How do we calculate the gradient of this function defined using the Riemannian logarithm on a Riemannian manifold?

Question

We consider the following function $\psi$ on an open subset $V\subset M,$ a Riemannian manifold of dimension $m,$ so that $\exp_p:U\to V$ is a diffeomorphism with its inverse $\log_p: V\to U$. Let $v\in T_pM$ be a fixed vector and define the function $\psi$ by:

$$\psi:V\to \mathbb{R}:=\psi(x):=\langle v,\log_p x\rangle_{T_pM}$$

We'd like to calculate its gradient $\nabla{\psi}(x).$ So I attempted this:

Let $\gamma:(-\epsilon,\epsilon)\to V \subset M$ be a smooth curve (not necessarily a geodesic) so that $\gamma(0)=x,\gamma'(0)=w\in T_xM.$ We calculate $D\psi_x(w)$ next and based on what we get, we'll then use: $D\psi_x(w)=\langle\nabla{\psi}(x),w\rangle_{T_xM},$ and perhaps use a normal chart that contains both $p,x.$ But I'm getting stuck here - any help, hint or solutions greatly appreciated!

So we go:

$$D\psi_x(w)=\frac{d}{dt}|_{t=0}\psi(\gamma(t))=\underbrace{\frac{d}{dt}|_{t=0}\langle v,\log_p{\gamma(t)}\rangle_{T_pM} }$$

This is where (see the underbrace) I'm getting stuck - I know that I can write the last quantity above as $\langle v,\frac{d}{dt}|_{t=0}\log_p{\gamma(t)}\rangle_{T_pM}.$ But I'm having trouble at calculating $,\frac{d}{dt}|_{t=0}\log_p{\gamma(t)}.$ Here I'd like to use the fact that $\gamma'(0)=w\in T_xM,$ and maybe some normal neighborhood that contains both $p,x$. But I've no idea how — any help, hint or solutions greatly appreciated!

Answer 1

In order to differentiate something containing the logarithm, it's best if you can differentiate the logarithm itself. And because you work in a neighborhood where the exponential map is a diffeomorphism, this is not substantially different from differentiating the exponential map.

The directional derivative of $\exp_p$ at $v\in T_pM$ in the direction $w\in T_vT_pM=T_pM$ is $J(1)$, where $J(t)$ is the Jacobi field along the geodesic $\gamma(t)=\exp_p(tv)$ with the initial data $J(0)=0$ and $D_tJ(0)=w$. That is, the derivative is more or less¹ the solution operator of the Jacobi equation.

This may feel like an unsatisfactory answer, because I've only translated the problem into another one. But this is both inevitable and useful: First, the behaviour of the gradient can depend greatly on the geometry so there is no simple formula. Second, Jacobi fields are often easier to understand (e.g. with comparison theorems) that more extended objects like the exponential map or the geodesic flow. Therefore getting stuck right where you are stuck is natural unless you have a specific kind of goal in mind with the calculation.

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¹ It's more natural to consider the Jacobi field as a pair $(J,D_tJ)$, and a solution operator maps $(0,w)$ to $(a,b)$ where you only care about $a$. One way to understand is to see the exponential map as a slice of the geodesic flow and see that the full solution operator of Jacobi fields is exactly the differential of the flow. The differential of the exponential map is the vertical-to-horizontal part of the whole differential. See e.g. these lecture notes for details.