For a closed Riemannian manifold $M$, must the set of points with non-unique closest points to a closed submanifold $S$ of $M$ be of 0 volume measure?

Question

Let $(M,g)$ be a closed (compact without boundary) Riemannian manifold of finite dimension, with the volume measure $\mu:= \mu(E):=\int_{E}dvol_g \forall E \in \mathcal{B}(M),$ the Borel sigma algebra of $M.$

Let $S\subset M$ be a closed embedded submanifold, so $dim(S)<dim(M).$

Let $P_S:=\{x\in M: \text{minimizer of the distance from $x$ to $S$ is *not* unique}\}= \{x\in M: \exists x_1\ne x_2, x_1, x_2\in S \text{ such that } d(x,S)=d(x,x_1)=d(x,x_2)\}$

One can intuitive think that $P_S$ is 'small' in some sense (to be made precise soon). However, this question and the answer provided there show that the complement of $P_S$ may not be open. Next one can ask if $P_S$ has $\mu$-measure zero, at least when $S$ is "nice" enough (Sorry for being vague here, but that's because I don't know what sufficient conditions on $S$ will be necessary...).

My questions are:

1. For $M, S$ as above, is $\mu(P_S)=0$ always?
2. If the answer is negative, is $\mu(P_S)=0$ under certain conditions on $(g,S)?$
3. Is the answer positive/negative when $M$ is non-compact?

Related questions: link.

Answer 1

More generally, for any closed subset $S$ of a complete manifold $M$, the set of points $x$ at whose minimal distance to $S$ is attained at more than point has measure $0$.

Indeed, consider the distance function $f:M\to\mathbb{R}$ given by $d_S(x)=d(S,x)$. Then $d_S$ is $1$-Lipschitz, so by Rademacher's theorem it is almost everywhere smooth. It will thus be enough to prove that:

Claim: If $f$ is smooth at some point $p\not\in S$, then there can be at most one point $s\in S$ minimizing distance to $p$.

Proof: First note that the derivative of $f$ at $p$, $df_p:T_pM\to\mathbb{R}$, has norm $1$ because $p$ is $1$-Lipschitz. Now, choose some $s\in S$ such that $d(p,s)=f(p)$ (that is, $s$ minimizes distance to $p$). Let $\gamma:\mathbb{R}\to M$ be a unit speed geodesic with $\gamma(0)=p,\gamma(f(p))=s$, and let $v=\gamma'(0)$. Then we have

$$df(v)=\left.\frac{d}{dt}\right|_{t=0}(f(\gamma(t)))\leq\left.\frac{d}{dt}\right|_{t=0}(d(\gamma(t),s))=\left.\frac{d}{dt}\right|_{t=0}(f(p)-t)=-1.$$

In fact $df(v)=-1$, because $f$ is $1$-Lipschitz so $df|_p$ is $1$-Lipschitz too. If there were another point $s'\in S$ at minimal distance from $p$, we would also have $df(v')=-1$ (where $v'$ is a tangent vector pointing to $s'$ as above). Moreover, we would have $v'\neq v$, as the points $s,s'$ are determined by the vectors $v,v'$ due to uniqueness of geodesics: $\exp(f(p)v)=s\neq s'=\exp(f(p)v')$. So it is enough to check that there is just one vector $v\in T_pM$ such that $df(v)\leq-1$. We prove that in the claim below.$\square$

Subclaim: For any Hilbert space $H$ over $\mathbb{R}$ and any linear map $L:H\to\mathbb{R}$ with $|L|\leq1$, there can be at most one unit vector $v$ such that $L(v)=-1$.

Proof of subclaim: Note that, as $L$ is $1$-Lipschitz, $|L(v)|$ is at most the distance from $v$ to ker$(L)$. But the codomain of $L$ has dimension $1$, so ker$(L)$ has codimension $\leq1$. Thus, there can be at most two unit vectors at distance $1$ of ker$(L)$: the two unit vectors orthogonal to ker$(L)$. Those are the only two unit vectors $v$ which may satisfy $L(v)=-1$, and as they are opposite, $L$ cannot be $-1$ in both of them.$\square$

Also, if $M$ is not complete (I assume we are still giving $M$ the length metric) then $P_S$ may have positive measure, e.g. let $M=\{(x,y)\in\mathbb{R}^2;x>0\text{ or }y>0\}$ and let $S$ be a curve contained in $\{(x,y)\in\mathbb{R}^2;x<0,0<y\leq|x|\}$ such the $\mathbb{R}^2$-distance from $(0,0)$ to $S$ is minimized at more than one point. Then, all points $(x,y)$ such that $x>0$ and $y\leq -x$ are contained in $P_S$.

Answer 2

My answers:

1. yes
2. does not apply, see 1.
3. I did not think sufficiently about it in order to make a precise statement, but I would guess, the answer is positive as well, probably with the same proof as for 1. The answer is certainly affirmative, if $S$ is compact and $M$ is complete. I have concerns, if $M$ is not complete.

A possibility to prove (or to sketch a proof of) answer 1 is to use the theory of the cut locus. This is worked out in all detail in the case that $S$ is a point (see e.g. Takashi Sakai, Riemannian Geometry, Chapter III, Section 4), but the relevant statements still hold for general closed embedded submanifold. Consider the normal exponential map $\exp^S$ whose domain is the normal bundle of $S$ in $M$ and whose target is $M$. Consider the set $\mathcal{I}_S$ of all vectors $X$ in the normal bundle such that $[0,1]\to M$, $t\mapsto \gamma_X(t):=\exp^S (t X)$ is a unique length minimizing geodesic without focal points. Let $\mathcal{C}_S$ be the boundary of $\mathcal{I}_S$ in the normal bundel of $S$ in $M$. This set is called the tangential cut locus of $S$ in $M$. It requires a bit of work to see that $\mathcal{C}_S$ has zero measure in this normal bundle.

The set you are interested in is a subset of $\exp^S(\mathcal{C}_S)$. As $\exp^S$ is absolutely continuous, it maps zero sets to zero sets.