MA4E0 Lie Groups

Home , Group homomorphism, Immersion (mathematics)

MA4E0 Lie Groups

December 5, 2012

Contents

0 Topological Groups 1

1 Manifolds 2

2 Lie Groups 3

3 The Lie Algebra 6

4 The Tangent space 8

5 The Lie Algebra of a Lie Group 10

6 Integrating Vector ﬁelds and the exponential map 13

7 Lie Subgroups 17

8 Continuous homomorphisms are smooth 22

9 Distributions and Frobenius Theorem 23

10 Lie Group topics 24

These notes are based on the 2012 MA4E0 Lie Groups course, taught by John Rawnsley, typeset by Matthew Egginton. No guarantee is given that they are accurate or applicable, but hopefully they will assist your study. Please report any errors, factual or typographical, to [email protected]

i MA4E0 Lie Groups Lecture Notes Autumn 2012

0 Topological Groups

Let G be a group and then write the group structure in terms of maps; multiplication becomes m : G × G → G −1 deﬁned by m(g1, g2) = g1g2 and inversion becomes i : G → G deﬁned by i(g) = g . If we suppose that there is a topology on G as a set given by a subset T ⊂ P (G) with the usual rules. We give G × G the product topology. Then we require that m and i are continuous maps. Then G with this topology is a topological group Examples include 1. G any group with the discrete topology

2. Rn with the Euclidean topology and m(x, y) = x + y and i(x) = −x 1 2 2 2 3. S = {(x, y) ∈ R : x + y = 1} = {z ∈ C : |z| = 1} with m(z1, z2) = z1z2 and i(z) =z ¯. Note that this is compact, connected and commutative

4. If G1,G2 are topological groups then G1 × G2 with the product topology is a topological group. 5. T k = S1 × ... × S1 We now state a ﬁrst result and prove it using the following few lemmas. Proposition 0.1 If G is a topological group which is connected then any open set containing the identity element generates G as a group, i.e. every element of G is a ﬁnite product of elements of the open set.

Lemma 0.2 If G is a topological group and g ∈ G then Lg(h) = gh = m(g, h) deﬁnes a continuous map G → G. In fact Lg is a homeomorphism.

Proof

Lg1 ◦ Lg2 (h) = g1g2h = Lg1g2 (h) and also Le(h) = eh = h = IdG(h) and so Lg ◦ Lg−1 = Le = IdG = Lg−1 ◦ Lg and so Lg is bijective and has a continuous inverse. Q.E.D.

Deﬁnition 0.3 If G is a group and A, B ⊂ G then we deﬁne 1. AB = {ab|a ∈ A, b ∈ B} 2. A−1 = {a−1|a ∈ A}

3. we inductively deﬁne Ak+1 = AkA = AAk for k ∈ N and A−k = (A−1)k Lemma 0.4 Suppose that A ⊂ G and then • if e ∈ A then Ak ⊂ Ak+1 • A−k = (Ak)−1 • If e ∈ A and A−1 = A then S Ak is a subgroup of G, called the subgroup of G generated by A. i∈Z Lemma 0.5 If G is a topological group and U is an open subset of G with e ∈ U then V = U ∩ U −1 is an open set with V = V −1 and e ∈ V .

Theorem 0.6 If G is a topological group and U is an open subset such that e ∈ U then ∪U k is an open subset of G and i fG is connected then ∪U k = G.

Proof U k is open since it can be written as [ Lg(U) g∈U k−1 and since Lg is a homeomorphism so Lg(U) is open hence the above is a union of open sets, and so is open. If we set V = U ∩ U −1 and note this is a subset of U and so V k ⊂ U k for all k and so ∪V k ⊂ ∪U k. V is open and H := ∪V k is a subgroup of G. H has cosets Lg(H) which are open sets and so G \ H = ∪gH6=H gH is open in G. Since G is connected we have G \ H = ∅. Since G ⊂ ∪U k ⊂ G we have G = ∪U k. Q.E.D. Examples include R? ⊃ R+ and the latter is an open subgroup and R? = R+ ∪ (−1)R+. and so R+ is closed as well. Any closed subgroup of R+, eg {1} or {2kk ∈ Z}.

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1 Manifolds

Deﬁnition 1.1 A chart (U, φ) of dimension n on an open set U of a topological space M is a homeomorphism φ of U onto an open set in Rn.   x1(p) We get functions x1, ..., xn on U such that, for p ∈ U φ(p) =  ... . These are called the coordinate xn(p) functions of the chart. Examples include M = Rn = U the identity map. This is called the canonical chart. One could instead have U to be an open set, and φ to be the inclusion map. Another example is S1 with the angle, or with stereographic projection as the chart. −1 If (U, φ) and (V, ψ) are two charts then ψ ◦ φ |φ(U∩V ) is the change of coordinates map between the two. This is a homeomorphism φ(U ∩ V ) to ψ(U ∩ V ).

Deﬁnition 1.2 Two charts (U, φ) and (V, ψ) of dimension n on a topological space M are compatible of class k −1 −1 k C if either U ∩ V = ∅ or if U ∩ V 6= ∅ then the maps ψ ◦ φ |φ(U∩V ) and φ ◦ ψ |ψ(U∩V ) are C as maps from open sets of Rn. Ck means, for k > 0 that all partial derivatives of orders up to k are continuous, for k = 0 that the functions are continuous, and for k = ω that the function is real analytic.

Example 1.1 INSERT FIGURE Two stereographic projections on the circle. Suppose we project from two y φ(x,y) antipodal points and set the centre of the circle to be the origin. Then we have that 1+x = 2 by considering 4−t2 4t −1 similar triangles. Thus we get that x = 4+t2 and y = 4+t2 , and thus we have found φ (t). Also we have that ψ(x,y) y −1 4 2 = 1−x and so ψ ◦ φ (t) = t by direct calculation. This is a map from φ(U ∩ V ) which is R \{0} since U = S1 \{(−1, 0))} and V = S1 \{(1, 0))} and so the change of coordinates map is C∞.

Exercise: Check other examples of charts are compatible for suitable k.

Deﬁnition 1.3 An atlas A of dimension n and class Ck on a topological space M is a collection of charts which are dimension n and pairwise compatible of class Ck such that the domains of the charts cover M. An atlas is said to be maximal if it contains all charts of dimension n which are Ck compatible with all of its charts.

Remark Every atlas is contained in a unique maximal atlas of the same dimension and class.

Example 1.2 On S1 we have two charts given by projections from two distinct points covering S1.

Deﬁnition 1.4 A manifold structure of dimension n and class Ck on a topological space M is a choice of maximal atlas of class Ck and dimension n.

A manifold structure can be speciﬁed by a non maximal atlas, e.g. the atlas of two charts on S1. We can then use charts from the maximal atlas when needed.

n n ∞ ω n Example 1.3 1. R with atlas of one chart {(R , IdR )} of dimension n and class C (C even). n 2. Any ﬁnite dimensional real vector space V by picking a basis b = (v1, ..., vn). We get a map φb : V → R P by φb(v) = (a1, ..., an) if v = aivi. V gets a topology by making φb a homeomorphism. Then (V, φb) is a chart of dimension n on V . This construction is independent of the choice of bases, b2 = (w1, ..., wn) and b1 = b. Then we have b2 = b1g where g is an n × n real matrix that is invertible. Then we have           a1 c1 a1 a1 c1 v = b  .  = b  .  = b g  .  and so  .  = g  .  and so φ = g ◦ φ and g is Ck for all k 1  .  2  .  1  .   .   .  b1 b2 an cn an an cn and so the charts are compatible and hence in the same maximal atlas.

n 2(x1,...,xn) 3. S the the n sphere and one can generalise stereographic projection. φ(x) = 1−x is projection onto n 0 the plane tangent at (1, 0, ..., 0) from U1 = {x ∈ S |x0 6= 1}.

We can consider charts from the implicit function theorem. Let F : Rn+k → Rk be a C∞ map then n+k n+k y ∈ F (R is called a regular value if for all x ∈ R with F (x) = y the derivative DxF is onto as a linear map Rn+k → Rk.

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Theorem 1.5 F −1(y) = {x ∈ Rn+k|F (x) = y} when y is a regular value is an n dimensional manifold. Charts can be obtained by taking n of the domain coordinates on a suitable open set and the remaining k domain coordinates are C∞ functions of these.

1 −1 2 2 2 2 Example 1.4 S = F (1) where F (x, y) = x + y . D(x,y)F (h, k) = 2xh + 2yk and since x + y = 1 x and y cannot simultaneously be zero and so the derivative is onto so 1 is a regular value. Then we solve x2 + y2 = 1 for either x and y. This gives x = p1 − y2 and we can do this likewise for Sn. Definition 1.6 A continuous map F : M → N between two manifolds is smooth if for any pair of charts (U, φ) and (V, ψ) of M and N respectively with f −1(V ) ∩ U 6= ∅ the map of open sets ψ ◦ f ◦ φ−1 : φ(f −1(V ) ∩ U) → ψ(V ) is C∞ as a map between open sets of Euclidean space. f is called a diffeomorphism if it is a homeomorphism and both f and f −1 are smooth. If such a map exists, then we say M and N are diffeomorphic. When N = R we refer to smooth maps as smooth functions. The set of all smooth functions on a manifold M is denoted by C∞(M). This is a real vector space and a ring with pointwise properties Definition 1.7 If M and N are manifolds of dimension m and n respectively then we can form a product chart (U × V, φ × ψ) from charts on M and N by

m n m+n φ × ψ(x, y) = (φ(x), ψ(y)) ∈ R × R = R and then this is a homeomorphism of U × V onto an open set of Rm+n. This construction gives an atlas on M × N which determines a product manifold structure of dimension m + n. For example T n = S1 × ... × S1 is a product manifold of dimension n and is compact. We generally assume that our manifolds are Hausdorﬀ topological spaces and, for Lie groups, we assume that the topology is second countable, i.e. there is a countable basis for the topology.

2 Lie Groups

Definition 2.1 A Lie group G is a set on which there are two structures: 1. G is a group 2. G is a C∞ manifold with Hausdorff, second countable topology such that m : G×G → G given by m(g, h) = gh and i : G → G given by i(g) = g−1 are both smooth maps Remark Sometimes one mapm ˜ (g, h) = g−1h is used in the definition. This is equivalent to the one given above. Example 2.1 1. G a finite group is a Lie group with the discrete topology

2. Rn is a Lie group under addition with the standard (canonical) smooth structure 3. S1 is a group under addition of angle or complex multiplication, and is a Lie group 4. If G and H are Lie groups then G × H is a Lie group 5. T n is a Lie group and is abelian and connected.

6. R? is a group under multiplication so is a 1 dimensional Lie group that is not connected

7. GL(n, R) is a Lie group. It is a group under matrix multiplication, and is an open set in Mn×n(R) since it is the preimage of R? under the determinant map. It has a global chart using the n2 matrix entries as coordinates, and the group operations are polynomials in these coordinates, and so are smooth.

8. GL(n, C) is a Lie group of dimension 2n2 9. The quaternions H. H = R1 + Ri + Rj + Rk where i2 = j2 = k2 = −1 and k = ij = −ji and so is a 4 skew ﬁeld. H = R as a real vector space. An element in H is written as q = q01 + q1i + q2j + q3k and 2 0 0 ? q¯ = q01 − q1i − q2j − q3k and qq¯ = ||q|| 1 and qq = qq . H = H \{0} under quaternion multiplication is a group with quadratic multiplication map and invers i(q) =q/ ˜ ||q||2 and so both are smooth and so it is a Lie group of dimension 4. 3 Quaternions of unit length for a subgroup Sp(1) = S as a manifold. Sp(1) is a manifold and the Euclidean coordinates are smooth functions when restricted to Sp(1) and so it is a Lie group of dimension 3.

