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> **Possible Duplicate:**  
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> [Must a surface obtained by exponentiating a plane in a tangent space of a Riemannian manifold be geodesically convex?](https://mathoverflow.net/questions/18108/must-a-surface-obtained-by-exponentiating-a-plane-in-a-tangent-space-of-a-riemann) 
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The one dimensional geodesic submanifolds of a given Riemannian manifold $(M,g)$ are just geodesics. So one can can wonder, how to construct 2-dimensional geodesic submanifolds. Lets first consider the following question:


Given any point $x\in M$ and a two dimensional subspace of $V\subset T\_xM$. Then the exponential map restricted to a sufficiently small ball around $0\in V$ gives an embedding of the open disc into $M$. When is it a geodesic submanifold?


Note that there are many spaces, that have this property at every point and at every tangent plane, like $\mathbb{S}^n,\mathbb{H}^n,\mathbb{R}^n$ and (if I am not mistaken) products of those. So one can also ask:


What properties must the metric $g$ have to ensure, that at every point $x\in M$ and at every two dimensional subspace $V\subset T\_xM$ the exponential map $B\_\varepsilon(0)\subset V\rightarrow M$ gives locally geodesic surfaces?


A "general" manifold should not have this property I think. It would be nice to have a simple and short counterexample.



geodesic 2-dimensional submanifolds of a Riemannian manifold
Your examples (i.e. space forms) are the only manifolds with the property that the exponential map sends 2-dimensional disks to totally geodesic surfaces. One way to see this is using Jacobi vector fields.


More precisely, let $X$ and $Y$ be two orthogonal vectors in some tangent space $T\_xM$ and assume that the exponential map sends (a neighbourhood of 0 of) the plane spanned by $X$ and $Y$ into a totally geodesic surface $S$. If you denote $\gamma\_s(t)=exp\_x(t(X\cos(s)+Y\sin(s)))$, then 
$$J:=\frac{\partial\gamma}{\partial s}$$
is a Jacobi field along the geodesic $\gamma\_0$ and thus satisfies (denoting $\gamma\_0$ by $\gamma$):
$$\nabla^2\_{\dot\gamma,\dot\gamma}J=-R\_{J,\dot\gamma}\dot\gamma.$$
Since $J$ and $\dot\gamma$ are tangent to $S$, which is totally geodesic, and moreover $R\_{J,\dot\gamma}\dot\gamma$ is orthogonal to $\dot\gamma$, we see that $R\_{J,\dot\gamma}\dot\gamma$ ahs to be proportional to $J$. A standard argument then shows that $M$ has constant sectional curvature (provided that $dim(M)>2$). I can give more details about this if you need.


