebraic groups induces a homomorphism of Lie algebras dϕe : g → h. Solution: Embedding H,→ GL(V ) for some ﬁnite-dimensional vector space V , we obtain a rational representation ϕ: G → GL(V ) and want to show that dϕe : g → End(V ) has image in h. Let v ∈ g, h ∈ H and f ∈ I(H) ⊂ A(G). ∗ Then (Ddϕ(v)f)(h) = dϕe(v)(λhf) = v(ϕ (λhf)) = 0, because (λhf)(ϕ(g)) = f(hϕ(g)) and f vanishes on H. Hence Ddϕ(v)(I(H)) = 0, which means that d%e(v) ∈ h. Algebra II — solutions to exercise sheet 7

31. Find a symmetric, non-degenerate bilinear form Γ on Kn such that T := SO(Γ) ∩ Dn is a torus of dimension m if n = 2m or n = 2m + 1. Prove that T is a maximal torus in SO(Γ): if H ⊆ SO(Γ) is abelian with T ⊆ H, then T = H.

0 1 Solution: Let Γ have entries 1 on the skew diagonal and 0 else; e.g. Γ = 1 0 for n = 2. Then Γ is obviously symmetric and non-degenerate. For n = 2m even, SO(Γ) ∩ D2m contains the diagonal matrices −1 −1 diag(x1, . . . , xm, x1 , . . . , xm ). If n = 2m + 1 is odd, then SO(Γ) ∩ D2m+1 contains the diagonal matrices −1 −1 diag(x1, . . . , xm, 1, x1 , . . . , xm ). In either case, these form a torus Dm in SO(Γ). n Now let g ∈ SO(Γ) such that gh = hg for all h ∈ SO(Γ) ∩ Dn. Each h acts on the standard basis vectors ei ∈ K by a character χi, i.e. h(ei) = χi(h) · ei. These characters are −1 −1 n = 2m : χ1 = x1, . . . , χm = xm, χm+1 = x1 , . . . , χ2m = xm , −1 −1 n = 2m + 1 : χ1 = x1, . . . , χm = xm, χm+1 = 1, χm+2 = x1 , . . . , χ2m+1 = xm . In either case, the characters are pairwise distinct. Hence, all weight spaces are one-dimensional. As g centralises SO(Γ) ∩ Dn, it must preserves all weight spaces, thus acts diagonally itself. But then g ∈ Dn.

n 32. Compute the orbits of the natural actions of GLn, Tn, Un, Dn on A and draw them for n = 2. Describe orbit closures as unions of orbits.

n n Solution: GLn acts on A with two orbits: 0 is a ﬁxed point, i.e. a closed orbit; A \{0} is an open orbit. i There are n + 1 orbits for the Tn-action whose closures are nested: O0 = {0} and Oi = A × {0}\ Oi−1 for i > 0; so that Oi = O0 ∪ ... ∪ Oi. The Un-orbit of a vector v = (v1, . . . , vn) is (∗,..., ∗, vi, 0,..., 0), where vi 6= 0 and vi+1 = ... = vn = 0. In i−1 particular, all orbits are of the form A × {(x, 0,..., 0)} and hence closed. n n n Dn acts on A with 2 orbits: any map t: {1, . . . , n} → {0, 1} determines a type of vectors v ∈ N by vi = 0 ⇐⇒ t(i) = 0. These types are preserved by the Dn-action and classify orbits. The closed orbit is {0}, corresponding to t = (0,..., 0); the orbit of type (1,..., 1), i.e. of vectors without zero components is open and dense. Given two orbits Ot and Os of types t and s, then Ot ⊆ Os if and only if s(i) = 0 ⇒ t(i) = 0 for all i = 1, . . . , n.

33. Show that the adjoint representation of SL2 has disconnected isotropy groups.

Solution: [The point is that some (not all) isotropy groups are disconnected. Note that both SL2 and sl2 are irreducible; this exercise shows that these properties are not enough to ensure irreducible isotropy.] −1 We consider the adjoint representation SL2 → GL(sl2), which is given by g · M = gMg for g ∈ SL2 and 0 1 M ∈ sl2, i.e. a traceless matrix. The isotropy group of M = 0 0 is F = {g ∈ SL2 | gM = Mg}. A quick matrix a b a b 2 computation gives c d ∈ F ⇐⇒ a = d, c = 0. Together with det(g) = 1, we get F = { 0 a | a, b ∈ K, a = 1}. 1 ∗ −1 ∗ This set decomposes into the two components 0 1 and 0 −1 . 34. Let G be a unipotent linear algebraic group and X an affine G-variety. Show that all orbits of G in X are closed. Solution: This is the theorem of Kostant–Rosenlicht (1961). The property even characterises unipotent groups. Let O = G · x be an orbit. By replacing X with O, we can assume that O is dense in X. Next, O is also open in X: the image of the map G → X, g 7→ g · x contains an open subset U ⊆ X (this is a general property of S morphisms between affine varieties) and thus G · x = g∈G g · U is open. Denote by Z := X \ O the closed complement. Then G acts on Z, hence on the ideal I(Z) ⊆ A(X). The latter action is locally finite, i.e. there exists a finite-dimensional, G-invariant subspace V ⊂ I(Z). As G is unipotent, the G representation G → GL(V ) has a fixed vector 0 6= f ∈ V (Kolchin). This means λg(f) = f, i.e. f(g·x) = f(x) for all g ∈ G. Hence f is constant on O, and thus constant on O = X. However, f(Z) = 0, so f = 0, a contradicton.

35. For a homogeneous ideal I ⊆ K[x0, . . . , xn], geometrically compare the varieties V (I) ⊆ An+1 and V (I) ⊆ Pn. Prove the homogeneous Nullstellensatz. n+1 n n+1 n Solution: Denote π : A → P , (a0, . . . , an) 7→ (a0 : ... : an), also X := V (I) ⊆ A and Y := V (I) ⊆ P . n+1 The crucial observation: x ∈ A satisfies f(x) = 0 for all f ∈ I (i.e. x ∈ X) if and only if f(π(x)) = f([x]) = 0 for all homogeneous f ∈ I, i.e. [x] ∈ Y . Hence X consists of lines through the origin, and all fibres of π : X \{0} → Y are K∗. In particular, π(X \{0}) = Y and π−1(Y ) ∪ {0} = X. [This is why X is called the affine cone of Y .]

