Let $f:\mathbb{R}^d \to \mathbb{R}$ be smooth with non-vanishing gradient and define $n: \mathbb{R}^d \to \mathbb{R}^d$ by
$$n(x)=\nabla f(x)/\|\nabla f(x) \|.$$
Let $J_n(x)$ be the Jacobian matrix of $n$.
Question
Does there exist a smooth real-valued function $v:\mathbb{R}^d \to \mathbb{R}$ such that the vector function
$$Z(x) = (J_n(x)+v(x)I)n(x)$$
has a symmetric Jacobian matrix, i.e.
$$\frac{\partial Z^i}{\partial x_j} = \frac{\partial Z^j}{\partial x_i}?$$
Context/Examples
Here $M=f^{-1}(\{0\})$ defines a smooth manifold and $P=I-nn^T$ is the orthogonal projection onto the tangent space of $M$. We wish to find scalar function $p$ such that
$$P(x) \nabla \log p(x) = J_n(x) n(x).$$
Note that $P$ leaves the RHS unchanged, so that
$$\nabla \log p(x) -J_n(x) n(x)$$
must be in the kernel of $P$ which is equal to the span of $\nabla f$. Thus,
$$\nabla \log p(x) =J_n(x)n(x) + c \nabla f(x).$$
For some scalar $c$. Let’s assume this is actually a scalar function, in the form $c = v(x)/\|\nabla f(x)\|$ so that we can rewrite the last displayed equation as
$$\nabla \log p(x) = [J_n(x) +v(x)I]n(x)=: Z(x)$$
and for the RHS to be a potential we need the “curl free” condition written above regarding the partials. So we wish to choose $v(x)$ such that this condition holds.
For the ellipse and ellipsoid, I have guessed solutions $p$ which satisfy $P\nabla \log p = J_n n$ so that I then could back out $v(x)$ (I can typeset these when I’m back at my desk later) but unfortunately I have not found a direct way of constructing or finding $v$ to then recover $p$ through gradient integration. Any help is appreciated, and please comment for corrections or clarifications.
Best Answer
We will show that if we add a multiple of $n$ to $J_n(n)$, then it is a gradient of some function. (Automatically, symmetric Jacobian matrix is followed).
Proof : Define $$ H = {\rm ln}\ |\nabla f|^2 $$ where $ \nabla f =f_i E_i, \ f_i=E_i(f)$ and $E_i = \frac{\partial }{ \partial x_i}$
Hence $E_q (H)=\frac{2\sum_i f_i f_{iq} }{|\nabla f|^2} $ so that $$ \nabla H = 2 {\rm Hess}\ (f)(n) \frac{1}{|\nabla f|} $$ where ${\rm Hess} (f) $ is a Hessian matrix of a function $f$.
Here note that $J_n(n) = n(n)$ so that \begin{align*} n(n) &= \frac{\nabla f}{|\nabla f|} ( \frac{ \nabla f}{|\nabla f|} ) \\& = \frac{1}{H} f_i f_{ij}E_j + \frac{1}{|\nabla f|} \nabla f (e^{-\frac{1}{2} H } ) \nabla f \\ & = {\rm Hess}\ (f)(n) \frac{1}{|\nabla f|} + \nabla f(e^{-\frac{1}{2} H} ) n \end{align*}
Hence $$ n(n) - \nabla f(e^{-\frac{1}{2} H} )n = \frac{1}{2} \nabla H $$