Random Points on a Group

Azimuth 2018-08-02

In Random Points on a Sphere (Part 1), we learned an interesting fact. You can take the unit sphere in $\mathbb{R}^n$ , randomly choose two points on it, and compute their distance. This gives a random variable, whose moments you can calculate.

And now the interesting part: when n = 1, 2 or 4, and seemingly in no other cases, all the even moments are integers.

These are the dimensions in which the spheres are groups. We can prove that the even moments are integers because they are differences of dimensions of certain representations of these groups. Rogier Brussee and Allen Knutson pointed out that if we want to broaden our line of investigation, we can look at other groups. So that’s what I’ll do today.

If we take a representation of a compact Lie group $G,$ we get a map from group into a space of square matrices. Since there is a standard metric on any space of square matrices, this lets us define the distance between two points on the group. This is different than the distance defined using the shortest geodesic in the group: instead, we’re taking a straight-line path in the larger space of matrices.

If we randomly choose two points on the group, we get a random variable, namely the distance between them. We can compute the moments of this random variable, and today I’ll prove that the even moments are all integers.

So, we get a sequence of integers from any representation $\rho$ of any compact Lie group $G.$ So far we’ve only studied groups that are spheres:

• The defining representation of $\mathrm{O}(1) \cong S^0$ on the real numbers $\mathbb{R}$ gives the powers of 2.

• The defining representation of $\mathrm{U}(1) \cong S^1$ on the complex numbers $\mathbb{C}$ gives the central binomial coefficients $\binom{2n}{n}.$

• The defining representation of $\mathrm{Sp}(1) \cong S^3$ on the quaternions $\mathbb{H}$ gives the Catalan numbers.

It could be fun to work out these sequences for other examples. Our proof that the even moments are integers will give a way to calculate these sequences, not by doing integrals over the group, but by counting certain ‘random walks in the Weyl chamber’ of the group. Unfortunately, we need to count walks in a certain weighted way that makes things a bit tricky for me.

But let’s see why the even moments are integers!

If our group representation is real or quaternionic, we can either turn it into a complex representation or adapt my argument below. So, let’s do the complex case.

Let $G$ be a compact Lie group with a unitary representation $\rho$ on $\mathbb{C}^n.$ This means we have a smooth map

$\rho \colon G \to \mathrm{End}(\mathbb{C}^n)$

where $\mathrm{End}(\mathbb{C}^n)$ is the algebra of $n \times n$ complex matrices, such that

$\rho(1) = 1$

$\rho(gh) = \rho(g) \rho(h)$

and

$\rho(g) \rho(g)^\dagger = 1$

where $A^\dagger$ is the conjugate transpose of the matrix $A.$

To define a distance between points on $G$ we’ll give $\mathrm{End}(\mathbb{C}^n)$ its metric

$\displaystyle{ d(A,B) = \sqrt{ \sum_{i,j} \left|A_{ij} - B_{ij}\right|^2} }$

This clearly makes $\mathrm{End}(\mathbb{C}^n)$ into a $2n^2$ -dimensional Euclidean space. But a better way to think about this metric is that it comes from the norm

$\displaystyle{ \|A\|^2 = \mathrm{tr}(AA^\dagger) = \sum_{i,j} |A_{ij}|^2 }$

where $\mathrm{tr}$ is the trace, or sum of the diagonal entries. We have

$d(A,B) = \|A - B\|$

I want to think about the distance between two randomly chosen points in the group, where ‘randomly chosen’ means with respect to normalized Haar measure: the unique translation-invariant probability Borel measure on the group. But because this measure and also the distance function are translation-invariant, we can equally well think about the distance between the identity 1 and one randomly chosen point $g$ in the group. So let’s work out this distance!

I really mean the distance between $\rho(g)$ and $\rho(1),$ so let’s compute that. Actually its square will be nicer, which is why we only consider even moments. We have

$\begin{array}{ccl} d(\rho(g),\rho(1))^2 &=& \|\rho(g) - \rho(1)\|^2 \\ \\ &=& \|\rho(g) - 1\|^2 \\ \\ &=& \mathrm{tr}\left((\rho(g) - 1)(\rho(g) - 1)^\dagger)\right) \\ \\ &=& \mathrm{tr}\left(\rho(g)\rho(g)^\dagger - \rho(g) - \rho(g)^\ast + 1\right) \\ \\ &=& \mathrm{tr}\left(2 - \rho(g) - \rho(g)^\dagger \right) \end{array}$

Now, any representation $\sigma$ of $G$ has a character

$\chi_\sigma \colon G \to \mathbb{C}$

defined by

$\chi_\sigma(g) = \mathrm{tr}(\sigma(g))$

and characters have many nice properties. So, we should rewrite the distance between $g$ and the identity using characters. We have our representation $\rho,$ whose character can be seen lurking in the formula we saw:

