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.