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...
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.
