Lecture 5: the Exponential Map and Ad

The exponential map Representations

Lecture 5: The exponential map and Ad

Daniel Bump

April 21, 2020 The exponential map Representations

The exponential map in previous lectures

Previously we considered a Lie subgroup G of GL(n, C). The Lie algebra g is then identiﬁed with a Lie subalgebra of gl(n, C), and X ∈ gl(n, C) is realized as a matrix. Then the exponential map ∞ X Xk X 7→ eX = exp(X) := k! k=0 is a map g → G with important properties.

The exponential map g → G has an intrinsic deﬁnition independent of any particular realization of G as a subgroup of GL(n, C), and we will discuss that today. The exponential map Representations

First order systems

The exponential map, introduced for closed Lie subgroups of GL(n, C) in earlier lectures or Chapter 5 of the book, can be deﬁned for a general Lie group G as a map Lie(G) −→ G.

One version of the fundamental theorem of first order systems of differential equations essentially says that given a system of coefficients ai(x1, ··· , xn) which we consider to be smooth n functions defined in a neighborhood of 0 in R , system of equations

0 xi(t) = ai x1(t),..., xn(t) , xi(0) = 0, (i = 1,..., n)

has a unique solution for t ∈ (−ε, ε) with some > 0. We cannot assert a solution for all t ∈ R since the solution might move off to inﬁnite in ﬁnite time. The exponential map Representations

Interpretation for vector ﬁelds

n Continuing to work in R , at any point x = (x1, ··· , xn) the derivations ∂/∂xi span the tangent space and we can interpret the ai as deﬁning a vector ﬁeld

n X ∂ ai(x1, ··· , xn) . ∂xi i=1 Then the fundamental theorem of first order systems says that n given a smooth vector field in a neighborhood U of 0 ∈ R then for small enough ε > 0 there exists a smooth parametrized path p :(−, ) → U tangent to the vector field, such that p(0) = 0.

Formulated this way, the result may be immediately transferred to a smooth manifold. We will apply this to left-invariant vector ﬁelds on a Lie group. The exponential map Representations

The exponential map

Let G be a Lie group and g its Lie algebra. Let X ∈ g. We interpret X as a left-invariant vector field. The fundamental theorem for first order systems only guarantees the existence of a curve p :(−ε, ) → G tangent to this vector field, but in this particular case we will show that p may be extended to all of R. Theorem Let G be a Lie group and g its Lie algebra. There exists a map exp : g −→ G that is a local homeomorphism in a neighborhood of the origin in g such that, for any X ∈ g, t −→ exp(tX) is an integral curve for the left-invariant vector field X. Moreover, exp (t + u)X = exp(tX) exp(uX). The exponential map Representations

Multiplicativity of the integral curve

Let X ∈ g. We know that for sufficiently small > 0 there exists an integral curve p :(−, ) −→ G for the left-invariant vector field X with p(0) = 1. We show first that if p :(a, b) −→ G is any integral curve for an open interval (a, b) containing 0, then

p(s) p(t) = p(s + t) when s, t, s + t ∈ (a, b).

Indeed, since X is invariant under left-translation, left-translation by p(s) takes an integral curve for the vector ﬁeld into another integral curve. Thus, t −→ p(s) p(t) and t −→ p(s + t) are both integral curves, with the same initial condition 0 −→ p(s). They are thus the same. The exponential map Representations

Bootstrapping the information

With this in mind, we show next that if p :(−a, a) −→ G is an integral curve for the left-invariant vector field X, then we may extend it to all of R. Of course, it is sufficient to show that we 3 3 may extend it to (− 2 a, 2 a). We extend it by the rule p(t) = p(a/2) p(t − a/2) when −a/2 6 t 6 3a/2 and p(t) = p(−a/2) p(t + a/2) when −3a/2 6 t 6 a/2, and it follows from p(s) p(t) = p(s + t) when s, t, s + t ∈ (a, b). that this definition is consistent on regions of overlap.

