Exponential maps

Idea

The exponential function of classical analysis given by the series,

(1)$\exp x \coloneqq \sum_{i = 0}^{\infty} \frac{x^i}{i!} ,$

is the solution of the differential equation

$f' = f$

with initial value $f(0) = 1$.

This classical function is defined on the real line (or the complex plane). To generalise it to other manifolds, we need two things:

• By dimensionalysis?, the argument of the function should be a tangent vector; so in the classical function $\mathbb{R} \to \mathbb{R}$, the source $\mathbb{R}$ is really the tangent space to the target $\mathbb{R}$ at the point $\exp 0 = 1$.
• We need a covariant derivative to tell us what $f'$ means.

So in the end we have, for any point $p$ on a differentiable manifold $M$ with an affine connection $\Del$, a map $\exp_p\colon T_p M \to M$, which is defined at least on a neighbourhood of $0$ in the tangent space $T_p M$.

Note that $p$ here comes from the initial value $\exp_p 0 = p$; we usually take $p = 1$ when we work in a Lie group, but otherwise we are really generalising the classical exponential function $x \mapsto p \exp x$; every solution to $f' = f$ takes this form.

Classically, there are some other functions called ‘exponential’; given any nonzero real (or complex) number $b$, the map $x \mapsto b^x$ (or even $x \mapsto p\, b^x$) is also an exponential map. Using the natural logarithm, we can define $b^x$ in terms of the natural exponential map $\exp$:

$b^x \coloneqq \exp (x \ln b) .$

So while $b$ is traditionally called the ‘base’, it is really the number $\ln b$ that matters, or even better the operation of multiplication by $\ln b$. This operation is an endomorphism of the real line (or complex plane), and every such endomorphism takes this form for some nonzero $b$ (and some branch of the natural logarithm, in the complex case). So we see that this generalised exponential map is simply the composite of the natural exponential map after a linear endomorphism.

Definition

Let $M$ be a differentiable manifold, let $\Del$ be an affine connection on $M$, and let $p$ be a point in $M$. Then by the general theory of differential equations, there is a unique maximally defined partial function $\exp_p$ from the tangent space $T_p M$ to $M$ such that:

• $\Del \exp_p = \exp_p$ and
• $\exp_p(0) = 1$.

This function is the natural exponential map on $M$ at $p$ relative to $\Del$. We have $\exp_p\colon U \to M$, where $U$ is some neighbourhood of $0$ in $T_p M$. If $M$ is complete? (relative to $\Del$), then $U$ will be all of $T_p M$.

Given any endomorphism $\phi\colon T_p M \to T_p M$, we can also consider the exponential map on $M$ at $p$ relative to $\Del$ with logarithmic base $\phi$, which is simply $x \mapsto \exp_p \phi(x)$. We say ‘logarithmic base’ since a classical exponential function with base $b$ corresponds to an exponential function whose logarithmic base is multiplication by $\ln b$.

Via geodesics

Recall that a geodesic is a curve on a manifold whose velocity? is constant (as measured along that curve relative to a given affine connection). Working naïvely, we may write

$\gamma' = v ,$

pretend that this is a differential equation for a function $\gamma\colon \mathbb{R} \to \mathbb{R}$, and take the solution

$\gamma(t) = p \exp t x ,$

where $p$ is given by the initial value $\gamma(0) = p$. We recognise this as being, morally, $\exp_p t x$. This suggests (although we need more work for a proof) the following result:

Let $M$ be a differentiable manifold, let $\Del$ be an affine connection on $M$, and let $p$ be a point in $M$. Given a tangent vector $x$ at $p$, there is a unique maximal geodesic $\gamma$ on $M$ tangent to $x$ at $p$. If $\gamma(1)$ is defined (which it will be whenever $M$ is complete? and may be in any case), we have $exp_p x = \gamma(1)$. In any case, we have $\exp_p (t x) = \gamma(t)$ for sufficiently small $t$.

In Riemannian manifolds

Let $M$ be a Riemannian manifold (or a pseudo-Riemannian manifold) and let $p$ be a point in $M$. Then $M$ may be equipped with the Levi-Civita connection $\Del_{lc}$, so we define the natural Riemannian exponential map on $M$ at $p$ to be the natural exponential map on $M$ at $p$ relative to $\Del_{lc}$.

In Lie groups

Note: this section is under repair.

