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

Let $G$ be a Lie group acting on $M$ by isometry, properly, but not necessarily freely. Under this action, denote by $O_p:=G.p,$ the $G$-orbit of $p.$ Fix $p^{*}\in M,$ and denote by $E$ the set $F_{p^{*}}:=\{p:\text{ there exists a unique minimizer of } d(p^{*}, ) \text{ in } O_p \}=\{p \in M: d(p^{*},p)=d(p^{*}, O_p), \text{ and } \forall r \ne p, r \in O_p, d(p^{*},r) > d(p^{*}, p)\}.$

My questions are:

(1) Is $\mu(M \setminus F_{p^{*}})=0?$ In words, must the union of orbits $F$ containing a unique minimizer of $d(p^{*},)$ be of full $\mu$-measure?

(2) What if $M$ is not closed, but just complete? (still without boundary). Is the above measure of $F$ still zero?

P.S. This question is related to this question I asked earlier, and in some sense, the current one is its dual question, because that earlier question concerned itself with fixing the embedded submanifold $S$ and asking if the set of points from where the distance attains a unique minimizer to $S$ had measure zero. This one interchanges the role of the point and submanifold in question - here we fix the point $p^{*}$ and ask if the union of all $G$-orbits (that are embedded submanifolds of $M$) where the distance function from $p^{*}$ has a unique minimizer?

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    $\begingroup$ Wouldn't this be not hold when $\mathbb{S}^1$ acts on $\mathbb{S}^2$ by rotations around some fixed axis, and if you let $p^*$ be a point in that axis? $\endgroup$
    – Saúl RM
    Commented Apr 16 at 1:03
  • $\begingroup$ @SaúlRM Apologies, but the axis of rotation doesn't intersect the manifold $S^2$ except for the two end points, right? So does this mean you're choosing one of these two end points? P.S. I may reply in a few hours, since it's very late where I'm writing this from. $\endgroup$ Commented Apr 16 at 1:12
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    $\begingroup$ Yes, I would choose one of the two intersections of the axis with $\mathbb{S}^2$ $\endgroup$
    – Saúl RM
    Commented Apr 16 at 1:22
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    $\begingroup$ @SaúlRM Thanks for confirming - okay let me think about it; my feeling is that for 'most' such $p^{*}\in M,$ it should be true - however, I'll sit down and think more... $\endgroup$ Commented Apr 16 at 1:24
  • $\begingroup$ It seems a statement similar to the question you asked earlier should be enough to prove that for almost all $p^*$ in $M$, your condition holds (this is equivalent by Fubini to saying that for almost all $p\in M$, the set of points $p^*$ for which $d(p^*,\cdot)$ is minimized at more than one point in $O_p$ has measure $0$) $\endgroup$
    – Saúl RM
    Commented Apr 16 at 1:35

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Perhaps this is not the strongest result one can get, but it is true that, if $M$ is a complete Riemannian manifold, then for almost all $p^*\in M$ the set $F_{p^*}$ you define in the question has measure $0$. Due to Fubini's theorem, we have:

For almost all $p^*\in M$ the set $M\setminus F_{p^*}$ has measure $0$.

iff

$$\mu\times\mu(\{(p,p^*)\in M\times M;\text{there is more than one minimizer of $d(p^*,\cdot)$ in }O_p\})=0.$$

iff

For almost all $p\in M$, the set $\{p^*\in M;\text{there is a unique minimizer of $d(p^*,\cdot)$ in }O_p\}$ has full measure in $M$.

But the last statement is in fact satisfied for all $p\in M$; indeed, $O_p$ is closed for all $p$, because the action of $G$ on $M$ is proper, so by the answer I gave to your other question, we are done.

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  • $\begingroup$ Sorry but could you please clarify your argument using Fubini's theorem on product measure spaces? I think the following is true, could you please check? If $F:=\{p\in M: \forall p'\in M, \text{ there exists a unique minimizer of distance from }p \text{ in } O_{p'} \}$, then $vol_g(M\setminus F)=0.$ This is because, if $p'\notin F,$ there's $p_1\in M$ so that $p\mapsto d(p,O_{p_1})$ is not differentiable at $p'.$ But since $p\mapsto d(p,O_{p_1})$ is a.e. differentiable, so $p'$ must be an element of the set of non-differentiable points of $p\mapsto d(p,O_{p_1})$, hence has measure $0.$ $\endgroup$ Commented Apr 22 at 14:50
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    $\begingroup$ Well, the set of points where $p\mapsto d(p,O_{p_1})$ is non-differentiable has measure $0$, but there are uncountably many candidates for $p_1$, so your argument does not need to work. I will write my Fubini argument $\endgroup$
    – Saúl RM
    Commented Apr 22 at 16:23
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    $\begingroup$ Thank you, yes indeed in my argument, the $p_1$ I chose depends on $p,$ so it doesn't go through. So $F$ may not have full measure, right (even after correcting my argument)? $F=\bigcap_{p^{*}}F_{p^{*}}$. Let me check your updated Fubini argument, thanks again! I think you took the product measure $\mu \times \mu$ in the line after the first 'iff'? $\endgroup$ Commented Apr 22 at 16:33
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    $\begingroup$ Thanks for the catch, I meant the product measure indeed. $\endgroup$
    – Saúl RM
    Commented Apr 22 at 19:53
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    $\begingroup$ Sorry, I was not careful enough when writing the answer. I will check that $\endgroup$
    – Saúl RM
    Commented Apr 22 at 20:02

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