Timeline for 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
Current License: CC BY-SA 4.0
11 events
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| Apr 22 at 20:05 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| Apr 22 at 20:02 | comment | added | Saúl RM | Sorry, I was not careful enough when writing the answer. I will check that | |
| Apr 22 at 19:56 | comment | added | Learning math | Thanks, another thing is that I think you meant that for almost all $p^{*}\in M, M\setminus F_{p^{*}}$ has measure $0?$ | |
| Apr 22 at 19:53 | comment | added | Saúl RM | Thanks for the catch, I meant the product measure indeed. | |
| Apr 22 at 19:52 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| Apr 22 at 16:33 | comment | added | Learning math | 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'? | |
| Apr 22 at 16:29 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| Apr 22 at 16:23 | comment | added | Saúl RM | 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 | |
| Apr 22 at 14:50 | comment | added | Learning math | 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 $0.$ | |
| Apr 16 at 2:17 | history | edited | Saúl RM | CC BY-SA 4.0 |
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| Apr 16 at 2:12 | history | answered | Saúl RM | CC BY-SA 4.0 |