Steady state density for BM on a compact manifold Background
Suppose we have a surface $M$ in $\mathbb{R}^D$ parameterized by $\theta \in \Omega \subset \mathbb{R}^d$ in the sense that (almost all) points $x$ of $M$ can be written as $\xi(\theta) =x$ for some smooth mapping $\xi$ with smooth inverse. The metric tensor $g$ can be induced from this mapping via $g=J^T J$ where $J$ is the Jacobian of $\xi$ (which is not square).
With the metric tensor in these local coordinates we can write down the Laplace-Beltrami operator and from there a system of SDEs defining Brownian motion on the manifold (after mapping back with $\xi$). Indeed, we have
$$d\theta_t = \frac12 \operatorname{div}_M g^{-1}(\theta) dt + \sqrt{g^{-1}(\theta)} dB_t$$ where $B$ is a BM in $\mathbb{R}^d$, and the manifold-divergence is used here, ie the $i$-th component is equal to $\frac{1}{\sqrt{\det g }} \partial_j (\sqrt{\det g} g^{ij})$ (summation convention is used here).
Now if we consider the Langevin SDE for some potential $U$,
$$ d\theta_t = (\operatorname{div}_M g^{-1} - g^{-1}\nabla U) dt+\sqrt{2g^{-1}}dB_t$$
then we can easily prove this has steady state
$$p_\infty(\theta) \propto e^{-U(\theta)} \sqrt{\det g(\theta)}$$
And if $U=0$ then this reduces to the SDE for a Brownian motion (missing the factor of $1/2$ in the coefficients, of course) that has steady state proportional to $\sqrt{\det g(\theta)}$. This is, of course, normalizable since $M$ is compact.
Question
What is the steady state in the higher dimensional extrinsic setting? How can we transform this PDF when the original mapping $\xi:\Omega \to M$ does not have a square Jacobian with non-zero determinant? I feel like I am missing something simple or subtle here. I was thinking maybe we can “lift” the parameters $\theta$ to $\mathbb{R}^D$ in some suitable fashion and get a new mapping $\tilde{\xi}$ that then has a square Jacobian from which we can apply the usual change of variables formula for PDFs, but I am unsure of how rigorous this can be done.
Any help is greatly appreciated and please correct me if I’ve made any mistakes in my post. Thanks.
Update
After some googling, I think the appropriate change of variables for the PDF in the higher dimensional setting is
$$f(x)=p(\xi^{-1}(x)) \sqrt{\det A(x)}$$
where
$$A(x)=(J\xi^{-1}(x))(J\xi^{-1} (x))^T$$
using the Jacobian of the inverse map $\xi^{-1}: M \to \Omega$ and $x\in \mathbb{R}^D$. I have checked this for the sphere, i.e. that $f$ is the steady state PDF for the SDE
$$dX_t=2c(X_t)n(X_t)dt+\sqrt{2} P(X_t)dW_t$$
where $W$ is a $D$-dimensional BM, $P=I-nn^T$ is the orthogonal projection onto the tangent space at a point on the sphere, $c=-\frac12 \operatorname{div}(n)$ is the mean curvature and $n=F/\|\nabla F \|$ is the normal vector where $M=F^{-1} (\{0\})$ is the implicit function defining $M$ (i.e. $F=x^2+y^2+z^2-1$ for the sphere). I have not been able to verify this same conjecture for a prolate spheroid yet.
It’d be great if someone could verify this, give a reference or a derivation. It seems I mainly need to know about change of variables for PDFs between coordinates of different dimensions, so I may open a new question with this specifically and stripped of the Brownian motion context, or edit this one, not sure what is preferred…
Update 2:
After doing some more work I have found the conjecture in the previous update is false, but is very close. Indeed, if we take
$$f(x)=\frac{1}{\|x\|}p(\xi^{-1}(x))\sqrt{J_{\xi^{-1}}(x) J_{\xi^{-1}}(x)^T}$$
then for the circle, ellipse, sphere, sphereoid, and ellipsoid, I have verified that $f$ satisfies $\mathscr{G}^*f=0$ where $\mathscr{G}$ is the infinitesimal generator of the aforementioned extrinsic SDE and $\mathscr{G}^*$ is its adjoint. Further $f$ satisfies
$$P(x)\nabla \log f(x)=J_n(x)n(x)$$
where $J_n$ is the Jacobian of the normal vector $n$. This is a sufficient and necessary condition for $\mathscr{G}^* h=0$. I believe it also works in the pinched sphere but I am still working out the verification (it is long and tedious).
There are some hypotheses on the inverse parameterization $\xi^{-1}$ that I have not been able to specify that seem to be required for this to work. It would be great if anyone more familiar with differential geometry could elucidate whether this "non-square" change of variable formula is readily in the literature somewhere (the closest to it I have seen is the coarea formula...)
Update 3
(Apologies for lengthening the question too much)
Let me summarize some other findings in the purely extrinsic viewpoint. Let $M=f^{-1}(\{0\})$ be our manifold defined via an implicit equation for a smooth $f:\mathbb{R}^D\to \mathbb{R}$ with non-vanishing gradient. Again let $n=\nabla f/\|\nabla f\|$ be the normal vector to $m$ defined everywhere in a neighborhood of $M$ in $\mathbb{R}^D$ and let $P=I-nn^T$ be the orthogonal projection to the tangent space of $M$ at $x$. Let $\mathscr{G}^*$ be the adjoint of the infinitesimal generator $\mathscr{G}$ of the SDE
$$dX=c(X)n(X)dt +P(X)dW.$$
Finally let $J_n$ be the Jacobian matrix of $n$.
Then

