geodesic 2-dimensional submanifolds of a Riemannian manifold - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-06-19T22:33:17Z http://mathoverflow.net/feeds/question/54635 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54635/geodesic-2-dimensional-submanifolds-of-a-riemannian-manifold geodesic 2-dimensional submanifolds of a Riemannian manifold HenrikRüping 2011-02-07T13:53:09Z 2011-02-07T14:53:33Z <blockquote> <p><strong>Possible Duplicate:</strong><br> <a href="http://mathoverflow.net/questions/18108/must-a-surface-obtained-by-exponentiating-a-plane-in-a-tangent-space-of-a-riemann" rel="nofollow">Must a surface obtained by exponentiating a plane in a tangent space of a Riemannian manifold be geodesically convex?</a> </p> </blockquote> <p>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:</p> <p>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?</p> <p>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:</p> <p>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?</p> <p>A "general" manifold should not have this property I think. It would be nice to have a simple and short counterexample.</p> http://mathoverflow.net/questions/54635/geodesic-2-dimensional-submanifolds-of-a-riemannian-manifold/54641#54641 Answer by Andrei Moroianu for geodesic 2-dimensional submanifolds of a Riemannian manifold Andrei Moroianu 2011-02-07T14:35:59Z 2011-02-07T14:53:33Z <p>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.</p> <p>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.</p>