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''](https://www.math.uni-bielefeld.de/%7Egrigor/nog.pdf) or [''Heat kernel bounds on hyperbolic space and Kleinian space''](https://londmathsoc.onlinelibrary.wiley.com/doi/epdf/10.1112/plms/s3-57.1.182)) 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.



Heat kernel on hyperbolic space of variable curvature
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


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


