Let $d\vec{X} = \vec{\mu}(\vec{X})dt +\sigma(\vec{X})d\vec{B}_t$ be a multidimensional SDE. It has infinitesimal generator $$\mathscr{L}f = \vec{\mu}^T \nabla f + \frac12 \operatorname{Tr}(\Sigma \nabla^2 f),$$ where $\Sigma = \sigma \sigma^T$ for smooth functions $f\in C^2$. The adjoint operator $\mathscr{L}^*$ is found by differentiating with respect to $t$ the integral $$\int f(y) p(t,x,y)dy$$ and using Ito's lemma, integration by parts and some regularity to get $$\int f(y) p_t(t,x,y)dy = \int \mathscr{L}f(y) p(t,x,y)dy=\int f(y) \mathscr{L}^* p(t,x,y)dy,$$ obtaining the Fokker-Planck equation $$p_t = \mathscr{L}^* p.$$ Explicitly, $$\mathscr{L}^* p = -\partial_i [p \mu^i]+\frac12 \partial_{ij}^2 [p\Sigma_{ij}],$$ where we have omitted the summation signs over $i$ and $j$ to de-clutter the notation.
Let $\tilde{p}(t,x,y) =p(t,x,y) \sqrt{\det g(y)}$ for a PD matrix field $g(y)$. Further let $$\tilde{\mathscr{L}} p := \frac{1}{\sqrt{\det g}} \mathscr{L}^* \tilde{p} = \frac{1}{\sqrt{\det g}}\mathscr{L}^* (p \sqrt{\det g}).$$
The above work shows that, with $\nu(dy) = \sqrt{\det g(y)} dy$, $$\int f(y) \tilde{\mathscr{L}} p(t,x,y) \nu(dy)=\int \mathscr{L}[f(y)] p(t,x,y) \nu(dy),$$ in other words the adjoint of $\mathscr{L}$ with respect to the volume measure is $\tilde{\mathscr{L}}$. Therefore, when $\mathscr{L} = \frac12 \Delta_g$, i.e. is one half the Laplace-Beltrami operator, we should get $\mathscr{L} = \tilde{\mathscr{L}}$, since it is well known that the Laplace-Beltrami operator is self-adjoint with reference to the volume measure. In this case $\mu =\frac12 \nabla_g \cdot g^{-1}$, the manifold divergence of the rows of $g^{-1}$, and $\Sigma = g^{-1}$, the inverse-metric tensor. Precisely, $$\nabla_g \cdot \vec{f} = \frac{1}{\sqrt \det g} \nabla \cdot (\sqrt{\det g}\vec{f})$$ for vector fields $\vec{f}$.
Questions:
Is this all correct?
How can I actually show the Laplace-Beltrami operator is self-adjoint in this context/language $\tilde{\mathscr{L}} = \mathscr{L}$? i.e.
$$\tilde{\mathscr{L}} f = -\frac{1}{\sqrt \det g} \partial_i \left[0.5 \nabla_g \cdot g^{i} \sqrt{\det g} f\right]+\frac{1}{2\sqrt{\det g}} \partial_{ij}^2 [g^{ij} \sqrt{\det g} f]$$ $$ = (0.5 \nabla_g \cdot g^{-1})^T \nabla f +\frac12 \operatorname{Tr}(g^{-1} \nabla^2 f)?$$ I am familiar with Christofell symbols $\Gamma_{ik}^r$, their relation to manifold divergence etc. My guess is to use Jacobi's formula for derivatives of determinants and some properties of $\Gamma_{ik}^r$, $g$, and $\nabla_g \cdot g^{-1}$ and simplify, simplify, simplify. I will try to write this out tonight.
Update: I think using the form $$\Delta_g f = \frac{1}{\sqrt{\det g}} \partial_i (\sqrt{\det g} g^{ij} \partial_j f)$$ makes it easy to directly show, integrating by parts twice, $$\int \Delta_g [f(y)] p(t,x,y) \nu(dy) = \int f(y) \Delta_g[p(t,x,y)] \nu(dy),$$ hence it follows that $\Delta_g$ is self-adjoint (w.r.t. to $\nu$) and thus we must also have $\Delta_g = 2\tilde{\mathscr{L}}$.
But, I would still like to see direct computations that show $\tilde{\mathscr{L}}=\frac12 \Delta_g$ (which I am trying to work out the idea I mentioned above).