We showed that the differential of the exponential map for a smooth manifold and is the identity at . 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 is **analytic** if is analytic for analytic vector fields . Using the real-analytic versions of the ODE theorem, it follows that is an analytic morphism.

So, make the above assumptions: analyticity of both the manifold and the connection. Now there is a small disk such that maps diffeomorphically onto a neighborhood containing . We will compute when is sufficiently small and (recall that we identify with its tangent spaces at each point).

First, we need to set some notation. If , then define a vector field on with . In other words, given , connect by a unique (up to rescaling parametrization) geodesic, and take parallel on this geodesic with .

Finally, given a vector field on , let be the Lie bracket operator . For any operator on a vector space, let

Theorem 1 (Helgason)Analyticity hypotheses as above, if , then for small enough,

(Note that I have abbreviated .)

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 , which we may assume is analytic in some neighborhood of —by shrinking if necessary, analytic on . In particular, it would be convenient if for we could get a formula for because by definition

**A formula for **

Now is analytic in , so we can write

To get the constants , we have to compute the derivatives at . Now so

Applying the previous formula to the analytic function and using induction yields

Here of course is regarded as an operator on smooth (or analytic) functions, because it sure isn’t a vector field.

Thus we have the and the formula

It follows that if is small

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

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.*

November 7, 2009 at 2:44 am

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

November 7, 2009 at 9:24 pm

[…] and that we have already shown […]

November 7, 2009 at 10:38 pm

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