The Exponential Map

Abstract

The exponential map, introduced for closed Lie subgroups of $\mathrm{GL}(n, \mathbb{C})$ in Chap. 5, can be defined for a general Lie group G as a map Lie(G) → G.

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The exponential map, introduced for closed Lie subgroups of $\mathrm{GL}(n, \mathbb{C})$ in Chap. 5, can be defined for a general Lie group G as a map Lie(G) → G.

We may consider a vector field (6.5) that is allowed to vary smoothly. By this we mean that we introduce a real parameter $\lambda \in (-\epsilon, \epsilon)$ for some ϵ > 0 and smooth functions $a_{i}: M \times (-\epsilon,\epsilon )\longrightarrow \mathbb{C}$ and consider a vector field, which in local coordinates is given by

$$\displaystyle{ (Xf)(m) =\sum _{ i=1}^{n}a_{ i}(m,\lambda ) \frac{\partial f} {\partial x_{i}}(m). }$$

(8.1)

FormalPara Proposition 8.1.

Suppose that M is a smooth manifold, m ∈ M, and X is a vector field on M. Then, for sufficiently small ϵ > 0, there exists a path p : $p : (-\epsilon, \epsilon)$ → M such that p(0) = m and $p_{{\ast}}(\mathrm{d}/\mathrm{d}t)(t) = X_{p(t)}$ for t ∈ (−ϵ, ϵ). Such a curve, on whatever interval it is defined, is uniquely determined. If the vector field X is allowed to depend on a parameter λ as in (8.1), then for small values of t, p(t) depends smoothly on λ.

Here we are regarding the interval (−ϵ, ϵ) as a manifold, and p _∗(d∕dt) is the image of the tangent vector d∕dt. We call such a curve an integral curve for the vector field.

FormalPara Proof.

In terms of local coordinates x ₁,…,x _n on M, the vector field X is

$$\displaystyle{\sum a_{i}(x_{1},\ldots,x_{n}) \frac{\partial } {\partial x_{i}},}$$

where the a _i are smooth functions in the coordinate neighborhood. If a path p(t) is specified, let us write x _i (t) for the x _i component of p(t), with the coordinates of m being $x_{1} = \cdots = x_{n} = 0$. Applying the tangent vector p _∗(t)(d∕dt)(t) to a function f ∈ C ^∞(G) gives

$$\displaystyle{ \frac{\mathrm{d}} {\mathrm{d}t}f\big(x_{1}(t),\ldots,x_{n}(t)\big) =\sum x_{i}^{\prime}(t) \frac{\partial f} {\partial x_{i}}\big(x_{1}(t),\ldots,x_{n}(t)\big).}$$

On the other hand, applying X _p(t) to the same f gives

$$\displaystyle{\sum _{i}a_{i}\big(x_{1}(t),\ldots,x_{n}(t)\big) \frac{\partial f} {\partial x_{i}}\big(x_{1}(t),\ldots,x_{n}(t)\big)\,,}$$

so we need a solution to the first-order system

$$\displaystyle{x_{i}^{\prime}(t) = a_{i}\big(x_{1}(t),\ldots,x_{n}(t)\big),\quad x_{i}(0) = 0,\quad (i = 1,\ldots,n).}$$

The existence of such a solution for sufficiently small |t|, and its uniqueness on whatever interval it does exist, is guaranteed by a standard result in the theory of ordinary differential equations, which may be found in most texts. See, for example, Ince [81], Chap. 3, particularly Sect. 3.3, for a rigorous treatment. The required Lipschitz condition follows from smoothness of the a _i. For the statement about continuously varying vector fields, one needs to know the corresponding fact about first-order systems, which is discussed in Sect. 3.31 of [81]. Here Ince imposes an assumption of analyticity on the dependence of the differential equation on λ, which he allows to be a complex parameter, because he wants to conclude analyticity of the solutions; if one weakens this assumption of analyticity to smoothness, one still gets smoothness of the solution. □

In general, the existence of the integral curve of a vector field is only guaranteed in a small segment (−ϵ, ϵ), as in Proposition 8.1. However, we will now see that, for left-invariant vector fields on a Lie group, the integral curve extends to all $\mathbb{R}$. This fact underlies the construction of the exponential map.

FormalPara Theorem 8.1.

Let G be a Lie group and $\mathfrak{g}$ its Lie algebra. There exists a map $\exp: \mathfrak{g}\longrightarrow G$ that is a local homeomorphism in a neighborhood of the origin in $\mathfrak{g}$ such that, for any $X \in \mathfrak{g}$ , t → exp (tX) is an integral curve for the left-invariant vector field X. Moreover, $\exp \big((t + u)X\big) =\exp (tX)\exp (uX)$ .

FormalPara Proof.

