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.