# Saddle points and negative scalar curvature

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...

• 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)? – Keshav Nov 24 '19 at 5:04
• @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? – L.F. Cavenaghi Nov 24 '19 at 5:09
• 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. – Keshav Nov 24 '19 at 5:14
• @Keshav, you are right. – 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.