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10. GL(n, H) is a Lie group of dimension 4n2 11. Orthogonal Group O(n). This is the group of operations that preserves the inner product on Rn. n O(n) = {g ∈ GL(n, R)|(gx) · (gy) = x.y∀x, y ∈ R }

Rewrite conditions so we can use the inverse function theorem to get charts. View x, y as column vectors, T T T T T T and so x · y = x y. Then (gx) · (gy) = x g gy ⇐⇒ x · (g gy − y) = 0 ⇐⇒ g gy − y = 0 ⇐⇒ g g = In. Thus T T O(n) = {g ∈ GL(n, R)|g g = In} = {g ∈ Mn×n(R)|g g = In} T T Set F (g) = g g for g deﬁned on Mn×n(R). What is the target. Note g g is not an arbitrary n × n matrix. T It is symmetric. Let Sn(R) = {A ∈ Mn×n(R)|A = A} and note that In ∈ Sn(R). Is In a regular value of F as a map Mn×n(R) → Sn(R). Take g ∈ O(n) and consider F (g + th) for h ∈ Mn×n(R). Then d D F (h) = F (g + th)| = gT h + hT g g dt t=0

For g ﬁxed, is every element of Sn(R) of this form for a suitable h. Take A ∈ Sn(R) and try to solve T T T 1 T 1 T g h + h g = A for some h or solve g h = 2 A and h g = 2 A, but these are the same as A = A . Thus 1 h = 2 gA. Thus DgF is surjective for every g ∈ O(n). −1 Since this holds forall g ∈ O(n), O(n) = F (In) and In is a regular value of F and hence O(n) gets a 2 2 1 manifold structure of dimension n − dim(Sn(R)) = n − 2 n(n − 1). Also O(n) has a group structure since T g1, g2 ∈ O(n) then (g1g2) g1g2 = In. The matrix entries are functions on O(n) and are smooth. Hence multiplication and inverses are smooth in manifold structure. Thus O(n) is a Lie group. O(1) = {±1}. a b b a Now consider O(2). If g = then a2 + c2 = 1 = b2 + d2 and ab + cd = 0. so ⊥ and c d d c −c b −c so is a multiple of , e.g. = λ and this gives λ2 = 1. Hence O(2) is the matrices of the a d a a −λc form with λ = ±1 and a2 + c2 = 1. Note that det(g) = λ and so λ depends continuously on g. c λa Hence O(2) has two connected components, with {g ∈ O(2)| det(g) = 1} as a subgroup and as a manifold is a circle. So O(2) is a disjoint union of two circles.

12. Special Linear Group SL(n, R) = {g ∈ GL(n, R)| det(g) = 1}. We look at det : Mn×n(R) → R. We have −1 SL(n, R) = det (1) but is it a regular value of det? If g ∈ SL(n, R) and h ∈ Mn×n(R) we need to compute d d d D det(h) = det(g + th)| = det(g(I + tg−1h))| = det(I + tg−1h)| = tr(g−1h) = n 6= 0 g dt t=0 dt n t=0 dt n t=0 and so 1 is a regular value of det so SL(n, R) is a Lie group of dimension n2 − 1. T 13. SO(n) = O(n)∩SL(n, R) = {g ∈ Mn×n(R)|g g = In, det(g) = 1}. For example SO(1) = {1} and SO(2) = a −c ∈ M ( )|a2 + c2 = 1 which is the same as O(2). Note that g ∈ SO(n) =⇒ det(g) = ±1 c a 2×2 R and hence SO(n) is an open set in O(n) since SO(n) = det−1(0, 2) viewing det as a function on O(n). det is a polynomial in the matrix entries so is smooth hence continuous and so det−1(0, 2) is open

Lemma 2.2 If G is a Lie group and H ⊂ G is a subgroup and an open set then H is a Lie group

Proof Take charts from G which intersect H and take (U ∩ H, φ|U∩H ) as a chart on H. Q.E.D.

If G is a Lie group this makes G0, the component containing the identity, into a Lie group. Also note that O(n)0 ⊂ O(n) and O(n) ⊃ SO(n) ⊃ SO(n)0 and also SO(2) = O(2)0 = SO(2)0. We can show this holds for all n by showing SO(n) is connected.

SO(n − 1) sits inside SO(n). Given g ∈ SO(n) with g viewed as n column vectors g = (x1, ..., xn) 0 . n−1 . so xi · xj = δij, and so xn ∈ S so rotate xn to the vector   this will rotate the other vectors 0 1 .  ...... 0   simultaneously and end up with a matrix of the form . . . .. Repeating we eventually get to . . . . 0 ... 0 1 SO(2) which is connected. Hence SO(n) is connected for all n.

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We have maps O(n) → Sn−1 by picking any column. Putting these maps together we get a map O(n) → ×nSn−1 which is continuous and injective and gives O(n) its usual topology. The image consists of the subset of this product of pairwise orthogonal vectors. The image is closed hence compact,so O(n) is a compact Lie group. So is SO(n).

14. Pseudo -orthogonal groups O(p, q) and SO(p, q). p+q Pp Pp+q T On R let Bp,q(x, y) = 1 xiyi − p+1 xiyi = x Ap,qy. We then deﬁne

O(p, q) = {g ∈ M(p+q)×(p+q)(R)|Bp,q(gx, gy) = Bp,q(x, y)}

T It is clear that g ∈ O(p, q) ⇐⇒ g Ap,qg = Ap,q and we can deﬁne F : M(p+q)×(p+q)(R) → Sp+q(R) by T −1 F (g) = g Ap+qg so O(p, q) = F (Ap,q) and is a regular value. This makes O(p, q) into a Lie group of 1 dimension 2 (p + q)(p + q − 1). Observe that O(p, 0) = O(p) and also deﬁne SO(p, q) = O(p, q) ∩ SL(p + q, R). a b We consider O(1, 1). Take an element here and write it as g = and then we get that a2 − c2 = 1, c d b −c b2 − d2 = −1 and ab − cd = 0 and so = −λ and with λ2 = 1. Now det(g) = −λ and so for −d a a c SO(1, 1) we have the form with a2 −c2 = 1. This is a hyperbolic function, and so SO(1, 1) has two c a connected components and O(1, 1) has four connected components. Now O(1, 1)0 corresponds to det(g) = 1 and a > 0 and then cosh t sinh t t 7→ sinh t cosh t

is a homomorphism of groups from R → O(1, 1)0 which is a diﬀeomorphism. This is an isomorphism of Lie groups. O(1, 3) is the Lorentz group, occurring in special relativity.

Deﬁnition 2.3 A homomorphism of Lie groups φ : H → G is a map which is a homomorphism of groups and a smooth map of manifolds. An isomorphism of Lie groups is a bijection which is a homomorphism of Lie groups, as well as its inverse.

15. An antisymmetric bilinear form can only be non singular when its dimension is even. So take R2n and set 0 0 0T 00 00T 0 0 −In T x y x y − x y Jn = and Ω(x, y) = x Jny. If x = 00 and y = 00 then Ω(x, y) = 0 00 00 0 In 0 x y x y − x y Remark If V is a real 2n dimensional vector space and w is an antisymmetric bilinear form on V which is non singular We deﬁne

Sp(2n, R) = {g ∈ GL(2n, R)|Ω(gx, gy) = Ω(x, y)}

T T T T T Then g ∈ Sp(2n, R) ⇐⇒ g Jng = Jn and set F (g) = g Jng and then (g Jng) = −g Jng and so T F maps M2n×2n(R) into A2n(R) = {A ∈ M2n×2n(R)|A = −A} which is a vector space of dimension 1 2 2n(2n − 1) = n(2n − 1). Then one can show Jn is a regular value of F and so Sp(2n, R) is a Lie group of dimension (2n)2 − n(2n − 1) = n(2n + 1). This group is called the real symplectic group

16. Complex versions GL(n, C) , etc. O(p, q, C) is the same as O(p + q, C) as one C bilinear form would have all positive signs as can multiply by i.

17. Unitary group On Cn one can form the Hermitian inner product hz, vi =z ¯T v. A complex linear map g : Cn → Cn is unitary if hgz, gvi = hz, vi for all z, v ∈ Cn. The set of unitary matrices is denoted by U(n). We can write the inner product as hz, vi =z ¯T (¯gT g)v and we denote g? =g ¯T . Then g ∈ U(n) ⇐⇒ ? g g = In. This is a group under matrix multiplication. ? ? ? ? ? Deﬁne F (g) = g g and then note that (g g) = g g and let Hn(C) = {A ∈ Mn×n(C)|A = A} which is a real vector space of dimension n2. If A = X + iY then A? == XT − iY T = X + iY , and so X = XT and Y = −Y T and so corresponds to symmetric and antisymmetric and so the dimension is the sum. 2 2 2 In is a regular value of F : Mn×n(C) → Hn(C) and so U(n) is a Lie group of dimension 2n − n = n . We can write g?g in terms of the columns of g. This is that the columns of g form an orthogonal basis of Cn so we get a map U(n) → ×nS2n−1 which is a homomorphism onto its image, which is n-tuples z1, ..., zn ∈ Cn with hzi, zji = 0 for i 6= j. The image is a closed set in ×nS2n−1 hence is compact, so U(n) is a compact Lie group.

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1 U(1) = {z ∈ C||z| = 1} = S a b If is in U(2) then |a|2 + |c|2 = 1 = |b|2 + |d|2 and ab¯ +cd ¯ = 0. Thus b = −λc¯ and d = λa¯ with c d a b a a b |λ| = 1 and so is determined by λ ∈ S1 and ∈ S3 with λ = det and so as a manifold c d c c d U(2) is diﬀeomorphic to S1 × S3. Note that g? =g ¯T and so det g? = detg ¯ = det g and so | det g| = 1. If we set

−1 SU(n) = U(n) ∩ SL(n, C) = det(1) by viewing det : U(n) → S1. Then one can check that 1 is a regular value of det and one gets SU(n) as a manifold of dimension n2 − 1 with matrix entries deﬁning smooth functions, and so is a Lie group.

SU(1) = {1}

a b We now consider SU(2). An element of U(2) is written with |a|2 + |c|2 = 1 = |b|2 + |d|2 and c d ab¯ +cd ¯ = 0. Then we get, using similar arguments to above, that a −λc¯ 2 2 3 1 U(2) = |a| + |c| = 1, |λ| = 1 ' S × S c λa¯

and note that det g = λ for g ∈ U(2) and so SU(2) is diffeomorphic to S3 and S3 is diffeomorphic to Sp(1) and so SU(2) and Sp(1) are diffeomorphic. 18. We here consider U(p, q) and SU(p, q). We define

p q X X ? hz, vip,q = z¯kvk − z¯p+kvp+k = z Ap,qv 1 1 and set ? U(p, q) = {g ∈ M(p+q)×(p+q)(C)|g Ap,qg = Ap,q} ? and observe that Ap,q is a regular value of F (g) = g Ap,qg and so U(p, q) is a Lie group of dimension (p + q)2. For g ∈ U(p, q) we have | det g| = 1 and 1 is a regular value of det : U(p, q) → S1 and so SU(p, q) is a Lie group of dimension (p + q)2 − 1. SU(1, 1) is the isometry group of the Poincar´edisc. It is three dimensional and related to SL(2, R). SU(2, 2) is the conformal group and is important in Penrose’s twister theory.

