Let $M$ be a complete Riemannian manifold and let $S_1,S_2 \subset M$ be totally geodesic submanifolds which are closed as subspaces of $M$.

Question: Is $S_1 \cap S_2$ a submanifold of $M$? Or is it at least locally path connected?

Notice that if we drop the condition that $S_1,S_2$ are totally geodesic, then $S_1 \cap S_2$ can be homeomorphic to any closed subset of $\mathbb{R}$ and so is far from being a manifold in general.


Interesting question. It turns out that each component of the intersection is always a closed, embedded smooth submanifold. Here's a sketch of a proof.

Let $p$ be a point in $S_1\cap S_2$, and let $(x^1,\dots,x^n)$ be Riemannian normal coordinates for $M$ on some geodesic ball $U$ containing $p$. Let $k_i = \dim S_i$ for $i=1,2$.

By properties of normal coordinates, the images of geodesics with initial velocities in $T_pS_1$ sweep out a linear slice $L_1\subset U$ in these coordinates, which is a $k_1$-dimensional embedded submanifold of $U$. Because $S_1$ is totally geodesic, this slice is contained in $S_1$. As a $k_1$-dimensional manifold contained in $S_1$, $L_1$ is actually an open subset of $S_1$. Since $S_1$ is closed in $M$, after shrinking $U$ if necessary, we can assume that $L_1 = S_1\cap U$.

Now do the same thing for $S_2$, obtaining a $k_2$-dimensional linear slice $L_2\subset U$ such that $L_2 = S_2\cap U$.

It follows that $S_1\cap S_2\cap U$ is equal to the intersection of the two linear slices $L_1$ and $L_2$, and therefore it is itself a linear slice of $U$. Since every point of $S_1\cap S_2$ has a coordinate neighborhood in which $S_1\cap S_2$ is a linear slice, it follows that $S_1\cap S_2$ is an embedded submanifold provided it has constant dimension. Since the argument above shows that the dimension is locally constant, it follows that each connected component will have contant dimension and thus will be an embedded submanifold. It is of course closed because it's the intersection of the closed subsets $S_1$ and $S_2$. $\square$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.