I initially asked on math.stackoverflow but have since come to understand this forum may be more appropriate, as this is indeed a question that arose in writing a research article.

Denote $g$ a symmetric positive definite matrix field (defined over, say, $\mathbb R^n$), and $g_{ij}$ its components in some basis (the same for all $g(x), x \in \mathbb R^n$). So $M$ is a smooth function that, to $x\in \mathbb R^n$, associates a positive definite matrix $g(x)$.

I am carrying out some computations, and they turn out to be valid only if:

$\sum_{s=1}^n g_{si}\partial_s g_{jk} - g_{sk}\partial_s g_{ji} = 0$

where $\partial_t$ denotes differentiation wrt to the $t$-th variable (components in the same basis).

I first thought this condition to be unreasonable. I now have the suspicion this may be some kind of metric compatibility condition, but I don't manage to make sense of the results I find on this topic. Understanding this constraint better could help me know if it is realistic to expect such an $g$ be produced in my circumstances, and perhaps how.

Another lead may be to look at this as a PDE. This resembles the divergence of some field, but I am unfortunately not an expert of this either. Perhaps this corresponds to a well-known PDE.

Per suggestion, for $n=2$. For $j=1$:

$g_{11} \partial_1 g_{11} - g_{21} \partial_2 g_{11} = 0$

$g_{11} \partial_1 g_{12} - g_{22} \partial_2 g_{11} = 0$

$g_{12} \partial_1 g_{12} - g_{22} \partial_2 g_{12} = 0$

For $j=2$:

$g_{11} \partial_1 g_{21} - g_{21} \partial_2 g_{21} = 0$

$g_{11} \partial_1 g_{22} - g_{22} \partial_2 g_{21} = 0$

$g_{12} \partial_1 g_{22} - g_{22} \partial_2 g_{22} = 0$

Despite that there are fewer terms than possible combinations, the following repeat:

$g_{11} \partial_1 g_{12}$ in eqs. (2) and (4).

$g_{22} \partial_2 g_{12}$ in eqs. (3) and (5).

I thought I might add some context. Originally, I am interested in a smooth function $F : \mathbb R^n \to \mathbb R^n$. If I want to impose $J_F(x) = g(F(x))M$ ($M$ a matrix allowing this to hold with $g$ spd) then, for all $i,k$,

$\partial_i F_k = \sum_s g_{ks}(F) M_{si}$

Now if we differentiate:

$\partial_{ij} F_k = \sum_{st} \partial_j F_t \partial_t g_{ks}(F) M_{si}$

$\partial_j F_t$ can be substituted using the previous relationship:

$\partial_{ij} F_k = \sum_{stu} g_{tu}(F) M_{uj} \partial_t g_{ks}(F) M_{si}$

rearranging the sums:

$\partial_{ij} F_k = \sum_{su} M_{uj}M_{si} \sum_t g_{tu}(F) \partial_t g_{ks}(F) $

Fixing $k$, the left-hand side is symmetric wrt $(i,j)$ thus so is the right-hand side. Thus:

$\sum_{su} M_{uj}M_{si} \sum_t g_{tu}(F) \partial_t g_{ks}(F) = \sum_{su} M_{uj}M_{si} \sum_t g_{ts}(F) \partial_t g_{ku}(F)$

i.e.:

$\sum_{su} M_{uj}M_{si} \sum_t (g_{tu}(F) \partial_t g_{ks}(F) - g_{ts}(F) \partial_t g_{ku}(F)) = 0$

In my application case, I can choose $M$ "arbitrary enough" (but perhaps there is some pattern I didn't see that makes some of these relations trivial). Thus I need that, for all $u,s$,

$\sum_t (g_{tu} \partial_t g_{ks} - g_{ts} \partial_t g_{ku})$

Maybe my trouble is I am writing $J_F(x) = g(F(x))M$ in a sense too restricted, and a more general framework in which $\partial_{ij} F \ne \partial_{ji} F$ would be more suited.

Also, following a slightly different path (imposing a different condition than $J_F = ...$), this condition does not appear as for all $x$, but is only necessary at a given $x_0$.