When a Riemannian manifold is of Hessian Type (i.e., a Riemannian manifold which its metric is Hessian)



When a Riemannian manifold is of Hessian Typ
First, the definition: A Riemannian $n$-manifold $(M^n,g)$ is *of Hessian type* if there exist $(n{+}1)$ functions $x^1,\ldots,x^n, u$ on $M$ such that $dx^1\wedge\cdots\wedge dx^n\not=0$ and such that
$$
g = \frac{\partial^2u}{\partial x^i\partial x^j} dx^idx^j
$$
(of course, the summation convention is in force for this formula). Note that the independence of the differentials $dx^i$ is needed in order to define the 'partial derivatives' in the formula. One says that $(M^n,g)$ is *locally of Hessian type* if each point of $M$ has an open neighborhood $U\subset M$ on which there exists a coordinate chart $x:U\to\mathbb{R}^n$ and a function $u\in C^\infty(U)$ such that the above formula holds on $U$.


Since metrics in dimension $n$ depend on $\tfrac12n(n{+}1)$ functions of $n$ variables and the data of a Hessian representation depends only on $(n{+}1)$ functions of $n$ variables, it is clear that, when $n>2$, not every metric is locally of Hessian type, and, in principle, such a set of criterion can be developed, but it's not trivial. Of course, as $n$ increases, the condition of being locally of Hessian type becomes more and more restrictive, even implying algebraic conditions on the Riemann curvature tensor once $n$ is sufficiently large.


However, when $n=2$, this is a determined problem. However, it is never elliptic, so one never gets elliptic regularity. The characteristic variety consists of $3$ points, so at least one of them has to be real. Depending on the sign of the Gauss curvature of the metric, one can sometimes formulate the problem as having $3$ real characteristics and sometimes one can't. Of course, in the real-analytic case, the problem is always solvable locally, so every real-analytic metric in dimension $2$ is locally of Hessian type.


