# For proper group action on closed Riemannian manifold, must the union of orbits with non-unique closest points to a given point be of 0 volume measure

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?

• 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? Commented Apr 16 at 1:03
• @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. Commented Apr 16 at 1:12
• Yes, I would choose one of the two intersections of the axis with $\mathbb{S}^2$ Commented Apr 16 at 1:22
• @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... Commented Apr 16 at 1:24
• 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 $0$) Commented Apr 16 at 1:35

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.

• 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 $0.$ Commented Apr 22 at 14:50
• 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 Commented Apr 22 at 16:23
• 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'? Commented Apr 22 at 16:33
• Thanks for the catch, I meant the product measure indeed. Commented Apr 22 at 19:53
• Sorry, I was not careful enough when writing the answer. I will check that Commented Apr 22 at 20:02