For the Nullstellensatz, let I ⊂ K[x0, . . . , xn] be a radical ideal with (x0, . . . , xn) 6⊆ I. We first consider its affine n+1 variety X = V (I) ⊆ A : the affine Nullstellensatz yields that X contains points different from the origin (I is contained in maximal ideals, and it cannot coincide with the homogeneous maximal ideal by assumption). From n the above, we then see that the projective variety Y = V (I) ⊆ P contains points. Algebra II — solutions to exercise sheet 8

2 36. Compute the orbits of the actions of GL3 on P , of GL4 on Gr(2, 4), and of 1 1 GL2 on P × P (diagonal action). Also compute isotropy groups for all orbits. 3 2 Solution: GL3 acts transitively on A \{0}, hence it also acts transitively on P . The isotropy group of the point ∗∗∗ (1 : 0 : 0) consists of the block matrices of type ◦∗∗ . ◦∗∗ 4 For the action of GL4 on Gr(2, 4), we consider an arbitrary subspace V ⊂ K with dim(V ) = 2. If v1, v2 is a basis 4 for V , then we can apply an automorphism of K , which maps v1 to e1 = (1, 0, 0, 0). Hence we can assume that 4 the basis is e1, v for some v ∈ K . The subgroup T4 ⊂ GL4 preserves e1, up to scalars. Using an appropriate triangular matrix, we can ﬁx e1 and map v to a vector of the form (0, ∗, ∗, ∗). At this point we use the subgroup 3 GL1 × GL3 ⊂ GL4, and with GL3 acting transitively on A \{0}, we can map (0, ∗, ∗, ∗) 7→ (0, 1, 0, 0) = e2. Hence GL4 acts transitively on Gr(2, 4). The isotropy group of the subspace V = (∗, ∗, 0, 0) consists of matrices whose lower left 2 × 2-block is zero. 1 2 1 The diagonal action of GL2 on P × P has two orbits: one is the diagonal ∆ := {(p, p) | p ∈ P } (this is a closed 1 1 orbit), the other is the open complement U := P × P \ ∆. It is obvious that ∆ is GL2-invariant. Moreover, ∼ 1 1 ∆ = P and GL2 acts transitively on P ; hence ∆ is an orbit. For a point ((x0 : x1), (y0 : y1)) ∈/ ∆, we have x0y1 − x1y0 6= 0, and applying an appropriate matrix multiplication, we get y1 −y0 · ((x : x ), (y : y )) = ((y x − y x : 0), (0 : y x − y x )) = ((1 : 0), (0 : 1)) ∈ 1 × 1. −x1 x0 0 1 0 1 1 0 0 1 1 0 0 1 P P 1 1 The isotropy group of any point in P × P is D2.

37. Show that any non-degenerate conic in P2 is isomorphic to P1. Deduce that homogeneous coordinate rings are not isomorphism invariants of projective varieties. 2 Solution: A conic in P is, by deﬁnition, the vanishing locus of some homogeneous polynomial of degree two: 2 2 2 a00x0 + a11x1 + a22x2 + a01x0x1 + a02x0x2 + a12x1x2 = 0 with aij ∈ K. This zero set can be also described in matrix form xtMx = 0, where the symmetric bilinear form M can be diagonalised to

    a00 a01/2 a02/2 c0 0 0 M = a01/2 a11 a12/2  0 c1 0  a02/2 a12/2 a22 0 0 c2 with ci either 0 or 1 (K is algebraically closed). This is given by a linear coordinate change, i.e. an automorphism 2 2 of P . However, if one of the ci = 0, then the quadric is degenerate. Hence every non-degenerate quadric in P is 2 2 2 1 2 2 2 isomorphic to V (x0 + x1 + x2). Moreover, P → P , (y0 : y1) 7→ (y0 : y0y1 : y1 ) is a morphism which is injective 2 2 2 1 ∼ 2 2 2 ∼ 2 2 2 and whose image is cut out by V (x0 − x1 + x2). Hence P = V (x0 − x1 + x2) = V (x0 + x1 + x2) using the 1 coordinate change x1 7→ ix1, and all non-degenerate conics are isomorphic to P . 1 For the second claim, note that the homogeneous coordinate ring of P is K[x, y], the polynomial ring in two variables (with its standard grading). However, the homogeneous coordinate ring of a quadric is e.g. S = K[x, y, z]/(x2 + y2 + z2). These rings are not isomorphic, for example because the associated aﬃne va- 2 2 2 3 rieties aren’t (V (x + y + z ) ⊂ A is singular).

38. Show that A1 and P1 are homeomorphic, but A2 and P2 are not. 1 1 1 Solution: As topological spaces, both A and P = A ∪ {∞} are sets of the same cardinality as K, with the cofinite topology. Hence they are homeomorphic. 2 2 When comparing A and P , let a curve be a closed, irreducible subset which is neither a point nor the whole 2 0 2 0 space. We claim that for any curve C ⊂ A , there exists another curve C ⊂ A with C ∩ C = ∅. (If C = V (f), 0 2 then C = V (f + 1) will work.) However, in P there are curves which intersect all other curves. (For example, a line V (x0) will work: if V (g) is a curve given by a homogeneous polynomial g ∈ K[x0, x1, x2], then either g ∈ (x0) 1 and V (g)∩V (x0) is even infinite, or g(0, x1, x2) 6= 0 and defines a finite set of points in P — this set is non-empty 2 2 because K is algebraically closed.) These are topological properties, so A and P cannot be homeomorphic. 39. For an affine, irreducible variety X and a point p ∈ X, show that the local ring of X at p is given by the localisation of A(X) at the maximal ideal Mp.

