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!



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