19. Sp(n) are called the quaternary unitary groups. Deﬁned like U(n) but using quaternions Hn.

T Sp(n) = {g ∈ Mn×n(H)|g¯ g = In}

−1 Now In is a regular value of F : Mn×n(H) → Hn(H) and Sp(n) = F (In). The dimension of Hn(H) is 1 2 n + 2 n(n − 1) and the dimension of Mn×n(H) is 4n and so the dimension of Sp(n) = n(2n + 1) which is the same dimension as that of Sp(2n, R). If g ∈ Sp(n) then the columns of g are unit vectors in Hn so are in S4n−1 and Sp(n) maps into ×nS4n−1 with image a closed set of pairwise orthogonal vectors, and so is compact. Now note O(n) ⊂ U(n) ⊂ Sp(n) since R ⊂ C ⊂ H. These are the compact exceptional groups, together with their det = 1 versions. There is no determinant function on H as it makes no sense on non commutative rings. ? There are ﬁve more compact simple groups G2,F4,E6,E7,E8. Sp(p, q) can also be deﬁned by g Ap,qg = Ap,q with g ∈ M(p+q)×(p+q)(H)

3 The Lie Algebra

Deﬁnition 3.1 A real Lie algebra is a real vector space V with binary operation [·, ·]: V × V → V satisfying 1. it is bilinear

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2. [x, y] = −[y, x] 3. [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 called the Jacobi identity

Observe that these relations are homogeneous.

Example 3.1 1. We can take any real vector space and set [x, y] = 0. This is the trivial Lie algebra, or also called the abelian Lie algebra

2. A an associate Lie algebra over R and set [a, b] = ab − ba. We make A into a Lie algebra, for example A = EndV the set of linear maps V → V . This is called the commutator bracket. It measures how non n zero the commutativity of the algebra is. For instance if V = R and then End(V ) = Mn×n(R) and so this is the Lie algebra with the commutator. 3. If (V, [, ]) is a Lie algebra then a subspace U of V becomes a Lie algebra so long as [x, y] ∈ U for x, y ∈ U. This is called a Lie subalgebra. For instance if A a commutator algebra, then End(A) is a Lie algebra. Denote by Der(A) the space of linear maps D : A → A which satisfy the Leibniz rule,

D(ab) = D(a)b + aD(b)

This is obviously a subspace of End(A) and we claim that it is a Lie subalgebra. Let D1,D2 ∈ Der(A). Then

[D1,D2](a, b) = D1(D2(ab)) − D2(D1(ab))

= D1(D2(a)b + aD2(b)) − D2(D1(a)b + aD1(b)) = ... =

[D1,D2](a)b + a[D1,D2](b)

and so is in Der(A).

Our main use of this last example is with A = C∞(M) for M a manifold. Let X (M) = Der(C∞(M)). If M = R and D ∈ X (M) then D(1) = D(1 × 1) = D(1) × 1 + 1 × D(1) = 2D(1) and so D(c) = 0 for any constant c ∈ R. If we take f ∈ C∞(R and a ∈ R then we can use Taylor expansion to get

0 2 f(x) = f(a) + (x − a)f (a) + (x − a) ga(x)

∞ where ga ∈ C (R). We can rearrange this and then check that it is smooth

0 2 D(f)(x) = 0 + D(x − a)f (a) + D((x − a) ga(x)) 0 2 = D(x − a)f (a) + D(x − a)(x − a)ga(x) + (x − a)D(x − a)ga(x) + (x − a) D(ga(x)) and if we set x = a we get D(f)(a) = D(x − a)(a)f 0(a) + 0 0 ∞ d d and so D(f)(a) = h(a)f (a) for some h ∈ C (R) for all a ∈ R and so D = h dx . If Di = hi dx for i = 1, 2 then d [D ,D ] = (h h0 − h h0 ) 1 2 1 2 2 1 dx and this is not the trivial Lie algebra. X (M) is the Lie algebra of vector fields on M and [x, y] is the Lie bracket of vector fields x, y ∈ X (M). If σ : M → M is a diffeomorphism and X ∈ X (M) then we can form an action by

(σ · X)(f) = (X(f ◦ σ)) ◦ σ−1

Then if X ∈ X (M) then σ · X ∈ X (M). ∞ If G is a Lie group deﬁne Lg : G → G for g ∈ G by Lg(h) = gh. This is a C map. Note Lg1 ◦ Lg2 = Lg1g2 −1 and Le = IdG and so Lg−1 = Lg and so is invertible. We say X ∈ X (M) is left invariant if Lg · X = X for all G G g ∈ G. Denote by X (G) the set of left invariant vector ﬁelds. Since X 7→ Lg · X is linear we have that X (G) G is a subspace of X (G). Note that Lg · ([X,Y ]) = [Lg · X,Lg · Y ] for all X,Y ∈ X (G) and so X (G) is a Lie subalgebra of X (G).

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4 The Tangent space

In Rn if we have a smooth curve γ :(a, b) → Rn we can associate a tangent vector to each point on γ by taking dγ(t) dt orγ ˙ (t). How though can we defineγ ˙ (t) for γ :(a, b) → M a smooth map? n d We can take a chart on M around a point on γ and take the tangent vector in R of φ ◦ γ, say dt (φ ◦ γ(t)) = n d v ∈ R . If we take another chart (V, ψ) around the same point then in general dt (ψ ◦ γ(t)) = w is different but −1 −1 −1 ψ ◦ γ(t) = ψ ◦ φ ◦ φ(γ(t)) = ψ ◦ φ (φ ◦ γ(t)) and so w = Dφ◦γ(t)(ψ ◦ φ )v. Fix a point x ∈ M and consider n objects of the form (U, φ, v) where (U, φ) is a chart about x and v ∈ R . Define an equivalence relation ∼x by −1 (U, φ, v) ∼x (V, ψ, w) if w = Dφ◦γ(t)(ψ ◦ φ )v.

Deﬁnition 4.1 The tangent space TxM is the set of equivalence classes of ∼x. We denote by [U, φ, v]x the equivalence class of (U, φ, v) at x.

This becomes a vector space by picking a chart (U, φ) at x and deﬁning

[U, φ, v]x + [U, φ, w]x = [U, φ, v + w]x and a[U, φ, v]x = [U, φ, av]x This is well deﬁned as the derivative of a linear map.

dγ(t) Deﬁnition 4.2 If γ :(a, b) → M is a smooth curve with x = γ(t0) we deﬁne γ˙ (t0) = dt |t=t0 by taking a chart (U, φ) around x and setting d(φ ◦ γ(t)) γ˙ (t0) = U, φ, |t=t0 dt x

Theorem 4.1 If M is a manifold of dimension n then for all x ∈ MTxM is a vector space of dimension n. If F : M → N is smooth then there is a natural linear map dxF : TxM → TF (x)N which satisﬁes the chain rule, namely if F : M → N and E : N → P and x ∈ M then

dx(E ◦ F ) = dF (x)E ◦ dxF

Furthermore if γ :(a, b) → M is a smooth curve then

· dγ(t)F (γ(t)) = (F ◦ γ) (t)

n Proof Fix x ∈ M and (U, φ) a chart around x. Deﬁne a map R → TxM by v 7→ [U, φ, v]x which is linear and −1 n also deﬁne [V, ψ, w]x 7→ Dψ(x)φ ◦ ψ (w) and this is also linear. Note that these are inverses and so R ' TxM if we pick a chart. Then −1 dxF [U, φ, v]x = [V, ψ, Dφ(x)(ψ ◦ F ◦ φ )(v)]F (x) which is independent of charts on M and N. This formula also works to show that the chain rule holds: d (F ◦ γ)·(t) = [V, ψ, ψ ◦ F ◦ γ(t)] dt F ◦γ(t) d = [V, ψ, ψ ◦ F ◦ φ−1 ◦ φ ◦ γ(t)] dt F ◦γ(t) d = [V, ψ, D ψ ◦ F ◦ φ−1 φ ◦ γ(t)] φ◦γ(t) dt F ◦γ(t) d = d F [U, φ, φ ◦ γ(t)] γ(t) dt γ(t) = dγ(t)F (γ ˙ (t))

Q.E.D. n n n dγ For example consider R with identity chart. This allows a canonical identiﬁcation TxR ' R and then dt is the same ordinary derivative.

∞ Deﬁnition 4.3 Let f ∈ C (M) and let x ∈ M. If X ∈ TxM then we set X(f) ∈ R to be equal to

−1 (Dφ(x)f ◦ φ )(v) if (U, φ) is a chart about x and X = [U, φ, v]x

This is the identiﬁcation of df : TxM → Tf(x)R with Tf(x)R = R, and write the X to the left, X(f) = df(X).

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∞ Proposition 4.4 Fixing x ∈ TxM the map f 7→ X(f) (C (M) → R) is linear and satisﬁes the product rule, namely X(fg) = X(f)g(x) + f(x)X(g)

Note that this should be clear since it is the directional derivative for a ﬁxed chart. Remark

1. This says X is a derivation at x from C∞(M) → R. ∞ 2. Suppose D : C (M) → R is linear and satisﬁes the product rule, then there is an x ∈ TxM such that D(f) = X(f) for all f ∈ C∞(M). To see this let (U, φ) be a chart about x such that the coordinate functions   D(f1) ∞  .  n xi. Then x1, ..., xn are the restrictions of globally deﬁned f1, ..., fn ∈ C (M). Then v =  .  ∈ R D(fn) and X = [U, φ, v]x.

∞ 3. This means that we could alternatively deﬁne TxM as the linear maps C (M) → R that satisfy the product rule.

Example 4.1 1. M an open set in a vector space V (for example GL(n, R) in Mn×n(R)). When we pick a basis b = {e1, ..., en} for V we get a chart (M, φb) as follows. Take x ∈ M and expand x in the basis ! x1(x) so that x = Pn x (x)e and put φ = . The x , ..., x are coordinate functinos of this chart. i=1 i i b . 1 n .xn(x) ! v1 M then becomes a manifold using the atlas {(M, φ )}. If v ∈ V then v = P v e and v = and b i i . .vn [M, φb, v]x ∈ TxM. The map v 7→ [M, φb, v]x is a linear isomorphism of vector spaces. This is independent ∞ d of the basis. The corresponding derivation at x we write as vx and for f ∈ C (M), vx(f) = dt f(x+tv)|t=0. 2. Suppose that M = F −1(c) for F : Rn+k → Rk smooth and c a regular value of f, and M given the manifold structure by the implicit function theorem. Take X ∈ TxM and any curve γ through x, say with γ(t0) = x n such that X =γ ˙ (t0). To do this pick a chart (U, φ) about x and then X = [U, φ, v]x for v ∈ R and put −1 d γ(t) = φ (φ(x) + (t − t0)v) and so dt φ(γ(t))|t=t0 = v. n+k 0 n+k Then we can view this as a curve in R and take its derivative at t0 to get a v ∈ R . But γ(t) ∈ M 0 0 n+k k for all t and so F (γ(t)) = c and hence (Dγ(t0)F )(v ) = 0 i.e. v ∈ KerDxF , and DxF : R → R n+k d and DxF is surjective so dim KerDxF = n. We have a map TxM → R deﬁned by X 7→ dt γ(t)|t=t0 which is linear and injective and maps TxM to Ker(DxF ) and so we can use Ker(DxF ) as a model for n n+1 −1 TxM. For example, S ⊂ R with F (x) = x · x. For x ∈ F (1) we have DxF (v) = 2x · v and so n+1 KerDxF = {v ∈ R |x · v = 0}, i.e. all vectors perpendicular to the radius vector x.

3. GL(n, R) has Mn×n(R) as the tangent space at In.

4. GL(n, C) has Mn×n(C) as the tangent space at In.

5. GL(n, H) has Mn×n(H) as the tangent space at In.

6. SL(n, R) = {g ∈ Mn×n(R) : det g = 1}. Now the tangent space at In is the kernel of DIn det = tr and so the space is {A ∈ Mn×n(R) : trA = 0}

7. SL(n, C) has TIn SL(n, C) = {A ∈ Mn×n(C) : trA = 0} T T 8. O(n) = {g ∈ Mn×n(R): g g = In}. Deﬁne F (g) = g g : Mn×n(R) → Sn(R) and now note that T T DIn F (A) = A + A and ker DIn F = {A ∈ Mn×n(R): A + A = 0} = An(R) ? 9. TIn U(n) = {A ∈ Mn×n(R): A + A = 0}

Recall we have diﬀeomorphisms Lg : G → G such that Lg(h) = gh with the properties that Lg1 ◦ Lg2 = Lg1g2 and Le = Id. Similarly for on the right with Rg(h) = hg but we have Rg1 ◦ Rg2 = Rg2g1 and Re = Id.

