Random Points on a Sphere (Part 2)

This is the tale of a mathematical adventure. Last time our hardy band of explorers discovered that if you randomly choose two points on the unit sphere in 1-, 2- or 4-dimensional space and look at the probability distribution of their distances, then the even moments of this probability distribution are always integers. I gave a proof using some group representation theory.

On the other hand, with the help of Mathematica, Greg Egan showed that we can work out these moments for a sphere in any dimension by actually doing the bloody integrals.

He looked at the nth moment of the distance for two randomly chosen points in the unit sphere in \mathbb{R}^d, and he got

\displaystyle{ \text{moment}(d,n) = \frac{2^{d+n-2} \Gamma(\frac{d}{2}) \Gamma(\frac{1}{2} (d+n-1))}{\sqrt{\pi} \, \Gamma(d+ \frac{n}{2} - 1)} }

This looks pretty scary, but you can simplify it using the relation between the gamma function and factorials. Remember, for integers we have

\Gamma(n) = (n-1)!

We also need to know \Gamma at half-integers, which we can get knowing

\Gamma(\frac{1}{2}) = \sqrt{\pi}

and

\Gamma(x + 1) = x \Gamma(x)

Using these we can express moment(d,n) in terms of factorials, but the details depend on whether d and n are even or odd.

I’m going to focus on the case where both the dimension d and the moment number n are even, so let

d = 2e, \; n = 2m

In this case we get

\text{moment}(2e,2m) = \displaystyle{\frac{\binom{2(e+m-1)} {m}}{\binom{e+m-1}{m}} }

Here ‘we’ means that Greg Egan did all the hard work:

From this formula

\text{moment}(2e,2m) = \displaystyle{\frac{\binom{2(e+m-1)} {m}}{\binom{e+m-1}{m}} }

you can show directly that the even moments in 4 dimensions are Catalan numbers:

\text{moment}(4,2m) = C_{m+1}

while in 2 dimensions they are binomial coefficients:

\mathrm{moment}(2,2m) = \displaystyle{ {2m \choose m} }

More precisely, they are ‘central’ binomial cofficients, forming the middle column of Pascal’s triangle:

1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, \dots



So, it seems that with some real work one can get vastly more informative results than with my argument using group representation theory. The only thing you don’t get, so far, is an appealing explanation of why the even moments are integral in dimensions 1, 2 and 4.

The computational approach also opens up a huge new realm of questions! For example, are there any dimensions other than 1, 2 and 4 where the even moments are all integral?

I was especially curious about dimension 8, where the octonions live. Remember, 1, 2 and 4 are the dimensions of the associative normed division algebras, but there’s also a nonassociative normed division algebra in dimension 8: the octonions.

The d = 8 row seemed to have a fairly high fraction of integer entries:


distance_between_points_on_unit_sphere_moments_cropped.jpg

I wondered if there were only finitely many entries in the 8th row that weren’t integers. Greg Egan did a calculation and replied:

The d=8 moments don’t seem to become all integers permanently at any point, but the non-integers become increasingly sparse.

He also got evidence suggesting that for any even dimension d, a large fraction of the even moments are integers. After some further conversation he found the nice way to think about this. Recall that

\text{moment}(2e,2m) = \displaystyle{\frac{\binom{2(e+m-1)} {m}}{\binom{e+m-1}{m}} }

If we let

r = e-1

then this moment is just

\text{moment}(2r+2,2m) = \displaystyle{\frac{\binom{2(m+r)}{m}}{\binom{m+r}{m}} }

so the question becomes: when is this an integer?

It’s good to think about this naively a bit. We can cancel out a bunch of stuff in that ratio of binomial coefficents and write it like this:

\displaystyle{ \text{moment}(2r+2,2m) = \frac{(2r+m+1) \cdots (2r+2m)}{(r+1) \cdots (r+m)} }

So when is this an integer? Let’s do the 8th moment in 4 dimensions:

\text{moment}(4,8) = \displaystyle{ \frac{7 \cdot 8 \cdot 9 \cdot 10 }{2 \cdot 3 \cdot 4 \cdot 5} }

This is an integer, namely the Catalan number 42: the Answer to the Ultimate Question of Life, the Universe, and Everything.  But apparently we had to be a bit ‘lucky’ to get an integer. For example, we needed the 10 on top to deal with the 5 on the bottom.

