Lecture 13: Lie Subgroups and Lie Subalgebras

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 ∈ 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 ∈ R = T1G, the left invariant vector d d ﬁeld corresponds to x = x dt ∈ T1G at a ∈ 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 |t=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 neighborhood of 0 ∈ TeG to a neighborhood of e ∈ G. As another consequence, we can prove Proposition 1.4. For any X,Y ∈ g, there is a smooth map Z :(−ε, ε) → g so that for all t ∈ (−ε, ε), exp(tX) exp(tY ) = exp(t(X + Y ) + t2Z(t)). Proof. Since exp is a diffeomorphism near 0 ∈ g, there is an ε > 0 so that the map ϕ :(−ε, ε) → g, t 7→ ϕ(t) = exp−1(exp(tX) exp(tY )) is smooth. Note that we can write ϕ 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 ϕ (0) = (d exp0) (γ ˙ e (0) +γ ˙ e (0)) = X + Y. Since ϕ(0) = 0, by Taylor’s theorem, ϕ(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 = {X ∈ g | expG(tX) ∈ H for all t ∈ R}. Example. (The special linear group) The special linear group is deﬁned as SL(n, R) = {X ∈ GL(n, R) : det X = 1}. 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 ∈ SL(n, R) if and only if Tr(A) = 0. We conclude sl(n, R) = {A ∈ gl(n, R) | Tr(A) = 0}. LECTURE 13: LIE SUBGROUPS AND LIE SUBALGEBRAS 3

Example. (The orthogonal group) Next let’s consider the orthogonal group T O(n) = {X ∈ GL(n, R): X X = In}. 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 ∈ o(n) if and only if AT = −A. So T o(n) = {A ∈ gl(n, R) | A + A = 0}, 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) = {X ∈ GL(n, R): X X = In, det X = 1} = O(n) ∩ SL(n, R). Its Lie algebra so(n) is the same as o(n).

Proof of theorem 2.1. If X ∈ h, then by the argument above, for any t ∈ R,

expG(tX) = expH (tX) ∈ H. Conversely suppose X 6∈ h. Consider the map ϕ : R × h → G, (t, Y ) 7→ exp(tX) exp(Y ).

Since d exp0 is the identity map, dϕ0,0(t, Y ) = tX + Y.

Since X 6∈ h, dϕ0,0 is injective. It follows that there exists a small ε > 0 and a neighborhood U of 0 in h such that ϕ maps (−ε, ε) × 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 {hj | j ∈ N} ⊂ 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 = {t ∈ R | exp(tX) ∈ hjU0}. We claim that Tj is a countable set. In fact, if |t − s| < ε and t, s ∈ Tj, then exp(t − s)X = exp(−sX) exp(tX) ∈ U. So exp(t − s)X = exp(Y ) for a unique Y ∈ U. It follows ϕ(t − s, 0) = ϕ(0,Y ). Since ϕ is injective, we conclude Y = 0 and t = s.

Now each Tj is a countable set. So one can ﬁnd t ∈ R such that t 6∈ Tj for all j. It follows that exp(tX) 6∈ ∪jhjU0 = H. So

X 6∈ {X ∈ g | expG(tX) ∈ H for all t ∈ R}. 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 ∈ G. In other words,

Vg = span{X1(g), ··· ,Xk(g)} gives us a k-dimensional distribution on G. Since [Xi,Xj] ∈ 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 ∈ H. Since h1 = Lh1 e ∈

H ∩ Lh1 H, and since H is maximal, we have Lh1 H ⊂ H. So in particular h1h2 = Lh1 h2 ∈ H. −1 Similarly, h ∈ H since L −1 (h1) = e ∈ 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 := {(e , e ) | t ∈ R} for some co-prime pair of integers (p, q). In particular, Then S1 × {0} and {0} × S1 are Lie subgroups. They are all isomorphic to S1. On the other hand, any one-dimensional subspace of g with “irrational angle” corresponds to a Lie subgroup of the form α it iαt H := {(e , e ) | t ∈ R}, where α is an irrational number. These are immersed submanifolds which are isomorphic to . α R Note H = T2, so they are not submanifolds. (In particular, we see that compact Lie groups may have noncompact subgroups.)

