Intuitive proof that the first $(n-2)$ coordinates on a sphere are uniform in a ball

Question

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.)

Answer 1

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 $\dim_\mathbb{R} \mathbb{C} = 2$.

Answer 2

Turning around the same ideas than previously exposed (I am not sure there exist two fundamentally different arguments), let me give a rough geometric proof.

Assume you project the uniform measure on $\mathbb{S}^{n-1}$ on $\mathbb{B}^{n-k}$. To compute the density of the projected measure with respect to Lebesgue measure, you have to take into account two factors :

• the size of the inverse image of the point, which is a sphere of dimension $k-1$ and radius $r=\sqrt{1-R^2}$, $R$ being the distance of the point from the center,
• the width $d\ell$ of the slice around the inverse image, which is slanted: $d\ell = \sqrt{dr^2+dR^2} = \frac1r dR$ (this is a two line computation that might seem magical, but you have to use the equation of a sphere somehow, right?)

You get a density proportional to $r^{k-1}/r$, so that for it to be constant you need to take $k=2$.

Answer 3

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.

Answer 4

In a comment Mark Meckes asked about an argument to reduce the $n\geq 4$ case to the $n=3$ case. Let $n\geq n'\geq 3.$ A smooth function on a convex set $B$ is constant if and only if the restriction to every $B\cap S,$ where $S$ is an $(n'-2)$-dimensional affine subspace, is constant. At least intuititvely, this means the pushforward measure on $B^{n-2}$ from the projection $S^{n-1}\to B^{n-2}$ is uniform if and only if there is a uniform conditional distribution for each $(n'-2)$-dimensional subspace. But fixing such a subspace corresponds to restricting to a scaled version of $S^{n'-1}\to B^{n'-2}.$ This implies that the results for $n$ and $n'$ are equivalent.

So if you accept Archimedes' theorem, you get the result for all $n.$ Alternatively you might find it more intuitive to prove the result for $n=4.$ This calculation is not too bad. "On the volumes of balls" by Blass and Schaunuel uses the parameterization $s_i=\tfrac 1 2 r_i^2$ to give a two-to-one map $B^2\times B^2\to B^4$ preserving the volume form $ds_1d\theta_1ds_2d\theta_2.$ Specifically, $(s_1,\theta_1,s_2,\theta_2)\to (s_1-s_2,\theta_1,s_2,\theta_2)$ for $s_1>s_2.$ For $s_1=1$ this defines a volume preserving map $S^1\times B^2\to S^3.$

Answer 5

Caveat: I gloss over constants, but it still holds.

Consider the basic case of projecting the 3-d sphere $S^2$ onto $B^1=I$. Let's answer the question what is the probability of a point (x,y,z) having z value between z and z+dz. If we use spherical coordinates, we have $\sin(\varphi_1)=z$. The probability of a "belt" is its area. The area is the width $d\varphi_1$ times the circumference of a circle at that "latitude" $\varphi_1$ which has radius $r=\cos(\varphi_1)$:

$P(z<z'<z+dz) = 2\pi r d\varphi_1= 2\pi \cos(\varphi_1)d\varphi_1$

But since $\sin(\varphi_1)=z$, taking differentials we have $\cos(\varphi_1)d\varphi_1=dz$. Intuitively, this is because the width of the belt has an angle with respect to the z axis, so for example close to the pole a small segment in z produces a very wide (though short) belt, in a way that just cancels out.

So

$P(z<z'<z+dz) = 2\pi \cos(\varphi_1)d\varphi_1 = 2\pi dz $ which is independent of z, so constant. We've reduced one spherical coordinate $\varphi_1$ to project to a lower dimensional ball.

Now, if we consider the n-dimensional sphere $S^{n-1}$ we can use similar coordinates, and reduce $\varphi_1$ again to find the belt of $(x_1, x_1+dx_1)$. This time, the radius of the (now not circle but) $S^{n-2}$ sphere is the same as before, $r=\cos(\varphi_1)$. Its surface is then proportional to $r^{n-2}=\cos^{n-2}(\varphi_1)$, and for the probability we need to multiply by the width which is again $\cos(\varphi_1)d\varphi_1=dx_1$.

So we have $P(x_1<x_1'<x_1+dx_1) = \cos^{n-2}(\varphi_1)d\varphi_1 = \cos^{n-3}(\varphi_1) dx_1 $

If $n \neq 3$ this probability now is dependent on $x_1$. But if we reduced it not to a single coordinate $x_1$, but to a shape which has the same density as a function of $x_1$ we could have a uniform distribution over that shape. The ball $B^{n-2}$ has precisely this density as a function of $x_1$. If we don't reduce the dimension by 2, but by something else, we would not have the right density for each belt in $x_1$. If you think about it, for reasons of symmetry if it holds for $x_1$ it follows that it's uniform on the ball, but I'm not going for a full proof here.

