Lie Group, Lie Algebra and their Representations Prof. Totaro ([email protected]) Typeset by Aaron Chan ([email protected]) Last update: June 8, 2010 Recommended Books: A. Kirillov - An introduction to Lie groups and Lie algebras J-P. Serre - Complex semisimple Lie algebra W. Fulton, J. Harris - Representation theory Kirillov is the closest to what we will cover, Fulton-Harris is longer but with lots of example, which provides a good way to understand representation theory. This course fit in especially well with Differential Geometry and Algebraic Topology. Definition 1 A Lie group is a group which is also a smooth manifold Example: (R,+) is a Lie group of dimension 1 S1 = {z ∈ C : |z| = 1} under multiplication Definition 2 The n-sphere Sn = {(x ,...,x )} ∈ Rn+1|x2+···+x2 = 1} is an n-manifold 0 n 0 n Many interesting Lie groups act on S2 Example: SO(3)=group of rotation in R3 (this is non-abelian) PGL(2,C) acts on S2 = C∪{∞} as Mobius transformation (cid:26)(cid:18) a 0 (cid:19)(cid:12) (cid:27) Here SO(3) ⊆ PGL(2,C) = GL(2,C)/ (cid:12)a ∈ C× 0 a (cid:12) z (cid:55)→ 2z, say, is in PGL(2,C) acting on S2 = C∪{∞} Examples of Lie groups • (Rn,+) any n ∈ N (or any finite dimensional real vector space) • R× = {x ∈ R|x (cid:54)= 0} under multiplication • C× = {x ∈ C|x (cid:54)= 0} under multiplication • GL(n,R) = {A ∈ M (R)|detA (cid:54)= 0} under mulitiplication n • GL(V) - General Linear group, where V is a finite dimensional vector space • SL(n,R) = {A ∈ M (R)|detA = 1} = {f : Rn → Rn linear and preserves volume} (Special n Linear group) • O(n) (Orthogonal group) • Sp(2n,R) (Sympletic group) • U(n) (Unitary group) • SU(n) (Special Unitary group) Remark: S0,S1,S3 are the only spheres that are also Lie groups 1 Orthogonal group O(n) = {A ∈ M (R)|AA(cid:62) = 1} n = {f : Rn → Rn|f linear and preserves distances } = {f : Rn → Rn|(cid:104)f(x),f(y)(cid:105) = (cid:104)x,y(cid:105) ∀x,y ∈ Rn} where the standard inner product on Rn is (cid:104)(x ,...,x ),(y ,...,y )(cid:105) = x y +x y +···+x y ∈ R 1 n 1 n 1 1 2 2 n n Elements of O(n) includes rotations and reflections Note that det is a homomorphism det : GL(n,R) → R× and this restricts to det : O(n) → {±1} since, A ∈ O(n) ⇒ 1 = det(1) = det(AA(cid:62)) = det(A)det(A(cid:62)) = det(A)2 Definition 3 Special orthogonal group SO(n) = {A ∈ M (R)|AA(cid:62) = 1,detA = 1} n Elements include rotations but not reflections (on Rn) SO(n) is a subgroup of index 2 in O(n). In fact, O(n) has 2 connected component, the one containing 1 is SO(n) Also note that SO(2)∼=S1 Symplectic group Sp(2n,R) = {f : R2n → R2n|w(x,y) = w(f(x),f(y))∀x,y ∈ R2n} where w is a non-degenerate alternating bilinear form on R2n: n w(cid:0)(q ,...,q ,p ,...,p ),(q(cid:48),...,q(cid:48),p(cid:48),...,p(cid:48) )(cid:1) = (cid:88)q p(cid:48) −p q(cid:48) 1 n 1 n 1 n 1 n i i i i i=1 for some choice of basis. Remark: Any non-degenerate alternating bilinear form w