When a Riemannian manifold is of Hessian Type (i.e., a Riemannian manifold which its metric is Hessian)
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1$\begingroup$ This doesn't seem to be a question. $\endgroup$– HJRWCommented Feb 19, 2013 at 14:26
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$\begingroup$ 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 $\endgroup$– user21574Commented Feb 19, 2013 at 14:36
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$\begingroup$ You can see some works of Hichin related to my previous comment $\endgroup$– user21574Commented Feb 19, 2013 at 14:37
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1$\begingroup$ I think HW means the question is too board. $\endgroup$– RalphCommented Feb 19, 2013 at 14:41
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$\begingroup$ 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?' $\endgroup$– HJRWCommented Feb 19, 2013 at 14:52
2 Answers
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.
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$\begingroup$ Nice answer. I have a question here if possible. Does the following condition make difference? $$\rm rank(\rm{Hess}\hspace{2pt}\it f-\alpha g\rm )\le1$$ Would you kindly see this question math.stackexchange.com/questions/2419564/… ? $\endgroup$– SemsemCommented Sep 9, 2017 at 19:15
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$\begingroup$ It's quite a few years later, and I suspect you have seen it already as it cites this answer, but John Armstrong and Shin-Ichi Amari were able to find the precise curvature obstructions for a given 4-metric to be Hessian. arxiv.org/pdf/1312.1103.pdf One of the conditions is equivalent to the fact that the Pontryagin forms vanish. The other is some mysterious cubic expression in the curvature. I was wondering if you recognize that cubic as anything familiar. $\endgroup$– Gabe KCommented Aug 17, 2018 at 17:08
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.
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4$\begingroup$ 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. $\endgroup$ Commented Feb 19, 2013 at 17:18
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$\begingroup$ Yes, exactly I am agree with @ Robert $\endgroup$– user21574Commented Feb 19, 2013 at 19:22