Lemma 4.5 TgG = deLg(TeG) since Lg(e) = g.

Deﬁnition 4.6 A vector ﬁeld on a manifold M is a choice of tangent vector at each point of M, i.e. for each x ∈ M we pick an Xx ∈ TxM.

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∞ Given a vector field X on M we can differentiate f ∈ C (M) to give a function x 7→ Xx(f), i.e. we can define a function X(f) by X(f)(x) = Xx(f) and it satisfies the Leibniz rule, since

X(f1f2)(x) = Xx(f1f2) = Xx(f1)f2(x) + f1Xx(f2(x)) = X(f1)f2(x) + f1(x)X(f2) = (X(f1)f2 + f1X(f2))(x) X is linear but X may not be a derivation since we do not know X(f) ∈ C∞(M). Definition 4.7 A vector field X on M is said to be smooth if X(f) ∈ C∞(M) for all f ∈ C∞(M). If X is smooth then X is a derivation. Conversely if D : C∞(M) → C∞(M) is a derivation then consider, for a given x ∈ M, the map f 7→ D(f)(x) ∞ from C (M) → R. Then it defines a derivation at x and so a tangent vector Xx ∈ TxM with D(f)(x) = Xx(f). Then D determines a vector field X which is smooth. Thus derivations of C∞(M) are smooth vector fields. Thus X (M) is now the vector space of smooth vector fields with the bracket of derivations. This is the link between the algebraic point of view (derivations) and the geometric point of view (tangent space). Definition 4.8 Let F : M → N be a smooth map of smooth manifolds. Vector fields X on M and Y on N are said to be F -related if dxF : TxM → TF (x)N is such that

dxF (Xx) = YF (x) for all x ∈ M.

Theorem 4.9 If Xi ∈ X (M) and Yi ∈ X (N ) for i = 1, 2 and F : M → N is a smooth map and Xi and Yi are F -related for i = 1, 2 then [X1,X2] is F -related to [Y1,Y2]. Proof Take f ∈ C∞(N ) Then dxF (Xi)x(f) = (Xi)x(f ◦ F ) and dxF (Xi)x = (Yi)F (x) and so (Yi)F (x)(f) = (Xi)x(f ◦ F ) and is the same as Yi(f)(F (x)) or Yi(f) ◦ F = Xi(f ◦ F ) (4.1) This is the derivation version of being F -related. Now we have

[X1,X2](f ◦ F ) = X1(X2(f ◦ F )) − X2(X1(f ◦ F ))

= X1(Y2(f) ◦ F ) − X2(Y1(f) ◦ F )

= Y1(Y2(f)) ◦ F − Y2(Y1(f)) ◦ F

= ([Y1,Y2](f)) ◦ F

Evaluating at x leads to dxF [X1,X2]x = [Y1,Y2]F (x) as required Q.E.D.

5 The Lie Algebra of a Lie Group

Deﬁnition 5.1 A vector ﬁeld X on a Lie group G is called left invariant if

dhLg(Xh) = Xgh for all g, h ∈ G.

Note that this means X is Lg related to X for all g ∈ G. Lemma 5.2 If X is a vector ﬁeld on a Lie group which is left invariant then X is smooth Proof Take f ∈ C∞(G) and then

X(f)(g) = Xg(f) = Xge(f) = (deLg(Xe))(f) = Xe(f ◦ Lg) and also note that (f ◦ Lg)(h) = f(Lg(h)) = f(gh) = f(m(g, h)) = (f ◦ m)(g, h) ∞ and so differentiating in h in the Xe direction leaves something smooth in g and so Xf ∈ C (G). Q.E.D. G G To find out how big X (G) is we compare it to TeG. Define ε : X (G) → TeG by ε(X) = Xe. This is linear.

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G Theorem 5.3 ε : X (G) → TeG deﬁned by ε(X) = Xe is a linear isomorphism.

Proof Let X ∈ ker ε and so Xe = 0. But Xg = deLgXe = 0 for all g and so X = 0 and so ε is injective. ˜ Now take ξ ∈ TeG and set ξh = deLh(ξ) for all h ∈ G. Then ˜ ˜ dhLg(ξh) = dhLg(deLh(ξ)) = de(Lg ◦ Lh)(ξ) = deLgh(ξ) = ξgh ˜ G ˜ ˜ Hence ξ is left invariant and so smooth by the above lemma. Thus ξ ∈ X (G) and ε(ξe) = ξe = deLe(ξ) = ξ and so ε is surjective. Q.E.D.

˜ Definition 5.4 The left invariant vector field ξ determined by ξ ∈ TeG is called the left invariant extension ˜ of ξ, with deLg(ξ) = ξ. ˜ G ˜ If ξ, η ∈ TeG then ξ, η˜ are in X (G) which is a Lie algebra so we can form [ξ, η˜] which is left invariant and so of ˜ G the form h for some h ∈ TeG with h = ε([ξ, η˜]). This makes ε : X (G) → TeG. We call TeG with this bracket the Lie algebra of G. Remark We could equally use right invariant vector fields and get a second bracket on TeG denoted by [·, ·]right ˆ ˆ with ξh = deRh(ξ) and [ξ, η]right = −[ξ, ηˆ]e Corollary 5.5 If G is an abelian Lie group then [ξ, η] = 0 Note the bracket can be zero for non abelian Lie groups. ˜ ˜ ˜ [ξ, η˜]e(f) = ξe(˜η(f)) − η˜e(ξ(f)) means [·, ·] only contains information about G near the identity. It gives the best information when G is connected.

Example 5.1 Examples of abelian Lie groups: Rn, R?, (0, ∞), S1, T n, C? = S1 × (0, ∞), Cn.

We need a technique for computing with vector ﬁelds. If (U, φ) is a chart with coordinates x1, ..., xn so n xi = ti ◦ φ where ti are Cartesian coordinates on R . −1 If f is a smooth function on the manifold we can form ∂f◦φ ◦ φ ∈ C∞(U) is a vector ﬁeld on U denoted ∂ti by ∂ ∂xi Pn i ∂ i ∞ Lemma 5.6 If X ∈ X (M) then X|U = a with a ∈ C (U). i=1 ∂xi

Example 5.2 GL(n, R) ⊂ Mn×n(R) is an open set. We have coordinates xij(g) = gij. since GL(n, R) is open d ˜ in a vector space, TIn GL(n, R) = Mn×n(R) as vector spaces. A ∈ Mn×n(R) corresponds to dt (In + tA)|t=0. A is the left invariant extension and then d d d A˜ = (d L )A = (d L ) (I + tA)| = L (I + tA)| = (g + tgA)| g In g In g dt n t=0 dt g n t=0 dt t=0 and applying this to coordinate functions gives d d X X X A˜ (x ) = (g + tgA) | = g + t g A | = g A = x (g)A g ij dt ij t=0 dt ij ik kj t=0 ik kj ik kj k k k ˜ P and so A(xij) = k xikAkj Then

[A,˜ B˜](xij) : = A˜(B˜(xij)) − B˜(A˜(xij)) X X = A˜( xikBkj) − B˜( xikAkj) k k X X = A(xik)Bkj − B˜(xik)Akj k k X X = xilAlkBkj − xilBlkAkj k,l k,l X X = xil (AlkBkj − BlkAkj) l k X = xil(AB − BA)lj l

= AB^− BA(xij) ˜ ˜ Using the lemma below we get [A, B] = AB^− BA and so the Lie algebra of GL(n, R) is Mn×n(R) with the Lie bracket as the commutator.

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Lemma 5.7 If X,Y ∈ X (M) and for an atlas of charts (U, φ) we have X(xi) = Y (xi) for the coordinates (x1, ..., xn) on U then X = Y .

Example 5.3 For all classical groups SL(n, R), O(n) etc, the same calculation works in the sense that TIn G is identiﬁed with ker DIn F for F mapping Mn×n(R) to a suitable set, and then if A ∈ ker DIn F then there is a 2 d 2 curve In + tA + O(t ) with A associated to dt In + tA + O(t )|t=0. The higher order terms do not change the argument, so all brackets are the commutators.

We denote Mn×n(R) with the commutator bracket by gl(n, R) and likewise for the others. If G and H are Lie groups , F : G → H is a homomorphism of Lie groups then F (eG) = eH and so deg F : TeG G → TeH H is there for a linear map from the Lie algebra of g to h.

Theorem 5.8 If F : G → H is a Lie group homomorphism then deG F : g → h is a Lie algebra homomorphism.

Proof Take ξ ∈ g and consider ξ˜ its left invariant extension: ˜ dgF ξg = dgF (deG Lg(ξ)) = deG (F ◦ Lg)(ξ) and then F ◦ Lg(h) = F (Lg(h)) = F (gh) = F (g)F (h) = LF (g)(F (h)) and then we get that

˜ ^ dgF (ξ) = deG (LF (g) ◦ F )(ξ) = deH LF (g)deG F (ξ) = deG F (ξ)F (g)

˜ ˜ and so ξ and de^G F (ξ) are F related. Take ξ, η ∈ g and then [ξ, η˜] is F related to [de^G F (ξ), de^G F (η)] and so

˜ dgF ([ξ, η˜]g) = [de^G F (ξ), de^G F (η)]F (g) and setting g = eG we get

deG F ([ξ, η]g) = [deG F (ξ), deG F (η)]h Q.E.D.

Deﬁnition 5.9 We denote deG F by F? : g → h.

Corollary 5.10 Homomorphic Lie groups have isomorphic Lie algebras,

However, the converse is not true. There are a few examples of this. R and S1 are abelian and one dimensional. Also SU(2) and SO(3) have isomorphic Lie algebras but different centres. U(1) × SU(2) and U(2) are diffeomorphic as manifolds but not isomorphic as groups, and have the same Lie algebras. If F : G → H and E : K → G are homomorphisms of Lie groups we get F? : g → h and E? : K → G and also F ◦ E : K → H is a homomorphism of Lie groups so we get (F ◦ E)? : k → h is a Lie algebra homomorphism. By the chain rule (F ◦ E)? = F? ◦ E?. Then IdG : G → G has derivative (IdG)? = Idg. Suppose G and H are compact connected Lie groups and F : G → H i as homomorphism such that F? : g → h is an isomorphism. By the inverse function theorem there exists a U ⊂ G and V ⊂ H both open and eG ∈ U, eH ∈ V and F a diffeomorphism of U into V . Hence F is an open map so F (G) is an open subgroup of H. H is connected so F (G) = H and so F is surjective. Let K = ker(F ). This is a normal subgroup of G which is closed in G. G is compact so K is compact. F is one to one on U and eG ∈ U and so K ∩ U = {eg} so K is discrete in the relative topology. The sets {{k}}k∈K then form an open covering of K, and so K is a finite group. Fix k ∈ K and consider the map G → K given by g 7→ gkg−1 ∈ K which is continuous, and so it has a connected image. Hence gkg−1 = k for all g ∈ G and so K ⊂ Z(G). For example there are no trivial homomorphisms from SO(3) → SU(2). To show this, suppose that there is one, and call it F . Then F? : SO(3) → SU(2) is a homomorphism of Lie algebras. F? is non zero otherwise F would be constant. Do some algebra and see that F? has to be surjective and so a linear isomorphism. Therefore F is onto with a finite kernel in the centre of SO(3) but the centre of SO(3) is {I3} so F is an isomorphism. This is not possible since Z(SU(2)){±I2}.

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6 Integrating Vector ﬁelds and the exponential map

Deﬁnition 6.1 Let X ∈ X (M) and x ∈ M then a smooth curve γ :(a, b) → M in M with γ(to) = x for some t0 ∈ (a, b) is said to be an integral curve through x ∈ M if γ˙ (t) = Xγ(t) for all t ∈ (a, b)

Theorem 6.2 If X ∈ X (M) and x ∈ M then there exists an ε > 0 and γ :(−ε, ε) → M a smooth curve with γ(0) = x and γ an integral curve of X. If γi :(a, b) → M are two integral curves of X with γ1(t0) = x = γ2(t0) then γ1(t) = γ2(t) for all t ∈ (a, b).