These are the closest illustrations I could find.

Answer 6

I discussed this with Rocco Chirivi' and the easiest explanation for us is the following. Writing $x=(x', x_{n-1}, x_n)$, the unit sphere $\mathbb S^{n-1}$ is a "surface" of revolution with the respect to the $x'$ plane and we parametrize as $x'=x',\ x_{n-1}=f(x') \sin \theta,\ x_n=f(x') \cos \theta$ with $x' \in B_{n-2}$ and $f(x')=\sqrt{1-\|x'\|^2}$.

By Guldino's theorem its area element is $d\sigma=f(x') \sqrt{1+\|\nabla f(x')\|^2} dx' d\theta=dx'd \theta$.

Integration in spherical coordinates also yields the result but perhaps in a more obscure way.

Answer 7

Fix integer ${ n \in \mathbb{Z} _{>0} }.$ Like here, consider positive part of ${ n-}$ball ${ (B ^n) ^{+} = \lbrace x \in \mathbb{R} ^{n} : \text{each } x _i > 0, \lVert x \rVert < 1 \rbrace }$ and positive part of ${ n -}$sphere ${ (S ^{n}) ^{+} = \lbrace x \in \mathbb{R} ^{n+1} : \text{each } x _i > 0, \lVert x \rVert = 1 \rbrace }.$

Consider bijection ${ \phi : (B ^n) ^{+} \to (S ^n) ^{+} }$ sending ${ (y _1, \ldots, y _n) \mapsto (z _1, \ldots, z _{n+1}) = (y _1, \ldots, y _n, \sqrt{1 - \sum _{1} ^{n} y _i ^2 } ) .}$
From ${ p = 2 }$ case of theorem ${ 1 }$ in the link, random vector ${ Z = \phi(Y) \in (S ^n) ^{+} }$ is uniform iff random vector ${ Y \in (B ^{n}) ^{+} }$ has density ${ f _{Y} (y) = \frac{\text{const}}{z _{n+1}} }$ ${ = \frac{\text{const}}{\sqrt{1 - \sum _{i=1} ^{n} y _i ^2 }} .}$

Consider the map ${ \psi : (B ^{n}) ^{+} \to (B ^{n-1}) ^{+} }$ sending ${ (y _1, \ldots, y _n) \mapsto (x _1, \ldots, x _{n-1}) = (y _1, \ldots, y _{n-1}) }.$
If ${ Y \in (B ^n) ^{+} }$ has density ${ f _{Y}(y) }$ then ${ X = \psi(Y) \in (B ^{n-1}) ^{+} }$ has density ${ f _{X} (x) = \int _{0} ^{\sqrt{1-\sum _{i=1} ^{n-1} x _i ^2} } f _{Y} (x _1, \ldots, x _{n-1}, u) \, du }.$

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If ${ Z \in (S ^n) ^{+} }$ is uniform, we saw the density of ${ Y = \phi ^{-1} (Z) \in (B ^n) ^{+} }$ is ${ f _{Y} (y) = \frac{\text{const}}{\sqrt{1 - \sum _{i=1} ^{n} y _i ^2 }} }$.
Hence the density of ${ X = \psi(Y) \in (B ^{n-1}) ^{+} }$ is ${ f _{X} (x) = \int _{0} ^{\sqrt{1-\sum _{i=1} ^{n-1} x _i ^2} } f _{Y} (x _1, \ldots, x _{n-1}, u) \, du }$ ${ = \int _{0} ^{L} \frac{\text{const}}{\sqrt{L ^2 - u ^2}}\, du }$ where ${ L = \sqrt{1-\sum _{i=1} ^{n-1} x _i ^2 } }.$
But integral ${ \int _{0} ^{L} \frac{du}{\sqrt{L ^2 - u ^2}} }$, on substituting ${ u = L \sin(\theta) ,}$ is the constant ${ \frac{\pi}{2} }.$ Hence above density of ${ X \in (B ^{n-1}) ^{+} }$ is constant. This gives:

Thm: Fix integer ${ n \geq 2 }.$ Say random vector ${ Z \in (S ^{n}) ^{+} }$ is uniform over ${ (S ^n) ^+ }.$ Then the random vector ${ X = (\psi \circ \phi ^{-1}) (Z) = (Z _1, \ldots, Z _{n-1}) }$ in ${ (B ^{n-1}) ^{+} }$ is uniform over ${ (B ^{n-1}) ^{+} .}$

Now by symmetry:

Thm: Fix integer ${ n \geq 2 }.$ Say random vector ${ Z }$ is uniform over ${ S ^n }.$ Then the random vector ${ X = (Z _1, \ldots, Z _{n-1}) }$ in ${ B ^{n-1} }$ is uniform over ${ B ^{n-1} .}$