The classical exponential function $\exp \colon \mathbb{R} \to \mathbb{R}^*$ or $\exp \colon \mathbb{C} \to \mathbb{C}^*$ satisfies the fundamental property:

Proposition

The function $\exp \colon \mathbb{C} \to \mathbb{C}^*$ is a homomorphism taking addition to multiplication:

$\exp(x + y) = \exp(x) \cdot \exp(y)$
Proof

A number of proofs may be given. One rests on the combinatorial binomial identity

$(x + y)^n = \sum_{j + k = n} \frac{n!}{j! k!} x^j y^k$

(which crucially depends on the fact that multiplication is commutative), whereupon

$\array{ \sum_{n \geq 0} \frac{(x+y)^n}{n!} & = & \sum_{n \geq 0} \sum_{j + k = n} \frac1{j!} \frac1{k!} x^j y^k \\ & = & (\sum_{j \geq 0} \frac{x^j}{j!}) \cdot (\sum_{k \geq 0} \frac{y^k}{k!}) \\ & = & \exp(x) \cdot \exp(y) }$

An alternative proof begins with the observation that $f = \exp$ is the solution to the system $f' = f$, $f(0) = 1$. For each $y$, the function $g_1 \colon x \mapsto f(x) f(y)$ is a solution to the system $g' = g$, $g(0) = f(y)$, as is the function $g_2 \colon x \mapsto f(x + y)$. Then by uniqueness of solutions to ordinary differential equations (over connected domains; see, e.g., here), $g_1 = g_2$, i.e., $f(x + y) = f(x)f(y)$ for all $x, y$.1

Let $M$ be Lie group and let $\mathfrak{g}$ be its Lie algebra $T_1 M$, the tangent space to the identity element $1$. Then $M$ may be equipped with the canonical left-invariant connection $\Del_l$ or the canonical right-invariant connection $\Del_r$. It turns out that the natural Riemannian exponential maps on $M$ at $1$ relative to $\Del_l$ and $\Del_r$ are the same; we define this to be the natural Lie exponential map on $M$ at the identity, denoted simply $\exp$. Several nice properties follow:

• $\exp$ is defined on all of $\mathfrak{g}$.
• $\exp \colon \mathfrak{g} \to G$ is a smooth map.
• If $\rho \colon \mathbb{R} \to \mathfrak{g}$ is a smooth homomorphism from the additive group $\mathbb{R}$ (i.e., if $\rho$ is an $\mathbb{R}$-linear map, uniquely determined by specifying $X = \rho(1)$), then $\exp \circ \rho_X \colon \mathbb{R} \to G$ is a smooth homomorphism.
• For $X, Y \in \mathfrak{g}$, if $[X, Y] = 0$, then the restriction of $\exp \colon \mathfrak{g} \to G$ to the subspace spanned by $X$ and $Y$ is a smooth homomorphism to $G$. In particular, $\exp \colon \mathfrak{g} \to G$ is a homomorphism if $\mathfrak{g}$ is abelian (e.g., if $G$ is a commutative Lie group).
• $\exp$ is surjective (a regular epimorphism) if $G$ is connected and compact (and also in some other situations, such as the classical cases where $G$ is $]0,\infty[$ or $\mathbb{C} \setminus \{0\}$). See this post by Terence Tao, Proposition 1; see also the first comment which indicates an alternative proof based on the fact that maximal tori in $G$ are all conjugate to one another. Note also that the exponential map might not be surjective if the compactness assumption is dropped, as in the case of $G = SL_2(\mathbb{R})$ or $SL_2(\mathbb{C})$, both of which are connected; see here for instance.
• If $G$ is compact, then it may be equipped with a Riemannian metric that is both left and right invariant (see Tao’s post linked in the previous remark); then the Lie exponential map is the same as the Riemannian exponential map at $1$.
• If $G$ is a matrix Lie group, then $\exp$ is given by the classical series formula (1).

(to be expanded on)

Logarithms

A logarithm is a local section of an exponential map.

1. A previous edit offered even more detail: “An alternative proof begins with the premise that each solution of the ordinary differential equation $g' = 0$ is locally constant. Suppose $c$ is a complex number. As $\exp' = \exp$, we find that $(\exp(x) \exp(c - x))' = \exp(x) \exp(c - x) + \exp(x) (-\exp(c-x)) = 0$. Hence, by the premise and the connectedness of the domain of $\exp$ (either ${\mathbb{R}}$ or ${\mathbb{C}}$), we obtain $\exp(x)\exp(c - x) = \exp(0)\exp(c)$. The initial condition $\exp(0) = 1$ then yields $\exp(x)\exp(c - x) = \exp(c)$. The result follows by setting $c = x + y$.”

Revised on January 2, 2015 16:06:16 by Urs Schreiber (127.0.0.1)