Connection between focal points and singularities of the normal exponential map

390 Views Asked by At

I am looking for nice references on focal points of Riemannian submanifolds. In particular, I would like to see a proof for the connection between focal points and singularities of the normal exponential map.

The statement should be something like this:

(actually it seems like what I am saying next is inaccurate. What I am really after is some kind of characterization of singular points of the normal exponential in terms of Jacobi fields, similar to the one which exists when $S$ is a single point).

Assume $S \subset M$ is a submanifold of a Riemannian manifold $M$. Then, the normal exponential map fails to be a diffeomorphism at $(p,v) \in NS$ if and only if $exp(p,v)$ is a focal point of $S$.

Of course, I mean for the statement to have some true content, so I am not using it as a definition of focal points, rather the one regarding Jacobi fields...

Here is one definition: (though I think I saw others which are probably equivalent, I am really not sure...)

Let $c:[0,a] \to M$ be a norml geodesic to $S$, i.e $c(0)=p \in S, \dot c(0) =v \in N_pS=(T_pS)^{\perp}$. We say $Y(t)$ is an $S$-Jacobi field along $c$, if it is a standard Jacobi field along $c$, which in addition satisfies:

1) $Y(t)$ is orthogonal to $c$: $\langle Y(t),\dot c(t) \rangle =0$

2) $Y(0) \in T_pS$

3) $\frac{DY}{Dt}(0)+A_{\dot c(0)}(Y(0)) \in (T_pS)^{\perp}$ where $A$ is the Weingarten map (I am not sure about its sign convention here).

A point $c(t_0)$ on $c$ is called a focal point of $S$ with respect to the geodesic $c$ if there is a non-trivial $S$-Jacobi field along $c$ which vanishes at $t=t_0$.

1

There are 1 best solutions below

0
On

The claim is false, but for a sort of stupid reason. Let's take $M$ to be the plane, and $S$ to consist of two circles: the unit circle $C$ at the origin, and the circle $Q$ of radius $1/4$ centered at $(-1/4, 0)$; note that $Q$ contains the origin.

Now at the point $( (0,0), 0)$ of $NS$, the exponential map is just lovely.

On the other hand, the point $(0,0) = exp( (0,0), 0)$ is a focal point of the circle $C$, and hence of $S$.

You might argue that my example isn't connected, but you can join the two circles by a pair of parallel vertical lines from the top of $Q$ to a point in the NW side of $C$, smoothed out at the joints, so the disconnectedness is not essential to this construction.

By the way, I expect a local version is true, something like "if $X$ is a "focal point of $S$ at $(p, v)$," then $exp$ is singular at $(p, v)$." But that version may be completely trivial -- a simple contortion of the definitions. By "a focal point of $S$ at $(p, v)$," I mean something like that it's a focal point of any submanifold consisting of any open subset $U$ of $S$ that contains $p$.