# connected complete totally geodesic sub manifold of $S^n$

Let $M$ and $N$ be manifolds with Riemannian metrics $g$ and $h$ respectively. A diffeomorphism $F: M\to N$ is an isometry if \begin{equation*} h_{F(x)}(T_x F(u), T_x F(v))=g_x(u,v) \end{equation*} for all $x\in M$ and $u,v\in T_x M$. A submanifold $\Sigma\subset M$ is called totally geodesic if for every $x\in\Sigma$ and every $v\in T_x\Sigma\subset T_x M$, the geodesic in $M$ with initial point $x$ and initial velocity is contained in $\Sigma$.

I want to show

i) The intersection of $S^n$ with any vector subspace (at least 2D) is a connected complete totally geodesic sub manifold.

Then deduce:

ii) Every connected, complete, totally geodesic submanifold of $S^n$ is the intersection of $S^n$ with vector subspace of $\mathbb{R}^{n+1}$

Here is my thought for (1): The first of my claim is that the closed totally geodesic subspaces of the sphere $S^n\subset\mathbb{R}^{n+1}$ are precisely the intersection of $S^n$ with linear subspace of $\mathbb{R}^{n+1}$. This follows from the description of the geodesics on $S^n$, according to Jurgen Jost's 4th Edn. So one would naturally look at how to determine the geodesics of $S^n$, in this case, the idea is to show the intersection of $S^n$ with any linear subspaces of $\mathbb{R}^{n+1}$ determined a great circle The orthogonal group $O(n+1)$ operates isometrically on $\mathbb{R}^{n+1}$, and since it maps $S^n$ into $S^n$, it also operate isometrically on $S^n$. Now let take a point $x\in S^n$ and a vector $v\in T_x S^n$. Let $E$ be the 2D plane through the origin of $\mathbb{R}^{n+1}$ containing $v$. We claim that the geodesic $\gamma_v$ through $x$ with tangent vector $v$ is the great circle (intersection of 2-plane with $S^n$ through the origin) through $x$ with tangent vector $v$ (parametrized by arc length). For this, let $S\in O(n+1)$ be the reflection across that 2-plane $E$. Together with $\gamma_v$, $S\gamma_v$ is also a geodesic through $x$ with tangent vector $v$. Now the uniqueness of the solution to geodesic equations implies $\gamma_v=S\gamma_v$, and thus the image of $\gamma_v$ is the great circle we want.

Hence we've shown that every tangent vector on the plane uniquely determines a connected complete totally geodesic sub manifold. Then what do I need to do for (2), i.e. to conclude every connected complete totally geodesic complete totally geodesic submanifold of $S^n$ is the intersection of $S^n$. I am lost, any correction on what I've written is appreciated.

• A vector subspace of one dimension cuts the sphere in two distinct points????? – Semsem Feb 26 '14 at 5:49
• sorry i don't understand what you mean…could you point out where in what i've written is not clear to you? I guess you mean the fact that $S$ is the reflection across the 2D plane means cutting the sphere in 2 distinct points. – math101 Feb 26 '14 at 5:55
• in point i: the intersection of $S^n$ and a vector subspace of dimension one is not connected – Semsem Feb 26 '14 at 5:58

First of all, the comment by Semsem is valid. If the vector space is one dimensional, then $V\cap \mathbb S^n$ is not connected. That is, statement (i) is correct only if you consider vector subspaces that are not one dimensional.
Let $M$ be a totally geodesic submanifold of $\mathbb S^n$. After some $O(n+1)$ on $\mathbb S^n$, we assume that the north pole $N=(0,\cdots, 0, 1)$ is in $M$. Let $L = T_NM$ and treat this space as a vector subspace in $\mathbb R^{n+1}$. Let $V = N \oplus L$. It is a vector subspace in $\mathbb R^{n+1}$.
Now we show $M = G := V \cap \mathbb S^n$ (Note that $V$ is at least two dimensional so $G$ is connected). First of all, let $x\in G$. Then there is a geodesic $\gamma$ such that $\gamma(0) = N$ and $\gamma(1) =x$. Then this is a portion of a great circle and thus $\dot \gamma (0) \in L$. As $M$ is totally geodesic, $x=\gamma(1)\in M$. Thus $G\subset M$.
On the other hand, let $q\in M$. Then as $M$ is connected and complete, there is a geodesic in $M$ joining $N$ and $q$. But this geodesic is also a geodesic in $\mathbb S^n$, as $M$ is totally geodesic. Thus this geodesic is a great circle and so are inside $G$ (As its tangent vector at $N$ is by definition in $L$). Thus $M\subset G$.
• Well, even if the subspace is $1$ dimensional, it is true that every geodesic of the sphere whose tangent vector is tangent to the intersction with the subspace is totally contained in the intersection: there are no such geodesics! – Mariano Suárez-Álvarez Feb 26 '14 at 7:11
• @algebra_geometry101: There is in general a definition of connectedness of a topological space. But in the case of a manifold, connectedness is the same as "for all $p$ and $q$ in $M$, there is a continuous paths joining $p$ and $q$". So for example, if $M$ is just a set of two points, then it is not connected in this sense. – user99914 Feb 27 '14 at 7:10