1 40. Show that Aut(P ) = PGL2. 1 ∼ 1 Solution: We already know that GL2 acts on P with kernel Z(GL2) = Gm. Hence PGL2 acts faithfully on P , and we have to show that all automorphisms of the projective line come from linear maps. 1 Let ϕ ∈ Aut(P ) be an arbitrary automorphism. Put (x0 : x1) := ϕ(1 : 0). If (x0 : x1) 6= (1 : 0), then x1 6= 0, and 0 1/x1 1 1 then matrix ψ := x −x satisﬁes ψϕ(1 : 0) = (1 : 0). Hence ψϕ induces an automorphism of P \{(1 : 0)} = A . 1 1 0 1 1 As A is an aﬃne variety, we have End(A ) = End(A(A )) = End(K[x]), so the automorphism must be of the a b form ψϕ = ax + b, i.e. a linear polynomial. Then θ := 0 1 maps (x : 1) 7→ (ax + b : 1) = ψϕ(x : 1) and −1 (1 : 0) 7→ (a : 0) = (1 : 0). Hence θ = ψϕ and ϕ = ψ θ ∈ PGL2. n [It is also true that Aut(P ) = PGLn+1 but the proof requires considerably harder methods.] Algebra II — solutions to exercise sheet 9

41. Show that varieties are compact in the Zariski topology: any open cover has a finite subcover. Solution: This holds true for any noetherian topological space, i.e. a space X where descending chains of closed S subsets stabilise or, equivalently, ascending chains of open subsets stabilise. If X = i∈I Ui is an open cover, S then choose a well-ordering on I and define Vj := i≤j Ui. Then the Vj form an ascending chain of open subsets. [In fact, a topological space is noetherian if and only if every open subset is compact.] Affine varieties are noetherian topological spaces because their affine coordinate rings are noetherian, so that asscending chains of ideals stabilise or, equivalently, descending chains of Zariski-closed subsets stabilise. The same reasoning works for projective varieties. For quasi-affine or quasi-projective varieties, we observe that a subspace of a noetherian topological space is still noetherian. 42. Show that a locally compact Hausdorff space X is compact if and only if the projection X × Y → Y is a closed map for all topological spaces Y . Solution: Let X be compact and take a closed subset Z ⊆ X × Y . We have to show that π(Z) ⊆ Y is closed, where π : X × Y → Y is the projection. Let y ∈ Y \ π(Z). Then (x, y) ∈/ Z for all x ∈ X. By the definition of the product topology, for each x ∈ X, there exist open neighbourhoods x ∈ Ux ⊆ X and y ∈ Vx ⊆ Y such S that Ux × Vx ⊆ X × Y \ Z. In particular, x∈X Ux = X. By compactness of X, there are finitely many points x1, . . . , xr such that X = Ux1 ∪ · · · ∪ Uxr . Let V := Vx1 ∩ · · · ∩ Vxr , then y ∈ V ⊆ Y open. By construction, V ⊂ Y \ π(Z), so that Y \ π(Z) is open, hence π(Z) closed. For the reverse implication, assume that X is not compact. Then we consider its one-point compactification Y := Xˆ = X ∪ {∞} — this is a compact space; Y is Hausdorff space since X is locally compact and Hausdorff. Moreover, let Z := {(x, x) | x ∈ X} ⊂ X × Xˆ be the diagonal. As X is Hausdorff, Z ⊂ X × X is closed; thus X is also closed in X × Xˆ. Then π(Z) = X ⊂ Xˆ, a closed subset in a compact Hausdorff space, hence compact itself. S The statement actually holds for arbitrary topological spaces X: for the reverse implication, let X = i∈I Ui be an open covering without any finite subcovering. Let now Y := X ∪ {ω}, topologised with the following open sets: all subsets of X; all sets containing {ω} ∪ (X \ U) where U is a finite union of the Ui. This is indeed closed under arbitrary unions and finite intersections. Moreover, by our assumption on the covering Ui, the subset X ⊂ Y is not closed, i.e. {ω} ⊂ Y is not open. Let Z be the closure of the diagonal ∆ := {(x, x) | x ∈ X} in X × Y . By hypothesis, its closure π(Z) is closed in Y , hence π(Z) = Y . Hence (x, ω) ∈ Z for some x ∈ X. By the definitions of closure, of the topology on Y and of the product topology, V × ({ω} ∪ (X \ Ui)) ∩ ∆ 6= ∅ for all i ∈ I and open sets x ∈ V ⊆ X. Hence V ∩ (X \ Ui) 6= ∅, i.e. V 6⊆ Ui for all V and i. But this would imply x∈ / Ui for all i ∈ I, contradicting that the Ui cover X.

43. Let X = V (y2 − x3) ⊂ A2 be the cuspidal cubic. Show that the map A1 → X, t 7→ (t2, t3) is regular, birational and a homeomorphism, but not an isomorphism of varieties. Extend this to an example of a morphism of projective varieties with the same properties. Solution: The map is given by polynomials, so it is regular. It is bijective, because for any point (x, y) ∈ X, we have y2 = x3 and (with K algebraically closed), there are two square roots of x, call them t and −t. Moreover, 6 3 2 3 1 t = x = y , regardless of sign. But exactly one of the two roots satisfies t = y. Next, A and X are infinite sets with the cofinite topology, hence every bijection between them is automatically a homeomorphism. 1 Finally, the two varieties are birational: K(A ) = Quot(K[t]) = K(t), i.e. the function field of the affine line is the field of rational functions in one variable. But A(X) = K[x, y]/(y2 − x3) = K[t2, t3]) has the same quotient field: Quot(K[t2, t3]) = K(t), as t3/t2 = t in Quot(A(X)). The two curves are not isomorphic because their affine coordinate rings aren’t. For example, the tangent space 2 ∗ 2 T0X = (M0/M0 ) is 2-dimensional, whereas all tangent spaces of A are 1-dimensional. 1 2 3 3 2 3 By homogenisation, we get an example of projective curves: P → V (x0x2 − x1), (t0 : t1) 7→ (t0 : t0t1 : t1). 44. Let A be a commutative ring with unit and N a finitely generated A-module. Show that a surjective module endomorphism N → N is an isomorphism. Solution: The theorem of Vasconcelos. Write f : N → N and use it to consider N as an A[X]-module, with X · n := f(n). Then XN = N by the assumption that f is surjective. Applying the Nakayama lemma (with I = (X) ⊂ A[X]) yields an element Y ∈ A[X] such that YN = 0 and Y ∈ 1 + (X), i.e. Y = 1 + XZ for some Z ∈ A[X]. Let n ∈ ker(f), then X · n = 0 and hence 0 = (1 + XZ)n = n + ZXn = n, so that f is indeed injective.

