Must a surface obtained by exponentiating a plane in a tangent space of a Riemannian manifold be geodesically convex?

Question

Perhaps this is basic knowledge in Riemannian geometry, but I can't seem to figure out the answer. Here is the precise statement of my question.

Let $M$ be a Riemannian manifold, $p$ a point in $M$. Let $R$ be small enough that $exp_p$ restricts to a diffeomorphism on the ball $B_R(0)$ of radius $R$ centered at the origin, and let $U_R$ be the intersection of $B_R(0)$ and any two dimensional plane through the origin in $T_p M$. Question: does there exist $R$ such that $exp_p(U_R)$ is geodesically convex, in the sense that for every two points of $exp_p(U_R)$ the unique geodesic segment connecting them lies entirely in $exp_p(U_R)$?

It would be really convenient for me if the answer is yes. If so, I am curious to know if the statement is still true if $P$ is replaced by a subspace of higher dimension, but I only need the result for planes.

Thanks!

Answer 1

No, a generic Riemannian metric does not have totally geodesic 2-dimensional submanifolds at all. The property that you ask for is very rare. For example, it implies that $R(X,Y)Y$ belongs to the linear span of $X$ and $Y$, for every $p\in M$ and every $X,Y\in T_pM$. This means point-wise constant sectional curvature if I remember correctly.

Answer 2

I don't think that's true. If the dimension of $M$ is bigger than that of $P$, then a necessary condition is that every tangent hyperplane $P\subset T_pM$ develops locally into a totally geodesic submanifold of $M$. This is not true for arbitrary manifolds. Some examples of manifolds in which this is true include those of constant sectional curvature. Symmetric spaces being a special subclass which verifies this conditions. Someone more awake can probably come up with a sufficient criterion.

I can't think of a very good example right now, but I am pretty sure that if take the 3-dimensional Riemannian Schwarzschild solution, start from a $r$-orthogonal plane outside of the apparent horizon, you'd get a counterexample.

If the dimension of $P$ is the same as dimension of $M$, however, then you should be okay as long as you make $U$ small enough.

Edit: Ah, for $P$ 2 dimensional and $M$ 3, by applying the Codazzi equations one sees that a necessary condition for locally developing the hyperplanes to totally geodesic submanifolds is that $Ric(X,Y) = 0$ whenever $g(X,Y) = 0$. This is obviously a very strong condition that is not satisfied by most manifolds.

Answer 3

No. Take the sphere with $p$ the north pole, and let $U$ be the neighborhood of $0$ in $T_p S^2$ on which $\exp_p$ takes to a diffeomorphism onto $S^2$ without the south pole. Let $\gamma$ be a geodesic through the north pole, and let $p_1$ and $p_2$ be two points on $\gamma$ in the southern hemisphere. Then the shortest geodesic segment passes through the south pole, hence leaves $U$.

The difficulty is that length minimization is a global phenomenon, whereas $\exp$ is only a local diffeomorphism. Do you have additional assumptions in your problem? e.g. non-negative curvature?