on Rn must have n even, and after a change of basis, such a form is given by above formula Example: Sp(2n,R) ⊂ SL(2n,R) Sp(2,R) = SL(2,R) = the group of area preserving linear maps Unitary group U(n) = {f : Cn → Cn linear and preserves distance} (= GL(n,C)∩O(2n)) Definition 4 The standard inner product on Cn is the nondegenerate positive definite Hermitian form n (cid:88) (cid:104)(z ,...,z ),(w ,...,w )(cid:105) = z w 1 n 1 n 1 1 i=1 Note that the length of a vector z ∈ Cn(= R2n) is (cid:107)z(cid:107) = (cid:112)(cid:104)z,z(cid:105) (as z = x y,zz = |z|2 = x2+y2) i 2 So, we have U(n) = {f : Cn → Cn linear |(cid:104)f(x),f(y)(cid:105) = (cid:104)x,y(cid:105) ∀x,y ∈ Cn} = {A ∈ GL(n,C)|AA∗ = 1} (A∗ = A(cid:62)) Special Unitary group The det of a unitary matrix gives a homomorphism det : U(n) → S1 ⊂ C× SU(n) = kerdet| U(n) Example U(1) = {z ∈ M (C)|zz = 1} = S1 1 (cid:26)(cid:18) (cid:19)(cid:12) (cid:27) SU(2) = a b (cid:12)(cid:12)a,b ∈ C,|a|2+|b|2 = 1 −b a (cid:12) Wee see that SU(2) is diffeomorphic to S3 = {(x ,x ,x ,x )|x2+x2+x2+x2 = 1} 0 1 2 3 0 1 2 3 Remark: S0,S1,S3 are the only sphere that are also Lie groups Some Basics of Smooth Manifold Definition 5 A subset M ⊆ Rn is called k-dimensional manifold (in Rn) if for every point x ∈ M, the following condition is satisfied: (M) There is an oen set U (cid:51) x, and open set V ⊂ Rn, and a diffeomorphism h : U → V s.t. h(U ∩M) = V ∩(Rk×{0}) = {x ∈ V|x = ···x = 0} k+1 n Definition 6 Let U ⊆ Rn,n > 0 be an open set, a smooth function f : U → R (or C∞) if all partial derivatives ∂r f (r ≥ 0) ∂ ···∂ xi1 xir are defined and continuous on U For U ⊆ Rm open, a smooth mapping f : U → Rn is a function s.t. f = (f ,...,f ), with f : U → R 1 m i smooth function For U,V ⊆ Rn, a diffeomorphism f : U → V of degree n is a smooth map (on Rn) with a smooth inverse The derivative df| of a smooth map f : U(⊆ Rm) → Rn at a point x ∈ U is a linear map Rm → Rn x given by matrix (∂fi) ∂xj ∂f1 ··· ∂f1 u  v  ∂x1 ∂xm 1 1  ... ... ...  ...  =  ...  ∂fn ··· ∂f1 um vn ∂x1 ∂x1 composite of the smooth maps is smooth, and d(g◦f)| = dg| ◦df| x f(x) x Theorem 7 (Inverse Function Theorem) Let U ⊂ Rn open, f : U → Rn be a smooth map. Suppose df| is an isomorphism, for some x ∈ U x Then ∃V ⊂ U, V (cid:51) x, s.t. f(V) is open and f is a diffeomorphism from V to f(V) Theorem 8 (Implicit Function Theorem) Let U ⊆ Rm open, f : U → Rn smooth map. Suppose df| is surjective at a point x ∈ U (n ≤ m) x Then ∃V ⊆ U,V (cid:51) x and diffeomorphism φ : W → V (W ⊆ Rn open), s.t. f(φ(x ,...,x )) = (x ,...,x ) 1 n 1 n 3 Definition 9 A submersion is a smooth map which has derivatives being surjective everywhere A smooth submanifold X ⊆ RN of dimension n is a subset s.t. ∀x ∈ X,∃ nbhd U (cid:51) x, U ⊆ RN and a submersion F : U → RN−n s.t. X ∩U = F−1(0) ⊆ U Example: Claim: The sphere Sn ⊆ Rn+1 is a smooth n-dimensional submanifold Proof Sn = F−1(0), where F : Rn+1 → R (x ,...,x ) (cid:55)→ x2+···+x2 −1 0 n 0 n We have to check that F is a submersion at points x ∈ Sn: dF = (2x ,2x ,...,2x ) (row matrix) 0 1 n This is surjective whenever (x ,...,x ) (cid:54)= (0,...,0) ⇒ surjective everywhere on Sn 0 n Example: X = {(x,y) ∈ R2|xy = a} is a smooth 1-dimensional submanifold for nonzero a ∈ R, but not for a = 0, here we have: F : R2 → R (x,y) (cid:55)→ xy−a Example: X = {x ∈ R|x2 = 0} is an 0-dimensional submanifold of R, but x2 : R → R is not a submersion at 0. To prove that X is a 0-dimensional submanifold, you have to notice that X = {x ∈ R|x = 0} Definition 10 The tangent space to a smooth n-dimensional submanifold X ⊆ RN at a point x ∈ X (if we describe X as X = F−1(V) for some submersion F : V → RN−n) is defined as: T X = ker(dF| : RN → RN−n) x x This is an n-dimensional linear subspace of RN Let X ⊂ RN be a smooth n-dimensional submanifold. A function f : X → R is smooth ⇔ near each point x ∈ X, f is the restriction of a smooth function on an open nbhd of X in RN (0-dimensional submanifold of RN = discrete subset) For submanifold X ⊆ RM, Y ⊆ RN (dim=m,n resp.) a smooth map f : X → Y has derivative df| : T X → T Y which is a linear map x x f(x) A diffeomorphism between 2 submanifolds is a smooth map with smooth inverse. Fact: (Hausdorff countable basis) Every smooth manifold is diffeomorphic to a submanifold of RN For submanifold X ⊆ RM,Y ⊆ RN (dim=m,n resp.), the product X ×Y ⊆ RM ×RN = RM+N is a smooth submanifold. It has the product topology Lie Group Definition 11 A Lie group G is a smooth manifold which is also a group s.t. multiplication : G×G → G (g,h) (cid:55)→ gh inverse : G → G g (cid:55)→ g−1 are smooth maps. We have a point 1 ∈ G (the identity) 4 Note that a Lie group need not be connected. (0-dimensional submanifold of RN=discrete subset) In particular, we can view any group (say countable) as a 0-dimensional Lie group. Lemma 12 Let G be a Lie group, G0 be the connected component of G containing 1. Then G0(cid:69)G and G/G0 is discrete (with the quotient topology) Proof multiplication : G×G → G is continuous, so it maps connected space G0×G0 onto connected subset of G, which contains 1. ⇒ G0×G0 (cid:16) G0 Likewise, inverse : G0 (cid:16) g0. Therefore, G0 ≤ G C : G → G To show G0(cid:69)G, need to show ∀g ∈ G the map g sends G0 to G0 x (cid:55)→ gxg−1 Have C smooth⇒continuous, and 1(cid:55)→1 g ⇒ C : G0 (cid:16) G0 g ⇒ G0(cid:69)G L : G → G We have, ∀g ∈ G a diffeomorphism g x (cid:55)→ gx (Can check that L is an inverse map, using that G is associative) g−1 Therefore, L (G0) = gG0 is the connected component of G containing g. g We know that G is the disjoint union of some of these left cosets gG and G/G0 is the set of cosets. To show that G/G0 has discrete topology. I have to show that each component gG0 is open in G. In fact, all connected component in any manifold are open subsets Lemma 13 Let G be a connected