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On a closed Riemannian manifold $M$, the heat kernel $k_t(x, y)$ of the Laplace-Beltrami operator (or more general of any generalized symmetric Laplace-type operator acting on sections of a vector bundle) admits an asymptotic expansion of the form $$ k_t(x, y) \sim \exp\left( -\frac{1}{4t}d(x, y)^2\right) \sum_{j=0}^\infty t^j \Phi_j(x, y) $$ where $d(x, y)$ denotes the Riemannian distance and $\Phi_j$ are appropriate smooth functions, not depending on $t$. This is meant in the sense that for each $N \in \mathbb{N}$, there exists a constant $C>0$ such that for all $x, y \in M$, we have $$ \left| k_t(x, y) - \chi(x, y)\exp\left( -\frac{1}{4t}d(x, y)^2\right) \sum_{j=0}^Nt^j \Phi_j(x, y) \right| < C t^{N+1}$$ where $\chi(x, y)$ is an appropriate cutoff function that is $\equiv 1$ near the diagonal.

In the case that $M$ is still compact but has a boundary, in many books there can be found an asymptotic expansion of the trace, but I could not find an asymptotic expansion of the kernel itself, uniform on $M \times M$. Is there such an expansion?

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You should specify a boundary condition for the heat equation, because the kernel will depend on it. // If the boundary is geodesic, then the double of $M$ can help (placing the second source of heat on the opposite point in the double), but in general this does not work. –  user53153 Feb 22 '13 at 16:07
    
Well, to start with, suppose that you have Neumann or Dirichlet boundary conditions and assume that the operator acts on functions. If one takes the double in the case of geodesic boundary, then the problem is quite simple, I know, but this is a quite heavy condition... –  Kofi Feb 22 '13 at 17:00

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