5

So, I've seen in a few places this method of calculating the heat kernel on a manifold given the kernel of its universal cover, through a so-called 'tiling method' as in section five of this paper (http://arxiv.org/pdf/1007.5467.pdf). Basically, you end up getting:

$$ \begin{equation} K_M(\textbf{x}, \textbf{y}, t) = \sum\limits_{g \in G} K_\tilde{M}( \tilde{\textbf{x}}, g \cdot \tilde{\textbf{y}}, t) \end{equation} $$

where $\tilde{M}$ is the universal cover of $M$ and $G$ is the covering group. I'm wondering if you can do a similar thing with a fundamental solution of the Laplace equation, which I'd really like to calculate for arbitrary Riemann surfaces (actually, compact hyperbolic surfaces).

specterhunter
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  • I'm not immensely satisfied but this as an answer BUT, it is true that the fundamental solution of the Laplace equation on a closed manifold (like the torus or the compact hyperbolic surfaces I'm interested in) satisfies:

    $$ \Delta G = \delta(x) - \frac{1}{|A|} $$

    where $A$ is the surface area of the manifold in question. Whereas the fundamental solution on the Euclidean or hyperbolic plane just satisfies:

    $$ \Delta G = \delta(x). $$

     – specterhunter Jan 29 '14 at 06:00
  • So, I guess you can't, in general for the Laplace equation, lift initial conditions and get a G-periodic solution to the same equation. – specterhunter Jan 29 '14 at 06:06

0 Answers0