Lie group, Then G is generated by a neighbourhood of 1 ∈ G Proof Let N be an neighbourhood of 1 ∈ G Let H ≤ G, generated by N ⇒ H open in G because ∀h ∈ H hN ⊆ H and hN is an open subset of G containing h In fact, H is also closed in G ifx ∈ G−H ⇒ xN ⊆ G−H (If xn = h ∈ H for some n ∈ N, then x = hn−1 #) So H is open and closed and contains 1 ⇒ H = G since G is connected ref.: Armstrong, Basic Topology Definition 14 A homomorphism f : G → H of Lie groups is a group homomorphism which is also smooth Lemma 15 Let f : G → H be a homomorphism of connected Lie groups. Suppose that df| : T G (cid:16) T H (1) 1 1 1 Then f : G (cid:16) H Proof By the Implicit Function Theorem, f maps some neighbourhood of 1 ∈ G onto some neighbourhood of 1 ∈ H, so f(G) contains the subgroup of H generated by this neighbourhood which is all of H because H is connected Example: f : R → S1 ⊆ C× i (cid:55)→ eit is a homomorphism of Lie groups (It’s smooth, and it’s a hom. because 5 f(s+t) = f(s)f(t)) Its derivative at 1 is (cid:12) d(eit)(cid:12) (cid:12) = ieit| = i (2) dt (cid:12) t=0 (cid:12) t=0 which is an isomorphism R = T R ∼= T S1 = iR ⊂ C 0 1 So lemma applies and indeed R (cid:16) S1 In fact, S1 ∼= R/Z where 2πZ = Z = kerf Definition 16 A closed Lie subgroup H of a Lie group G is a closed submanifold of G which is a subgroup of G Note that such a subgroup H is a Lie group. Indeed, multi: H ×H → H is just the restriction of multi: G×G → G so it is also smooth, likewise for inverses. Use this to prove that the classical groups actually are Lie groups Example: GL(n,R). This is an open subset of M R = Rn2 so it is a smooth n2-dimensional manifold. Multi- n plication of matrices is smooth (in fact, polynomial or mapping smooth)    a11 ··· a1n b11 ··· b1n (cid:32)a b +a b +... ···(cid:33)  ... ... ...  ... ... ...  = 11 11 ...12 21 ... a ··· a b ··· b n1 nn n1 nn is a smooth function. Inverse is a polynomial in entries of given matrix A and in 1/detA which is a smooth function of GL(n,R) = {A | detA (cid:54)= 0}. For example, (cid:18) (cid:19)−1 (cid:18) (cid:19) a b 1 d −b = c d ad−bc −c a SL(n,R) = {A ∈ GL(nR) | detA = 1} This is a closed Lie subgroup of GL(n,R). Clearly it is a closed subgroup To show: SL(n,R) is a smooth submanifold of dimension n2−1. It suffices to check that SL(n,R) is a smooth submanifold near 1 ∈ G(n,R) using left translation (see notes for pictures) It suffices to show that det : GL(n,R) → R× is a submersion near 1. Todothis, weseehowdetchangesasyoumovefrom1 ∈ GL(n,R). SolookatA = 1+(cid:15)B, B ∈ M R. n We solve the equation detA = 1(mod (cid:15)2) We compute: (cid:32) (cid:32) (cid:33)(cid:33) b ··· 11 det(1+(cid:15)B) = det 1+(cid:15) ... ... = (1+(cid:15)b )···(1+(cid:15)b ) (mod (cid:15)2) 11 nn = 1+(cid:15)(b +···+b ) (mod (cid:15)2) 11 nn ⇒ ker(d(det)| ) = {B ∈ M R | tr(B) = 0} 1 n This is a codimension 1 subspace of M R so det is a submersion at 1 ∈ GL(n,R), so SL(n,R) is a n closed Lie subgroup , and sl(n) = T SL(n,R) = {B ∈ M R | tr(B) = 0} 1 n gl(n) = M R = T GL(n,R) n 1 6 Example: Orthogonal group O(n). Again this is a closed subgroup of GL(n,R). To show that it is a smooth submanifold it suffice to check that near 1 ∈ GL(n,R). So we differentiate these equations for O(n) ⊂ GL(n,R) So, for B ∈ gl(n) we compute where is : (1+(cid:15)B)(1+(cid:15)B)t = 1 (mod (cid:15)2) (3) (1+(cid:15)B)(1+(cid:15)B)t = 1+(cid:15)(B+Bt) (mod (cid:15)2) (4) F : GL(n,R) → Rn2 We have O(n) = F−1(1) for some smooth mapping and we have computing ker(dF) = {B ∈ gl(n) | B+Bt = 0} n(n−1) ⇒ dimR(ker(dF)) = dim(zero diagonal matrix) = 2 SowewouldliketosaythatO(n)isthefibreofasmoothmapGL(n,R) → Rn2−(n(n−1)/2) = Rn(n+1)/2 Can we define O(n) using only n(n+1)/2 equations? Yes, since ∀A ∈ GL(n,R),AAt is symmetric So AAt = 1 reduces to n(n+1)/2 equations. SoO(n)isasmoothsubmanifoldofdimensionn(n−1)/2inGL(n,R)andhenceaclosedLiesubgroup. Also so(n) = T O(n) = {B ∈ gl(n)|Bt = −B} 1 Example: Unitary group U(n) ⊂ GL(n,C). We just show that it is a smooth (real) submanifold of GL(n,C) near 1 Differentiate the equation for U(n) ⊂ GL(n,C) at 1: Write A = 1+(cid:15)B, B ∈ gl(n,C) = M C n Solve (1+(cid:15)B)(1+(cid:15)B)∗ = 1 (mod (cid:15)2) (1+(cid:15)B)(1+(cid:15)B)∗ = 1+(cid:15)(B+B∗) (mod (cid:15)2) So U(n) = F−1(1) ⊂ GL(n,C) where ker(dF| ) = {B ∈ gl(n,C) | B∗ = −B} = {skew hermitian matrices} = i{hermitian matrices} and 1 gl(n,C) = {hermitian}+{skew-hermitian} (cid:18) (cid:19) ia z Skew-hermitian matrix is a,b ∈ R,z ∈ C −z¯ ib So dimR(ker(dF|1)) = (1/2)dimRgl(n,C) = n2 So I would like to define U(n) ⊂ GL(n,C) by exactly 2n2−n2 = n2 real equations. Indeed, for any A ∈ GL(n,C), AA∗ is always hermitian (since (AB)∗ = B∗A∗). So AA∗ = 1 reduces to only n2 real equations (say that the element of AA∗ above diagonal are zeroes and the elements on diagonal, which are real =1) So U(n) =fibre of submersion GL(n,C) → Rn2 so it is a closed Lie subgroup of GL(n,C) Definition 17 For A ∈ M (K), where K = R or C, define the exponential of A by: n (cid:88)∞ An exp(A) = ∈ M K (5) n n! n=0 To check that this series converges, define the norm: (cid:107)A(cid:107) := sup (cid:107)Ax(cid:107) (6) (cid:107)x(cid:107)=1,x∈Rn 7 Clearly (cid:107)AB(cid:107) ≤ (cid:107)A(cid:107).(cid:107)B(cid:107) ⇒ ∀A ∈ M (K), (cid:107)An(cid:107) ≤ (cid:107)A(cid:107)n and this series converges in R ∀(cid:107)A(cid:107). So the series of matrices n n! n! converges absolutely. Easy that exp : M R → M R is smooth and exp : M C → M C is complex analytic. n n n n Also, for (cid:107)A(cid:107) < 1 define the logarithm (cid:88)∞ An log(1+A) = (−1)n+1 (7) n n=1 This converges for (cid:107)A(cid:107) < 1. ⇒ log is defined on the open ball of radius 1 around 1 ∈ M (K) n Theorem 18 (1) For x in some neighbourhood of 0 ∈ M K, log(exp(x)) = x. n For X with (cid:107)X −1(cid:107) < 1, exp(log(X)) = X (2) exp(x) = 1+x+···. That is exp(0) = 1 and dexp| = id 0 MnK (3) If xy = yx in M K, then exp(x+y) = exp(x)exp(y) In particular, exp(x)exp(−x) = 1 for any n x ∈ M K. So exp(x) ∈ GL(n,K) n (4) Forafixedx ∈ M K,defineasmoothmapR → GL(n,K)byt (cid:55)→ exp(tx). Thenexp((s+t)x) = n exp(sx)exp(tx) ∀s,t ∈ K. In other wrods, t (cid:55)→ exp(tx) is a homomorphism of Lie groups. (5) The exponential map commutes with conjugation and transpose. That is exp(A × A−1) = Aexp(x)A−1 and exp(x)t = exp(xt) Proof (1) follows from the fact that log(exp(x)) = x for x ∈ R, so that is true as an identity of formal power series. So it works for a matrix X (2) - (3) Try to compute exp(x)exp(y) for any x,y ∈ M (K) n exp(x)exp(y) = (1+x+x2/2+···)(1+y+y2/2+···) (8) = 1+(x+y)+(x2/2+xy+y2/2)+··· (9) and exp(x+y) = 1+(x+y)+(x+y)2/2+··· (10) = 1+(x+y)+(x2+xy+yx+y2)/2+··· (11) If yx = yx, then exp(x + y) = exp(x)exp(y) is an identity of power series in commmuting variables, say because it’s true for x,y ∈ R (4) follows from (3) because for any x ∈ M K, and any s,t ∈ K sx and tx commute. So exp(sx+ n tx) = exp(sx)exp(tx) (5) These follow from the power series for exp, using that (AxA−1)n = AxnA−1, and likelwise (xt)n = (xn)t Definition 19 A one-parameter subgroup of a Lie group G is a homomorphism R → G of Lie groups The theorem gives, for any x ∈ gl(n,R), a one-parameter subbgroup of GL(n,R), R → GL(n,R) with tangent vector at 0 is x ∈ gl(n,R) = T GL(n,R) 1 8 Theorem 20 For every classical group G ⊆ GL(n,K) (to be listed), G is a closed Lie subgroup of GL(n,K). In fact, if we let g = T G, then exp gives diffeomorphism, for some neighbourhood U of 1 in GL(n,K) 1 and u of 0 in gl(n,K), U ∩G (cid:29)log(cid:29) u∩g exp The classical groups: (1) Compact (real) groups: SO(n),U(n),SU(n),Sp(n) (2) GL(n,K), SL(n,K), SO(n,K), O(n,K); for K = R or C (3) Real Lie group: Sp(2n,R) (4) Complex Lie gorup: Sp(2n,C) Example: O(n,C) = subgroup of GL(n,C) preserving the symmetric C-bilinear form: (cid:88) < (z ,...,z ),(w ,...,w ) >= z w 1 n 1 n i i Sp(2n,C) = subgroup of GL(2n,C) preserving the standard C-symplectic (i.e. alternating nondegen- erate) form: w((z ,...,z ),(w ,...,w )) = (z w −z w )+(z w −z w )+··· 1 2n 1 2n 1 n+1 n+1 1 2 n+2 n+2 2 Compact symplectic group Sp(n):=subgroup of GL(n,H) preserving distance on Hn = R4n. Here the quaternions H = R1⊕Ri⊕Rj ⊕Rk determined by i2 = k2 = j2 = −1 and ij = k (etc.). Say we define an H-vector space V to be a right H-module, va ∈ V for a ∈ H. For example Hn = {(z ,...,z )(cid:62) | z ∈ H} is an H-vector space. 