# Ito drift term of extrinsic Brownian motion aka Mean curvature of intersection of hypersurfaces

Suppose we have manifold in the form $$M=f^{-1}(\{\vec{0}\})$$, where $$f:\mathbb{R}^d\to \mathbb{R}^p$$ where $$p=D-d$$, and $$f \in C^{\infty}$$ ands its Jacobian, $$J_f(x)$$ has full rank on $$M$$. Here, we take the convention that the $$i$$-th row of $$J_f(x)$$ is $$\nabla f^i(x)^T$$. The orthogonal projection of a point $$y\in \mathbb{R}^D$$ to the tangent space $$TM_x$$ of $$M$$ at $$x$$ is given by $$P(x)=I_{D\times D} -N(x)^TN(x),$$ where $$N(x)$$ is the matrix obtained by orthonormalizing the rows of $$J_f(x)$$. See my previous question for a discussion of this.

The Stratonovich SDE of Brownian motion on the manifold $$M\subset \mathbb{R}^D$$ driven by a $$D$$-dimensional Brownian motion $$B_t$$ is simply $$dX_t = P(X_t)\circ dB_t.$$

For a variety of purposes (simulation, studying Fokker-Planck equation) I have the following Question What is the Ito form of the above SDE?

Attempt Obviously, we start by applying the conversion formula. In this case, the $$i$$-th component of the drift term in the Ito SDE is given by $$\mu^i(x)=\frac12 \sum_{k=1}^D \sum_{j=1}^D P_{kj}(x)\frac{\partial }{\partial x_k} P_{ij}(x).$$ Well some simplification are obvious. Let $$A(x)=N(x)^T N(x)$$. After applying the definition of $$P_{ij}(x)$$, simplifying the partial, and using properties of $$\delta_{ij}$$, we obtain $$2 \mu^{i}(x)=-\operatorname{div} A_{i \cdot} (x)+\sum_{k=1}^D \sum_{j=1}^D A_{kj}(x) \frac{\partial A_{ij}(x)}{\partial x_k}.$$

The first term is the (negative of the) divergence of the $$i$$-th row of $$A=N^TN$$. The second term I cannot simplify any further with anything clever.

Special case and connection to mean curvature of hypersurfaces:

I do know if $$p=1$$, then things simplify a bit. We have instead $$P(x)=I-n(x)n(x)^T$$ where $$n(x)=\nabla f(x)/\|\nabla f(x)\|$$ is the unit normal vector to $$M$$ at $$x$$. The computations then simplify to $$\mu^i(x) = -\frac12 \operatorname{div}(n(x)) n^i(x),$$ if I am not mistaken. In this case, then $$\mu(x)=c(x)n(x),$$ where $$c(x)=-\frac12 \operatorname{div}(n(x))$$ is actually the mean curvature of $$M$$. We would be tempted to guess then that in the general case, the drift should contain the mean curvature of $$M$$ as a factor. Perhaps something like $$N(x)^T c(x)$$? Here, $$N$$ is the $$p\times D$$ matrix defined above while $$c$$ would be a $$p\times 1$$ vector. Not sure if this is along the right path, but it leads me to ask: what is the mean curvature of a manifold defined by a set of implicit equations $$f^1(x)=0,\dotsc, f^p(x)=0?$$

Summary questions Does anyone have any idea how to simplify the above double summation? Asked in another manner, what is the mean curvature of $$M$$ in this case?

We claim $$\mu = N^T[q+c],$$ where the $$r$$-th component of the vector $$c$$ is given by the mean-curvature vector in the direction of the normal vectors $$c^r = -\frac12 \operatorname{div}(n_r),$$ $$n_r$$ is the transpose of the $$r$$-th row of $$N$$, and $$q^r = \frac12 \operatorname{Tr}(N J_{n_r} N^T),$$ for $$r=1,2\dotsc, p$$.
The proof is relatively straightforward actually. One continues from the expression obtained in the OP $$2 \mu^{i}(x)=-\operatorname{div} A_{i \cdot} (x)+\sum_{k=1}^D \sum_{j=1}^D A_{kj}(x) \frac{\partial A_{ij}(x)}{\partial x_k}.$$ Plug in the expression for the entries of $$A=N^T N$$, apply product rule, one term will cancel after everything and we will be left with $$\sum_{k=1}^D \sum_{j=1}^D \sum_{r=1}^p \sum_{l=1}^p N_{lk} N_{ri} N_{lj} \partial_k [N_{rj}]-(N^T c)^i.$$ The first term is exactly, after a little thought, $$(N^T q)^i$$, as defined above. This completes our sketch.
This will be an incomplete answer, but I believe I found the correct expression. I have verified it for many examples but have yet to completely derive it in general. Nevertheless, the drift, I claim, is given by $$\mu(x)=N(x)^Tc(x)$$ where $$c(x)$$ is the vector of mean curvatures in the direction of the (ortho)normal vectors to the manifold making up the rows of $$N$$, i.e. $$c^i(x)=-0.5 \operatorname{div}(n_i(x))$$ where $$n_i$$ is the $$i$$-th row of $$N$$ written as a column vector.