Proof Take a chart (U, φ) with x ∈ U. Then

Xφ−1(y) = [U, φ, v(y)]φ−1(y) for y ∈ φ(U) a vector valued function v on φ(U) ⊂ Rn. X is smooth implies taht v is smooth on φ(U). n d γ˙ (t) = [U, φ, Γ(t)]γ(t). with Γ(t) a curve in R . which is dt φ(γ(t)) and so we haveγ ˙ (t)Xγ(t). Thus d φ(γ(t)) = v(φ(γ(t))) (6.1) dt and for γ(t) to pass through x at t0 we need

φ(γ(t0)) = φ(x) and so φ(γ(t)) is a solution of an ODE with initial value conditions at t0 and so the existence and uniqueness theorem for ODES gives existence of integral curves for at least a short time. Q.E.D. Note there is no explicit t in the RHS of (6.1) so it is autonomous. Hence φ(γ(t)) is a solution implies that φ(γ(t + a)) is a solution and so we can always ﬁnd an integral curve through a given point at a convenient value of t. so if γ1(t1) = x = γ2(t2) then we can shift one so thatγ ˜2(t) = γ2(t + t2 − t1) is an integral curve with γ˜2(t1) = x and so γ1(t) =γ ˜2(t) for common values of t near t1. Thus you can get a large solution form combining γ1 and γ2

Definition 6.3 An integral curve γ of a vector field X is called a maximal integral curve if any integral curve passing through a point of γ is just γ restricted to a subinterval (after shifting parameters to agree at a common point). X is said to be complete if the interval of definition of any maximal integral curve is (−∞, ∞).

2 ∂ Example 6.1 Consider R with X(x,y) = ∂x . Then v = (x, y) = (1, 0). If (x(t), y(t)) is an integral curve then (x ˙(t), y˙(t)) = (1, 0) and we want the integral cuve through (x0, y0) at t0. This gives the integral cuve as (t − t0 + x0, y0) for t ∈ (−∞, ∞) so X is a complete vector ﬁeld.

If we were on R \{0} with the same X then the integral curves have the same formula but if y0 = 0 then t = t0 − x0 gives (0, 0) which is not allowed. We have two integral curves for either t > t0 − x0 or t < t0 − x0. On Rn \{0} there are many complete vector ﬁelds. Theorem 6.4 If G is a Lie group then all left invariant vector ﬁelds are complete

Lemma 6.5 If M is a manifold and X ∈ X (M) and γ :(a, b) → M is an integral curve then for c ∈ R we have γ˜(t) = γ(t + c) is an integral curve on (a − c, b − c).

Proof Done before Q.E.D.

Lemma 6.6 Let X be a left invariant vector ﬁeld on a Lie group G. If γ :(a, b) → G is an integral curve through g at t = t0 and h ∈ G then γ¯(t) = hγ(t) is an integral curve on (a, b) passing through hg at t = t0. Proof d d γ¯(t) = hγ(t) = L γ(t) = d L (γ ˙ (t)) = X dt dt h γ(t) h Lh(γ(t))=Xγ¯(t) and soγ ¯ is an integral curve and γ¯(t0) = hγ(t0) = hg Q.E.D.

Lemma 6.7 Let X be a left invariant vector field on a Lie group G and γX be the maximal integral curve of X with γX (0) = e. Then γX is defined on (−∞, ∞) and satisfies γX (s)γX (t) = γX (s + t) for all s, t.

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Proof We prove that γX (s)γX (t) = γX (s + t) for those s, t where both sides are defined. i.e. if γX is defined (a, b) then we take s, t, s+t ∈ (a, b). Fix s ∈ (a, b) and consider γ1(t) = γX (s)γX (t) on (a, b) and γ2(t) = γX (s+t) on (a − s, b − s) for t such that t, s + t ∈ (a, b). By the previous lemmas, γ1 and γ2 are integral curves and at t = 0 both pass through γX (s) so γ1(t) = γ2(t) for all t ∈ (a, b) ∪ (a − s, b − s), i.e. t, s + t ∈ (a, b). Now suppose b < ∞, and we shift by an amount s ∈ (a, 0) then b − s > b and intervals (a, b) and (0, b − a) −1 overlap and we have a common solution on the overlap γX (s)γX (s + t) and γX (t). The new solution is defined on (a, b − a) which is a contradiction, so b = ∞. The case a = −∞ is similar. Q.E.D.

Proof (of theorem 6.4) Let X be a left invariant vector ﬁeld and γX (t) the maximal integral curve through e at t = 0. Then lemmas 6.5, 6.6 imply that gγX (t − t0) is the integral curve through g at time t0. The property γX (s)γX (t) = γX (t + s) says γX : R → G is a homomorphism of groups and γX a smooth curve implies that γX : R → G is a homomorphism of Lie groups. Q.E.D.

Deﬁnition 6.8 A Lie group homomorphism σ : R → G is called a 1-parameter subgroup of G, and is denoted 1-PSG. If G is a Lie group and σ is a continuous homomorphism then we say σ is a continuous 1-PSG.

Example 6.2 If G = R then we are looking for continuous maps σ : R → R with σ(s) + σ(t) = σ(s + t) for all s, t ∈ R. We have that σ(0) = 0. Now, for n ∈ Z we have σ(n + 1) = σ(n) + σ(1) and hence by induction m mn σ(n) = nσ(1). We can apply this argument to σ(nx) to get σ(nx) = nσ(x). Then nσ( n x) = σ( n x) = mσ(x) m m and so σ( n x) = n σ(x) and since σ is continuous and rationals are dense in R we get that σ(x) = xσ(1) for all x ∈ R. This gives that σ is linear and hence smooth. We can also show that σ˙ (0) = σ(1). Theorem 6.9 The following three sets are in bijection for a Lie group G. 1. The set of 1-PSG of a Lie group G. 2. The maximal integral curves of left invariant vector ﬁelds 3. The Lie algebra g.

Proof To map 2 to 1 note γX is a 1-PSG. To map 1 to 3 note σ 7→ σ˙ (0) gives us a map. To map 3 to 2 note ξ 7→ γξ˜ is one. Then we show the action of these in turn gives the identity, so they are all bijections. ˜ Given σ a 1-PSG set ξ =σ ˙ (0) and form γξ˜ It is enough to show that σ is the integral curve of ξ through e. It should be clear that σ(0) = e. Then d d d σ˙ (t) = σ(t + s)| = σ(t)σ(s)| = d L σ(s)| = d L ξ = ξ˜ ds s=0 ds s=0 e σ(t) ds s=0 e σ(t) σ(t) ˜ and so σ is an integral curve of ξ so σ = γξ˜ by uniqueness. Q.E.D.

According to the theorem, γξ˜(1) ∈ G is determined by ξ ∈ g.

Deﬁnition 6.10 We deﬁne expG : g → G by expG ξ = γξ˜(1). This is called the exponential map of G.

Proposition 6.11 expG : g → G is a smooth map Proof A smooth family of ODEs (jointly smooth in parameters and manifold variables) has solutions depending ˜ smoothly on the parameters. Hence the ODE for ξ depends linearly on ξ. Picking a basis ξ1, ..., ξk for g we have ξ = a1ξ1 + ... + akξk and so we are looking for solutions of ˜ ˜ ˜ γ˙ a(t) = ξγ(t) = (a1ξ1 + ... + akξk)γa(t) where a = (a1, ..., ak) Q.E.D.

Proposition 6.12

γξ˜(s) = exp(sξ)

Proof Consider t 7→ γξ˜(st) and note this is smooth and γξ˜(st1)γξ˜(st2) = γξ˜(s(t1 + t2)) and so γξ˜(st) = γη˜(t) for some η and d d η =γ ˙ (0) = γ (st)| = (st)| γ˙ (0) = sξ ξ˜ dt ξ˜ t=0 dt t=0 ξ˜ and therefore γ (xt) = γ (t) and then set t = 1 to conclude Q.E.D. ξ˜ sξe

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Corollary 6.13 exp(sξ) exp(tξ) = exp((t + s)ξ)

It is NOT true that exp ξ exp η = exp(ξ + η) in general. ˜ Corollary 6.14 ξ has integral curve through g at time t0 given by g exp((t − t0)ξ)

Example 6.3 1. For Rn, and 1-PSG σ has the form σ(v) = vσ(1) and so exp v = v under the standard n n identity T0R = R ? ? d 2. G = R with g = R and then T1R = R with the identiﬁcation dt (1 + tv)|t=0 ↔ v For ξ ∈ R we want a left invariant extension to R?. We use the coordinate functions x from R? ⊂ R, and ˜ d we then have ξx = f(x) dx and we want to know when this is left invariant. d ξ˜ = d L ξ = d L (1 + tξ)| x 1 x 1 x dt t=0 d = L (1 + tξ)| dt x t=0 d = (x + txξ)| dt t=0 = (xξ)x

and so f(x) = xξ. ˜ The integral curve σ(t) through 1 at time t = 0 will satisfy σ˙ (t) = ξσ(t) and be given by a function x(t) and so x˙(t) = x(t)ξ and so x(t) = Aeξt with x(0) = A = 1 and so exp ξ = x(1) = eξ.

3. Suppose G = GL(n, R) and g = Mn×n(R). Then ξ ∈ Mn×n(R) corresponds with the tangent vector d dt (In + tξ)|t=0 at In ∈ GL(n, R). We have d d ξ˜ = d L (I + tξ)| = (g + tgξ)| = gξ g In g dt n t=0 dt t=0 g where the over-line means directional derivative.

If g(t) is the integral curve through In at t = 0 then it satisﬁes ˜ g˙(t) = ξg(t) = g(t)ξg(t)

and g˙(t) as a tangent vector is g˙(t)g(t) and the derivative is that of a function. Thus g˙(t) = g(t)ξ. given a matrix ξ we form ξ2, ξ3, ... and hence we form

∞ X ξk eξ = I + n k! k=1

d tξ P∞ tk−1ξk tξ which is a smooth function in ξ with dt e = 0 + k=1 k k! = e ξ and so d d d (g(t)e−tξ) = (g(t))e−tξ + g(t) (e−tξ) = g(t)ξe−tξ + g(t)(−e−tξξ) = 0 dt dt dt and so g(t)e−tξ is a constant A and so g(t) = Aetξ and for t = 0 we have I = A and so exp ξ = eξ. n GL(n,R) ξ ξ 4. We can show that, for example, if ξ ∈ o(n) then e ∈ O(n) and so expO(n) ξ = e .

Theorem 6.15 If F : G → H is a homomorphism of Lie groups and F? : g → h is the corresponding homomorphism of Lie algebras then F (expG ξ) = expH (F?ξ)

Proof Consider t 7→ F (expG(tξ)) which is a 1-PSG of H, hence of the form expH (tη) for some η ∈ h given by d η = F (exp (tξ))| = F ξ dt G t=0 ? Q.E.D.

Corollary 6.16 Example 4 above. Observe that if we have a matrix group, the inclusion map O(n) ,→ GL(n, R) is a homomorphism of Lie groups whose derivative is the inclusion o(n) ,→ gl(n, R) of Lie algebras. Hence exp ξ = exp ξ for ξ ∈ o(n). O(n) GL(n,R)

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? ξ det? ξ Example 6.4 det : GL(n, R) → R is a homomorphism of Lie groups and det? = tr hence det(e ) = e = etrξ

exp : g → G is a smooth map with 0 ∈ g and e ∈ G with exp(0) = e. Then

d0 exp : T0g → TeG is a linear map. However T0g = g (using directional derivatives) and also TeG = g and so we can view d0 exp : g → g as a linear map. Then we have the following.

