3
$\begingroup$

Let $(M^n,g), n \geq 3$ be a closed Riemannian manifold. Assume that there is a function $\phi : M \to \mathbb{R}$ of class $C^2$ such that $\phi$ has a saddle point. Then, is necessarily true that in this point $M$ has negative scalar curvature?

It seems true, but I just have an intuition about it, like saddle points have a two plane with negative sectional curvature...

$\endgroup$
  • $\begingroup$ Do you mean the graph manifold of $\phi$ or something? Otherwise, can't you take a regular saddle-point-function on $M = \mathbb R^3$ (or the flat torus if it needs to be closed)? $\endgroup$ – Keshav Nov 24 '19 at 5:04
  • $\begingroup$ @Keshav, not the graph, I mean $(M,g).$ I was wondering if one can conclude anything about the curvature of $g$ given the existence of such function. Do you have an explicit example for a function with saddle points on $M = \mathbb{R}^3$ and the flat standard metric? $\endgroup$ – L.F. Cavenaghi Nov 24 '19 at 5:09
  • 1
    $\begingroup$ What's wrong with e.g. $f(x,y,z) = x^2 - y^2 - z^2$ or similar? I'm assuming by saddle point you mean (as on wiki) a critical point which is not a local extremum. $\endgroup$ – Keshav Nov 24 '19 at 5:14
  • $\begingroup$ @Keshav, you are right. $\endgroup$ – L.F. Cavenaghi Nov 24 '19 at 5:16
5
$\begingroup$

Take any closed Riemannian manifold $N$ equippped with a function $g \colon N \to \mathbb{R}$ which has a saddle point at $y_0 \in N$, and assume that $N$ has scalar curvature $R_0 < 0$ at $y_0$. Scale the round metric on $S^2$ so that it has constant scalar curvature $S > 0$, and choose a base point $x_0 \in S^2$ and a smooth function $f \colon S^2 \to \mathbb{R}$ so that $df_{x_0}$ has rank 1.

Now set $M = S^2 \times N$ and define $\phi \colon M \to \mathbb{R}$ by $\phi(x, y) = f(x) + g(y)$. Then $\phi$ has a saddle point at $(x_0, y_0)$, but the scalar curvature of $M$ at this point is $R_0 + S$; choosing the scale factor $S$ large enough will ensure this is positive.

| cite | improve this answer | |
$\endgroup$

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.