6
$\begingroup$

First, a bit of notation (I'm assuming Einstein's convention on summation of index):

$\text{Hessian}(f):=\nabla^2(f):=\nabla\nabla(f)=\left(\partial_i\partial_j(f)-\Gamma_{ij}^k\partial_k(f)\right)dx^i\otimes dx^j$

Where $\nabla$ is the Levi-Civita connection on a riemannian manifold $(M,g)$ and $\Gamma$ are the Christoffel symbols. By definition and a bit of computation, it is easy to note that $\nabla^2(f)\in\ \text{TM}^*\otimes\text{TM}^*$, as $g$, the metric tensor. My questions are:

As it never been analyzed the function $f:\nabla^2(f)=g$ for a general manifold (that is: we can associate to a function $f$ a metric, but can we do the opposite in general?) How? Has this $f$ any kind of application? If so, where?

I would really appreciate any kind of suggestion, expecially books or articles (I've found something, like https://arxiv.org/pdf/1312.1103.pdf, but is a bit too advanced for me, talking about transformations I haven't studied yet).

Thanks in advance

$\endgroup$
2
  • $\begingroup$ You need some conditions on the function so that it's Hessian is positive. $\endgroup$
    – user99914
    Mar 8, 2018 at 14:24
  • $\begingroup$ @JohnMa I'm not looking for conditions under which a function's hessian is a metric, the opposite: under which condition a metric is an hessian? $\endgroup$
    – Caffeine
    Mar 8, 2018 at 14:42

1 Answer 1

4
$\begingroup$

If $\nabla^2 f = g$ then we have $\nabla^3 f = 0$ (because $\nabla g=0$), so the Ricci identity tells us that $$R(\cdot, \cdot,\nabla f,\cdot) = 0.$$ Thus, for many choices of $g$ we can immediately rule out finding such an $f$: for example, if $g$ has all non-zero sectional curvatures in some neighbourhood, we can only satisfy this equation if $\nabla f=0$ there, but then of course $\nabla^2 f = 0\ne g.$

It might be possible to engineer interesting pairs $(f,g)$ together to avoid this problem—some solutions exist, like the obvious $f=\frac12 |x|^2$ on $(\mathbb R^n,dx_1^2+\cdots+dx_n^2)$—but I would expect that the obstruction due to curvature is fairly generic if you just pick a random $g$.

A particularly concrete case is when $M$ is compact: then we have $\int_M \Delta f = 0,$ which contradicts $\Delta f = \operatorname{tr}_g \nabla^2 f = n>0,$ so there are no solutions. This idea should also rule out solutions whenever $M$ contains a compact minimal submanifold.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.