# geodesic 2-dimensional submanifolds of a Riemannian manifold [duplicate]

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.

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## marked as duplicate by HenrikRüping, S. Carnahan♦Feb 7 '11 at 15:06

Such $g$ must have constant curvature (assuming $\dim\ge 3$ of course), see mathoverflow.net/questions/18108. Products do not work. – Sergei Ivanov Feb 7 '11 at 14:17
currently I am willing to close this as a duplicate. thanks to both of you anyway. – HenrikRüping Feb 7 '11 at 15:01

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.

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While I was writing my answer, Sergei noticed that the question was a more or less a duplicate... – Andrei Moroianu Feb 7 '11 at 14:40
A typo: it has constant sectional curvature. – Sergei Ivanov Feb 7 '11 at 14:48
Sure, thanks Sergei. – Andrei Moroianu Feb 7 '11 at 14:53