I am working with the heat kernel on the hyperbolic space explicitly (as you may guess by my previous questions) and I got the desired results when the curvature is $-\kappa=-1$. Now I am trying to do the same for a fixed but arbitrary curvature $-\kappa<0$, so I need to generalize the explicit formulas for the heat kernel that I got in $\mathbb{H}^n(1)$ (see ''The Heat Kernel on Hyperbolic Space'' or ''Heat kernel bounds on hyperbolic space and Kleinian space'') to a more general $\mathbb{H}^n(\kappa)$.

I achieve to get a general formula when $n$ is odd, that is $$ p_{n}(\rho, t)=\frac{(-1)^{m}}{2^{m} \pi^{m}} \frac{1}{(4 \pi t)^{\frac{1}{2}}}\left(\frac{\kappa}{\sinh (\kappa\rho)} \frac{\partial}{\partial \rho}\right)^{m} e^{-\kappa^2m^{2} t-\frac{\rho^{2}}{4 t}} $$ with $n=2m +1$, but I did it by just adding $\kappa$ in the formula and then cheking that it satisfies the equation, that is now $$\frac{\partial^2}{\partial \rho^2}p_{n}(\rho, t)+(n-1)\kappa \coth(\kappa \rho)\frac{\partial}{\partial \rho}p_{n}(\rho, t)-\frac{\partial}{\partial t}p_{n}(\rho, t).$$ When I try to do the same to even $n$, it fails because I can not compute the derivatives so easily.

Could someone help me find a general formula when $n=2m+2$? I have some intuition but I can not check if they are, in fact, the fundamental solution that I want.


1 Answer 1


The sectional curvature scales like the inverse of the metric. So fixing a coordinate system on $\mathbb{H}^n(1)$, with metric $g$, the scaled metric $\kappa^{-1} g$ has sectional curvature $-\kappa$.

If $u(t,x)$ solves the heat equation you have $$ u_t = \Delta_g u \iff \kappa u_t = \kappa \Delta_g u \iff \kappa u_t = \Delta_{\kappa^{-1} g} u $$ and so $u(\kappa t,x)$ solves the heat equation for the $-\kappa$ curvature.

The function $\rho$ being the geodesic distance scales like $\kappa^{1/2}$: that is $\rho_g = \kappa^{1/2} \rho_{\kappa^{-1} g}$.

And the volume form scales like $\mathrm{dvol}_g = \kappa^{n/2} \mathrm{dvol}_{\kappa^{-1} g}$.

This tells you that if you write $\tilde{p}$ for the heat kernel when the section curvature equals $-\kappa$, and $p$ for the heat kernel when the sectional curvature equals $-1$, you should have

$$ \tilde{p}_n(\rho,t) = \kappa^{n/2} p_n(\kappa^{1/2}\rho, \kappa t)$$

So first: your formula for the odd dimensional case is wrong. Where you have $\kappa$ you should have $\kappa^{1/2}$ instead (unless you are actually looking at the case where sectional curvature equals $-\kappa^2$). The correct formula should be

$$ \tilde{p}_{n}(\rho, t)=\frac{(-1)^{m}}{2^{m} \pi^{m}} \frac{1}{(4 \pi t)^{\frac{1}{2}}}\left(\frac{\kappa^{1/2}}{\sinh (\kappa^{1/2}\rho)} \frac{\partial}{\partial \rho}\right)^{m} e^{-\kappa m^{2} t-\frac{\rho^{2}}{4 t}} $$

In the even case (n = 2m+2) you should have

$$ \tilde{p}_n(\rho, t) = \frac{(-1)^m}{2^{m+5/2} \pi^{m+3/2}} \kappa^{-1/2} t^{-3/2} e^{-\frac{(2m+1)^2}{4} \kappa t} \left( \frac{\kappa^{1/2}}{\sinh \kappa^{1/2}\rho} \partial_\rho \right)^m \int_{\kappa^{1/2}\rho}^\infty \frac{s e^{- s^2/(4\kappa t)}}{(\cosh s - \cosh \kappa^{1/2}\rho)^{1/2}} ds $$

  • $\begingroup$ You're right. I was confused between $\kappa$ and $\kappa^2$. And the answer is very clear and helpful. Thank you. $\endgroup$
    – MathqA
    Nov 23, 2021 at 18:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.