It seems plausible that our chances of getting an integer increase as the moment gets big compared to the dimension. For example, try the 4th moment in dimension 10:

\text{moment}(10,4) = \displaystyle{ \frac{11 \cdot 12}{5 \cdot 6}}

This not an integer, because we’re just not multiplying enough numbers to handle the prime 5 in the denominator. The 6th moment in dimension 10 is also not an integer. But if we try the 8th moment, we get lucky:

\text{moment}(10,8) = \displaystyle{ \frac{13 \cdot 14 \cdot 15 \cdot 16}{5 \cdot 6 \cdot 7 \cdot 8}}

This is an integer! We’ve got enough in the numerator to handle everything in the denominator.

Greg posted a question about this on MathOverflow:

• Greg Egan, When does doubling the size of a set multiply the number of subsets by an integer?, 9 July 2018.

He got a very nice answer from a mysterious figure named Lucia, who pointed out relevant results from this interesting paper:

• Carl Pomerance, Divisors of the middle binomial coefficient, American Mathematical Monthly 122 (2015), 636–644.

Using these, Lucia proved a result that implies the following:

Theorem. If we fix a sphere of some even dimension, and look at the even moments of the probability distribution of distances between randomly chosen points on that sphere, from the 2nd moment to the (2m)th, the fraction of these that are integers approaches 1 as m → ∞.

On the other hand, Lucia also believes Pomerance’s techniques can be used to prove a result that would imply this:

Conjecture. If we fix a sphere of some even dimension > 4, and consider the even moments of the probability distribution of distances between randomly chosen points on that sphere, infinitely many of these are not integers.

In summary: we’re seeing a more or less typical rabbit-hole in mathematics. We started by trying to understand how noncommutative quaternions are on average. We figured that out, but we got sidetracked by thinking about how far points on a sphere are on average. We started calculating, we got interested in moments of the probability distribution of distances, we noticed that the Catalan numbers show up, and we got pulled into some representation theory and number theory!

I wouldn’t say our results are earth-shaking, but we definitely had fun and learned a thing or two. One thing at least is clear. In pure math, at least, it pays to follow the ideas wherever they lead. Math isn’t really divided into different branches—it’s all connected!

Afterword

Oh, and one more thing. Remember how this quest started with John D. Cook numerically computing the average of |xy - yx| over unit quaternions? Well, he went on and numerically computed the average of |(xy)z - x(yz)| over unit octonions!

• John D. Cook, How close is octonion multiplication to being associative?, 9 July 2018.

He showed the average is about 1.095, and he created this histogram:

Later, Greg Egan computed the exact value! It’s

\displaystyle{ \frac{147456}{42875 \pi} \approx 1.0947335878 \dots }

On Twitter, Christopher D. Long, whose handle is @octonion, pointed out the hidden beauty of this answer—it equals

\displaystyle{ \frac{2^{14}3^2}{5^3 7^3 \pi} }

Nice! Here’s how Greg did this calculation:

• Greg Egan, The average associator, 12 July 2018.

Details

If you want more details on the proof of this:

Theorem. If we fix a sphere of some even dimension, and look at the even moments of the probability distribution of distances between randomly chosen points on that sphere, from the 2nd moment to the (2m)th, the fraction of these that are integers approaches 1 as m → ∞.

you should read Greg Egan’s question on Mathoverflow, Lucia’s reply, and Pomerance’s paper. Here is Greg’s question:

For natural numbers m, r, consider the ratio of the number of subsets of size m taken from a set of size 2(m+r) to the number of subsets of the same size taken from a set of size m+r:

\displaystyle{ R(m,r)=\frac{\binom{2(m+r)}{m}}{\binom{m+r}{m}} }

For r=0 we have the central binomial coefficients, which of course are all integers:

