If $L$ is elliptic, is $\partial_t - L$ hypoelliptic?

Question

If $L$ is the Beltrami Laplacian on $(M,g)$, then it is known that $P = \partial_t - L$ is hypoelliptic. Can the same be said in general, when $L$ is arbitrary elliptic (of order $2$, if it helps)? Is it possible to draw conclusions about hypoellipticity by looking at the symbol of the operator? (For simplicity, I consider $L$ to act on complex-valued functions, not on sections in arbitrary vector bundles.)

Answer 1

No, the Schrödinger operator $i\Delta$ is a counterexample. If you assume that $L$ is self-adjoint (and some technical assumptions) then you get that $L$ generates an analytic semigroup and therefore $e^{-tL}$ is smoothing. For details see for instance the book by Renardy and Rogers.