Theorem 6.17 d0 exp = Idg Proof Given ξ ∈ g there is a curve in g, say γ(t) with γ(0) = 0 andγ ˙ (0) = ξ, for instance γ(t) = tξ. Then d d d expγ ˙ (0) = exp(γ(t))| = exp(tξ)| = ξ ∈ T G 0 dt t=0 dt t=0 e i.e. d0 exp ξ = ξ. Q.E.D.

Corollary 6.18 There are hence neighbourhoods U of 0 ∈ g and V of e ∈ G such that V = expG U and exp U → V is a diﬀeomorphism.

Proof Inverse function theorem since Idg is an isomorphism Q.E.D.

Corollary 6.19 If G is connected then every element is a ﬁnite product of exponentials, i.e. g ∈ G =⇒ g = exp ξ1... exp ξN for ξi ∈ g. Proof Any open set around e generates G. Take one of the form V = exp U where exp : U → V is a diﬀeomorphism. Q.E.D.

Corollary 6.20 If G is a connected Lie group and Fi : G → H for i = 1, 2 are homomorphisms of Lie groups such that Fi? : g → h are equal, i.e. F1? = F2? then F1 = F2

Proof Take g ∈ G then by the previous corollary g = exp ξ1... exp ξN and so

F1(g) = F1(exp(ξ1)...F1(exp(ξN ))

= exp(F1?(ξ1))... exp(F1?(ξN ))

= exp(F2?(ξ1))... exp(F2?(ξN ))

= F2(exp(ξ1)...F2(exp(ξN ))

= F2(g) Q.E.D. Let exp : U → V be a diffeomorphism and exp−1 : V → U be the inverse function. Then this is a diffeomorphism from a neighbourhood e 3 V ⊂ G into U ⊂ g. If we pick a basis ξ1, ..., ξN for g then for g ∈ V we have −1 PN exp g = i=1 xi(g)ξi and this will define a smooth function on V which are the components of a smooth map into Rn, ψ : V → Rn which is a diffeomorphism onto an open set in Rn.(V, ψ) is a chart on G compatible with the atlas and these charts depend on the choice of open set U in g and the basis of g. These are all compatible. If g ∈ G and then g ∈ gV so we can left translate to any element. Then (gV, ψ ◦ Lg−1 ) will be a chart around g. These are all compatible, so we get an atlas for G built from the exponential map (problem sheet 6). Any real vector space is a group under addition so the Lie algebra of a Lie group is itself a Lie group. When then is exp : g → G a homomorphism of Lie groups, i.e. exp(ξ + η) = exp(ξ) exp(η).

Proposition 6.21 Let G be connected Lie group and then exp : g → G is a homomorphism of Lie groups if and only if G is abelian.

Proof Suppose exp : g → G is a homomorphism. Then exp(ξ) exp(η) = exp(ξ + η). G is connected and so G is generated by exponentials and so G is abelian. Now suppose that G is abelian and connected. Then t 7→ exp(tξ) exp(tη) is a 1-PSG of G therefore exp(tξ) exp(tη) = exp(tζ) for some ζ ∈ g. We have d ζ = exp(tξ) exp(tη)| = ξ + η dt t=0 using the lemma below. Thus exp is a homomorphism of Lie groups. Q.E.D.

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Lemma 6.22 If f :(a, b) → M is given by a function F :(a, b) × ... × (a, b) → M where f(t) = F (t, t, ..., t) and PN 0 ∈ (a, b) then f(0) = i=1 fi(0) where fi(t) = F (0, ..., t, ..., 0) with the t in the i-th spot. Proof The chain rule for f = F ◦ ∆ where ∆ is the diagonal map, ∆(t) = (t, t, ..., t). Q.E.D.

Corollary 6.23 IF g ∈ G is a connected and abelian Lie group then exp : g → G is surjective.

Proof IF g ∈ G then g = exp(ξ1)... exp(ξN ) = exp(ξ1 + ... + ξN ) Q.E.D. Examples of connected abelian Lie groups are Rn,T n, Rk × T n−k of dimension n. Theorem 6.24 Up to isomorphism, the above list are the only connected abelian Lie groups.

Lemma 6.25 Let Γ ⊂ Rn be an additive subgroup which is closed and for which there is an open ball B around n 0 with Γ ∩ B = {0}. Then either Γ = {0} or there are k linearly independent vectors v1, ..., vk ∈ R such that Pk Γ = { i=1 mivi|mi ∈ Z} Proof We proceed by induction. Take n = 1. Then if Γ 6= {0} there exists a non zero element v ∈ Γ of minimal length. Since Γ is closed and ||v|| ≥ R (radius of B), we claim that Γ = Zv. We know that 0 6= w ∈ Γ then w = λv for λ ∈ R and assume that λ > 0, and then we can write λ = p + r for p ∈ N and 0 ≤ r < 1. Then v ∈ Γ =⇒ pv ∈ Γ and therefore w − pv ∈ Γ and now ||w − pv|| = r||v|| < ||v|| and so, since v was minimal, we have that r = 0 and so w = pv ∈ Zv. The general case. Suppose we know the result for n − 1 and Γ 6= {0}. Then there is some non-zero v1 ∈ Γ of minimal length. Then Rv1 ⊃ Rv1 ∩ Γ is the case of n = 1 and so Rv1 ∩ Γ = Zv1. n Consider R /Rv1 ⊃ Γ/Zv1. We need to show that Γ/Zv1 satisfies the same assumptions as Γ. If it doesn’t then there is a non zero element sequence in Γ/Zv1 tending to 0. Using coset representation there is a sequence n n γm ∈ Γ such that γm 6∈ Zv1 and γm + Zv1 → 0 in R /Rv1, i.e. γm + Rv1 → 0 in R /Rv1. To tend to zero in the quotient Euclidean space means we can find a sequence in lm ∈ Z such that γm −lmv1 → n 0 in R . Let lm = hm + rm for some hm ∈ Z, and rm ∈ [0, 1]. Then γm − lmv1 = (γm − hmv1) − rmv1. Then sequence rm ∈ [0, 1) ⊂ [0, 1] and the latter is compact and so rm has a convergent subsequence. Hence we can suppose we have a sequence γm and coset representation γm such that there is a sequence rm ∈ [0, 1] convergent n n with γm − rmv1 → 0 in R , but γm ∈ Rv1. Let r = lim rm. γm is convergent in R to rv1. Γ is a closed subgroup n of R and so rv1 ∈ Γ. Therefore rv1 ∈ Zv1 ∩ [0, 1]v1 and so r = 0 or r = 1. n If r = 0 then γm → 0 in R . n If r = 1 then γm − v1 → 0 in R and both are sequence in Γ. Since Γ ∩ B = {0} there is an M such that m ≥ M implies that γm = 0 or γm − v1 = 0. But γm is a representative of a non-zero coset. 0 n 0 n Hence there exists a B in R /Rv1 with B ∩ Γ/Zv1 = {0}. Then by induction there exist v2, ..., vk in R /Rv1 such that v2 + Rv1, ..., vk + Rv1 are linearly independent and integer combinations of v2 + Zv1, ..., vk + Zvk give Γ/Zv1. Q.E.D. Proof (of theorem 6.24) G connected abelian implies that exp : g → G is a homomorphism and exp is a surjective map. Let Γ = ker exp which is a subgroup of g which is closed and since exp is a diffeomorphism near 0 ∈ g then there exists an open U ⊂ g with 0 ∈ U and Γ ∩ U = {0}. It is enough to study such subgroups γ of Rn. We thus know from the above lemma that there exist v1, ..., vk linearly independent such that Γ = Zv1 + ... + v . Put g = span {v , ..., v } which has dimension k. Pick g ⊂ g such that g = g ⊕ g and then Γ ⊂ g and Z k 1 R 1 k 2 1 2 1 exp : g2 → G will be an isomorphism onto its image G2 ⊂ G. Put G1 = exp(g1) which is a subgroup of G and ∼ n−k ∼ ∼ k k ∼ k k ∼ k n−k G = G1 × G2. G2 = R and G1 = g/Γ = R /Z = (R/Z) = T and so G = T × R . Q.E.D.

Corollary 6.26 Any connected compact Abelian Lie group is isomorphic to a torus.

7 Lie Subgroups

Deﬁnition 7.1 A smooth map of manifolds f : M → N is an immersion if

dxf : TxM → Tf(x)N has zero kernel, i.e. it is injective, for all x ∈ M. A smooth map of manifolds f : M → N is a submersion if

dxf : TxM → Tf(x)N is surjective for all x ∈ M. A smooth map of manifolds f : M → N is a diﬀeomorphism if it is a bijection and both f and f −1 are smooth.

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A diﬀeomorphism is both an immersion and a submersion. Something which is both an immersion and a submersion is locally a diﬀeomorphism. Locally an immersion looks like an axis in a product, e.g. M ,→ M × N . A submersion is locally a projection, e.g. M × N → M.

Deﬁnition 7.2 A submanifold of a manifold M is a smooth map f : M → N which is an injective immersion, i.e. both f and df are injective.

Thus crossings are not allowed in a submanifold. We often view N ⊂ M and TxN ⊂ TxM. Deﬁnition 7.3 A submanifold f : N → M is called embedded if f is a homeomorphism to f(N ) with the relative topology.

Example 7.1 1. Linear subspaces of a ﬁnite dimensional vector space are embedded submanifolds

2. R → T 2 by t 7→ (eiat, eibt) is an immersion so long as (a, b) 6= (0, 0) and injective when there is no common iat ibt 2πn 2πm period for e and e . The periods of these are t1 = a and t2 = b and a common period would be b iat ibt when t1 = t2. For this we would need a ∈ Q. When it isn’t, then t 7→ (e , e ) is injective. This is a submanifold and sits R as a subgroup of T 2, a line of irrational slope. This is dense.

3. To see S1 as a subgroup of T n we send z 7→ (zm1 , ..., zmn ). So long as there are no common divisors of 1 n m1, ..., mn then this will be injective and an immersion. Thus S is a submanifold of T and is embedded.

Deﬁnition 7.4 If G is a Lie group then a Lie submanifold f : H → G is a smooth map of manifolds where H is a Lie group and f is a homomorphism.

The above uses the properties, in Lie group terms of 1. H and G are Lie groups. 2. f : H → G is a homomorphism of Lie groups. 3. f is injective.

4. dhf : ThH → Tf(h)G is injective.

Given ξ ∈ h, deH Lhξ ∈ ThH and all arise in this way. Then observe that

0 0 0 0 0 (f ◦ LH )(h ) = f(hh ) = f(h)f(h ) = Lf(h)f(h ) = (Lf(h) ◦ f)(h ) and so we get that

dhf(deH Lhξ) = deH (f ◦ Lh)(ξ) = deH (Lf(h) ◦ f)(ξ) = deG Lf(h)deH f(ξ) = deG Lf(h)(f?(ξ))

Both of deG Lf(h) and deH Lh are invertible and so dhf is injective if and only if f? is injective. Thu f is an immersion if and only if f? : g → h is injective.

Proposition 7.5 A homomorphism of Lie groups f : H → G is an immersion of manifolds if and only if f and f? are injective.

This then characterises Lie subgroups as homomorphisms f with both f and f? injective. Then f(H) is a subgroup of G and f?(h) is a Lie subalgebra of g.

Theorem 7.6 If f : H → G is a Lie subgroup then

f?(h) = {ξ ∈ g| expG(tξ) ∈ f(H) for t in some open interval }

Proof Firstly f?(h) = {ξ ∈ g|...} since if η ∈ h put ξ = f?(η) and then expG(tξ) = expG(tf?(η)) = f(expH (tη)) ∈ f(H) for all t. We prove the converse, suppose ξ ∈ g has expG(tξ) ∈ f(H) for all t ∈ (a, b). For t ∈ (a, b) there exists h(t) ∈ H with f(h(t)) = expG(tξ). 0 0 We show that h(t) is a smooth curve in H. We do this by (locally ) ﬁnding a solution map for f(h ) = expG ξ 0 for ξ near ξ in f?(h). f?(h) is a linear subspace of g so we can ﬁnd a subspace m ⊂ g such that g = f?(h) ⊕ m and then H × m is a manifold of the same dimension as G.

