Timeline for Intuitive proof that the first $(n-2)$ coordinates on a sphere are uniform in a ball
Current License: CC BY-SA 3.0
10 events
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| Apr 18, 2017 at 5:45 | history | edited | Michael Hardy | CC BY-SA 3.0 |
added 7 characters in body
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| Jul 26, 2010 at 21:15 | comment | added | Mark Meckes | Actually your original interpretation, "if a sequence of points is uniformly distributed in $S^{n−1}$, then its projection is uniformly distributed in $B^{n−2}$", is exactly right, in an appropriate asymptotic sense. | |
| Jul 26, 2010 at 18:52 | comment | added | Mark Meckes | I've added a sentence clarifying that the pull-back construction is what I meant. | |
| Jul 26, 2010 at 16:55 | comment | added | Dick Palais | Thanks, Peter, Joel, Mark, and Greg. This was the kind of answer I was hoping for. That the pull-back metric was meant makes clear the relation to the result of Archimedes. And having taught measure theory so many times, perhaps I should have thought of the pull-back construction, but the wording really confused me. When I first read the question, it seemed to me that what you were saying was that if a sequence of points is uniformly distributed in $S^{n-1}$, then its projection is uniformly distributed in $B^{n-2}$. It is a nice question, but perhaps it should be edited for clarity. | |
| Jul 26, 2010 at 14:52 | comment | added | Greg Kuperberg | Hi Dick! The short answer is that the axis projection $f:(x,y,z) \mapsto z$ preserves measure in the sense that the area of $f^{-1}(S)$ is proportional to the length of $S$. (Which in more erudite terms is the push-forward as Joel says.) You can in any case put the circle or torus factor back in any of these moment maps to get a same-dimensional measure-preserving map in the opposite direction. E.g., from the cylinder to the sphere in the original case of Archimedes. This last map has to be in that direction, because otherwise where would you send the poles? | |
| Jul 26, 2010 at 13:14 | comment | added | Mark Meckes | Your difficulty in interpreting the question may be my fault for stating the question in language familiar to probabilists, when I wanted an answer in terms of geometry. | |
| Jul 25, 2010 at 11:39 | comment | added | Joel Fine | Grr, browser bug means I can't delete and replace the above comment. The formula for the push-forward measure that I mistyped there should read $f_*\mu(U) = \mu(f^{-1}(U))$. | |
| Jul 25, 2010 at 11:36 | comment | added | Joel Fine | Note you can always push measures forward but you can't in general pull them back. Given a map $f \colon (X,\mu) \to Y$, define the push-forward measure on Y by the formula $\f_*\mu(U) = \mu(f^{-1}(U))$. Given a measure $\nu$ on $Y$ you might try to define the pull back by $f^*\nu(V) = \nu(f(V))$. But this won't always be a measure; eg $f$ may send disjoint sets to overlapping ones, so $f^*\nu$ may not be additive. This is the case for the projection $\pi \colon S^2 \to I$ we are talking about here. So really it only makes sense to talk of $\pi$ being measure preserving in one direction. | |
| Jul 25, 2010 at 11:17 | comment | added | Peter LeFanu Lumsdaine | This is just a confusion in the use of “measure-preserving”, I think. The sense in which it holds here is for inverse images: for a subset $U \subseteq I$, we have $\mu(p^-1(U)) = \mu(U)$. As you point out, the analogous property doesn't hold for forward images. I've seen “measure-preserving” used for both of these properties in the past, iirc; surely there ought to be some terminology to distinguish the two? | |
| Jul 25, 2010 at 10:29 | history | answered | Dick Palais | CC BY-SA 2.5 |