Let $X \in \mathfrak{g}$. We know that for sufficiently small ϵ > 0 there exists an integral curve p : (−ϵ, ϵ) → 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

$$ { p(s)\,p(t)} = {p(s + t)}\,\mathrm{when}\,s,t,s + t \in (a,b). $$

(8.2)

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.

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 $\mathbb{R}$. Of course, it is sufficient to show that we may extend it to $(-\frac{3} {2}a, \frac{3} {2}a)$. We extend it by the rule $p(t) = p(a/2)\,p(t - a/2)$ when $-a/2 \leqslant t \leqslant 3a/2$ and $p(t) = p(-a/2)\,p(t + a/2)$ when $-3a/2 \leqslant t \leqslant a/2$, and it follows from (8.2) that this definition is consistent on regions of overlap.

Now define $\exp: \mathfrak{g}\longrightarrow G$ as follows. Let $X \in \mathfrak{g}$, and let $p: \mathbb{R}\longrightarrow G$ 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 $u \in \mathbb{R}$, then $t\mapsto p(tu)$ is an integral curve for uX, so exp(uX) = p(u).

The exponential map is a smooth map, at least for X near the origin in $\mathfrak{g}$, by the last statement in Proposition 8.1. Identifying the tangent space at the origin in the vector space $\mathfrak{g}$ with $\mathfrak{g}$ itself, exp induces a map $T_{0}(\mathfrak{g})\longrightarrow T_{e}(G)$ (that is $\mathfrak{g}\longrightarrow \mathfrak{g}$), and this map is the identity map by construction. Thus, the Jacobian of exp is nonzero and, by the Inverse Function Theorem, exp is a local homeomorphism near 0. □

We also denote exp(X) as e^X for $X \in \mathfrak{g}$.

FormalPara Remark 8.1.

If $G = \mathrm{GL}(n, \mathbb{C})$, then as we explained in Chap. 7, Proposition 7.2 allows us to identify the Lie algebra of G with $\mathrm{Mat}_{n}(\mathbb{C})$. We observe that the definition of $\exp: \mathrm{Mat}_{n}(\mathbb{C})\longrightarrow \mathrm{GL}(n, \mathbb{C})$ by a series in (5.2) agrees with the definition in Theorem 8.1. This is because $t\longmapsto \exp (tX)$ with either definition is an integral curve for the same left-invariant vector field, and the uniqueness of such an integral curve follows from Proposition 8.1.

FormalPara Proposition 8.2.

Let G, H be Lie groups and let $\mathfrak{g}$, $\mathfrak{h}$ be their respective Lie algebras. Let f : G → H be a homomorphism. Then the following diagram is commutative:

FormalPara Proof.

It is clear from the definitions that f takes an integral curve for a left-invariant vector field X on G to an integral curve for df(X), and the statement follows. □

A representation of a Lie algebra $\mathfrak{g}$ over a field F is a Lie algebra homomorphism $\pi: \mathfrak{g}\longrightarrow \mathrm{End}(V )$, where V is an F-vector space, or more generally a vector space over a field E containing F, and End(V ) is given the Lie algebra structure that it inherits from its structure as an associative algebra. Thus,

$$\displaystyle{\pi ([x,y]) = \pi (x)\,\pi (y) - \pi (y)\,\pi (x).}$$

We may sometimes find it convenient to denote π(x)v as just xv for $x \in \mathfrak{g}$ and v ∈ V. We may think of $(x,v)\mapsto xv = \pi (x)v$ as a multiplication. If V is a vector space, given a map $\mathfrak{g} \times V \longrightarrow V$ denoted $(x,v)\mapsto xv$ such that $x\mapsto \pi (x)$ is a representation, where π(x) : V → V is the endomorphism v → xv, then we call V a $\mathfrak{g}$-module. A homomorphism ϕ : U → V of $\mathfrak{g}$-modules is an F-linear map satisfying ϕ(xv) = x ϕ(v).

FormalPara Example 8.1.

If π : G → GL(V ) is a representation, where V is a real or complex vector space, then the Lie algebra of GL(V ) is End(V ), so the differential Lie(π) : Lie(G) → End(V ), defined by Proposition 7.3, is a Lie algebra representation.

By the universal property of $U(\mathfrak{g})$ in Theorem 10.1, A Lie algebra representation $\pi: \mathfrak{g}\longrightarrow \mathrm{End}(V )$ extends to a ring homomorphism $U(\mathfrak{g})\longrightarrow \mathrm{End}(V )$, which we continue to denote as π.

If $\mathfrak{g}$ is a Lie algebra over a field F, we get a homomorphism $\mathrm{ad}: \mathfrak{g}\longrightarrow \mathrm{End}(\mathfrak{g})$, called the adjoint map, defined by ad(x)y = [x,y]. We give $\mathrm{End}(\mathfrak{g})$ the Lie algebra structure it inherits as an associative ring. We have

$$\displaystyle{ \mathrm{ad}(x)([y,z]) = [\mathrm{ad}(x)(y),z] + [y,\mathrm{ad}(x)(z)] }$$

(8.3)

since, by the Jacobi identity, both sides equal $[x,[y,z]] = [[x,y],z] + [y,[x,z]]$. This means that ad(x) is a derivation of $\mathfrak{g}$.