Pick t0 ∈ (a, b) and consider Ψt0 : H × m → G given by

Ψt0 (h, ζ) = expG(t0ξ)f(h) expG(ζ)

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and note Ψt0 is smooth. d d Ψ (η, ρ) = Ψ (exp sη, sρ)| (eH ,0) t0 ds t0 H s=0 d = exp (t ξ)f(exp (sη))exp (sρ))| ds G 0 H G s=0 = d L (f (η) + ρ)) eG expG(t0ξ) ? therefore d Ψ is surjective onto T G and the domain is h ⊕ m which has the same dimension so (eH ,0) t0 expG(t0ξ d(eH ,0)Ψt0 is a linear isomorphism.

Therefore Ψt0 is a diﬀeomorphism near (eH , 0) and therefore there exist open sets U1 ⊂ H and U2 ⊂ m such that if V = Ψt0 (U1 × U2) then V is open in G, eH ∈ U1, s ∈ U2 and expG(tξ) ∈ V and Ψt0 is a diﬀeomorphism of U1 × U2 onto V . −1 −1 Ψt0 : V → U1×U2 will be smooth and we put Φt0 = π1◦Ψt0 and then Φt0 is a smooth map G ⊃ V → U1 ⊂ H. 0 0 0 0 0 0 There is an interval around t0 ∈ (a , b ) such that (a , b ) ⊂ (a, b) and expG(tξ) ∈ V . Then t ∈ (a , b ) we have 0 0 h(t) = Φt0 (expG(tξ)) and so h(t) is smooth on (a , b ). t0 is arbitrary and so h(t) is smooth on (a, b). ˙ Pick c ∈ (a, b) such that h(c) ∈ Th(c)H and so η = dh(c)Lh(c)−1 (h(c)) ∈ h and then ˙ f?(η) = f?(dh(c)Lh(c)−1 (h(c))) = (df(h(c))Lf(h(c)−1)) ◦ (ff(h(c))f)(h(c)) but ˙ · d d f(h(c)) = f(h(c)) = (exp (tξ))|t=c = de L (ξ) f(h(c)) dt G G expG(cξ) −1 −1 and f(h(c) ) = f(h(c)) = expG(−cξ) and so

f?(η) = ξ

Q.E.D.

This means that f?(h) = {ξ ∈ g| expG(tξ) ∈ f(H)∀t} and f(expH (tη)) = exp tf?(η). This means expH is determined by f? and expG hence an atlas for H is determined by f? and the topology of H. Proposition 7.7 Let H be a subgroup of a Lie group G and h ⊂ g a subspace such that we have open sets U of s ∈ g and V of eG ∈ G with expG : U → V a diﬀeomorphism such that expG(U ∩ h) = V ∩ H. Then H with the −1 relative topology has an atlas built using expG on V ∩ H and left translating making H into a Lie subgroup of G.

Lemma 7.8 Let G be a Lie group, ξ, η ∈ g, then there exists a ζ(t) ∈ g for |t| < δ for some δ > 0 such that

exp(tξ) exp(tη) = exp(tξ + tη + ζ(t)) for |t| < δ, and ζ(t)/t2 → 0 as t → 0.

Proof Take U ⊂ g, 0 ∈ U such that exp : U → V = exp(U) is a diﬀeomorphism then choose δ such that exp(tξ) exp(tη) ∈ V for |t| < δ. Then set

ζ(t) = exp−1(exp(tξ) exp(tη) − tξ − tη) for |t| < δ. Then ζ is a smooth function of t and ζ(0) = 0. Also

exp(tξ) exp(tη) = exp(tξ + tη + ζ(t)) and diﬀerentiating at t = 0 gives ξ + η = ξ + η + ζ˙(t) and so ζ˙(0) = 0. ζ is smooth in t around t = 0 and so we have a Taylor expansion with remainder 1 ζ(t) = ζ(0) + tζ˙(0) + t2R(t) 2

ζ(t) where R(t) is bounded as t → 0. clearly R(t) = t2 . Q.E.D.

Corollary 7.9 If ξ, η ∈ g then

ξ η N exp(ξ + η) = lim exp exp N→∞ N N

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1 Proof When N ≥ δ we have ξ η ξ η 1 exp exp = exp + + ζ N N N N N and so ξ η N 1 exp exp = exp ξ + η + Nζ N N N 1 and Nζ N → 0 as N → ∞. Q.E.D.

Theorem 7.10 Let H be a closed subgroup of a Lie group G then H with the relative topology has a unique manifold structure making the inclusion map i : H →,→ G a Lie subgroup.

Proof Take a subset of g which would be a Lie algebra of H if the result were true, i.e. set

h = {ξ ∈ g| expG(tξ) ∈ H∀t ∈ R} We show h is a linear subspace: Take ξ ∈ h adn c ∈ R, and consider t 7→ expG(t(cξ)) = expG((ct)ξ) ∈ H for all t and so cξ ∈ h. Take ξ, η ∈ h and consider ξ η N exp exp ∈ H for all N G N G N and this converges in G as N → ∞. Since H is closed the limit expG(ξ + η) is in H. Replacing ξ by tξ and η by tη results in expG(tξ + tη) ∈ H for all t and so ξ + η ∈ h. We now prove that conditions for proposition 7.7 hold for this subspace, i.e. there exists open sets U ⊂ g and V = expG(U) such that expG : U → V is a diffeomorphism and expG(U ∩ h) = expG(U) ∩ H. We proceed by contradiction. Suppose no such U exists, i.e. we never have expG(U ∩ h) = expG(U) ∩ H for a U ⊂ g. Fix W ⊂ g where exp : W → exp (W ) is a diffeomorphism, and fix a norm on g. Consider open balls B 1 (0) G G k for large enough k ∈ to ensure B 1 (0) ⊂ W . We have N k

exp (B 1 (0) ∩ h) 6= exp (B 1 (0)) ∩ H G k G k

If ξ ∈ B 1 (0) ∩ h then exp ξ ∈ exp (B 1 (0)) ∩ H and so exp (B 1 (0)) ∩ H is strictly larger than exp (B 1 (0) ∩ h) k G G k G k G k Let hk ∈ H ∩ exp (B 1 (0)), but hk 6∈ exp (B 1 (0) ∩ h). Then hk = exp ξk for ξk ∈ B 1 (0) and so (ξk) → 0 G k G k G k in g as k → ∞ and therefore (hk) → e in G. We claim that hk 6∈ expG(W ∩ h). Otherwise hk = expG ηk with ηk ∈ W ∩ h and hk = expG ξk with ξk ∈ B 1 (0) ⊂ W . Due to bijectivity of exp on W we have ηk = ξk and so ηk ∈ B 1 (0)∩h and so hk ∈ exp (B 1 (0)∩h) k G k G k which contradicts our choice of hk, and so the claim holds. Thus we have a sequence hk in H with (hk) → e in G and an open set W1 = W ∩ h containing 0 in h but hk 6∈ expG(W1) for all k large. Pick a subspace m ⊂ g supplementary to h in g making g = h ⊕ m. Deﬁne α : h × m → G by

α(ξ, η) = expG(ξ) expG(η) Then α is smooth and has derivative at (0, 0) given by

d(0,0)α(ξ, η) = ξ + η which is the identification of h ⊕ m with g and so is a linear isomorphism. Hence α is a diffeomorphism near (0, 0) so we have open sets Wh and Wm around the origin in h and m respectively so that α is a diffeomorphism of Wh × Wm with α(Wh × Wm) which is an open set in G containing e. We claim(to prove the claim above!!) that we can shrink Wm until H and expG(Wm) meet only at e. Suppose not, then we can find a sequence ηk in Wm so that expG ηk ∈ H \{e} and ηk → 0. Put tk = ||ηk|| > 0 and then ηk/tk is a vector of norm 1 for each k and hence we have a sequence ηk/tk in the unit sphere in Wm which is compact, and so this sequence has a convergent subsequence. We throw away the other terms and relabel the subsequence ηk/tk. Let η be the limit of this sequence. Let t > 0, nk(t) = bt/tkc and so

t/tk − 1 ≤ nk(t) ≤ t/tk and thus we have that t − tk ≤ tknk(t) ≤ t

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but since tk → 0 as k → ∞ we have tknk(t) → t as k → ∞. Multiplying the left side by ηk/tk and the right side by η we obtain

nk(t)ηk → tη as k → ∞. Applying expG we have that

lim expG(nk(t)ηk) = expG(tη) k→∞ and so n lim (expG(ηk))k (t) = expG(tη) k→∞ n but expG(ηk) ∈ H =⇒ (expG(ηk))k (t) ∈ H =⇒ expG(tη) ∈ H and also from group property of H, expG(−tη) ∈ H for t > 0. Thus expG(tη) ∈ H for t ∈ R and thus ηinH but since η 6= 0 and η ∈ m we have a contradiction. Thus the claim holds, i.e. we can shrink Wm so that expG(Wm) ∩ H = {e}. Now to conclude the proof of the ﬁrst claim (using the second claim!!). Since hk → e in G and expG(Wh ×Wm) is open and contains e then hk ∈ expG(Wh × Wm) for all k ≥ k0, and hence

0 0 hk = expG(ξk) expG(ηk)

0 0 0 0 0 with ξk ∈ Wh and ηk ∈ Wm, but hk 6∈ expG W1 and so ηk 6= 0 for all k. Also expG ηk ∈ H since hk, expG ξk ∈ H 0 0 and then we have expG ηk ∈ expG Wm ∩ H = {e} and so ηk = 0 This is a contradiction. Thus the first claim is established, i.e. we can find U open in g and V = expG U such that expG : U → V is a diffeomorphism and expG(U ∩ h) = V ∩ H. By proposition 7.7 we then have a manifold structure on H making it into a Lie group with Lie algebra h. Then i : H,→ G is a Lie subgroup. Q.E.D.

Corollary 7.11 H has Lie algebra h = {ξ ∈ g| expG(tξ) ∈ H∀t} The following is an example of how this is used Theorem 7.12 If F : G → H is a homomorphism of Lie groups then

ker(F ) = {g ∈ G|F (g) = eH } is a Lie subgroup of G with Lie algebra given by ker(F?)

−1 Proof ker(F ) = F ({eH }) and by continuity of F and Hausdorﬀ property of H and G we have this set is closed, and it is also a subgroup of G. Thus it is a Lie subgroup of G with Lie algebra

{ξ ∈ g| expG(tξ) ∈ ker(F )∀t} = {ξ ∈ g|F (expG(tξ)) = eH ∀t} but since F (expG(tξ)) = expH (tF?(ξ)) we have F?(ξ) = 0 as required. Q.E.D.

? Example 7.2 det : GL(n, R) → R and then ker det = SL(n, R) and det? = tr and so sl(n, R) = {ξ ∈ gl(n, R)|trξ = 0} Theorem 7.13 If σ : G → G is an automorphism, then the ﬁxed point set Gσ = {g ∈ G|σ(g) = g} is a Lie subgroup with Lie algebra gσ?

Example 7.3 1. G = GL(n, R) with σ(g) = (g−1)T is an automorphism and Gσ = O(n), and gσ? = {ξ ∈ T Mn×n(R)| − ξ = ξ} = An(R). −1 2. G = GL(n, C) with σ(g) = (g )T then Gσ = U(n). 3. R2n = Rn ⊕ Rn and G = GL(2n, R) and write 2n × 2n matrices as 2 × 2 matrices of n × n blocks. 0 −In 2 −1 Jn = and so Jn = −In and deﬁne σ1(g) = JngJn , which is an inner automorphism. Then In 0 U −V Gσ1 = { |U + iV ∈ GL(n, )} and so GL(n, ) is a Lie subgroup of GL(2n, ). If instead VU C C R −1 T −1 σ2 σ2(g) = Jn(g ) Jn then G = Sp(2n, R).

