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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!

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Wouldn't you get a counterexample by taking $M=\mathbb{R}^2$ and $U$ any nonconvex neighborhood of the origin? –  Jonas Meyer Mar 13 '10 at 22:05
    
Arg, you're right. I want $U$ to be convex as well - I'll fix the question. –  Paul Siegel Mar 13 '10 at 22:16
    
Also, if it helps, I'm happy to assume that $U$ is a small ball. –  Paul Siegel Mar 13 '10 at 22:18
    
Jonas is right, of course. See en.wikipedia.org/wiki/Gauss's_lemma_(Riemannian_geometry) The memory is a bit vague but this should extend to your result for very small and strictly convex $U$. If I find a reference I will post it, meanwhile also see en.wikipedia.org/wiki/Normal_coordinates and en.wikipedia.org/wiki/Exponential_map –  Will Jagy Mar 13 '10 at 22:35
    
I have tried using the Gauss lemma to prove that the answer is 'yes', but I just don't see how to force a geodesic into a potentially high codimension surface. Can you elaborate? –  Paul Siegel Mar 13 '10 at 23:01
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up vote 12 down vote accepted

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.

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Ack. This ruins my evening :( Thanks, though! –  Paul Siegel Mar 13 '10 at 23:27
    
Sergei -- your last sentence if a bit confusing since it seems to say that any surface (which is a totally geodesic submanifold of itself) has constant Gaussian curvature. –  algori Mar 14 '10 at 3:07
    
Could not believe in your global condition (for every point in M). May be I misunderstand something. –  Petya Mar 14 '10 at 3:33
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@Petya. "Point-wise constant sectional curvature" means that sectional curvature at fixed point does not depend on direction. Schur's theorem says that if dimension is $\ge3$ then point-wise constant sectional curvature is equivalent to constant curvature. (The argument is local; it only use Bianchi identity.) –  Anton Petrunin Mar 14 '10 at 3:57
    
I see know (misread)! –  Petya Mar 14 '10 at 4:26
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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.

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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?

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Yes, but your $U$ is not small. –  Jonas Meyer Mar 13 '10 at 22:44
    
Small enough geodesic balls are geodesically convex, but the question is about staying in the image of the plane if $U$ is small and convex. –  Jonas Meyer Mar 13 '10 at 22:46
    
True, Jonas. There should exist a small $U$ for which Paul's convexity conclusion holds, following Will's suggestion of the Gauss lemma. My point is that if you just choose a neighborhood $U$ at the beginning, the property may not hold. –  Tom LaGatta Mar 13 '10 at 22:47
    
Good point. However for my purposes I only need the result for $U$ sufficiently small. I'll try to rephrase the question to make this more clear. –  Paul Siegel Mar 13 '10 at 22:48
    
I just removed my temporary downvote (it was for sorting purposes). –  Jonas Meyer Mar 13 '10 at 23:38
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