45. Find subgroups of SL2 such that the quotient is respectively projective, affine or strictly quasi-affine. ∼ 1 ∼ Solution: The quotient SL2/(T2 ∩ SL2) = P is a projective variety; SL2/(D2 ∩ SL2) = PSL2 is an affine variety; ∼ 2 SL2/U2 = A \{0} is quasi-affine but not affine. Algebra II — solutions to exercise sheet 10

46. Compute the dimensions of Grassmannians Gr(d, n) and ﬂag manifolds Fl(n).

Solution: We already know that Fl(n) = GLn/Tn is a homogeneous space. This implies that tangent spaces at the 2 flag manifold are given by quotients TeGLn/TeTn, hence dim(Fl(n)) = dim(gln) − dim(tn) = n − n(n + 1)/2 = n(n − 1)/2, where tn ⊂ gln is the subset of all upper triangular matrices. The Grassmannian Gr(d, n) has a transitive action by GLn, so Gr(d, n) = GLn/H where H is the stabiliser of some d-dimensional subspace U ⊂ Kn. Taking U to be the span of the first d standard basis vectors of n AB K , we find H = { 0 D | A ∈ GLd,B ∈ M(d × n − d),D ∈ GLn−d}. Again, we compute the dimension as 2 2 2 2 dim(Gr(d, n)) = dim(gln) − dim(TeH) = n − (d + d(n − d) + (n − d) ) = nd − d = d(n − d). 47. Show that two irreducible varieties X and Y are birational, i.e. their function fields are isomorphic: K(X) ∼= K(Y ), if and only if there exist affine open subsets ∼ UX ⊆ X and UY ⊆ Y which are isomorphic: UX = UY .

Solution: For any open affine subset ∅ 6= U ⊆ X, we have Quot(A(U)) = K(X). Hence UX ⊆ X and UY ⊆ Y with UX =∼ UY implies K(X) =∼ Quot(A(UX )) =∼ Quot(A(UY )) =∼ K(Y ). Now, assume K(X) =∼ K(Y ). Pick an open affine subset V ⊆ X; this gives K(X) =∼ K(V ). Choose an isomorphism ϕ: K(V ) → K(Y ). Let f1, . . . , fr ∈ A(V ) be generators of the affine coordinate ring of Y . Then the ϕ(fi) are rational functions on Y , and there is an open subset U ⊆ Y on which all ϕ(fi) are regular. Shrink U, if necessary, to ensure it is affine. Then ϕ induces a morphism ϕ: A(V ) → A(U). This map of K-algebras must be injective, since the induced map of quotient fields is. Hence we obtain a dominant morphism α: U → V of affine varieties. Repeating this argument for ϕ−1 : K(Y ) → K(X) yields a dominant morphism β : V → U 0 for some open, affine subset U 0 ⊆ Y . The morphisms α and β are inverse on an open subset.

p 48. Assume char(K) = p > 0. Show that the map Gm → Gm, x 7→ x is a bijective morphism of aﬃne algebraic groups, but is not an isomorphism. Solution: The map is obviously a group homomorphism, due to characteristic p. It is a morphism of varieties, since it is evidently given by a polynomial. It is surjective, because K is algebraically closed (every element of the ﬁeld has roots of any order). It is injective because of xp = yp ⇐⇒ (x − y)p = 0 ⇐⇒ x − y = 0. However, it is not an isomorphism: variety morphisms Gm → Gm are the same as endomorphisms of the K-algebra −1 p ∗ −1 − p A(Gm) = K[T,T ). The map F : Gm → Gm,F (x) = x induces F : K[T,T ] → K[T,T 1], f(T ) 7→ f(T ). This ring homomorphism is not surjective!

49. Let G be a linear algebraic group, acting on a quasi-projective variety X. Show that orbits of minimal dimension are closed; in particular, closed orbits exist. (Hint: you can use the statement of exercise 25.)

Solution: Let x ∈ X and Z := G · x be the closure of the orbit of x. Then Z is a G-invariant, closed subset of X. We already know that O := G · x ⊆ Z is open. Assume the orbit is not closed. Then Z \ O is closed in Z and G-invariant, hence quasi-projective and a union of orbits. Therefore, dim(Z \ O) < dim(Z) and since dimensions are non-negative and ﬁnite, orbits of minimal dimension must be closed.

50. Let G be a connected projective algebraic group. Show that G is commutative. Solution: We consider the two morphisms f, p: G × G → G with f(g, h) = ghg−1h−1 and p(g, h) = g. Let U ⊂ G be an open aﬃne subset of G such that e ∈ U. Then p(G × G \ f −1(U)) ⊂ G is a closed subset not containing e. Thus there is an open subset V ⊂ G such that p−1(V ) ⊆ f −1(U), hence f(p−1(V )) ⊆ U. Let g ∈ V ; then f({g} × G) = f(p−1(g)) ⊆ U. However, with G proper and connected, this set is the singleton {e} since f(g, e) = e. Thus f(p−1(V )) = {e}, but p−1(V ) ⊂ G × G is a dense subset, and so f(G × G) = {e}, i.e. G is commutative. −1 Sketch of an alternative proof if char(K) = 0: for any g ∈ G, let cg : G → G, h 7→ ghg . Its derivative at e gives an automorphism Ad(g) := dce : TeG → TeG. In particular, we get a morphism Ad: G → GL(TeG). With G proper and connected, its image must be the identity of TeG. Like in the aﬃne case, TeG is a Lie algebra and we now know that it is abelian. G connected and char(K) = 0 imply that G is commutative. Algebra II — solutions to exercise sheet 11

51. Find Borel subgroups in SO4, Sp4,T4 and U4. 4 4 Solution: Recall that GL4/T T4 =∼ Fl(K ), the variety of full flags in K . Given the subgroup SO4 ⊂ GL4, n 0 1 2 i we consider the subset of isotropic flags Fl0(K , β) = {V ( V ( V | dim(V ) = i, β|V i×V i = 0} where 4 4 n β : K × K → K is the bilinear form. This is a closed subset, so that Fl0(K , β) is a projective variety. SO4 acts transitively on isotropic flags (similar proof as with GLn and full flags). Using the bilinear form β from 0 1 1 2 1 3 Exercise 31, a particular isotropic flag is given by V = 0 ⊂ V = Ke ⊂ V = Ke + Ke . From g ∈ GL4 with ge1 = e1, ge3 = e3 and gtβg = β, we get

0 0 0 1 1 ∗ 1 ∗ 0 0 1 0 0 1 0 ∗ β =   , g =   0 1 0 0 0 0 1 ∗ 1 0 0 0 0 0 0 1 and in particular the stabiliser group in SO4 is solvable. T4 and U4 are solvable and connected, hence their Borel groups are the full group in each case.