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8 Continuous homomorphisms are smooth

The basic idea is to look at continuous 1-PSG of Rn. These are linear σ(t) = tv with v = σ(1), and hence smooth. Lemma 8.1 If G is a Lie group and H is a subgroup then the topological closure H¯ is a subgroup.

Proof Take g, h ∈ H¯ and suppose gk → g and hk → h for sequences hk, gk ∈ H. Then

gh = m(g, h) = lim m(gk, hk) k→∞ and m(gk, hk) ∈ H and so gh ∈ H¯ . Continuity of the inverse gives i(H¯ ) ⊂ H¯ so H¯ is a subgroup. Q.E.D.

Lemma 8.2 If σ : R → G is a continuous 1-PSG then σ(R) is a connected closed abelian Lie subgroup of G.

Proof By the previous lemma, it is a closed subgroup, and hence a Lie subgroup. σ(R) is connected henve σ(R) is connected. σ(R) is abelian and so σ(R) is abelian. Q.E.D.

Corollary 8.3 If σ : R → G is a continuous 1-PSG then σ is a continuous 1-PSG of σ(R). Thus it is enough to prove that σ is smooth when G is a connected abelian Lie group.

Theorem 8.4 Any continuous 1-PSG of a Lie subgroup is smooth

Proof If G is connected and abelian then G is isomorphic to Rk × T n−k for some k and n = dim G. Since maps into products are continuous or smooth or homomorphism if and only if the components are continuous or smooth or homomorphisms we need only consider the cases where G = R and G = S1. Continuous homomorphisms R → R are smooth since they are linear. for R → S1 let σ : R → S1 ⊂ C be continuous and σ(s) + σ(t) = σ(s + t) as a complex valued function. Use the result from complex function theory so there exists a unique g : R → R continuous with σ(t) = eig(t) with g(0) = 0. We then have ei(g(s)+g(t)) = eig(s+t) and so g(s + t) − g(s) − g(t) ∈ 2πZ so we have a continuous map R × R → 2πZ and so the image must be connected which must contain the image of (0, 0) which is 0. Hence g(s + t) − g(s) − g(t) = 0 for all s, t. This implies g(t) = tg(1) so σ is smooth. Q.E.D.

Theorem 8.5 If G and H are Lie groups and F : G → H is a continuous homomorphism then F is smooth.

Proof Let expG(tξ) be a smooth 1-PSG of G. Then F (expG(tξ)) is a continuous 1-PSG of H and by theorem 8.4 is smooth. We claim that F is smooth on G if and only if it is smooth on an open set V around eH . If F is smooth on V and g ∈ G then gV is an open set around g. If g0 ∈ gV then g−1g0 ∈ V and

0 −1 0 −1 0 0 F (g ) = F (gg g ) = F (g)F (g g ) = LF (g) ◦ F ◦ Lg−1 (g ) which is a composition of smooth maps hence is smooth. Thus the claim is established. n Take a basis ξ1, ..., xın for g and consider α : R → G deﬁned by α(t1, ..., tn) = expG(t1ξ1)... expG(tnξn). Let n h = (h1, ..., hn) ∈ R and then d d α(0 + th)| = exp (th ξ )... exp (th ξ )| = h ξ + ... + h ξ dt t=0 dt G 1 1 G n n t=0 1 1 n n n and so d0α is the identiﬁcation of R with g given by a basis, therefore d0α is a linear isomorphism and hence α is n onto a neighbourhood V of eG in G. For (t1, ..., tn) ∈ U,

F (α(t1, ..., tn)) = F (expG(t1ξ1)... expG(tnξn)) = F (expG(t1ξ1))...F (expG(tnξn)) which is smooth in (t1, ..., tn) by theorem 8.4. Therefore F is smooth on V , and hence smooth everywhere. Q.E.D.

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9 Distributions and Frobenius Theorem

Deﬁnition 9.1 1.A k-dimensional tangent distribution D on a manifold M is a k-dimensional subspace Dx ⊂ TxM for each x ∈ M.

2. We say D is smooth if each point in M has a neighbourhood U with k-smooth vector ﬁelds X1, ..., Xk ∈ X (U) with (X1)x, ..., (Xk)x forming a basis at each x ∈ U for Dx.

3. A vector field X on M is said to belong to D if Xx ∈ Dx for all x ∈ M. 4. D is said to be involutive if the set of vector fields belonging to D is a Lie subalgebra of X (M). 5. The set of smooth vector fields belonging to D is denoted by Γ(D).

We can rewrite involutive to be [Γ(D), Γ(D)] ⊂ Γ(D).

n n n Example 9.1 1. M = R and put Dx = {X ∈ TxR |X(tk+1) = ... = X(tn) = 0} = {X ∈ TxR |X = Pk ∂ ∞ n ∂ ∞ n ∂ ai |x}. Then Γ(D) = C ( ) + ... + C ( ) and so D is smooth and involutive. i=1 ∂ti R ∂t1 R ∂tk

2. G a Lie group, h ⊂ g a k-dimensional linear subspace. Put Dg = deLg(h) ⊂ TgG. Then D is a k- ˜ ˜ dimensional distribution and if ξ1, ..., ξk is a basis for h then Dg = span(ξ1,g, ..., ξk,g) and so D is smooth. ˜ ˜ ˜ ˜ ˜ We have ξi ∈ Γ(D) and so D is involutive so [ξi, ξj] ∈ Γ(D) but [^ξi, ξj] = [ξi, ξj] and so [ξi, ξj] ∈ h so h is a Lie subalgebra of g. The converse is also true.

Hence left invariant involutive smooth distributions are in bijection with Lie subalgebras.

Deﬁnition 9.2 Let D be a smooth k-dimensional tangent distribution on M, then an integral submanifold for D is a k-dimensional submanifold (N , φ) with φ : N → M with N a k-dimensional connected manifold and φ an immersion of N into M such that dyφ(TyN ) = Dφ(y) for all y ∈ N . Note that we have a notion of maximal integral submanifold, analogous to maximal integral curves of a vector ﬁeld.

Proposition 9.3 If D is a smooth distribution on M such that for each x ∈ M there is an integral submanifold φ : N → M with x ∈ φ(N ) then D is involutive.

Proof Let X,Y ∈ Γ(D), x ∈ M, φ : N → M with y ∈ N and φ(y) = x and (, φ) an integral submanifold. Then ˜ ˜ ˜ ˜ ˜ ˜ there exists X, Y on N with dy0 φ(Xy0 ) = Xφ(y0) and dy0 φ(Yy0 ) = Yφ(y0) and so near y, X and (Y ) are φ-related ˜ ˜ to X and Y respectively. This implies [X, Y ] and [X,Y ] are φ-related and therefore [X,Y ]x = [X,Y ]φ(Y ) = dyφ([X,˜ Y˜ ]y) ∈ Dx. Since x is arbitrary, [X,Y ] ∈ Γ(D) Q.E.D.

Theorem 9.4 (Frobenius) Let D be a smooth involutive k-dimensional tangent distribution on a manifold M and then each x ∈ M lies on a locally unique integral submanifold.

Deﬁnition 9.5 A maximal integral submanifold (or leaf) (N , φ) of an involutive distribution D on a manifold M is a connected integral submanifold whose image is not a proper subset of any other integral submanifold.

Theorem 9.6 If G is a Lie group, h ⊂ g a Lie subalgebra then there is a Lie subgroup (H, φ) of G with deH φ(TeH H) = h and deH φ : TeH H → h an isomorphism of Lie algebras. Proof If G is a Lie group, h ⊂ g a Lie subalgebra, and D the left invariant extension of h then D is involutive so we let (H, φ) be the corresponding leaf of D with eG ∈ φ(H). We know φ : H → G is smooth, injective with injective differential. We need to show H is a group in a way which is smooth and making φ a Lie group homomorphism. First step is to show φ(H) is a subgroup of G. Take g ∈ φ(H). The left invariance of D means left translates of integral manifolds are integral manifolds. Consider (H,Lg−1 ◦ φ) which is a submanifold of G and an integral submanifold. Also e = Lg−1 (g) ∈ Lg−1 (φ(H)) so Lg−1 (φ(H)) = φ(H) hence φ(H) is a subgroup of G. We make H into a group by defining φ(h1h2) = φ(h1)φ(h2). We claim this makes H into a Lie subgroup. Pick a supplement m for h in g, so g = h ⊕ m and define Φ : H × m → G by

Φ(h, ξ) = φ(h) exp ξ

Then we have d d d Φ(X, η) = Φ(γ(t), 0 + tη)| = φ(γ(t)) exp(tη)| = d φ(x) + d L η (h,0) dt t=0 dt t=0 h e φ(h)

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and both maps have zero kernel so d(h,0)Φ has zero kernel as a linear map ThH ⊕ m → g and hence d(h,0)Φ is an isomorphism. Hence by inverse function theorem there is a neighbourhood U h of h in H, W h of 0 in m and V h of φ(h) in G with V h = Φ(U h × W h) and Φ : U h × W h → V h a diﬀeomorphism. −1 h h h h h h h Let Φ on V be given by maps ψ1 : V → U and ψ2 : V → W . These are smooth maps and h 0 −1 0 −1 0 0 ψ1 (φ(h ) exp 0) is the ﬁrst component of Φ (φ(h ) exp 0) = Φ (Φ(h , 0)) = (h , 0) and so

h 0 0 ψ1 (φ(h )) = h

0 h 0 0 for h ∈ U . Let h1, h2 ∈ H then for h1 near h1 and h2 near h2 then

0 0 0 0 0 0 h1h2 0 0 h1h2 0 0 h1h2 = ψ1 (φ(h1h2)) = ψ1 (φ(h1)φ(h2)) and this is a composition of smooth functions. Then

0 0 0 0 h1h2 0 0 mH (h1, h2) = ψ1 (mH (φ(h1), φ(h2)))

h h −1 h1h2 on (U × U ) ∩ (mG ◦ φ × φ) (U ). This is an open set and contains h1h2 so mH is smooth on H × H.

Similarly iH : H → H is smooth. Let φ(eH ) = e and so deH (TeH H) = h = De. Q.E.D. An application of this is that every Lie algebra (real, ﬁnite dimensional) is the Lie algebra of a Lie group, up to isomorphism. The method to show this uses representation theory.

Deﬁnition 9.7 A representation of a Lie algebra g is a homomorphism σ : g → gl(n, R) where n is called the dimension of the representation.

σ([ξ, η]) = σ(ξ)σ(η) − σ(η)σ(ξ)

Note gl(n, R) = Mn×n(R) ⊂ Mn×n(C). Deﬁnition 9.8 A representation is faithful if ker σ = {0}.

Thus for a faithful representation g is isomorphic to a Lie subalgebra of gl(n, R). Theorem 9.9 (Ado) Every ﬁnite dimensional real Lie algebra has a faithful representation.

A consequence of this is every ﬁnite dimensional Lie algebra g is a Lie subalgebra of gl(n, R) for some n. Hence there is a Lie subgroup φ : H → GL(n, R) whose Lie algebra h is isomorphic to g. Thus H is a Lie group whose Lie algebra is g. Lie subgroups of GL(n, R) are called linear Lie groups. For example all classical matrix groups are linear. Also note SL(2n, R) has the same Lie algebra as many non-linear Lie groups.

10 Lie Group topics

• Actions of Lie groups. G acts on M becomes a map σ : G × M → M by σ(g, x) = g · x is a smooth map. G is a symmetry group of M. For example rotations are the symmetry group of sphere. O(n + 1) acting on Sn is the symmetry group of spherical geometry. This is an example of a homogeneous geometry. Sn = O(n + 1)/O(n). • Integration on Lie groups • Representation theory • Internal symmetry groups.

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