Revised to more specific question after realizing some steps

Let $N(x)$ be a $K\times D$ matrix function of $x\in \mathbb{R}^D$ with orthonormal rows, so that $N(x)N(x)^T=I_{K\times K}$. Let $\vec{n}_k$ be the $k$-th row of $N$, written as a column vector, and let $D[\vec{n}_k]$ be the Jacobian matrix of $\vec{n}_k$. Consider the matrix equation $$P(x) \nabla \log p(x) = N(x)^T \vec{q}(x)+\sum_{k=1}^K D[\vec{n}_k(x)]\vec{n}_k(x),$$ where the vector $\vec{q}(x)$ has $r$-th component $$q^r(x) = \operatorname{Tr}(N(x) D[n_r(x)] N^T(x)),$$ for $r=1,2,\dotsc, K$ and where $P(x)=I_{D\times D} - N(x)^T N(x)$ is the orthogonal projection of $\mathbb{R}^D$ onto the tangent space $T_x M$ at $x$ of some manifold $M$, and the unknown $p(x)$ is a nice enough function of $x\in \mathbb{R}^D$.

Question: Can we solve this equation explicitly for the unknown function $p(x)$?

Some simplifications and easy cases: When $K=1$, and $\operatorname{Tr}(n(x)^T D[n(x)] n(x)) n=0$ we have $P(x)\nabla \log p(x) = D[n]n$, I know how to proceed, see this answer. This more general case is giving me some trouble however.

In general, notice that applying $P(x)$ to both sides of the original equation and using the fact that $P(x)^2=P(x)$, that $D[\vec{n}_k(x)]\vec{n}_k(x)$ already lies on $T_x M$, and that (suppresing notation on $x$) \begin{align*} PN^T q &= (I-N^TN)N^T q\\ & = N^Tq -N^T N N^Tq\\ & = N^Tq - N^T I_{p\times p} q\\ & = N^Tq-N^Tq=\vec{0} \end{align*} where we have used the orthogonality of $N$, i.e. $NN^T=I_{K\times K}$, we obtain the equation $$P \nabla \log p = \sum_{k=1}^K D[\vec{n}_k]\vec{n}_k.$$ It follows that $$\nabla \log p = \sum_{k=1}^K \left(D[\vec{n}_k] +c_kI_{D\times D} \right)\vec{n}_k$$ for some scalar functions $c_k$ (since $\{n_1,\dotsc, n_K\}$ is an orthonormal basis for the kernel of $P$.

It remains to show the RHS is a gradient, analogous to the case $K=1$. I think it is possible to generalize the argument for $K=1$ but the exact details are escaping me at the moment, so I would appreciate any tips or ideas.

Update 2/26/2023: When $N$ comes from orthonormalizing a Jacobian matrix $D[f]$ with full rank on $f^{-1}(\{0\})$ of some smooth function, and we also assume that the rows of $D[f]$ are already orthogonal then we can directly use the previous method linked above with the appropriate changes. We obtain in this case that $\sum_{k=1}^K D[\vec{n}_k]\vec{n}_k = \frac12 \nabla \left[\sum_{k=1}^K H_k\right] - \sum_{k=1}^K \mu_k \vec{n}_k$ and then we get that $$\nabla \log p(x) = \nabla \left[\sum_{k=1}^K \frac12 H_k\right]$$ $$ = \nabla \left[\log \prod_{k=1}^K \|\nabla f_k\|\right],$$ hence $p(x) = \prod_{k=1}^K \|\nabla f_k(x)\|$. Here $H_k = 2\log \|\nabla f_k\|$ and $$\mu_k = \frac{1}{\|\nabla f_k\|} n_k^T (\nabla^2 f_k)n_k,$$ and $n_k = \nabla f_k / \|\nabla f_k\|$. I have omitted a bit of the details but if anyone wants to see them I can add them in later when I have more time.

In general, my conjecture is the solution is $p(x) = \sqrt{\det J_f(x) J_f(x)^T}$, where $J_f := D[f]$.



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