A curious speculation

Consider a matrix A chosen uniformly at random from the group SU(2). (Precisely, A is a random variable distributed according to Haar measure.) The expected value of Tr(A) is zero, because A \mapsto -A is a measure preserving symmetry of SU(2). The expected value of Tr(A)^2 is {1}. (A nice exercise.) In general, the expected value of Tr(A)^{2n+1} is zero, and the expected value of Tr(A)^{2n} is the Catalan number \binom{2n}{n}/(n+1). This is the sequence that starts {1}, {1}, {2}, {5}, {14}, {42}, …, where that first {1} is the expected value of Tr(A)^0=1.

I learned that fact from Hyperelliptic curves, L-polynomials, and random matrices, a very interesting
paper whose main substance I will completely ignore. That paper also discusses what happens when you replace SU(2) with SO(2). The SO(2) case is so easy we can work it out in detail: A must be of the form \left( \begin{smallmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{smallmatrix} \right), and \theta is chosen uniformly at random from the interval {} [0, 2 \pi]. So Tr(A)=2 \cos \theta = e^{i \theta} + e^{- i \theta} and the expected value of Tr(A)^{2n} is

\frac{1}{2 \pi} \int_{0}^{2 \pi} \left( e^{i \theta} + e^{- i \theta} \right)^{2n} d \theta =  \frac{1}{2 \pi} \sum_{k=0}^{2n} \binom{2n}{k} \int_{0}^{2 \pi} e^{(n-2k) i \theta} d \theta = \binom{2n}{n}.

Now, combinatorialists know that the Catalan numbers are closely related to properties of the symmetric groups S_n. And the sequence \binom{2n}{n} is known to play the same role for the symmetry groups of the hypercubes. In fact, we call the symmetric groups the “Coxeter groups of type A” and the symmetry groups of the hypercubes the “Coxeter groups of type B“. There is a growing philosophy that the sequence \binom{2n}{n} should be thought of as the Catalan numbers of type B.

Now, there is one more infinite sequence of Coxeter groups — the groups of type D. The Catalan numbers of type D are \frac{3n-2}{n}\binom{2n-2}{n-1}; the first few are 1, 4, 14, 50, …. Is there some random variable X such that E(X^{2n+1})=0 and E(X^{2n})=\frac{3n-2}{n}\binom{2n-2}{n-1}? And, if so, can we view
X as the pushforward of some kind of Haar measure?

UPDATE: I think I lose. Let’s write \Delta_n = \frac{3n-2}{n}\binom{2n-2}{n-1}. Then

\lim_{n \to \infty} \Delta_{n+1}/\Delta_{n} = 4.

I think that should implies that |X| \leq 2, with probability {1}.
But then, if we want the expected value of X^2 to be 4, then we need to have X = \pm 2 with probability 1. And that implies E(X^4)=16.

UPDATE: Well, this idea seems dead for now (although I’d be curious to here if anyone can revive it.) I encourage readers to check out the comments, though: Greg Kuperberg and Michael Lugo both bring interesting perspectives.

14 thoughts on “A curious speculation

  1. Your equation is, indeed, what I want the guess to be.

    I’ve been being sloppy about indexing because these things tend to work themselves out in my experience. But, for the record:
    \binom{2n+2}{n+1}/(n+2) is the Catalan number of type $A_{n}$, and is E(X^{2n+2}).
    \binom{2n}{n} is the Catalan number of type B_n, and is E(X^{2n}).
    \frac{3n-2}{n} \binom{2n-2}{n-1} is the Catalan number of type D_n, and is what I had hoped would be E(X^{2n-2}).

    So

  2. There is a possible salvage for both the update and Lior’s question. The update uses Lior’s suggested numbering and concludes that the construction is impossible. But with a different definition, you can use the original numbering. Let d_n be the terms of the sequence with the recurrence relation

    d_{n+1}/d_n = 2(3n+1)(2n-1)/(3n-2)/(n+1).

    Then you can let d_0 = 1 and instead d_{-1} is undefined. You come to the same conclusion if you express David’s formula in terms of the Gamma function, because the poles at 0 cancel.

    The type A and type B sequences are not just moments of random variables. They are also dimensions of invariant spaces of V^{tensor n} for some representation V of some compact group G. In all three cases, V has to be two-dimensional. So the question is whether you can find a subgroup G of U(2) to make it work in the D case. This would be an even stricter generalization than what David was asking for. But I think that the group G does not exist in this case. It cannot be a finite group, because then d_n would grow as Theta(2^n), instead of as Theta(2^n/sqrt(n)). But the only other choice is O(2) or the normalizer of the Cartan circle in SU(2), which give the same numbers as O(2) and I don’t think that they are the right numbers.

    That does not show that the random variable alone does not exist, nor even the Haar measure interpretation, just that the character interpretation does not exist. On the other hand having a Haar measure interpretation without some relation to characters is superficial, because Haar measure by itself is measure-o-morphic to uniform measure on an interval.

  3. There is fourth infinite family of Coxeter groups, but most dihedral groups don’t come from lattices.

  4. I’ll give an alternative proof that there’s no such random variable, which doesn’t seem all that enlightening to me but might trip something in somebody’s brain. I’ll show that there’s no random variable with zeroth through fourth moments equal to 1, 0, 4, 0, 14. A sequence of numbers a_0, a_1, a_2, … is a sequence of moments of some random variable if and only if the Hankel matrices of that sequence, which have the (i, j)th entry equal to a_{i+j}, are all positive semidefinite. (This is the problem of moments.) But the 3-by-3 Hankel matrix

    1 0 4
    0 4 0
    4 0 14

    has determinant -8, so is not positive semidefinite.

  5. Michael’s argument can be made more efficiently by pointing out that E(X^2)^2 should be less than E(X^4) by Cauchy-Schwartz (or the power mean inequality.) Notice that the determinant is block diagonal, with one block 4 and the other \left( \begin{smallmatrix} 1 & 4 \\ 4 & 14 \end{smallmatrix} \right). So the important comparison is between 4^2 and {14}.

    I thought I checked that, but I must have gotten the inequality backward. So thanks!

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