# Gaussian Curvature of Exponentiated 2-Planes

Consider a Riemannian manifold $M$ with sectional curvatures $K\ge 0$ and let $\Pi$ be a 2-plane in the tangent space of $M$ at a point $p$. In a small enough neighborhood $U$ of 0 the exponential map will be a diffeomorphism and $S=\text{exp}(U\cap \Pi)$ will be an open surface in $M$. The Gaussian curvature of this surface at $p$ is the sectional curvature of $M$ corresponding to the 2-plane $\Pi$.

My question is this: can anything be said about the Gaussian curvature of $S$ at points other than $p$? For example, is it still nonnegative?

The answer is "no". Say, there is a non-negatively curved Riemannian metric on $\mathbb R^3$ with a point $p\in M$ and a sectional direction $\Pi$ such that the surface $S$ has negaive curvature at the points arbitrary close to $p$.
First notice that $S$ is a ruled surface, in particular it has nonpositive Gauss curvature (= product of principle curvatures) at any point. Note further that generic ruled surface has negaive Gauss curvature at most of the points. Therefore if the ambient space has zero curvature at such point say $q$ then $S$ has negative curvature at $q$.
It remains to construct a metric which has vanishing curvature and most of the points and such that the surface is generic arbitrary close to $p$. Such a metric can be found among product metrics $\mathbb R\times (\mathbb R^2,g)$; the metric $g$ has mostly vanishing curvature; it has a sequence of islands of positive curvature which converge to the projection of $p$ in $\mathbb R^2$. Any choice of plane $\Pi$ which is not horizontal and not vertical will do the job.
• Anton, what if I add the condition that the manifold $M$ is compact? I'm specifically interested in the case of Riemannian symmetric spaces of compact type. Nov 15, 2014 at 6:51
• The example is local --- compactness makes no difference. It is $C^\infty$-smooth, but I do not see an analytic example of that type. Nov 15, 2014 at 17:43