When $\int_M \exp(-d_M(x,y)^2/t) dvol(y)$ becomes constant for a Riemannian manifold $M$?

Question

Let $(M,g)$ be a closed and connected Riemannian manifold. $d_M$ is its geodesic metric and $dvol_M$ is its standard volume measure. For each $t>0$, define a map $f:M\rightarrow\mathbb{R}_{>0}$ in the following way:

$$f_t(x):=\int_M e^{-\frac{d_M(x,y)^2}{t}} dvol_M(y).$$

Then, when this $f_t$ becomes constant map? Obviously it becomes constant if $M$ is symmetric, like $S^n$. But do we have more rich characterization?

This seems a simple question, so it might be a classical one. But I'm not expert in differential geometry. Is there any related reference about this question?

Answer 1

Here are a few comments:

1) I like the example of a cylinder, slightly more complicated than $S^n$. This satisfies the constancy in the problem. It is symmetric in having a transitive group of isometries, but the group of isometries is not doubly transitive.

2) We can solve the problem if for any points $p$ and $q$, we can exhibit an isometry $f$ taking $p$ to $q$. And there is an obvious candidate for such an isometry: Choose a geodesic $\gamma$ from $p$ to $q$. Then parallel transport along $\gamma$ takes the tangent vectors at $p$ to tangent vectors at $q$, and takes small geodesics starting at $p$ to small geodesics starting at $q$. This creates a map $f:B(p,r)\rightarrow B(q,r)$, where $r$ is the minimum of the radii of injectivity at $p$ and $q$. We need to prove first that $f$ is an isometry on its domain, and then that $f$ can be extended to the whole space.

3) To prove that $f$ is an isometry, it suffices to show that the minimal ball $S$ containing $p_1$ and $p_2$ has the same volume as the minimal ball containing $f(p_1)$ and $f(p_2)$. Then, by David Speyer's reformulation of the hypothesis, the two balls have the same radii, and the same diameters, and the same distance from $p_1$ to $p_2$ as from $f(p_1)$ to $f(p_2)$. This may be useful because if $S$ has (e.g.) twice the volume of $f(S)$, then we can identify a decreasing sequence of balls $S_i$ where $S_i$ has twice the volume of $f(S_i)$, so we can identify a particular point where the isometry fails.

Answer 2

A partial answer: Let $u(x,s)$ be the function such that $$\mathrm{Vol}{\Big \{} y \in M: s_1 \leq d(x,y)^2 \leq s_2 {\Big \}} = \int_{s=s_1}^{s_2} u(x,s) ds.$$ Then your $f_t(x)$ is the Laplace transform of $u(x,s)$. Since a function is determined by its Laplace transform, $f_t$ is constant in $x$ if and only if $u(x,s)$ is constant in $x$. It seems a bit more natural to work with the function $v(x,r)$ such that $$\mathrm{Vol}{\Big \{} y \in M: r_1 \leq d(x,y) \leq r_2 {\Big \}} = \int_{r=r_1}^{r_2} v(x,r) dr.$$ These are related by $v(x,r) = 2 r u(x,r^2)$ (just substitute $s=r^2$ in the integral), so an equivalent condition is that $v(x,r)$ is constant in $x$.

I couldn't find any references on this condition, though, and I didn't easily see how to make such a space that isn't symmetric.