Also

$$\displaystyle{ \mathrm{ad}(x)\,\mathrm{ad}(y) -\mathrm{ad}(y)\,\mathrm{ad}(x) = \mathrm{ad}\big([x,y]\big) }$$

(8.4)

since applying either side to $z \in \mathfrak{g}$ gives $[x,[y,z]] - [y,[x,z]] = [[x,y],z]$ by the Jacobi identity. So $\mathrm{ad}: \mathfrak{g}\longrightarrow \mathrm{End}(\mathfrak{g})$ is a Lie algebra representation.

We next explain the geometric origin of ad. To begin with, representations of Lie algebras arise naturally from representations of Lie groups. Suppose that G is a Lie group and $\mathfrak{g}$ is its Lie algebra. If V is a vector space over $\mathbb{R}$ or $\mathbb{C}$, any Lie group homomorphism π : G → GL(V ) induces a Lie algebra homomorphism $\mathfrak{g}\longrightarrow \mathrm{End}(V )$ by Proposition 7.3; that is, a real or complex representation.

In particular, G acts on itself by conjugation, and so it acts on $\mathfrak{g} = T_{e}(G)$. This representation is called the adjoint representation and is denoted $\mathrm{Ad}: G\longrightarrow \mathrm{GL}(\mathfrak{g})$. We show next that the differential of Ad is ad. That is:

FormalPara Theorem 8.2.

Let G be a Lie group, $\mathfrak{g}$ its Lie algebra, and $\mathrm{Ad}: G\longrightarrow \mathrm{GL}(\mathfrak{g})$ the adjoint representation. Then the Lie group representation $\mathfrak{g}\longrightarrow \mathrm{End}(\mathfrak{g})$ corresponding to Ad by Proposition 7.3 is ad .

FormalPara Proof.

It will be most convenient for us to think of elements of the Lie algebra as tangent vectors at the identity or as local derivations of the local ring there. Let $X,Y \in \mathfrak{g}$. If f ∈ C ^∞(G), define $c(g)f(h) = f({g}^{-1}hg)$. Then our definitions of the adjoint representation amount to

$$\displaystyle{\big(\mathrm{Ad}(g)Y \big)f = Y \big(c({g}^{-1})f\big).}$$

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. Therefore, under the representation of $\mathfrak{g}$ in Proposition 7.3, X maps Y to the local derivation at the identity

$$\displaystyle{f\longmapsto \frac{\mathrm{d}} {\mathrm{d}t}\,\big(\mathrm{Ad}({\mathrm{e}}^{tX})Y \big)f\,\Big\vert _{ t=0} = \frac{\mathrm{d}} {\mathrm{d}t} \frac{\mathrm{d}} {\mathrm{d}u}f({\mathrm{e}}^{tX}{\mathrm{e}}^{uY }{\mathrm{e}}^{-tX})\,\Big\vert _{ t=u=0}.}$$

By the chain rule, if F(t ₁,t ₂) is a function of two real variables,

$$\displaystyle{ \frac{\mathrm{d}} {\mathrm{d}t}F(t,t)\,\Big\vert _{t=0} = \frac{\partial F} {\partial t_{1}}(0,0) + \frac{\partial F} {\partial t_{2}}(0,0). }$$

(8.5)

Applying this, with u fixed to $F(t_{1},t_{2}) = f({\mathrm{e}}^{t_{1}X}{\mathrm{e}}^{uY }{\mathrm{e}}^{-t_{2}X})$, our last expression equals

$$\displaystyle{ \frac{\mathrm{d}} {\mathrm{d}u} \frac{\mathrm{d}} {\mathrm{d}t}f({\mathrm{e}}^{tX}\,{\mathrm{e}}^{uY })\,\Big\vert _{ t=u=0} - \frac{\mathrm{d}} {\mathrm{d}u} \frac{\mathrm{d}} {\mathrm{d}t}f({\mathrm{e}}^{uY }\,{\mathrm{e}}^{tX})\,\Big\vert _{ t=u=0} = XY f(1) - Y Xf(1).}$$

This is, of course, the same as the effect of [X,Y ] = ad(X)Y. □

Exercises

FormalPara Exercise 8.1.

Show that the exponential map $\mathfrak{s}\mathfrak{u}(2) \rightarrow \mathrm{SU}(2)$ is surjective, but the exponential map $\mathfrak{s}\mathfrak{l}(2, \mathbb{R}) \rightarrow \mathrm{SL}(2, \mathbb{R})$ is not.