\displaystyle{ R(m,0)=\binom{2m}{m} }

For r=1 we have the Catalan numbers, which again are integers:

\displaystyle{ R(m,1)=\frac{\binom{2(m+1)}{m}}{m+1}=\frac{(2(m+1))!}{m!(m+2)!(m+1)}}
            \displaystyle{ = \frac{(2(m+1))!}{(m+2)!(m+1)!}=C_{m+1}}

However, for any fixed r\ge 2, while R(m,r) seems to be mostly integral, it is not exclusively so. For example, with m ranging from 0 to 20000, the number of times R(m,r) is an integer for r= 2,3,4,5 are 19583, 19485, 18566, and 18312 respectively.

I am seeking general criteria for R(m,r) to be an integer.

Edited to add:

We can write:

\displaystyle{ R(m,r) = \prod_{k=1}^m{\frac{m+2r+k}{r+k}} }

So the denominator is the product of m consecutive numbers r+1, \ldots, m+r, while the numerator is the product of m consecutive numbers m+2r+1,\ldots,2m+2r. So there is a gap of r between the last of the numbers in the denominator and the first of the numbers in the numerator.

Lucia replied:

Put n=m+r, and then we can write R(m,r) more conveniently as

\displaystyle{ R(m,r) = \frac{(2n)!}{m! (n+r)!} \frac{m! r!}{n!} = \frac{\binom{2n}{n} }{\binom{n+r}{r}}. }

So the question essentially becomes one about which numbers n+k for k=1, \ldots, r divide the middle binomial coefficient \binom{2n}{n}. Obviously when k=1, n+1 always divides the middle binomial coefficient, but what about other values of k? This is treated in a lovely Monthly article of Pomerance:

• Carl Pomerance, Divisors of the middle binomial coefficient, American Mathematical Monthly 122 (2015), 636–644.

Pomerance shows that for any k \ge 2 there are infinitely many integers with n+k not dividing \binom{2n}{n}, but the set of integers n for which n+k does divide \binom{2n}{n} has density 1. So for any fixed r, for a density 1 set of values of n one has that (n+1), \ldots, (n+k) all divide \binom{2n}{n}, which means that their lcm must divide \binom{2n}{n}. But one can check without too much difficulty that the lcm of n+1, \ldots, n+k is a multiple of \binom{n+k}{k}, and so for fixed r one deduces that R(m,r) is an integer for a set of values m with density 1. (Actually, Pomerance mentions explicitly in (5) of his paper that (n+1)(n+2)\cdots (n+k) divides \binom{2n}{n} for a set of full density.)

I haven’t quite shown that R(m,r) is not an integer infinitely often for r\ge 2, but I think this can be deduced from Pomerance’s paper (by modifying his Theorem 1).

I highly recommend Pomerance’s paper—you don’t need to care much about which integers divide

\displaystyle{ \binom{2n}{n} }

to find it interesting, because it’s full of clever ideas and nice observations.

This entry was posted on Thursday, July 12th, 2018 at 12:00 pm and is filed under mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

24 Responses to Random Points on a Sphere (Part 2)

  1. kenneth abbott says:
    12 July, 2018 at 1:27 pm

    Much simpler result, but still very nice.. if you have a sphere in n-dimensional Euclidean Space and you insist that the surface area=the volume, then its radius is an Integer. And the integer is n.

    Reply
    • John Baez says:
      12 July, 2018 at 3:06 pm

      Nice. This must follow from

      \displaystyle{ \text{volume} = \text{surface area} \int_0^R \left(\frac{r}{R}\right)^n \, dr }

      which says that the volume of the sphere of radius R is the integral of the surface areas of spherical shells with radii going from 0 to R. The area of the spherical shell of radius r is

      \displaystyle{ \text{surface area} \left(\frac{r}{R}\right)^n }

      where

      \text{surface area}

      is the area of the sphere of radius R.

