LECTURE 13: LIE SUBGROUPS AND LIE SUBALGEBRAS 1. The Exponential Map Let G be any Lie group, g its Lie algebra. In PSet 4 you will prove that any left invariant X vector field X 2 g is complete. Let φt : G ! G be the flow generated by X. Definition 1.1. The exponential map of G is the map X exp : g ! G; X 7! φ1 (e): X tX tX X It is not hard to see φts = φs . So exp(tX) = φ1 (e) = φt (e): ∗ Example. For G = R , we can identify T1G = R. For any x 2 R = T1G, the left invariant vector d d field corresponds to x = x dt 2 T1G at a 2 G is X(a) = ax dt . The integral curve of X starting at tx X x e = 1 is γe(t) = e . So we get follows exp(x) = φ1 (e) = γe(1) = e . Similarly one can show, for G = (S1; ·), 1 1 ix exp : iR = TeS ! S ; exp(ix) = e and for G = (Rn; +), n n n exp : R = T0R ! R ; exp(x) = x and for G = GL(n; R), A2 A3 exp : gl(n; ) ! GL(n; ); exp(A) = eA = I + A + + + ··· : R R 2! 3! Lemma 1.2. exp : g ! G is a smooth map, and if we identify T0g with g, d exp0 = Id: Proof. Consider the map Φ:e R × G × g ! G × g; (t; g; X) 7! (g · exp(tX);X): One can check that this is the flow on G × g corresponding to the left invariant vector field (X; 0) on G × g, thus it is smooth. It follows that exp is smooth, since it is the composition Φe π1 g ,! R × G × g −! G × g −! G: X X d Since exp(tX) = φt (e) = γe (t), we get dt jt=0 exp(tX) = X. On the other hand, d d(Xt) exp ◦tX = (d exp)0 = (d exp)0X: dt t=0 dt We conclude that (d exp)0 equals to the identity map. Since (d exp)0 is bijective, we have 1 2 LECTURE 13: LIE SUBGROUPS AND LIE SUBALGEBRAS Corollary 1.3. exp is a local diffeomorphism near 0, i.e. it is a diffeomorphism from a neighbor- hood of 0 2 TeG to a neighborhood of e 2 G. As another consequence, we can prove Proposition 1.4. For any X; Y 2 g, there is a smooth map Z :(−"; ") ! g so that for all t 2 (−"; "), exp(tX) exp(tY ) = exp(t(X + Y ) + t2Z(t)): Proof. Since exp is a diffeomorphism near 0 2 g, there is an " > 0 so that the map ' :(−"; ") ! g; t 7! '(t) = exp−1(exp(tX) exp(tY )) is smooth. Note that we can write ' as the composition X Y −1 γe ×γe µ exp R −! G × G −! G −! R: According to lemma 1.2 in lecture 12, dµe;e(X; Y ) = X + Y . It follows 0 −1 X Y ' (0) = (d exp0) (_γe (0) +γ _ e (0)) = X + Y: Since '(0) = 0, by Taylor's theorem, '(t) = t(X + Y ) + t2Z(t) for some smooth function Z. Remark. The mysterious function Z is explicitly given by the Baker-Campbell-Hausdorff formula: 1 t t2 Z(t) = [X; Y ] + ([X; [X; Y ]] − [Y; [X; Y ]]) + [X; [Y; [X; Y ]]] + ··· 2 12 24 2. Lie subgroups v.s. Lie subalgebras Recall that a Lie subgroup H of G is an immersed submanifold which is also a Lie group with respect to the induced group operation. Note that we don't require H to be a smooth submanifold. Suppose H is a Lie subgroup of G, and h be the Lie algebra of H. Since dιH : h ! g is injective, we can think of h as a Lie subalgebra of g. Note that any integral curve of a left-invariant vector field in H is automatically an integral curve of G (with initial vector in TeH), so the exponential map expH : h ! H is exactly the restriction of expG : g ! G onto h. Theorem 2.1. Suppose H is a Lie subgroup of G. Then as a Lie subalgebra of g, h = fX 2 g j expG(tX) 2 H for all t 2 Rg: Example. (The special linear group) The special linear group is defined as SL(n; R) = fX 2 GL(n; R) : det X = 1g: It is a Lie subgroup of GL(n; R). To determine its Lie algebra sl(n; R), we notice that det eA = eTr(A): So for an n × n matrix A, eA 2 SL(n; R) if and only if Tr(A) = 0. We conclude sl(n; R) = fA 2 gl(n; R) j Tr(A) = 0g: LECTURE 13: LIE SUBGROUPS AND LIE SUBALGEBRAS 3 Example. (The orthogonal group) Next let's consider the orthogonal group T O(n) = fX 2 GL(n; R): X X = Ing: n(n+1) This is a compact 2 dimensional Lie subgroup of GL(n; R). To figure out its Lie algebra o(n), we note that (eA)T = eAT , so tA T tA tAT −tA (e ) e = In () e = e : Since exp is locally bijective, we conclude that A 2 o(n) if and only if AT = −A. So T o(n) = fA 2 gl(n; R) j A + A = 0g; which is the space of n × n skew-symmetric matrices. Notice that O(n) is not connected. It consists of two connected components, and the connected component of identity is the called the special orthogonal groups T SO(n) = fX 2 GL(n; R): X X = In; det X = 1g = O(n) \ SL(n; R): Its Lie algebra so(n) is the same as o(n). Proof of theorem 2.1. If X 2 h, then by the argument above, for any t 2 R, expG(tX) = expH (tX) 2 H: Conversely suppose X 62 h. Consider the map ' : R × h ! G; (t; Y ) 7! exp(tX) exp(Y ): Since d exp0 is the identity map, d'0;0(t; Y ) = tX + Y: Since X 62 h, d'0;0 is injective. It follows that there exists a small " > 0 and a neighborhood U of 0 in h such that ' maps (−"; ") × U injectively into G. Shrinking U if necessary, we may assume that expH maps U diffeomorphically onto a neighborhood U of e in H. Choose a smaller −1 neighborhood U0 of e in H such that U0 U0 ⊂ U. We pick a countable collection fhj j j 2 Ng ⊂ H such that hjU0 cover H. (This is always possible since H is the union of countable many compact sets.) For each j denote Tj = ft 2 R j exp(tX) 2 hjU0g. We claim that Tj is a countable set. In fact, if jt − sj < " and t; s 2 Tj, then exp(t − s)X = exp(−sX) exp(tX) 2 U: So exp(t − s)X = exp(Y ) for a unique Y 2 U. It follows '(t − s; 0) = '(0;Y ). Since ' is injective, we conclude Y = 0 and t = s. Now each Tj is a countable set. So one can find t 2 R such that t 62 Tj for all j. It follows that exp(tX) 62 [jhjU0 = H. So X 62 fX 2 g j expG(tX) 2 H for all t 2 Rg: This completes the proof. Conversely, any Lie subalgebra gives rise to some Lie subgroup: 4 LECTURE 13: LIE SUBGROUPS AND LIE SUBALGEBRAS Theorem 2.2. If h is a Lie subalgebra of g, then there is a unique connected Lie subgroup H of G with Lie algebra h. 0 Proof. Let X1; ··· ;Xk be a basis of h ⊂ g. Since Xis are left invariant vector fields on G, linearly independent at e, they are linearly independent at all g 2 G. In other words, Vg = spanfX1(g); ··· ;Xk(g)g gives us a k-dimensional distribution on G. Since [Xi;Xj] 2 h for all i; j, V is integrable. By Frobenius theorem, there is a unique maximal connected integral manifold of V through e. Denote this by H. To show that H is a subgroup, note that V is a left invariant distribution. So the left translation of any integral manifold is an integral manifold. Now suppose h1; h2 2 H. Since h1 = Lh1 e 2 H \ Lh1 H, and since H is maximal, we have Lh1 H ⊂ H. So in particular h1h2 = Lh1 h2 2 H. −1 Similarly, h 2 H since L −1 (h1) = e 2 H implies L −1 H ⊂ H. It follows that H is a subgroup 1 h1 h1 of G. Since the group operations on H are the restriction of group operations on G, they are smooth. So H is a Lie group. For uniqueness, let K be another connected Lie subgroup of G with Lie algebra h. Then K is also an integral manifold of V. So we have K ⊂ H. Since TeK = TeH the inclusion has to be a local diffeomorphism. In other words, K coincide with H near e. Since any connected Lie group is generated by any open set containing e, we conclude that K = H. Example. Consider G = T2 = S1 × S1. We know g = R2 endowed with trivial Lie bracket. The Lie subalgebras of g are one-dimensional subspaces of g. Obviously, any one-dimensional subspace of g with \rational angle" corresponds to a Lie subgroup p;q ipt iqt H := f(e ; e ) j t 2 Rg for some co-prime pair of integers (p; q). In particular, Then S1 × f0g and f0g × S1 are Lie subgroups.