*

*$\mathscr{G}^* p = 0$ if and only if $P(x) \nabla \log p = J_n(x) n(x)$.

Since $P J_n n = J_n n$ it follows that $\nabla \log p -J_n n$ is in the kernel of $P$, which we can easily compute as
$$\ker P(x) = \operatorname{span} \left\{\nabla f(x)\right\}$$
where $f_{x_D}$ is the partial of $f$ with respect to $x_D$, for all $x$ such that $f_{x_D}$ is non-zero at $x$. Thus
$$\nabla \log p(x) = J_n(x) n(x) + v\nabla f(x),$$
for some real $v\in \mathbb{R}$. Now we need to be able to do gradient integration in order to recover $\log p$, which is where I get stuck now. In the case of the circle, sphere, torus and cylinder, we have $J_n(x) n(x)=\vec{0}$ identically, and should have $p=1$ as one solution. Any ideas on how to proceed further?
 A: Here is a solution, although it is somewhat restricted to the case of hypersurfaces and does not establish (at least not immediately obvious to my geometric deficient mind) a transformation from the intrinsic to extrinsic setting. But here it goes.
Let $f:\mathbb{R}^D\to \mathbb{R}$ be a $C^2$ function with non-vanishing gradient on the level set of $0$, so that it defines a hypersurface $M=f^{-1}(\{0\})$. Let $n=\nabla f/\|\nabla f\|$ be the unit normal vector to $M$ defined everywhere in a neighborhood of $M$ and let $P=I-nn^T$ be the orthogonal projection onto the tangent space of $M$ at $x$. Let $\mathscr{G}$ be the infinitesimal generator of the process solving the SDE
$$dX = c(X)n(X)dt+P(X)dW,$$
where $W$ is a $D$-dimensional Brownian motion, $c(x)=-\frac12 \operatorname{div} n(x)$. Let $\mathscr{G}^*$ be the adjoint of $\mathscr{G}$, so that the Fokker-Planck equation for $X$ is $\partial_t u(t,x) = \mathscr{G}^* u(t,x)$. The steady-state density $p(x)$ for reversible diffusions satisfies
$$\mathscr{G}^*p =0.$$
We claim $p=\|\nabla f\|$ solves this steady-state equation. Here are the steps. First, we can show
$$\mathscr{G}^* p \iff P(x)\nabla \log p(x) = J_n(x)n(x)$$
where $J_n(x)$ is the Jacobian of $n$, i.e.
$$J_n(x)_{ij}=\frac{\partial n^i}{\partial x_j}.$$
Since, $M$ is a $D-1$ hypersurface, it is easy to prove that
$$\ker P(x)=\operatorname{span}\{\nabla f(x)\},$$
and together with the fact that $J_n n$ already lies on the tangent space, we have that $p$ must satisfy
$$\nabla \log p(x) = J_n(x)n(x)+\nu(x)n(x).$$
Now let $H=\log \|\nabla f\|^2$. We can show that
$$\nabla H = \frac{2}{\|\nabla f\|} (\nabla^2 f)n,$$
and that
$$J_n n = \frac12 \nabla H - \mu n$$
where
$$\mu = \frac{1}{\|\nabla f\|} n^T (\nabla^2 f)n.$$
The only way for the RHS of $\nabla \log p = J_n n +\nu n$ to be a gradient is if $\nu = \mu$. It follows that $p=\|\nabla f\|$, or more precisely,
$$p(x) \propto \|\nabla f(x)\|,$$
solves the steady-state equation.
Please comment for corrections, clarifications, or suggestions.
