I am interested in results regarding transnormal foliations on a Riemannian (smooth, connected and complete) manifold $(M,g)$. More specifically, a smooth function $f:M\to{\bf R}$ is called transnormal if there exists a smooth function $b:f(M)\to{\bf R}$ such that $\|\nabla f\|^2=b\circ f$. From [Wan], this induces a particularly nice decomposition of $M$ into level sets of $f$.
I'm interested in a special case of this decomposition. I take $M$ to be closed, and $L$ is a Riemannian submanifold which is isometrically embedded and totally geodesic (actually, $L$ is the fix-point set of an isometric involution on $M$). $\phi$ denotes the squared-distance function to $L$: $$\forall y\in M,\;\phi(y)=\min_{x\in L}\mathrm{dist}(x,y)^2.$$ I will let $T=\max\phi$ and $N=\{y\in M\mid0<\phi(y)<T\}$. I consider $h:{}]0,T[{}\to{\bf R}$ to be a smooth non-decreasing diffeomorphism defined by $h(x)=\tan\left(\frac{\pi}{T}(x-T/2)\right)$. Then, if $b(x)=4(1+x^2)^2h^{-1}(x)^2$, we have: $$\|\nabla\psi\|=b\circ\psi$$ on $N$, provided that $\psi$ is indeed smooth on $N$ (by using that $\sqrt{\phi}$ is solution to the Eikonal equation wherever it is smooth). In that case, $\psi$ is transnormal and I can apply the results in [Wan] to obtain the corresponding decomposition on $N$ where the focal submanifolds $V^\pm$ are empty. I would now like to extend the decomposition to the whole $M$. I already know that $L$ is a submanifold.
- Is it true that $\psi$ is smooth on $N$?
- Does the foliation on $N$ extend to one on $M$?
As stated, this cannot be true. An example I have in mind is if $L$ is the lowest-height point on a torus lying vertically on the table. There are index one critical points for the distance function, and so the function $\psi:N\to{\bf R}$ cannot be smooth, for it should not have any critical values. However, in this example, $L$ is not $Fix(\tau)$ for some isometric involution $\tau:M\to M$. If $M$ is a surface and $L=Fix(\tau)$, then there is not much room for anything, and I can convince myself 'visually' that the result holds. I know it holds in the following cases:
- $M={\bf CP}^2$, with $f=\mathrm{dist}(-,{\bf RP}^2)$. Then each leaf is a lens space $L(4,1)$, and there are two singular leaves ${\bf RP}^2$ and $\{z_0^2+z_1^2+z_2^2=0\}$. Both are codimension $2$ leaves.
- $M={\bf CP}^1\times{\bf CP}^1$ with $f=\mathrm{dist}(-,{\bf RP}^1\times{\bf RP}^1)$. There are two singular leaves, one being ${\bf RP}^1\times{\bf RP}^1$ and the other being $4$ points. This time, there is a codimension $2$ leaf and a codimension $4$ one.
Does anybody have any result in this direction?
Assuming I have more conditions on $(M,g)$ and on the involution $\tau$, maybe there are more things to be said? E.g. where is the squared-distance function smooth, and where is it not smooth?
Edit: According to [Pra], the squared-distance function to $L$ is smooth on $M\smallsetminus\mathrm{Cut}(L)$, where $\mathrm{Cut}(L)$ is the cut locus of $L$. If I could show that $\mathrm{Cut}(L)$ is the set of points furthest away from $L$, i.e. $\mathrm{Cut}(L)=\phi^{-1}(T)$, then $\psi$ is indeed smooth (and it is even smooth on $\phi^{-1}({}]-\infty,T[{})$).
Is any result known whether $\mathrm{Cut}(L)$ is equal to $\phi^{-1}(T)$ in the case where $L$ is $\mathrm{Fix}(\tau)$? Is it a submanifold?
