MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
This doesn't seem to be a question. – HJRW Feb 19 '13 at 14:26
This is very important queston because for studying local geometry of the moduli space of special lagrangian submanifolds we need to Riemannian manifolds of Hessian type which are useful for the study of monge ampere equations – baba ab dad Feb 19 '13 at 14:36
You can see some works of Hichin related to my previous comment – baba ab dad Feb 19 '13 at 14:37
I think HW means the question is too board. – Ralph Feb 19 '13 at 14:41
I meant that, grammatically, it was not a question. I don't want to penalise non-native English speakers, but I didn't understand what the question was. Since the edit, I think I do, viz, 'When is a Riemannian manifold of Hessian type?' – HJRW Feb 19 '13 at 14:52
up vote 7 down vote accepted

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.

share|cite|improve this answer

I might be misunderstanding the question, but I believe that if there are functions $h$ and $k$ so that $$ \text{Hess}_h = kg, $$ then $(M,g)$ must be (at least locally) a warped product $(a,b) \times_f N^{n-1}$. This follows from integrating along flowlines of $\nabla h$, to compare the induced metrics on different level sets of $h$.

I don't know if this is was the first proof of this result, but the result I've stated above is proven (and discussed a bit more than I have here) in Cheeger-Colding's paper "Lower Bounds on Ricci Curvature and Almost Rigidity of Warped Products" on p 192-194 in this copy of the paper.

share|cite|improve this answer
I think that you are misunderstanding the question. I believe that your $\mathrm{Hess}_h$ is actually $\nabla^2 h$, where $\nabla$ is the Levi-Civita connection of $g$. That is not the Hessian with respect to a coordinate system, as I described it above. In your equation, $g$ actually appears on both sides of the equation, whereas, in mine, it only appears on $1$ side and the $(n{+}1)$ unknowns all appear on the other side. – Robert Bryant Feb 19 '13 at 17:18
Yes, exactly I am agree with @ Robert – baba ab dad Feb 19 '13 at 19:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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