Now define exp : g −→ G as follows. Let X ∈ g, and let p : R −→ G be an integral curve for the left-invariant vector field X with p(0) = 0. We define exp(X) = p(1). We note that if u ∈ R, then t 7→ p(tu) is an integral curve for uX, so exp(uX) = p(u). The exponential map Representations

Compatibility with the GL(n, C) case

If G = GL(n, C), then we identify g = Matn(C). If X ∈ Matn(C), we interpret X as the left invariant vector ﬁeld (derivation of C∞(G)) d (Xf )(g) = f (getX)| , dt t=0 where etX is the matrix exponential map:

∞ X Xk exp(X) := k! k=0

It is easy to see that the path etX is tangent to this vector field, so this definition of the exponential agrees with the new definition. The exponential map Representations

Compatibility for homorphisms

Proposition Let G, H be Lie groups and let g, h be their respective Lie algebras. Let f : G → H be a homomorphism. Then the following diagram is commutative: df g h

exp exp f G H

Here df : g → h is the differential of the smooth map f at the identity. It is a Lie algebra homomorphism by Proposition 7.3 (covered in class last Thursday). See Proposition 8.2 for the easy proof. The exponential map Representations

Representations Lie groups and Lie algebras

If G is a compact Lie group, a smoothing argument together with the Peter-Weyl theorem implies that any (continuous) ﬁnite-dimensional representation π : G → GL(V) is smooth. See Theorem 8.2. In any case, we’ll assume that any representation of the Lie group G is ﬁnite-dimensional and smooth. Then by Proposition 7.3, the differential of π at the identity is a Lie algebra homomorphism g = Lie(G) → End(V), that is, a representation of the Lie algebra g.

This means that

dπ([X, Y]) = dπ(X) dπ(Y) − dπ(Y) dπ(X). The exponential map Representations

The Adjoint representation of G

There is one immedately available representation of G, namely the adjoint representation. The group G acts on itself by conjugation, and this action ﬁxes the identity, so it acts on the tangent space at 1, which is g.

Ad : G → GL(g) The exponential map Representations

The adjoint representation of a Lie algebra

Let g be any Lie algebra. Then we have a representation ad : g → End(g), deﬁned by:

ad(x)y = [x, y].

To see that this is a representation, note that the required identity ad([x, y])(z) = ad(x) ad(y)z − ad(y) ad(x)z means [[x, y], z] = [x, [y, z]] − [y, [x, z]] which is one way of rearranging the Jacobi identity.

We will show that if g is the Lie algebra of G, then the differential of Ad is ad. The exponential map Representations

Reminder of Chapter 7

Let M be a manifold, a ∈ M.

Let Oa be the ring of germs of smooth functions at a. Thus an element of this ring is an equivalence class of smooth functions defined near a, where two functions are identified if they agree on some sufficiently small neighborhood of a.

One deﬁnition of the tangent space Ta(M) is as the space of derivations Y : O1 → C, maps that satisfy

Y(f1f2) = Y(f1)f2(a) + f1(a)Y(f2).

Let us relate this to our definition of a vector field Y as a derivation of C∞(M). Given a vector field Y, define Ya(f ) = (Yf )(a). This only depends on the germ of f at a, so Ya is a tangent vector by the previous definition. The exponential map Representations

The differential of ad is Ad

Assume that g is the Lie algebra of a Lie group G. Theorem The differential of Ad : G → GL(g) is ad.

If f ∈ C∞(G), deﬁne c(g)f (h) = f (g−1hg). Then our deﬁnitions of the adjoint representation amount to

Ad(g)Yf = Yc(g−1)f .

To compute the differential of Ad, note that the path t −→ exp(tX) in G is tangent to the identity at t = 0 with tangent vector X. The exponential map Representations

The differential of ad is Ad (continued)

Therefore, under the representation of g in Proposition ??, X maps Y to the local derivation at the identity d d d f 7−→ Ad(etX)Yf = f (etXeuY e−tX) . dt t=0 dt du t=u=0

By the chain rule, if F(t1, t2) is a function of two real variables, d ∂F ∂F F(t, t) = (0, 0) + (0, 0). dt t=0 ∂t1 ∂t2

t X uY −t X Applying this, with u ﬁxed to F(t1, t2) = f (e 1 e e 2 ) gives d d d d f (etX euY ) − f (euY etX) = XYf (1) − YXf (1). du dt t=u=0 du dt t=u=0 This is, of course, the same as the effect of [X, Y] = ad(X)Y.