52. Find all parabolic subgroups P with T4 ⊆ P ⊆ GL4. Solution: There are 8 such parabolic subgroups, there are given by block matrices of the formats ∗∗∗∗ ∗∗∗∗ ∗∗∗∗ ∗∗∗∗ ∗∗∗∗ ∗∗∗∗ ∗∗∗∗ ∗∗∗∗ ∗∗∗∗ ∗∗∗∗ ∗∗∗ ∗∗∗ ∗∗∗ ∗∗∗∗ ∗∗∗∗ ∗∗∗ GL4 = ∗∗∗∗ , ∗∗ , ∗∗∗ , ∗∗ , ∗∗∗ , ∗∗ , ∗∗∗ , ∗∗ = T4. ∗∗∗∗ ∗ ∗ ∗∗ ∗∗∗ ∗∗ ∗ ∗

53. Let the connected linear algebraic group G act on a quasi-projective variety X with finitely many orbits. Show that every irreducible G-invariant subset in X is the closure of a G-orbit. Find a counterexample for an action with infinitely many orbits. Solution: We know that orbits in general are quasi-projective and open in their closure. An irreducible G- invariant subset Z ⊆ X is a union of orbits. Since there are only finitely many orbits, Z contains an orbit O as an open subset; the closure O must then be Z. 1 Counterexamples abound. (For a trivial one, take G = 1, Z = X = A .) 54. Find a connected linear algebraic group G and a maximal solvable subgroup U ⊂ G such that U is disconnected.

∗ 0 0 ∗ Solution: This is problem 17 in disguise: we take G = GL2 and U := N(D2) = 0 ∗ ∪ ∗ 0 . Then U is obviously disconnected, it is solvable by problem 17. It is a matrix calculation to show that the only subgroup of G strictly containing U is G.

55. Classify all root systems in the Euclidean plane E := R2. Algebra II — solutions to exercise sheet 12

56. Compute the radicals R(GLn),R(SLn),R(Un).

Solution: On the one hand, R(GLn) is the identity component of the intersection over all Borel subgroups. With − − the Borel subgroups B := Tn and B of upper/lower triangular matrices, we ﬁnd R(GLn) ⊆ B ∩ B = Dn. On ∗ the other hand, R(GLn) is the maximal normal, solvable, connected subgroup: we get R(GLn) = K · In from 1 1 a 0 0 1 b a−b R(GLn) normal: for n = 2, we have 1 0 0 b 1 −1 = 0 a ∈ R(GL2) ⊆ D2, hence a = b. The same computation for SLn shows R(SLn) = 1, i.e. SLn is semisimple. R(Un) = Un because Un is unipotent, hence solvable; connected; normal in itself.

∼ r 57. Let char(K) = 0 and U a commutative, unipotent group. Show that U = Ga for some r ∈ N. Solution: One way is to use exponential and logarithm for matrices (introduced in the classification of connected, 1-dimensional groups). Consider U ⊆ Un for some n (Kolchin). The map log: U → n is then polynomial, i.e. a morphism of affine varieties, where n is an additive subgroup of nilpotent matrices. As U is commutative, the logaritm is even a group homomorphism, and its inverse is exponential. Now n is a K-vector space, so in particular r isomorphic to Ga for some r ∈ N. 58. Let G be a finite group and A a trivial G-module, i.e. an abelian group which has the trivial left G-action. Show that H1(G, A) = Hom(G, A). Solution: This is an obvious application of the definitions: note that trivial G-action implies B1(G, A) = 0, so 1 1 that H (G, A) = Z (G, A) = Hom(G, A), since the 1-cocycle condition collapses ϕ(g1g2) = ϕ(g1) + g1(ϕ(g2)) = ϕ(g1) + ϕ(g2). 59. Let ϕ: G → H be a surjective morphism of linear algebraic groups, and T ⊆ G a maximal torus (or a maximal connected normal unipotent subgroup, respectively). Show that ϕ(T ) ⊆ H has the same property. Solution: Humphreys §21.3

60. Let (E, Φ) be a root system. A base of (E, Φ) is a subset S ⊆ Φ such that P S is a basis for E and every root β ∈ Φ can be written as β = α∈S mαα with either all mα ≥ 0 or all mα ≤ 0. The elements of S are also called simple roots, and their non-negative linear combinations positive roots and denoted Φ+. Draw bases and positive roots for the root systems in E = R2, of exercise 55. Algebra II — solutions to exercise sheet 13

61. Let G be a connected algebraic group, B ⊆ G a Borel subgroup and V a rational G-module. Show that the invariant subspaces coincide: V G = V B. Solution: Trivially, V G ⊆ V B . Let v ∈ V B . We look at the morphism G → G · v, g 7→ gv. Because of v ∈ V B , this map factors through G/B → G · v. However, G/B is projective and the orbit G · v is aﬃne; both varieties are connected. Hence, G/B → G · v is constant, and v ∈ V G.

62. Let G be a linear algebraic group. Show that R(G)u = Ru(G), i.e. the unipotent part of the radical is the unipotent radical. Solution: Springer 7.6.3

63. For a linear algebraic group G, show that G/R(G) is semisimple and that G/Ru(G) is reductive. Solution: We know that the surjective group homomorphism π : G → G/R(G) maps Borel subgroups to Borel T ◦ subgroups. Then R(G) = ( B B) , by a characterisation of the radical. Hence G/R(G) has trivial radical by

\ ◦ π(B) ⊂ π(R(G)). π(B)

