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It is a classical fact that if $(x_1,\ldots,x_n)$ is a random vector uniformly distributed on the sphere $S^{n-1} \subseteq \mathbb{R}^n$, then the random vector $(x_1,\ldots,x_{n-2})$ is uniformly distributed in the unit ball $B_{n-2} = \{ (y_1,\ldots,y_{n-2}) \mid \sum_{i=1}^{n-2} y_i^2 \le 1\} \subseteq \mathbb{R}^{n-2}$. In measure-theoretic language, the pull-back of volume measure on $B_{n-2}$ via the coordinate projection $S^{n-1} \to B_{n-2}$, $(x_1,\ldots,x_n) \mapsto (x_1,\ldots,x_{n-2})$ is Hausdorff measure on $S^{n-1}$ (up to normalization). Apparently the $n=3$ case was known to Archimedes.

Is there an intuitive geometric proof of this, that in particular explains why you drop 2 coordinates, as opposed to 1 or 3 or ...? Or even some heuristic that explains the 2?

I already know reasonably slick probabilistic proofs of this result, including a version for $\ell_p$ norms when $p$ is an integer and you project onto the first $n-p$ coordinates (using the right distribution on the $\ell_p$ sphere, which is not surface area except for $p=1,2$), but as far as I can see they just make it look like a coincidence that things turn out this way. (And as far as I know, maybe it is.)

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My intuition is that you just drop one dimension since the "dimension of $B_{n-2}$ is $n-2$ and the dimension of $S^{n-1}$ is $n-1$. –  robin girard Jul 23 '10 at 21:25
    
@Robin: I deliberately used the word "coordinate" instead of "dimension" for that reason. –  Mark Meckes Jul 24 '10 at 0:10
    
A couple people have emailed me to ask about the $\ell_p$ version I mentioned. It's Corollary 4 in this paper: arxiv.org/abs/math/0503650 –  Mark Meckes Mar 14 '13 at 23:30

2 Answers 2

One viewpoint, which is a bit gauche for your construction but valid, is that the general result is a corollary of Archimedes' theorem, that the projection from a 2-sphere $S^2$ to an interval $I$ is measure-preserving. Whether or not you view it as a coincidence, Archimedes' theorem has an important generalization. Namely, $S^2$ is the simplest example of a projective toric variety, and the coordinate projection is its toric moment map. The moment map of any projective toric variety is measure preserving. For instance, the moment map from $\mathbb{C}P^n$ to the $n$-simplex shows you that the Fubini-Study volume of the former is $\pi^n/n!$. You might also recognize this as the volume of the unit ball $B_{2n}$. There is a simple symplectic map from $B_{2n}$ to $\mathbb{C}P^n$ which is 1-to-1 in the interior and quotients the boundary to $\mathbb{C}P^{n-1}$. (I learned/realized these facts in an old discussion with Doug Ravenel and Yael Karshon.)

So you could say that the original relation has a good explanation in complex and symplectic geometry, and that the explanation has been disguised a bit in real geometry. Moreover, that 2 arises because $\dim_\mathbb{R} \mathbb{C} = 2$.

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When Richard Thomas taught me toric geometry he proved Delzant's theorem - on the correspondence between toric symplectic manifolds and Delzant polytopes - using Archimedes's Theorem. I give a brief version of his argument in an answer to this question (please excuse the shameless self-promotion!): mathoverflow.net/questions/4982/look-into-delzant-polytope –  Joel Fine Jul 24 '10 at 6:34
    
Your self-promotion needs no excuse. As long we've started that, I used Archimedes' theorem and other moment maps in a paper on numerical quadrature. front.math.ucdavis.edu/math/0405366 –  Greg Kuperberg Jul 24 '10 at 19:07
    
I'd be happier with your first observation, that the $n\ge 4$ case follows from the $n=3$ case, if I knew a nice intuitive geometric argument for that reduction, that didn't include anything that looked like a happy numerical coincidence. Even so, it's a very good point and does address the question "Why the 2?" at least halfway. ("Because 3-1=2.") –  Mark Meckes Jul 26 '10 at 13:19
    
I'd be happier with the rest of your answer if I knew anything about projective toric varieties; now I guess I have an excuse to go learn something about them. Maybe if I did I'd consider this as nice an answer as I'd hoped for in the first place. In any case, it's nice to see that this is part of a larger story. –  Mark Meckes Jul 26 '10 at 13:22

I find myself very confused by all this, and I suspect I must be missing something very important, and I am hoping someone (Greg?) can set me straight.

Let's just consider the classic case n=3, so we are projecting from the 2-sphere S onto its projection on the $x$-axis, i.e., the interval $I = [-1,1]$ using the map $(x,y,z) \mapsto x$.

The Archimedes projection that has been mentioned several times is very different---it is the projection of S to the right circular cylinder C tangent to the sphere along its equator. I agree that this is measure (i.e., area) preserving. (Who am I to argue with Archimedes?) On the other hand, the projection mentioned by the OP is dimension reducing so we seem to be comparing the area of a region with length of its projection.

Now the projection of S onto I can't be measure preserving can it? First of all it doesn't seem to be dimensionally correct. Consider for example a small spherical cap of radius $r$ centered at $(0,0,1)$. To first order its area is $\pi r^2$. However, its projection is the interval $[-r,r]$ which has length $2r$. How can these quantities be proportional? Even worse, if we apply a rotation, the area of the spherical cap stays constant, but the area of its projection varies wildly.

It sounds to me like I must be somehow misinterpreting the original question, but I haven't been able to re-interpret it in a way consistent with its wording.

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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? –  Peter LeFanu Lumsdaine Jul 25 '10 at 11:17
    
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. –  Joel Fine Jul 25 '10 at 11:36
    
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 $f_*\mu(U) = \mu(f^{-1}(U))$. –  Joel Fine Jul 25 '10 at 11:39
    
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. –  Mark Meckes Jul 26 '10 at 13:14
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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? –  Greg Kuperberg Jul 26 '10 at 14:52

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