Let $M$ be a differentiable manifold that is also a metric space $(M,d)$, equipped with the topology induced by the metric distance $d$; further, assume that $d$ is $C^2$ on $M\times M$. Now, let $D:M\times M\to[0,+\infty )$ be the squared metric distance, i.e. $D(p,q):=(d(p,q))^2$.
Is it true that $D$ is also a divergence?
Non-negativity and positivity are trivial but I can't prove that
For every $p\in M$, $D(p,p+dp)$ is a positive-definite quadratic form for infinitesimal displacements $dp$ from $p$.
This holds by definition if the manifold is a Riemannian manifold $(M,g)$ and $d$ is the geodesic distance induced by $g$. In general, though, I can only prove that the quadratic form is positive-semidefinite (as here).