0 00 0 For the statement about G/Ru(G), show that if an extension 1 → G → G → G → 1 the outer groups G and G00 are unipotent, then so is the middle one G. (This follows from Jordan decomposition.) Now the preimage of unipotent and normal subgroup U¯ ⊆ G/Ru(G) in G is Ru(G) ⊆ U ⊆ G. Then U is normal in G and also unipotent (from the extension 1 → Ru(G) → U → U/Ru(G) → 1, hence Ru(G) = U. 64. Let G be a group with subgroups H,N ⊆ G. Show that the following notions are equivalent — then G = N o H is called a semi-direct product of H by N: (1) There is a short exact sequence 1 → N −→ι G −→π H → 1 admitting a section σ : H → G, i.e. πσ = idH . (2) N is normal in G, and NH = G and N ∩ H = 1. (3) There is a homomorphism α: H → Aut(N) and G is isomorphic to the group 0 0 0 0 N oα H deﬁned by (n, h) · (n , h ) := (nα(h)(n ), hh ) on the set N × H. Also show that the existance of a retraction %: G → N, i.e. %ι = idN , is equivalent to a splitting G ∼= N × H of G as a direct product. Solution: (1) ⇐⇒ (2): Given the split extension, then N = ker(ι, hence is normal in G. Given g ∈ G, then πσπ(g) = π(g), so σπ(g) = gn for some n ∈ N; altogether g = n−1σπ(g) ∈ NH. Finally, if g ∈ N ∩ σ(H), then g = σ(h), hence 1 = π(g) = π(σ(h)) = h; so N ∩ σ(H) = 1. The reverse implication is proved similarly. (1) ⇒ (3): The homomorphism α: H → Aut(N) is deﬁned as follows: for h ∈ H and n ∈ N, let α(h)(n) = σ(h)nσ(h)−1. Then π(α(h)(n)) = πσ(h) · π(n) · πσ(h)−1 = h · 1 · h−1 = 1, hence α(h)(n) ∈ N.

65. Show that the following subsets deﬁne root systems of rank n: n+1 {ej − ei | i, j ∈ {1, . . . , n + 1}, i 6= j} ⊂ Q (called type An) n {±ej ± ei | i, j ∈ {1, . . . , n}, i < j} ⊂ Q (called type Dn) Algebra II — solutions to exercise sheet 14

i 0 66. Show that centraliser of 0 −i ∈ PGL2(C) is disconnected. 1 0 Solution: In fact, the centraliser of A := 0 −1 in SL2 is disconnected (th exercise follows from this). a b 1 0 d −b ad+bc −2ab Now M := c d 0 −1 −c a = 2cd −(ad+bc) , so M = A ⇐⇒ ad + bc = 1, ab = cd = 0, leading to

a 0 0 b CA(SL2) = 0 1/a ∪ −1/b 0 , which is a disjoint union.

67. For B ⊆ G a Borel subgroup of a connected linear algebraic group G, show that Z(B) = Z(G).

0 Solution: Trivially, Z(B) = CB (B) ⊆ CG(B) ⊆ CG(G) = Z(G). Let g ∈ Z(G). Then g ∈ B for some Borel subgroup B0 ⊆ G (because Borel groups cover G). Since Borel subgroups are conjugated, there is an h ∈ G with B0 = hBh−1. As g is central, we have g = h−1gh ∈ B.

68. Prove directly for G = SLn and for G = SOn (with the bilinear form from Problem 31) that G is covered by Borel subgroups and that maximal tori coincide with their centralisers.

Solution: G = SLn with B = Tn ∩ SLn (upper triangular matrices of determinant 1). If M ∈ SLn is any −1 matrix, then there is a base change matrix g ∈ GLn such that gMg is upper triangular (this is a weak form of the Gauss algorithm). Moreover, we can assume that det(B) = 1 (because K is algebraically closed). Hence gMg−1 ∈ B or, equivalently, M ∈ g−1Bg, a Borel subgroup of G. T = Dn ∩ SLn (diagonal matrices of determinant 1) is a maximal torus in G. Let d = diag(d1, . . . , dn) ∈ T and g ∈ G. Then gd is obtained from g by multiplying the i-th row of g by di, whereas in dg the i-th column of g is multiplied by di. If gij 6= 0, then we get digij = dj gij , hence gij = 0 unless di = dj . If g ∈ CG(T ), then we have dg = gd for all d ∈ T , hence g is a diagonal matrix.

G = SOn with B = Tn ∩ SOn a Borel subgroup. Here, any matrix can be brought into upper triangular form using the Gram-Schmidt algorithm. Hence Borel subgroups again cover G. The argument for CG(T ) = T proceeds as above (we can work with diagonal tori by choosing an appropriate symmetric bilinear form).

69. Let G be a connected linear algebraic group with a maximal torus and B be the set of Borel subgroups of G with its natural T -action. Show that there is a bijection between the ﬁxed point set BT and the Weyl group W (G).

T −1 Solution: We let W = NG(T )/CG(T ) act on B by n · B := nBn . In order to show that this is well- T defined, we first prove that CG(T ) ⊆ B∈BT B. As CG(T ) is connected (all torus centralisers are) and nilpotent (because it has the unique maximal torus T ), it is solvable and hence contained in some B ∈ BT . For another B0 = gBg−1 ∈ BT , we have maximal tori T, gT g−1 ⊆ B0. Hence these are conjugated to each other under some 0 0 0 −1 0−1 0 −1 0−1 0 b ∈ B . Then CG(T ) = C(b gT g b ) = b gCG(T )g b ⊆ B . Let B,B0 ∈ BT , i.e. T ⊆ B,B0 ⊆ G. By conjugacy of Borel subgroups, there is g ∈ G with B = gB0g−1. Then T ⊆ B and gT g−1 ⊆ B are maximal tori, so by conjugacy of maximal tori in B, there is b ∈ B with −1 −1 0 −1 0 bgT g b = T . Hence n := bg ∈ NG(T ). Then also nB n = B, so n ∈ W maps B 7→ B. Hence W acts transitively on BT . T −1 Assume now that some B ∈ B is fixed under W . So we want to show that nBn = B implies n ∈ CG(T ). But −1 nBn = B implies n ∈ NG(B) = B by the theorem about normalisers of Borel subgroups, so n ∈ B ∩ NG(T ) = NB (T ) = CB (T ) ⊆ CG(T ), where the last equality follows from the semi-direct decomposition B = Bu o T and Jordan decomposition.

70. Show that the following subsets deﬁne root systems of rank n: n {±ei | i ∈ {1, . . . , n}} ∪ {±ej ± ei | i, j ∈ {1, . . . , n}, i < j} ⊂ Q (type Bn) n {±2ei | i ∈ {1, . . . , n}} ∪ {±ej ± ei | i, j ∈ {1, . . . , n}, i < j} ⊂ Q (type Cn) 5 About the exam

Topics in descending importance.

Calculations with matrices and matrix groups For example, for any explicitly deﬁned matrix group G, you could check that it is a linear algebraic group, compute is centre, commutator subgroup, Lie algebra etc. Or check for properties such as solvable, unipotent, Borel etc. Exercises along these lines have been 6, 11, 16, 17, 18, 32, 36, 51, 52, 68.