Let G be a Lie group, and H a subgroup. If H is not only an immersed submanifold, but a (regular) submanifold, then one can show that H must be a closed subset in G. Deﬁnition 2.3. A subgroup H of G is said to be a closed subgroup if it is closed in H.

In what follows we will prove the following very powerful theorem. Theorem 2.4 (E. Cartan’s closed subgroup theorem). Any closed subgroup H of a Lie group G is a Lie subgroup (and thus a submanifold) of G. LECTURE 13: LIE SUBGROUPS AND LIE SUBALGEBRAS 5

Example. O(n), SL(n, R) are Lie groups, since they are closed subgroups, and thus Lie subgroups of GL(n, R). Suppose H is a closed subgroup of G. Let

h = {X ∈ g | exp(tX) ∈ H for all t ∈ R}. We will need the following lemmas: Lemma 2.5. h is a linear subspace of g.

Proof. Clearly h is closed under scalar multiplication. We have seen that there exists smooth function Z :(−ε, ε) → g so that exp(tX) exp(tY ) = exp(t(X + Y ) + t2Z(t)). It follows tX tY n exp(t(X + Y )) = lim exp( ) exp( ) . n→∞ n n So by closedness of H, h is also closed under addition.

Lemma 2.6. Suppose X1,X2, ··· be a sequence of nonzero elements in g so that

(1) Xi → 0 as i → ∞. (2) exp(Xi) ∈ H for all i. Xi (3) limi→∞ = X ∈ g. |Xi| Then X ∈ h.

t t Proof. For any ﬁxed t 6= 0, we take ni = [ ] be the integer part of . It follows that |Xi| |Xi|

ni exp(tX) = lim exp(niXi) = lim exp(Xi) ∈ H. i→∞ i→∞ Lemma 2.7. exp : g → G maps a neighborhood of 0 in h bijectively to a neighborhood of e in H.

Proof. Take a vector subspace h0 of g so that g = h ⊕ h0. Let Φ : g = h ⊕ h0 → G be the map Φ(X + Y ) = exp(X) exp(Y ).

Then as we have seen, dΦe(X + Y ) = X + Y . So Φ is a local diﬀeomorphism from g to G. Since exp |h = Φ|h, to prove the lemma, it is enough to prove that Φ maps a neighborhood of 0 in h bijectively to a neighborhood of e in H. 0 Suppose the lemma is false, then we can ﬁnd a sequence of vectors Xi +Yi ∈ h⊕h with Yi 6= 0 so that Xi + Yi → 0 and Φ(Xi + Yi) ∈ H. Since exp(Xi) ∈ H, we must have exp(Yi) ∈ H for all i. 0 We let Y be a limit point of Yi s. Then according to the previous lemma, Y ∈ h. Since Y ∈ h0, |Yi| we must have Y = 0, which is a contradiction since by construction, |Y | = 1. 6 LECTURE 13: LIE SUBGROUPS AND LIE SUBALGEBRAS

Proof of Cartan’s closed subgroup theorem. According to the previous lemma, one can ﬁnd a neighborhood U of e in G and a neighborhood V of 0 in g so that exp−1 : U → V is a diﬀeomorphism, and so that exp−1(U ∩ H) = V ∩ h. It follows that (exp−1, U, V ) is a chart on G which makes H a submanifold near e. For any other point h ∈ H, we can use left translation to get such a chart. As an immediate consequence, we get Corollary 2.8. If ϕ : G → H is Lie group homomorphism, then kerϕ is a Lie subgroup of G. Another very interesting consequence of Cartan’s theorem is Corollary 2.9. Every continuous homomorphism of Lie groups is smooth.

Proof. Let φ : G → H be a continuous homomorphism, then Γφ = {(g, φ(g)) | g ∈ G} is a closed subgroup, and thus a Lie subgroup of G × H. It follows that the projection

i π1 p :Γφ ,→ G × H −→ G is bijective and smooth. Thus dp is bijective near (eG, eH ), which implies that p is local diffeomorphism near (eG, eH ). Since it is a group, p is local diffeomorphism everywhere by translation. −1 Since p is bijective, it has to be a global diffeomorphism. Thus φ = π2 ◦ p is smooth.