Helgason’s formula for the differential of the exponential map November 6, 2009Posted by Akhil Mathew in differential geometry, MaBloWriMo.
Tags: analytic manifolds, exponential map, Sigurdur Helgason
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.