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

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.)

• 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 $n-1$. Jul 23, 2010 at 21:25
• @Robin: I deliberately used the word "coordinate" instead of "dimension" for that reason. Jul 24, 2010 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 Mar 14, 2013 at 23:30

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

• 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 Jul 24, 2010 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. arxiv.org/abs/math/0405366 Jul 24, 2010 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.") Jul 26, 2010 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. Jul 26, 2010 at 13:22

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 $k=2$.

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.

• 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 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. Jul 25, 2010 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 $f_*\mu(U) = \mu(f^{-1}(U))$. Jul 25, 2010 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. Jul 26, 2010 at 13:14
• 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 14:52

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 $S^1\times B^2\to S^3.$

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

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

$$P(z which is independent of z, so constant. We've reduced one spherical coordinate $$\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

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.   • I seems that tiff format is not supported (that's probably why the image the you've linked to was not displayed). I have tried to convert it to png and also changed size from i.stack.imgur.com/JBwRz.png to i.stack.imgur.com/JBwRzm.png - if you're satisfied with the result, you can probably omit the original link to the tiff file. Jan 17, 2020 at 14:54

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.

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 $${ f _{X} (x) = \int _{0} ^{\sqrt{1-\sum _{i=1} ^{n-1} x _i ^2} } f _{Y} (x _1, \ldots, x _{n-1}, u) \, du }.$$

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 $${ B ^{n-1} .}$$