Properties of linear algebraic groups, orbits and homogeneous spaces There have been some recurring arguments that you should be able to apply as well: embedding a group in GLn; Chevalley lemma; Jordan decomposition; the structure of diagonalisable and connected solvable groups; Borel ﬁxed point theorem (and a map from a connected, projective variety to an aﬃne variety is constant); conjugacy of Borel subgroups and of maximal tori. Exercises 14, 15, 46, 49, 53, 61, 67, 69.

Basic algebraic geometry Aﬃne geometry: Zariski topology, singular points, tangent spaces. Projective geometry: projective space, Grassmannians, proper morphisms. Exercises: 13, 35, 43, 47.

Lie algebras and adjoint representation

You should know the deﬁnitions of the classical groups (GLn, SLn, SOn, Spn) and of their Lie groups. If the non-standard bilinear forms are needed, I will list them; you don’t have to memorise these. Exercises 22, 27, 30. Matriculation number 1 2 3 4 5 6 7 8 P

Exam Algebra II

K is an algebraically closed ﬁeld. All algebras and varieties are deﬁned over K.

2. Prove that a connected linear algebraic group consisting of semisimple elements is a torus. Show by example that the connectedness assumption is necessary.

3. Let G be a connected solvable linear algebraic group. Does G have a unique maximal torus? (Proof or counterexample.)

4. Compute the Lie algebra sl2 of SL2 from ﬁrst principles, choose a basis and compute the matrix of ad(g): sl2 → sl2 for arbitrary g ∈ SL2 in the chosen basis.

6. Compute the roots of the group Sp4. You can use that this group has rank 2.

7. Let B ⊆ G be a Borel subgroup of a connected linear algebraic group G. Show that the centres coincide: Z(B) = Z(G).

8. Let G = GLn and B ⊂ G a Borel subgroup. Let W be the Weyl group of G. S a) Show G = w∈W BwB. (Explain ﬁrst that BwB is a well-deﬁned subset of G.) b) Show that the union is disjoint. Solutions for exam

1. An affine transformation of A2 is a bijection A2 −→∼ A2 mapping affine lines to affine lines. a) Show that the set G of affine transformations of A2 is a linear algebraic group. b) Describe a maximal torus in G. 2 Solution: a) First solution: It is clear that G is a group. Let us check first that G ⊂ Aut(A ) (automorphisms ∗ 2 as an affine variety). For this, we have to check g (f) = fg ∈ A(A ) = K[x, y]. If deg(f) = 1, i.e. f = ax + by + c for some a, b, c ∈ K, then we know g∗(f) is again a polynomial of degree 1 — this is condition that G preserves affines lines (i.e. zero sets V (ax + by + c)). Since g∗ is obviously a ring automorphism (of the ring of all K- 2 ∗ 2 valued functions on A , for example), we see that g preserves A(A ), as every polynomial is generated from ∗ 2 linear forms. In particular, we have g ∈ Aut(A ) = Aut(K[x, y]), but the above argument shows even more: ∗ g ∈ Aut(K + Kx + Ky) = GL3. 2 Second solution: If ϕ ∈ G is arbitrary with ϕ(O) 6= O (where O = (0, 0) ∈ A is a fixed origin), then we consider 2 2 the translation v : A → A , x 7→ x − ϕ(O). Translations are affine transformations, so vϕ ∈ G with vϕ(O) = O. 2 2 Hence vϕ is an affine transformation of A fixing O, hence it is a linear automorphism of K , i.e. vϕ ∈ GL2. 2 ∼ O There are two natural subgroups in G: translations Ga = V ⊂ G and linear transformations GL2 = G ⊂ G O (the subgroup fixing O). Then V ∩ GL2 = 1 and VG = G by the above argument. An immediate computation O 2 shows that G is a normal subgroup of G: for g ∈ G and v(x) = x + v the translation along v ∈ A , we have g−1vg(x) = g−1(g(x) + v) = x + g−1(v), which is again a translation. 0 Altogether we see that the abstract group G has a semi-direct product decomposition G = G o V . Since O ∼ ∼ 2 G = GL2 and v = Ga are affine varieties,so is their product (which is the underlying set of G). The composition is given by linear functions, so in particular is polynomial. ∼ 2 b) Here we use the semi-direct decomposition from above. Since Ga is unipotent, so is V = Ga. Tori consist of semisimple elements, hence any torus in G has to be disjoint from V . We find that a maximal torus in G is a maximal torus in GL2, for example D2.

2. Prove that a connected linear algebraic group consisting of semisimple elements is a torus. Show by example that the connectedness assumption is necessary.

Solution: Let G = Gs be connected. We have to show that G is commutative (because a torus is, by deﬁnition, a connected commutative group all of whose elements are semisimple). Let B ⊆ G be a Borel subgroup. Then B is connected solvable, so has a semi-direct product decomposition B = Bu n T where T ⊆ B ⊆ G is a maximal torus. From the assumption G = Gs we know Bu = 1, so B = T . In particular, B is commutative and hence nilpotent. This forces G = B. (In the course, we have seen that if a Borel subgroup is nilpotent (or normal), then it is the whole group.) Connectedness is necessary: any ﬁnite group consists of semisimple elements (this uses char(K) = 0 and Jordan decomposition).

3. Let G be a connected solvable linear algebraic group. Does G have a unique maximal torus? (Proof or counterexample.) Solution: G does not necessarily have only one maximal torus. (In fact, we have shown that connected nilpotent groups are characterised by having a unique maximal torus.) A counterexample can be found in B := T2 ∩ SL2: this is a Borel subgroup of SL2, so is conncted solvable. It has : x 0 1 1 1 −1 x 1/x−x the standard maximal torus T = D2 ∩ SL2 = 0 1/x but another torus is 0 1 T 0 1 = 0 1/x .