      Reply
      • kenneth abbott says:
        13 July, 2018 at 12:15 pm

        Yes. I’m pretty sure that in general surface area=d/dr(volume) so then when you require surface are=volume you get a surprise differential equation d/dr(volume)=volume
        What can this possibly mean?
        I like to think of these spheres as “holographic spheres” ;-)

        Reply
      • Greg Egan says:
        13 July, 2018 at 1:10 pm

        I think this should generalise to the radius of the inscribed sphere of any polytope whose faces all make contact with that sphere. It certainly holds for n-cubes (e.g. a cube of half-side 3 has volume and surface area both equal to 216), but it should also be true that the regular icosahedron whose surface area equals its volume must have an inscribed sphere of radius 3.

        Reply
  2. John Baez says:
    12 July, 2018 at 8:45 pm

    Over on Twitter, Nithilher the Colourless wrote approximately:

    Yes, it appears that the numbers for even dimensions d=2e have all bounded denominators with maximal prime number \le 2e-3. Odd dimensions d=2e-1 have denominators growing like O(n^{e-1}). Sorry, new for me, but probably trivial for you.

    The numbers he’s talking about are the moments

    \text{moment}(d,n)

    In the case where not only the dimension d but also the moment number n is even we should be able to prove his guess using the formula

    \displaystyle{ \text{moment}(2r+2,2m) = \frac{(2r+m+1) \cdots (2r+2m)}{(r+1) \cdots (r+m)} }

    Anyone want to take a try?

    There must be a somewhat similar formula when n is odd.

    Reply
  3. Graham Jones says:
    12 July, 2018 at 9:26 pm

    Just in case no one has noticed, if s has the P_n distribution from Part 1, and x=s^2/4, then x has a Beta distribution. I think you might be asking something like: when does \text{Beta}(n/2,(n-1)/2) have all moments integral?

    Reply
    • John Baez says:
      12 July, 2018 at 9:36 pm

      Nice! Since I don’t know much about Beta distributions, this doesn’t help me. But it could help someone who knew about them, and there’s a chance someone has already studied the problem in this guise!

      Reply
    • Greg Egan says:
      12 July, 2018 at 11:23 pm

      I’m not sure that’s true. The substitution x=s^2/4, and the right choice of parameters \alpha, \beta for the Beta distribution will make x^{\alpha-1} (1-x)^{\beta-1} proportional to s^{n-2} (4-s^2)^{\frac{n-3}{2}}, but when you do a change of variable the probability density function also picks up a factor from the derivative of the transformation.

      Reply
    • John Baez says:
      13 July, 2018 at 4:56 pm

      I believe Greg’s objection is correct. It makes me happy in a perverse way, because sometime I may want to do a literature search to see which of our results are new, and this may excuse me from having to read the literature on moments of Beta distributions.

      Reply
    • Graham Jones says:
      13 July, 2018 at 7:18 pm

      ds = x^{-1/2} dx so it still has the form of a beta distribution. No?

      Reply
    • Greg Egan says:
      13 July, 2018 at 10:13 pm

      Graham’s right, you can tweak the \alpha parameter to account for the derivative and match the probability distribution, so the relevant Beta distribution becomes \text{Beta}((n-1)/2,(n-1)/2).

      There is one remaining catch, though: the change of variables means that the original moments now come from powers, not of x, but of 4x.

      So the moments of these Beta distributions need to be multiplied by powers of 4 to make them integral. For example, to get the Catalan numbers, you need to multiply the mth moments of the Beta(3/2,3/2) distribution by 4^m.

      Reply
    • Graham Jones says:
      14 July, 2018 at 7:16 am

      I think everything is easier if you look at squared Euclidean distances. For example: Consider one point at (1,0,0…0), and a random point at (x1,x2,…xn). Let x = x1 and D be the sum of squares of x2,…xn. The squared distance is (1-x)^2 + D = 1-2x+x^2+D = 2-2x. The squared distance to the ‘mirror’ point (-x1,x2,…xn) is (1+x)^2 + D = 2+2x. So the squared distance has a symmetric distribution in [0,4]. In particular its mean is 2 for all dimensions. (Greg’s formula shows this too of course.) As for the factors of 4, just look at spheres with diameter 1 instead of radius 1!

