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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.

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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

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

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  • $\begingroup$ 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

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