4. Compute the Lie algebra sl2 of SL2 from ﬁrst principles, choose a basis and compute the matrix of ad(g): sl2 → sl2 for arbitrary g ∈ SL2 in the chosen basis. 4 Solution: By deﬁnition, SL2 = V (x1x4 − x2x3 − 1) ⊂ A . We want to compute the tangent space at the point e = (1, 0, 0, 1). For a variety cut out by a single polynomial f and a point p, the tangent space at p is given by P the vanishing locus of i ∂f/∂xi(p) xi = 0. In our situation, we get the linear equation x1 + x4 = 0, i.e. the Lie algebra TeSL2 = sl2 = {M ∈ M(2,K) | tr(A) = M} is given by traceless matrices. 1 0 0 1 0 0 A (very typical) basis for sl2 consists of the three matrices H = 0 −1 ,X = 0 0 ,Y = 1 0 . a b −1 A group element g = c d acts on a tangent vector M ∈ sl2 by ad(g)(M) = gMg . Hence we get

a b 1 0 d −b ad+bc −2ab ad(g)(H) = c d 0 −1 −c a = −2cd −bc−ad = (ad + bc)H − 2abX − 2cdY ; 2 ad(g)(X) = a b 0 1 d −b = −ac a = −acH + a2X − c2Y ; c d 0 0 −c a −c2 ac 2 ad(g)(Y ) = a b 0 0 d −b = bd −b = bdH − b2X + d2Y. c d 1 0 −c a d2 −bd Hence the matrix of ad(g) in the basis (H,X,Y ) is

ad + bc −ac bd   −2ab a2 −b2 . −2cd −c2 d2

5. Let B be a non-degenerate symmetric bilinear form on K2n. Show that the set of full isotropic flags V 0 ⊂ V 1 ⊂ V 2 ⊂ ... ⊂ K2n (i.e. dim(V i) = i and B|V i×V i = 0 for all i) is a projective variety. 2n Solution: We write Fn for the set of isotropic flags in K . Because B is non-degenerate, the maximal dimension 2n · 2n of an isotropic subspace in K is n. Now a flag V ∈ Fn is partial flag (of ordinary subspaces) in K ; the latter : Tn ∗ is given by the partial flag manifold Xn = GL2n/P where the parabolic subgroup has block form P = 0 ∗ . In particular, Xn is a projective variety. The subset Fn ⊂ Xn is closed: B is a bilinear form, so isotropicity is a polynomial condition. Hence, Fn is projective.

6. Compute the roots of the group Sp4. You can use that this group has rank 2. 1 0 0 I t Solution: With I = 0 1 and J = −I 0 , the symplectic group is deﬁned by Sp4 = {g ∈ GL4 | g Jg = J}. −1 −1 It is an easy computation that diag(x, y, x , y ) ∈ Sp4, hence T := D4 ∩ Sp4 is a 2-dimensional torus, so is a maximal torus (because we can use that Sp4 has rank 2). As Sp4 is given by a bilinear form, its Lie algebra is

t AB t t t sp4 = {X ∈ M(4,K) | X J + JX = 0} = { CD | B = B,C = C,D = −A }.

In particular, A = a1 a2 ,D = 0 d ,B = b1 b2 ,C = c1 c2 . a3 a4 −d 0 b2 b3 c2 c3 −1 The adjoint action of Sp4 (and hence that of T ) on sp4 is given by ad(g)(X) = gXg . We explicitly compute

     −1  x 0 0 0 a1 a2 b1 b2 x 0 0 0 −1 0 y 0 0  a3 a4 b2 b3   0 y 0 0 ad(t)(X) =  −1      0 0 x 0  c1 c2 −a1 −a3   0 0 x 0 −1 0 0 0 y c2 c3 −a2 −a4 0 0 0 y  −1 2  a1 xy a2 x b1 xyb2 −1 2  x ya3 a4 xyb2 y b3  =  −2 −1 −1 −1   x c1 x y c2 −a1 −x ya3 −1 −1 −2 −1 x y c2 y c3 −xy a2 −a4 and we are now looking for 8 eigenvectors relative to characters (i.e. functions xmyn). They are easy to see:

character weight eigenvector (all other matrix entries 0) 1 0 a1 = 0 or a4 = 0 −1 xy e1 − e2 a2 = 1 −1 x y e2 − e1 a3 = 1 2 x 2e1 b1 = 1 xy e1 + e2 b2 = 1 2 y 2e2 b3 = 1 −2 x −2e1 c1 = 1 −1 −1 x y −e1 − e2 c2 = 1 −2 y −2e2 c3 = 1

Therefore, the roots (i.e. the non-zero weights) are ±2e1, ±2e2, ±(e1 − e2), ±(e1 + e2).

7. Let B ⊆ G be a Borel subgroup of a connected linear algebraic group G. Show that the centres coincide: Z(B) = Z(G). Solution: This is exercise 67.

8. Let G = GLn and B ⊂ G a Borel subgroup. Let W be the Weyl group of G. S a) Show G = w∈W BwB. (Explain ﬁrst that BwB is a well-deﬁned subset of G.) b) Show that the union is disjoint.

Solution: a) We choose the standard torus and Borel subgroup: T = Dn ⊂ B = Tn ⊂ GLn. The Weyl group of GLn is the permutation group W = NG(T )/CG(T ) = NG(T )/T =∼ Sn. In particular, from CG(T ) = T (which has been an exercise for T = Dn ⊂ GLn and is easy to check again) and T ⊂ B, we see that nB = ntB for n ∈ NG(T ) and t ∈ T . Note that W is the subgroup of permutation matrices in GLn. (This is a special feature of GLn — for other reductive groups, the Weyl group is not necessarily a subgroup in a canonical way!) One way to see GLn = BWB is via the Gauss algorithm: bringing an arbitrary matrix g ∈ GLn in row echelon (hence upper triangular) form starts with (g | In) and uses row transformations to end in (U | L) with Lg = U where U is an upper triangular matrix and L is a lower triangular matrix, i.e. U, Lt ∈ B. However, in general one has to apply a permutation (e.g. if the top left entry of g is zero); i.e. there is a permutation matrix w1 ∈ W such that the row algorithm works on w1g (using w1g means that we permute rows of g). We record that for any t −1 g ∈ G, there are w1 ∈ W, L ∈ B ,U ∈ B with wg = L U. −1 −1 −1 −1 We arrive at g = w1 L U. Now there is a unique permutation matrix w2 ∈ W such that w1 L w2 ∈ B −1 — i.e. we undo the row permutations of w1 by appropriate column permutations of w2. Eventually, we get −1 −1 −1 0 0 −1 −1 −1 g = w1 L w2w2 U = U wU ∈ BwB where U := w1 L w2 ∈ B by construction and w := w2 . 0 0 −1 −1 0 b) If BwB ∩ Bw B 6= ∅, then there exist b1, b2, b3, b4 ∈ B such that b1wb2 = b3w b4, hence b3 b1wb2b4 = w and w0 ∈ BwB. This immediately implies Bw0B ⊆ BwB. The other inclusion follows by symmetry.