1 n i GL(n,H) := {invertible H-linear maps Hn → Hm} ⊆ M (H) n Warning: det only defined for matrices uses a commutative ring. Why are O(n),U(n),Sp(n) compact? O(n) = { matrix with column i = A(e ) | A(e ),...,A(e ) ∈ Rn orthonormal} ⊆ M R = Rn2 i 1 n n is a closed bounded subset and hence compact U(n) = GL(n,C)∩O(2n) which is closed subset of O(2n) hence compact. Sp(n) = GL(n,H)∩O(4n) ⊂ GL(4n,R) a closed subset of O(4n), so Sp(n) is compact Proof of Theorem 20 in a few cases: SL(n,R): Claim that: for x ∈ gl(n,R), near 0, exp(x) ∈ SL(n,R) ⇔ x ∈ sl(n,R) := {x ∈ gl(n) | tr(x) = 0}. Use Jordan canonical form: For any x ∈ M C, x is conjugate (over C) to an upper-triangular matrix. n  ea1 ∗  So exp(x) is conjugate (over C) to  ... .   0 ean In particular, detexp(x) = ea1···ean (12) = ea1+···an (13) = exp(tr(x)) (14) 9 So, exp(x) ∈ SL(n,R) ⇔ detexp(x) = 1 (15) ⇔ exp(tr(x)) = 1 ⇔ tr(x) ∈ 2πiZ (16) For x near 0, this happens ⇔tr(x) = 0 Definition 21 vector field V on a smooth manifold M assigns to every point p ∈ M a tangent vector v ∈ T M s.t. p p in any coordinate chart, it has the form n (cid:88) ∂ v = f (p) (17) i ∂x i i=1 where f ,...,f are smooth functions M → R (see picture) 1 n Here we write ∂ ,..., ∂ for the standard basis to T Rn for every p ∈ Rn ∂x1 ∂xn p Two ways to think of tangent vectors at p ∈ M: (1) A smooth curve c : R → M has a tangent vector c(cid:48)(t) ∈ T M c(t) (2) Differentiate a smooth function F on M in the direction of tangent vector X ∈ T M at point p p (one definition: pick a curve c with c(cid:48)(0) = X and then define X(f) = d| f(c(t))) dt t=0 We can identify T M with the space of “derivation at p”, X : C∞(M) → R, R-linear, s.t. X(fg) = p f(p)X(g)+X(f)g(p) ∈ R In particular, in some coordinates, ∂ (cid:12)(cid:12) ,... ∂ (cid:12)(cid:12) are derivation at p ∂x1 p ∂xn p Theorem 22 (Existence and Uniqueness for ODEs) Let M be a smooth manifold, X a vector field on M, p ∈ M. Then ∀a < 0,b > 0, ∃at most one curve c : (a,b) → M s.t. c(0) = p and c(cid:48)(t) = X ∈ T M c(t) c(t) Also, c(t) exists on some open interval around 0, the maximal interval might or might not be R. If M is compact then c(t) is defined ∀t ∈ R Theorem 23 Let G be a Lie group, x ∈ T G. Then ∃! one parameter subgroup f : R → G s.t. f(cid:48)(0) = x 1 Proof (see picture) Suppose we have such a f. We know that ∀t,t ∈ R, f(t+t ) = f(f )f(t) ∈ G 0 0 0 For t ∈ R, and think if t near 0. Then f(t+t ) = L f(t) ∈ G 0 0 f(t0) Differentiate this w.r.t. t at t = 0 gives: f(cid:48)(t ) = dL (x) ∈ T G, since f(cid:48)(0) = x ∈ T G so define a left-invariant vector field X on G by: 0 f(t0) f(x0) 1 ∀g ∈ G, take the tangent vector X := (dL )(x) ∈ T G g g g So f(t) must be the unique solution to the ODE: f(0) = 1 ∈ G and f(cid:48)(t) = X ∈ T G ∀t ∈ f(t) f(t) (a,b) ⊆ R One checks that a solution to the ODE is a one-parameter subgroup. Suppose we have defined f : [0,T] → G with f(s+t) = f(s)f(t) for s,t,s+t ∈ [0,T]. Then we can define f on [T,2T] by f(T +t) = f(T)f(t) for t ∈ [0,T]. (see picture) Repeat process. Definition 24 Let G be a Lie group. Then the exponential map exp : g → G (where g = T G) is defined by 1 exp(x) = f(1) (18) where f : R → G is the unique one-parameter subgroup with f(cid:48)(0) = x ∈ g (This is smooth, by theorems on ODEs) 10