$d(\rho(g),\rho(1))^2 = \mathrm{tr}\left(2 - \rho(g) - \rho(g)^\dagger \right)$

But there’s another representation lurking here, the dual

$\rho^\ast \colon G \to \mathrm{End}(\mathbb{C}^n)$

given by

$\rho^\ast(g)_{ij} = \overline{\rho(g)_{ij}}$

This is a fairly lowbrow way of defining the dual representation, good only for unitary representations on $\mathbb{C}^n,$ but it works well for us here, because it lets us instantly see

$\mathrm{tr}(\rho(g)^\dagger) = \mathrm{tr}(\rho^\ast(g)) = \chi_{\rho^\ast}(g)$

This is useful because it lets us write our distance squared

$d(\rho(g),\rho(1))^2 = \mathrm{tr}\left(2 - \rho(g) - \rho(g)^\dagger \right)$

in terms of characters:

$d(\rho(g),\rho(1))^2 = 2n - \chi_\rho(g) - \chi_{\rho^\ast}(g)$

So, the distance squared is an integral linear combination of characters. (The constant function 1 is the character of the 1-dimensional trivial representation.)

And this does the job: it shows that all the even moments of our distance squared function are integers!

Why? Because of these two facts:

1) If you take an integral linear combination of characters, and raise it to a power, you get another integral linear combination of characters.

2) If you take an integral linear combination of characters, and integrate it over $G,$ you get an integer.

I feel like explaining these facts a bit further, because they’re part of a very beautiful branch of math, called character theory, which every mathematician should know. So here’s a quick intro to character theory for beginners. It’s not as elegant as I could make it; it’s not as simple as I could make it: I’ll try to strike a balance here.

There’s an abelian group $R(G)$ consisting of formal differences of isomorphism classes of representations of $G$ , mod the relation

$[\rho] + [\sigma] = [\rho \oplus \sigma]$

Elements of $R(G)$ are called virtual representations of $G.$ Unlike actual representations we can subtract them. We can also add them, and the above formula relates addition in $R(G)$ to direct sums of representations.

We can also multiply them, by saying

$[\rho] [\sigma] = [\rho \otimes \sigma]$

and decreeing that multiplication distributes over addition and subtraction. This makes $R(G)$ into a ring, called the representation ring of $G.$

There’s a map

$\chi \colon R(G) \to C(G)$

where $C(G)$ is the ring of continuous complex-valued functions on $G.$ This map sends each finite-dimensional representation $\rho$ to its character $\chi_\rho.$ This map is one-to-one because we know a representation up to isomorphism if we know its character. This map is also a ring homomorphism, since

$\chi_{\rho \oplus \sigma} = \chi_\rho + \chi_\sigma$

and

$\chi_{\rho \otimes \sigma} = \chi_\rho \chi_\sigma$

These facts are easy to check directly.

We can integrate continuous complex-valued functions on $G,$ so we get a map

$\displaystyle{\int} \colon C(G) \to \mathbb{C}$

The first non-obvious fact in character theory is that we can compute inner products of characters as follows:

$\displaystyle{\int} \overline{\chi_\sigma} \chi_\rho = \dim(\mathrm{hom}(\sigma,\rho))$

where the expression at right is the dimension of the space of ‘intertwining operators’, or morphisms of representations, between the representation $\sigma$ and the representation $\rho.$

What matters most for us now is that this inner product is an integer. In particular, if $\chi_\rho$ is the character of any representation,

$\displaystyle{\int} \chi_\rho$

is an integer because we can take $\sigma$ to be the trivial representation in the previous formula, giving $\chi_\sigma = 1.$

Thus, the map

$R(G) \stackrel{\chi}{\longrightarrow} C(G) \stackrel{\int}{\longrightarrow} \mathbb{C}$

actually takes values in $\mathbb{Z}.$

Now, our distance squared function

$2n - \chi_\rho - \chi_{\rho^\ast} \in C(G)$

is actually the image under $\chi$ of an element of the representation ring, namely

$2n - [\rho] - [\rho^\ast]$

So the same is true for any of its powers—and when we integrate any of these powers we get an integer!

This stuff may seem abstract, but if you’re good at tensoring representations of some group, like $\mathrm{SU}(3),$ you should be able to use it to compute the even moments of the distance function on this group more efficiently than using the brute-force direct approach. Instead of complicated integrals we wind up doing combinatorics.

I would like to know what sequence of integers we get for $\mathrm{SU}(3).$ A much easier, less thrilling but still interesting example is $\mathrm{SO}(3).$ This is the 3-dimensional real projective space $\mathbb{R}\mathrm{P}^3,$ which we can think of as embedded in the 9-dimensional space of $3\times 3$ real matrices. It’s sort of cool that I could now work out the even moments of the distance function on this space by hand! But I haven’t done it yet.