Let $(M,g)$ be a noncompact Riemannian manifold whose isometry group acts transitively on $M$, i.e. a (not necessarily normal) homogeneous space. Let $e_{\lambda}(x,y)$ be the integral kernel of $f \mapsto \int_{0}^{\lambda} dE_{\nu}(f)$ where $dE_{\nu}$ is the spectral measure of the (non-negative) Laplacian associated to $(M,g)$. Is there a relatively simple proof that

$$ \int_M |\nabla_x e_{\lambda}(x,y)|^2_g\, dy = \int_M e_{\lambda}(x,y) \cdot \Delta_x e_{\lambda}(x,y)\, dy \ ?$$

Note that the differentiation is in $x$ and the integration is in $y$. In particular, it's surely false for most non-homogeneous manifolds.

I ask for `relatively simple' because I know of a proof using the expression of the Laplacian as the generator of the heat semi-group. That proof seems to be overkill.


Your Answer

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

Browse other questions tagged or ask your own question.