When are heat kernels only dependent on the distance?

Question

"All" the examples of heat kernels in circulation are only dependent on the distance between the space variables rather than on the space variables themselves, i.e. $$K(t;x,y) = K(t;d(x,y)).$$

Think to the heat kernel on the Euclidean $d$-dimensional space, on the $d$-dimensional hyperbolic space, or even to the heat kernel associated to the laplacian acting on modular forms of integer weight on the upper half plane.

While wondering why this was true I stumbled across this article, where, at the top of page $2$, the author observes that this happens because of the homogeneousness of the space. This seems to be a very reasonable statement, but how to prove it?

We claim that if $X$ is a homogeneous Riemannian manifold, i.e. if the group of isometries $\mathrm{Isom}(X)$ of $X$ acts transitively on the set of points of $X$, then, for $x,y,x',y'\in X$ s.t. $d(x,y) = d(x',y')$ we have $$K_X(t;x,y) = K_X(t;x',y').$$

Since the heat kernel is isometry invariant, it would be enough to produce a $\varphi\in\mathrm{Isom}(X)$ such that $\varphi(x) = x'$ and $\varphi(y) = y'$ to get the statement. But this seems to me a stronger requirement than what the transitiveness of $\mathrm{Isom}(X)$ ensures, specifically this is the $2$-transitivity. What am I missing?

Moreover I would be grateful if somebody could point to me an example of a manifold which is known not to have a heat kernel purely dependent on the distance.

Thanks!

Answer 1

This might be more a comment than an answer and I only adress your second question but here goes. The heat kernel will almost never depend only on the distance. For example, think of a manifold $M$ on which there are two distinct parts: a flat part (where you have Euclidean volume growth), and a very thin, cylindrical part (where the volume growth is much smaller than Euclidean). Now for $i=1,2$, think of a heat source at a point $q_i$ heating up a point $p_i$ at distance $\text{dist}(p_i,q_i) = d$, where $p_1,q_1$ are in the thin, cylindrical part and $p_2,q_2$ in the flat part. It is physically obvious that for a same amount of time, at least for small times, the heat from $q_2$ in the flat part of $M$ will be more spread out than the heat from $q_1$ in the thin part. So the point $p_1$ will be heated much more rapidly from the heat source $q_1$ than the point $p_2$ from the heat source $q_2$. Thus for $k(t,x,y)$ the heat kernel on that space, we can't have $k(t,p_1,q_1) = k(t,p_2,q_2)$ for small times. The same picture but from a Brownian movement point of view might be even more convincing. To make this rigorous, one could use the various bounds on the heat kernel in terms of the local volume growth of balls.

Answer 2

Getting back to your first question, the key ingredient is the uniqueness of the heat kernel and the invariance of the Laplace-Beltrami operator with respect to isometries. Let $G := Isom(M)$, then both functions $K(t,x,y)$ and $K(t,gx,gy)$ for any $g \in G$ solve the heat equation with the correct initial condition. By the uniqueness we thus have $K(t,x,y) = K(t,gx,gy)$ for all $g \in G$, the desired dependence on the distance.