      Reply
    • Graham Jones says:
      14 July, 2018 at 7:45 am

      Thinking about this a bit more I suspected that there would be a relationship between Dirichlet distributions on the unit simplex and uniform distributions on spheres. Google found this
      http://alea.impa.br/articles/v7/07-17.pdf
      for me, and Lemma 2.3 is a more general version of what I guessed. Those Rademacher distributed epsilons sound scary but they are just random + or – signs. The case p=2 is the one we’re talking about here.

      Reply
  4. Todd Trimble says:
    13 July, 2018 at 2:49 pm

    I’ll just mention that “Lucia” is the nom de plume of a pretty well-known number theorist with impeccable credentials. I’m contractually obligated as a MathOverflow moderator not to say who, but it’s a he.

    Reply
  5. Blake Stacey says:
    13 July, 2018 at 4:37 pm

    This is great stuff! I’ll have to clear some time (and find some mental focus) to explore it myself. I love unexpected appearances of (\mathbb{R}, \mathbb{C}, \mathbb{H}) or (\mathbb{C}, \mathbb{H}, \mathbb{O}), all the more so after I found one of my own.

    Reply
  6. John Baez says:
    13 July, 2018 at 5:13 pm

    By the way, it would be nice to settle this:

    Conjecture. In dimensions 1, 2 and 4, and only in these dimensions, all the even moments of the probability distribution of distances between two independently and uniformly distributed points are integers.

    It’s only the italicized phrase that’s left to show, and we can probably do that by looking at the 4th moment. If the dimension is even, say d = 2e, we have a nice formula for the moment:

    \text{moment}(2e,4) = \frac{(2e+1)(2e+2)}{e(e+1)} = 2\frac{2e+1}{e} = 4 + \frac{2}{e}

    and this is an integer only if e is 1 or 2, i.e. d is 2 or 4.

    So, we only need to handle all the odd dimensions. We’ve been sort of sidestepping those. We can probably take the formula

    \displaystyle{ \text{moment}(d,n) = \frac{2^{d+n-2} \Gamma(\frac{d}{2}) \Gamma(\frac{1}{2} (d+n-1))}{\sqrt{\pi} \, \Gamma(d+ \frac{n}{2} - 1)} }

    and rewrite it in terms of factorials or binomial coefficients when d is odd and n is even.

    Reply
  7. John Baez says:
    13 July, 2018 at 10:15 pm

    On MathOverflow, James Propp pushed the conversation in another direction:

    The 2mth moment of the (random) area of the triangle whose vertices are three independent, uniformly distributed random points on the unit circle appears to be ((3m)!/(m!)^3)/16^m. Can anyone prove this? Better yet, can anyone give a conceptual explanation for why this moment should be rational? (If this observation is not new, references would be appreciated.)

    Iosif Pinelis said that for every natural number n, the nth moment is

    \displaystyle{ \mathrm{E}(A^n) =\frac1{\pi^2}\int_{[0,\pi]^2}(2\sin u_1\,\sin u_2\,\sin(u_1+u_2))^n\,du_1\,du_2\ }

    Reply
  8. kenneth abbott says:
    14 July, 2018 at 12:34 pm

    Of course, this whole thing uses distance defined by the 2-metric.. d^2=x1^2+x2^2+…+xn^2 which is just a member of the metric family.. d^q=x1^q+x2^q+…+xn^q for any positive integer q. And there are many other metrics. I wonder how the results would play in different metrics?

    Reply

You can use Markdown or HTML in your comments. You can also use LaTeX, like this: $latex E = m c^2 $. The word 'latex' comes right after the first dollar sign, with a space after it.

Fill in your details below or click an icon to log in:

Gravatar
WordPress.com Logo

You are commenting using your WordPress.com account. ( Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. ( Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. ( Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. ( Log Out /  Change )

w
Cancel

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.