There is a notion of "line" on a Riemannian manifold. We say that a geodesic $\gamma$ on a complete Riemannian manifold is a line if it minimizes the distance between any two of its points.
The claim is:
For a complete Riemannian manifold with sectional curvature positive (not necessarily constant), there is no line.
The proof should not be so difficult, but I am having a little trouble. I imagine that the solution follows from the following lines:
Suppose that the claim is false. Take two points of such geodesic (line) $\gamma$, lets say $p,q$. Then the distance between them is the less possible, so the second formula of the variation of energy should be strictly positive, i.e., $I_a(V) = \int_{t_0}^{t_1}\{\langle V',V'\rangle - \langle R(\gamma',V)\gamma',V\rangle\}dt > 0.$ (Here, $V$ is a variational proper vector field along $\gamma.$)
The second term is clearly the curvature for an appropriate $V$. (We can take $V$ orthonormal for every $t.$)
But I don't know how to conclude.
I do appreciate any hints or comments.
Thanks.
This is a corollary of Toponogov-Cheeger-Gromoll Splitting Theorem: If $M$ is a connected complete Riemannian manifold of nonnegative Ricci curvature containing a line then $M$ is isometric to a Riemannian direct product ${\mathbb R}\times N$. Now, if $N$ is not a point then $M$ cannot have positive sectional curvature (it is a nice exercise to work out by yourself: Identify which directions have zero sectional curvature). Note that Toponogov proved the splitting theorem in the context of sectional curvature; Cheeger and Gromoll improved the result in the Ricci curvature setting.