Nature of a certain invariant on smooth field of positive definite matrices

Question

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.

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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).

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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 $x_0$.

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Update 3. If we take the condition punctually at some $x$. Posing $G_{x} : y \mapsto g(x) y$, the Jacobian is $g(x)$, then

$\partial_i (g_{jk} \circ G_x) = \sum_s g_{si} \partial_s g_{jk}(G_x)$

The condition then writes, in these terms, $\partial_i(g_{jk} \circ G) = \partial_k(g_{ji} \circ G_x)$. Moreover as $g$ is symmetric, all index permutations are allowed.

In summary, this amounts to saying the $3$-tensor $\nabla (g\circ G_x)$ is symmetric.

I am still not clear on the significance of this (g eating itself has a symmetric "Jacobian tensor") but this seems like a step forward.

Answer 1

I don't see where $g$ being SPD is used? The content of this question boils down to: given $F:\mathbb{R}^n\to\mathbb{R}^n$, I want to know whether there is a $n\times n$-matrix valued function $A:\mathbb{R}^n \to \mathbb{R}^{n\times n}$ such that $\partial_i F_j = A_{ij}\circ F$.

This would require the integrability condition that $$ \partial_k \partial_i F_j = \partial_i \partial_k F_j \implies \sum_{\ell} A_{k\ell} \partial_\ell A_{ij} - A_{i\ell} \partial_\ell A_{kj} = 0 $$ and has nothing to do with whether $A$ admits a factorization $A = M g$ where $M$ is a constant matrix and $g:\mathbb{R}^n\to\mathbb{R}^{n\times n}$ is SPD. (In other words, the existence of factorizations has nothing to do with this equation that you derived.)

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As to "solving" the PDE that you derived: my advice is "don't". Instead, argue as follows:

An immediate consequence of the assumption that $\partial F = A\circ F$ is that you can integrate this expression along paths; assuming $A$ is Lipschitz continuous, by Picard-Lindelof you have the following result:

If $x_1, x_2\in \mathbb{R}^n$ are such that $F(x_1) = F(x_2)$, then for any $y\in \mathbb{R}^n$ we have $F(y + x_2 - x_1) = F(y)$.

Proof: let $\gamma$ be a smooth curve from $x_1$ to $y$, let $\eta = \gamma + x_2 - x_1$ which is a curve from $x_2$ to $y + x_2 - x_1$. Observe that $\gamma' = \eta'$ since they differ by a constant. Thus $F\circ \gamma$ and $F\circ \eta$ solve the same ODE with the same initial conditions, and hence take equal values. q.e.d.

Thus there exists a subgroup $\Gamma$ of the translation group on $\mathbb{R}^n$ such that $F$ factors through a quotient $$ F = G\circ \Pi, \quad \mathbb{R}^n \overset{\Pi}{\to} \mathbb{R}^n / \Gamma \overset{G}{\to} \mathbb{R}^n $$ with $G$ being injective.

It turns out that this is also sufficient: suppose $F$ factors through a quotient as above, then $d\Pi$ is a (constant) projection map, and since $G$ is injective it is invertible as a function, in particular, we can set $$ A = (dG \circ G^{-1}) \cdot d\Pi $$ and this gives a representation.

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Now we can return to the factorization problem. Let's treat the generic case where $dF$ is surjective at one point. In this case, both $M$ and $g$ have to be full rank. Luckily $g$ is also assumed to be SPD, and so no problem. But as $M$ is constant, this means that $dF$ is full rank everywhere, so that $F$ is a local diffeomorphism, and the subgroup $\Gamma$ has to be discrete.

In fact, $\Gamma$ has to be the empty group. If the factorization exists, then we can linearly transform the domain $\mathbb{R}^n$ by the matrix $M$ so that for the transformed function $\tilde{F}$ we have $d\tilde{F} = g\circ \tilde{F}$. Taking the dot product against a fixed vector $v$ we have

$$ \partial_v (\tilde{F} \cdot v) > 0 $$

using the SPD condition, which shows that $\tilde{F}$ cannot be periodic in any direction.

Additionally, the assumption that $d\tilde{F}$ is symmetric means that the antisymmetric part of $d\tilde{F} = 0$, which means that $\tilde{F}$ can be written as $d\phi$ for some function $\phi:\mathbb{R}^n\to\mathbb{R}$.

So we have the following conclusion:

Thm: If a smooth function $F$ with $dF$ surjective at some point $x_0$ satisfies the decomposition $dF = M\cdot g(F)$ where $M$ is constant, and $g$ is SPD, then $M$ is full rank and $F(x) = \tilde{F}(Mx)$ where $\tilde{F} = d\phi$ for some strictly convex function $\phi$.

The converse is also obviously true.