# Establish if a manifold has non negative sectional curvature

Consider $(M,g)$ a riemannian manifold of dimension $n>2$ ($g$ is the riemannian metric). By definition, $(M,g)$ has non negative sectional curvature if for every $p\in M$ and every $u,v\in T_pM$ linearly independent it results $K(u,v)\ge 0$, where $K(u,v)$ is the sectional curvature computed on the plane spanned by $\{u,v\}$.

Now, for $p\in M$ consider a local moving frame $\displaystyle{\frac{\partial}{\partial x_1},\dots,\frac{\partial}{\partial x_n}}$ on a open neighborhood $U$ of $p$, that is, for every $q\in U$, $\{\displaystyle{\frac{\partial}{\partial x_1}(q),\dots,\frac{\partial}{\partial x_n}(q)}\}$ is a basis of $T_qM$. Clearly we can cover $M$ with such open sets $U$ on which we have local moving frames.

This is my question: in order to prove that $(M,g)$ has non negative sectional curvature, is it enough to prove that on every $U$ and for every $q\in U$ it results $\displaystyle{K(\frac{\partial}{\partial x_i}(q),\frac{\partial}{\partial x_j}(q))\ge 0}$ for every $i,j=1,\dots,n$?

This is essentially a linear algebra question. One consider $R$ a symmetric bilinear form on $\bigwedge^2 TM$. That is,

$$R(v_1\wedge v_2 , v_3\wedge v_4) := \langle R(v_1, v_2) v_4, v_3\rangle$$

Note any basis $w_1, \cdots w_n$ of $TM$ induces a basis $\left\{ w_i \wedge w_j : i<j\right\}$ of $\bigwedge^2 TM$. So your question is essentially:

If the diagonal entries of a symmetric matrix is non-negative, does it implies that the matrix is nonnegative?

The answer is quite clearly no, for example the matrix

$$\begin{bmatrix} 1 & -2 \\ -2 &1\end{bmatrix}$$

has eigenvalue $3$, and $-1$.

• (+1), but don't you want $2$ off the diagonal...? :) – Andrew D. Hwang Sep 30 '16 at 10:13