Given a Riemannian manifold $(M,g)$ (can be taken parallelizable if needed) a second order differential operator $\mathcal{L}$ can be expressed in a local chart as
$$ \mathcal{L}f(x)= \left(a^{ij}(x) \partial_i \partial_j+ b^j(x) \partial_j +c(x)\right)f(x) $$ The operator is said uniformly elliptic (w.r. to $g$) if, given an atlas $\{(U_\alpha,(x^i)_\alpha)\}$ exists a constant $C>0$ such that in any chart
$$ a^{ij}(x) \xi_i\xi_j \ge C g^{ij}(x) \xi_i\xi_j $$ It is immediate that a uniformly elliptic operator is elliptic.
Let now $u(t,x)$ a classical solution of the Cauchy problem
\begin{align} \label{parabolic}(\partial_t -\mathcal{L}) u(t,x)&=0\\ u(0,x)&=f(x) \end{align} Where $f: M \rightarrow \mathbb{R}$ Is a smooth function. If we assume the coefficients to be smooth functions it is true that $\mathcal{L}u(t,x)$ is a solution for the Cauchy problem? i.e has $\mathcal{L}u(t,x)$ the same regularity of $u(t,x)$?
EDIT:
If the manifold is compact the result should follow. Indeed it is possible to prove that for compact manifolds any elliptic operator is the generator of a Feller semi-group. For a dense classe of functions the semigroup can be written as $e^{tA}$. By the Kolmogorov backward formula. for any such function $e^{tA}f$ is a solution of the PDE.
If the function is taken to be smooth of course $u(t,x):=e^{tA}f(x)$ is smooth in $t$ and in $x$.
I am not abke to characterize in any useful way the set of functions for which this works.