Let $M$ be a Riemannian manifold and $E \to M$ be a Hermitian bundle. If $\nabla$ is a Hermitian connection on $E$, one may define the Laplacian $L = \nabla^* \nabla$, and then consider its Friedrichs extension, that we shall also denote $L$. The "heat" semigroup $t \mapsto \exp(-tL)$ has a smooth "heat" kernel $h$ such that $h(t,x,y) : E_x \to E_y$ is a linear map for every $(t, x, y) \in (0, \infty) \times M \times M$ (where $E_x$ is the fiber over $x \in M$).
- Is it possible that $h(t,x,y) = 0$ for some $(t,x,y)$?
- If yes, is it possible that $h(t,x,y) = 0$ for some $(t,x) \in (0, \infty) \times M$ and $y$ in some open subset $U_{t,x}$?
I am mostly interested in the case $E = M \times \mathbb C$ and $\nabla = \mathrm d + A \wedge$ (with $A$ an imaginary $1$-form).
I know that the answer is negative for the "usual" Laplace-Beltrami operator (corresponding to $A=0$), but the proof of this makes use of the parabolic maximum principle, which in turn depends on $h$ having real (in fact non-negative) values. When $A$ is imaginary, though, $h$ may take complex values, and most techniques from the case $A=0$ are lost.
