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Let $f\in C^{\infty}(\mathbb{R}^D, \mathbb{R}^K)$ be a smooth function whose Jacobian matrix $Df$ has full rank on $M=\{x\in \mathbb{R}^D: f(x)=\vec{0}\}$. We take the convention that the rows of the Jacobian are the transposes of the gradients of the components of $f$. Let $N(x)$ be the matrix obtained by orthonormalizing the rows of of $Df(x)$, so that $N(x)N(x)^T=I_{K\times K}$.

Question: Is it always true that $\operatorname{Tr}(N D[n_r] N^T)=0$?

When $K=1$, we can prove this is indeed true. In this case, we have $n=\nabla f/\|\nabla f\|$ and we just have to show $n^T D[n]n=0$. It is relatively straightforward to show that $$D[n]n = \frac{1}{\|\nabla f\|} (\nabla^2 f) n- \frac{1}{\|\nabla f\|}\left(n^T (\nabla^2 f)n\right)n.$$ Multiplying this on the right by $n^T$ gives $$n^T D[n]n = \frac{1}{\|\nabla f\|} n^T(\nabla^2 f) n- \frac{1}{\|\nabla f\|}n^T\left(n^T (\nabla^2 f)n\right)n.$$ $$=:\frac{1}{\|\nabla f\|} v- \frac{1}{\|\nabla f\|}n^Tvn.$$ where we have written $v:=n^T(\nabla^2 f)n$ for convenience. Now, $v$ is just a scalar, so we can rewrite the second term as $$\frac{1}{\|\nabla \|} vn^T n = \frac{v}{\|\nabla f\|},$$ since $n^Tn=1$ due to $n$ being a unit-normal vector. Thus we obtain $$n^T D[n]n = \frac{1}{\|\nabla f\|} v - \frac{1}{\|\nabla f\|} v=0,$$ as claimed.

When $K>1$, it is a lot more difficult to follow one's nose here...

Update 8/31/2023 Fix $r$ and $r'$ in $\{1,2,\dotsc, K\}$. Using linearity and product rule, we can show that, since $n_r^Tn_{r'} = \delta_{rr'}$ $$D[n_r]^Tn_{r'}=-D[n_{r'}]^Tn_r,$$ hence $D[n_r]^T n_r = \vec{0}$. It follows that $$(ND[n_r]N^T)_{ij} = -n_r^T D[n_i]n_j,$$ hence $$\operatorname{Tr}(ND[n_r]N^T) = -\sum_{j=1}^K n_r^T D[n_j]n_j$$ $$=-\sum_{j\neq r} n_r^T D[n_j]n_j,$$ since $D[n_r]^T n_r=\vec{0}$, hence we know $n_r^T D[n_r]n_r = (D[n_r]^T n_r)^T n_r = \vec{0}^T n_r = 0$. But it is difficult for me to see anything else.

Update 9/1/2023 Let's assume there exists a vector $v(x)$ such that $$P(x)v(x) = \sum_{k=1}^K D[n_k(x)]n_k(x),$$ where $P(x)=I-N(x)^T N(x)$ is the orthogonal projection onto $T_xM$ the tangent space of $M$ at $x$. Then from the previous update, we have with $r=i$ and $r'=k$, $$n_k^TD[n_i] = -n_i^T D[n_k]$$ hence multiplying on the right by $n_k$ gives $$n_k^T D[n_i] n_k = -n_i^T D[n_k]n_k,$$ so that summing from $k=1$ to $K$ gives $$\operatorname{Tr}(N D[n_i] N^T) = -n_i^T \left(\sum_{k=1}^K D[n_k]n_k\right)$$ $$ = -n_i^T Pv,$$ but $Pv$ lies in $T_xM$ and obviously $n_i \perp u$ for any $u\in T_xM$. Thus $-n_i^T Pv = 0$ hence this trace term is zero, $$\operatorname{Tr}(ND[n_i]N^T)=0,$$ for $i=1,2,\dotsc, K$.

So, now my questions are:

  1. taking the assumption for granted, is this argument sound and valid?
  2. Is the assumption true? I have essentially already asked it in this question and just made the connection today between the two. There $v=\nabla \log p$ where $p$ solves the steady-state Fokker-Planck equation of the process satisfying $dX_t = P(X_t)\circ dB_t$. I have proven it in the case $K=1$, in which then $p$ is proportional to $\|\nabla f\|$.
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