I recently gave an answer to this question, that I'm requesting you to check the correctness of.
"Let $f(x)=\text{dist}_M^2(p,x)$, and $p \in M$ (Riemannian manifold) is fixed. Show that $\text{grad}(f)=-2\exp^{-1}_x(p)$".
I feel that my intuition is correct, but there are two assumptions (equations (1) and (2) below) couldn't be fully justified, although I think they're correct but can't justify. Could you please check these assumptions and thus deduce if they were correct or not? Feel free to criticize.
It's been a while I did Riemannian geometry in depth, so my quesitions might seem rudimentary to many. here's my attempt at a proof, but I still need to think why (1) and (2) below are correct. Here's the outline of the proof:
Steps:
- I'll define what $exp, log :=exp^{-1}$ are in $\mathbb{R}^m.$
- Prove the formula: $\text{grad}(f)=-2\exp^{-1}_x(p)$ for $\mathbb{R}^m$ first.
- Use the normal coordinates to carry over the $log^{\mathbb{R}^m}$ to $log^{M},$ where $log^{*}$ denotes the $log$ maps for *, i.e. inverses of the $exp$ maps in the corresponding spaces. The assumptions (1) and (2) below are made at this very step and I'm a bit unsure if they're correct...
Steps 1) and 2):
Let's prove the formula in $\mathbb{R}^n$ first. Here $d^2(y,z)=||y-z||^2, exp_{a}b=a+b, log_{a}b=b-a.$ Take $z=0, g(y):=||y||^2, \nabla{g}(y)=2y=-2(-y)=-2log_{y}(0)\equiv -2exp_{y}^{-1}(0).$ Therefore, we established the formula in the Euclidean case.
Step 3):
Now let's work in the case of general Riemannian manifold $(M,g), dim(M)=m.$. Let $p\in(M,g),$ with its exponential map $exp^{M},$ log map $log^{M}.$ Pick a normal coordinate chart $(U,\phi)$ around $p$ so that $U$ is a $\delta$-uniformly normal neighborhood of $p,$ i.e. $U\subset B(x,\delta) \forall x\in U$ (such a $\delta>0$ exists, see for example, John Lee's book "Riemannian Manifolds: Introduction to curvature", Lemma 5.12, P.78). Let $W:=\phi(U)\subset \mathbb{R}^m$. [So here essentially we pick an ordered orthonormal basis $(E_i)_{1\le i\le m},$ of $T_pM$ and $E:\mathbb{R}^m\to T_pM:= (x_1\dots x_m)\mapsto \sum_i{x_iE_i},$ and then define $\phi:U\to W:= \phi(x):=E^{-1}log^{M}(x)=:(x_1\dots x_m).$]
Note that (I'm sure it's correct, but why exactly? By Gauss' lemma or radial isometry in a uniformly normal neighborhood?!)
$$log^{\mathbb{R}^m}\circ \phi = D\phi \circ log^{M}\dots (1)$$
Assuming the above is correct, the rest is easy:
Letting $y:=\phi(x), d^2(x,p)=||y||^2, ||.||$ meaning the Euclidean norm. Therefore,
$$D\phi_{x}(\nabla_{x}{d^2(x,p)})=\nabla_{y}||y||^2 \dots (2)$$
(why? Is this because: $d^2(x,p)=||y||^2\implies$ their differentials correspond, and hence their gradients do? So essentially $d(\phi\circ f)=D\phi(df)?$)
Assuming the above is correct,
$D\phi_{x}(\nabla_{x}{d^2(x,p)}) =-2log_{y}(0)$ $ \implies \nabla_{x}{d^2(x,p)} $ $= D\phi_{x}^{-1}(-2log_{y}(0))$ $= -2 D\phi_{x}^{-1} (log_{y}(0))$ (linearity) $= -2 log_{\phi^{-1}(y)}{\phi^{-1}(0)}=-2log_{x}(p)$ (applying (1) above, assuming correct)
This proves it.