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...

  • $\begingroup$ 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
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    $\begingroup$ 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

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.

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