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exponential map

Exponential maps

Idea

The exponential function of classical analysis given by the series,

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

is the solution of the differential equation

f=f f' = f

with initial value f(0)=1f(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 exp0=1\exp 0 = 1.
  • We need a covariant derivative to tell us what ff' means.

So in the end we have, for any point pp on a differentiable manifold MM with an affine connection \Del, a map exp p:T pMM\exp_p\colon T_p M \to M, which is defined at least on a neighbourhood of 00 in the tangent space T pMT_p M.

Note that pp here comes from the initial value exp p0=p\exp_p 0 = p; we usually take p=1p = 1 when we work in a Lie group, but otherwise we are really generalising the classical exponential function xpexpxx \mapsto p \exp x; every solution to f=ff' = f takes this form.

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

b xexp(xlnb). b^x \coloneqq \exp (x \ln b) .

So while bb is traditionally called the ‘base’, it is really the number lnb\ln b that matters, or even better the operation of multiplication by lnb\ln b. This operation is an endomorphism of the real line (or complex plane), and every such endomorphism takes this form for some nonzero bb (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 MM be a differentiable manifold, let \Del be an affine connection on MM, and let pp be a point in MM. Then by the general theory of differential equations, there is a unique maximally? defined partial function exp p\exp_p from the tangent space T pMT_p M to MM such that:

  • exp p=exp p\Del \exp_p = \exp_p and
  • exp p(0)=1\exp_p(0) = 1.

This function is the natural exponential map on MM at pp relative to \Del. We have exp p:UM\exp_p\colon U \to M, where UU is some neighbourhood of 00 in T pMT_p M. If MM is complete? (relative to \Del), then UU will be all of T pMT_p M.

Given any endomorphism ϕ:T pMT pM\phi\colon T_p M \to T_p M, we can also consider the exponential map on MM at pp relative to \Del with logarithmic base ϕ\phi, which is simply xexp pϕ(x)x \mapsto \exp_p \phi(x). We say ‘logarithmic base’ since a classical exponential function with base bb corresponds to an exponential function whose logarithmic base is multiplication by lnb\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

γ=v, \gamma' = v ,

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

γ(t)=pexptx, \gamma(t) = p \exp t x ,

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

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

In Riemannian manifolds

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

In Lie groups

Note: this section is under repair.

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

Proposition

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

exp(x+y)=exp(x)exp(y)\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= j+k=nn!j!k!x jy k(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

n0(x+y) nn! = n0 j+k=n1j!1k!x jy k = ( j0x jj!)( k0y kk!) = exp(x)exp(y)\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=expf = \exp is the solution to the system f=ff' = f, f(0)=1f(0) = 1. For each yy, the function g 1:xf(x)f(y)g_1 \colon x \mapsto f(x) f(y) is a solution to the system g=gg' = g, g(0)=f(y)g(0) = f(y), as is the function g 2:xf(x+y)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 2g_1 = g_2, i.e., f(x+y)=f(x)f(y)f(x + y) = f(x)f(y) for all x,yx, y.1

Let MM be Lie group and let 𝔤\mathfrak{g} be its Lie algebra T 1MT_1 M, the tangent space to the identity element 11. Then MM may be equipped with the canonical left-invariant connection l\Del_l or the canonical right-invariant connection r\Del_r. It turns out that the natural Riemannian exponential maps on MM at 11 relative to l\Del_l and r\Del_r are the same; we define this to be the natural Lie exponential map on MM at the identity, denoted simply exp\exp. Several nice properties follow:

  • exp\exp is defined on all of 𝔤\mathfrak{g}.
  • exp:𝔤G\exp \colon \mathfrak{g} \to G is a smooth map.
  • If ρ:𝔤\rho \colon \mathbb{R} \to \mathfrak{g} is a smooth homomorphism from the additive group \mathbb{R} (i.e., if ρ\rho is an \mathbb{R}-linear map, uniquely determined by specifying X=ρ(1)X = \rho(1)), then expρ X:G\exp \circ \rho_X \colon \mathbb{R} \to G is a smooth homomorphism.
  • For X,Y𝔤X, Y \in \mathfrak{g}, if [X,Y]=0[X, Y] = 0, then the restriction of exp:𝔤G\exp \colon \mathfrak{g} \to G to the subspace spanned by XX and YY is a smooth homomorphism to GG. In particular, exp:𝔤G\exp \colon \mathfrak{g} \to G is a homomorphism if 𝔤\mathfrak{g} is abelian (e.g., if GG is a commutative Lie group).
  • exp\exp is surjective (a regular epimorphism) if GG is connected and compact (and also in some other situations, such as the classical cases where GG is ]0,[]0,\infty[ or {0}\mathbb{C} \setminus \{0\}).
  • If GG is compact, then it may be equipped with a Riemannian metric; then the Lie exponential map is the same as the Riemannian exponential map at 11.
  • If GG is a matrix Lie group, then exp\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=0g' = 0 is locally constant. Suppose cc is a complex number. As exp=exp\exp' = \exp, we find that (exp(x)exp(cx))=exp(x)exp(cx)+exp(x)(exp(cx))=0(\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\exp (either {\mathbb{R}} or {\mathbb{C}}), we obtain exp(x)exp(cx)=exp(0)exp(c)\exp(x)\exp(c - x) = \exp(0)\exp(c). The initial condition exp(0)=1\exp(0) = 1 then yields exp(x)exp(cx)=exp(c)\exp(x)\exp(c - x) = \exp(c). The result follows by setting c=x+yc = x + y.”

Revised on April 18, 2014 05:42:53 by Todd Trimble (107.204.62.194)