Helgason’s formula for the differential of the exponential map

Helgason’s formula for the differential of the exponential map November 6, 2009

Posted by Akhil Mathew in differential geometry, MaBloWriMo.
Tags: analytic manifolds, exponential map, Sigurdur Helgason
trackback

We showed that the differential of the exponential map ${\exp_p: T_p(M) \rightarrow M}$ for ${M}$ a smooth manifold and ${p \in M}$ is the identity at ${0 \in T_p(M)}$ . In the case of analytic manifolds, it is possible to say somewhat more. First of all, if we’re working with real-analytic manifolds, we can say that a connection ${\nabla}$ is analytic if ${\nabla_XY}$ is analytic for analytic vector fields ${X,Y}$ . Using the real-analytic versions of the ODE theorem, it follows that ${\exp_p}$ is an analytic morphism.

So, make the above assumptions: analyticity of both the manifold and the connection. Now there is a small disk ${V_p \subset T_p(M)}$ such that ${\exp_p}$ maps ${V_p}$ diffeomorphically onto a neighborhood ${U \subset M}$ containing ${p}$ . We will compute ${d(\exp_p)_{X}(Y)}$ when ${X \in V_p}$ is sufficiently small and ${Y \in T_p(M)}$ (recall that we identify ${T_p(M)}$ with its tangent spaces at each point).

First, we need to set some notation. If ${Z \in V_p}$ , then define a vector field ${Z^*}$ on ${U}$ with ${Z^* = \exp_{p*}(Z)}$ . In other words, given ${q \in U}$ , connect ${p,q}$ by a unique (up to rescaling parametrization) geodesic, and take ${Z^*}$ parallel on this geodesic with ${Z^*(p) = Z}$ .

Finally, given a vector field ${W}$ on ${U}$ , let ${\theta(W)}$ be the Lie bracket operator ${\Gamma(U) \rightarrow \Gamma(U)}$ . For any operator on a vector space, let

$\displaystyle (1-e^{-A})/A := \sum_{m=0}^{\infty} \frac{(-A)^m}{(m+1)!}.$

Theorem 1 (Helgason)

Analyticity hypotheses as above, if ${X,Y \in T_p(M)}$ , then for ${t \in \mathbb{R}}$ small enough, $\displaystyle \boxed{ (d \exp)_{tX}(Y) = \left( \frac{ 1 - e^{\theta( - tX^* )}}{\theta(tX^*)} (Y^*) \right)_{\exp(tX)}.}$

(Note that I have abbreviated ${\exp := \exp_p}$ .)

This is clearly a messy formula, but I will try to motivate the proof as best as I can. Source: Helgason, Differential Geometry, Lie Groups, and Symmetric Spaces.

First of all, since both quantities here are tangent vectors, we should try to apply them on a function ${f: U \rightarrow \mathbb{R}}$ , which we may assume is analytic in some neighborhood of ${p}$ —by shrinking ${U}$ if necessary, analytic on ${U}$ . In particular, it would be convenient if for ${Z \in V_p}$ we could get a formula for ${f(\exp(Z))}$ because by definition

$\displaystyle (d \exp)_{tX}(Y) f = \frac{d}{du}|_{u=0} f( \exp(tX + uY)).$

A formula for ${f(\exp(tZ))}$

Now ${f(\exp(tZ))}$ is analytic in ${t}$ , so we can write

$\displaystyle f(\exp(tZ)) = \sum a_n \frac{ t^n}{n!}.$

To get the constants ${a_n}$ , we have to compute the derivatives at ${0}$ . Now ${Z^*(\exp(tZ)) = \frac{d}{dt} \exp(tZ)}$ so

$\displaystyle \frac{d}{dt} f(\exp(tZ)) = (Z^* f)(\exp(tZ)).$

Applying the previous formula to the analytic function ${Z^{*(n-1)}f}$ and using induction yields

$\displaystyle \frac{d^n}{dt^n } f(\exp(tZ)) = (Z^{*n} f)(\exp(tZ)),$

Here of course ${Z^{*n}}$ is regarded as an operator on smooth (or analytic) functions, because it sure isn’t a vector field.

Thus we have the ${a_n}$ and the formula

$\displaystyle \boxed{ f(\exp(tZ)) = \sum (Z^{*n} f )(p) \frac{ t^n}{n!} .}$

It follows that if ${t}$ is small

$\displaystyle (d \exp)_{tX}(Y) f = \frac{d}{du}|_{u=0} \sum_{n=0}^{\infty} \frac{1}{n!} (( tX^* + uY^*)^n f) (p).$

If we expand this out, then only the terms with ${u}$ occuring to the power 1 will remain when we differentiate with respect to ${u}$ at ${u=0}$ . The end result is

$\displaystyle \boxed{(d \exp)_{tX}(Y) f = \sum_{n=1}^{\infty} \frac{t^{n-1}}{n!} ( X^{*(n-1)} Y^* + X^{*(n-2)} Y^* X^* + \dots )f(p).}$

I think this is a good place to stop for today. Tomorrow, we’ll show how some algebraic manipulation with Lie brackets can turn this into Helgason’s formula.

Mildly corrected.

Comments»

1. liuxiaochuan - November 7, 2009: Hi guys, I just wrote a post, making a plan to learn maths in the next afew months. I am wodering if you are interested in joining me. I also set up a group at http://groups.google.com/group/maths-learning

Reply
Akhil Mathew - November 7, 2009: One of my current aims is to learn Riemannian geometry (though unfortunately I haven’t yet looked at a good reference for this; I plan to look at some of the ones recommended on MO in the near future). I was actually thinking of saying something about de Rham cohomology before the end of this month.

I’ve subscribed.

Reply
Akhil Mathew - November 7, 2009: Also, thanks for the link!
2. Helgason’s formula II « Delta Epsilons - November 7, 2009: […] and that we have already shown […]

Reply
3. gbideas - November 7, 2009: I probably could google it. But in order to say hello, I thought, I just ask here: How do you do the formulas? Are they supported by every wordpress blog? Or is this an upgrade option like video upload.

Have a great day.

GB

Reply
Akhil Mathew - November 7, 2009: Yes, the way to get LaTeX is to type “$”, then “latex”, and then the code (and then $ again); there’s also the command \displaystyle to get displayed math. The easiest way I know to do this quickly is Luca Trevisan’s LaTeX to WordPress python script

Edit: the official WordPress guide is at http://en.support.wordpress.com/latex/.

Reply

	Proof that a group r… on Basics of group representation…
	Jahnke on The Hahn-Banach theorem and tw…
	pageanu on Some unsolved problems
	Ingo Blechschmidt on Generic freeness II
	kreditvergleich deut… on Divisibility theorems for grou…

Delta Epsilons