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Let $O_n$ be the (real) orthogonal group of $n$ by $n$ matrices. I am interested in the following sequence which showed up in a calculation I was doing $$a_k = \int_{O_n} (\text{Tr } X)^k dX$$ where the integral is taken with respect to the normalized Haar measure on $O_n.$ I noticed some work of Dianconis et al. covers a more general set of moments, but essentially the formulas I saw only seem to hold for small $k.$ For the case of $O_2$ the sequence looks like $a_{2k} = \frac{1}{2}\binom{2k}{k},$ which can be calculated by hand since $O_2$ has some special structure.

I was hoping there would be some general known formula (since, if I recall correctly, this calculates the multiplicity the trivial representation in a tensor power the standard representation of $O_n$), or that it might be in the OEIS, but I don't know exactly what to look for, and don't know enough small terms to search.

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up vote 13 down vote accepted

Pastur and Vasilchuk have extended the result of Diaconis and Evans for $a_{2k}$ from $2k\leq n/2$ to $2k\leq n-1$:

$$a_{2k}=\pi^{-1/2}2^{k}\Gamma(k+1/2)\;\;\text{for}\;\;2k\leq n-1\quad\quad[*]$$

As suggested by Liviu Nicolaescu, for small $n$ you might directly integrate over the probability distribution of the eigenvalues in $O_n$, which you can find here. (This is the socalled "circular real ensemble" of random matrices.)

$\bullet$ For $n=2$ one reproduces the result $$a_{2k}=\frac{1}{\pi}\int_0^{\pi/2}dx\,(2\cos x)^{2k}=\tfrac{1}{2}(2k)!(k!)^{-2}$$ you quoted above, which agrees with $[*]$ for $k=1$ and $k=2$. (For $k=3$ it gives $10$ instead of $15$.)

$\bullet$ For $n=3$ one obtains $$a_{2k}=\frac{1}{2\pi}\int_0^{2\pi}dx\,(1-\cos x)(1+2\cos x)^{2k}$$ which evaluates to $$a_2=1,\;\; a_4=3,\;\; a_6=15,\;\; a_8=91,\;\;a_{10}=603,\;\;a_{12}=4213$$ in agreement with the series of Qiaochu Yuan's answer. The formula $[*]$ for $a_{2k}$ holds for $k\leq 3$ (for $k=4$ it gives $105$ instead of $91$).

$\bullet$ For $n=4$ one finds $$a_{2k}=\frac{1}{8\pi^2}\int_0^{2\pi}dx\int_0^{2\pi}dy\,(\cos x-\cos y)^2(2\cos x+2\cos y)^{2k}$$ $$\quad\quad+\frac{1}{2\pi}\int_0^{2\pi}dz\,(1-\cos^2 z)(2\cos z)^{2k}$$ which evaluates to $$a_2=1,\;\;a_4=3,\;\;a_6=15,\;\;a_8=105,\;\;a_{10}=903,\;\;a_{12}=8778.$$ Now the formula $[*]$ for $a_{2k}$ holds for $k\leq 4$ (for $k=5$ it gives $945$ instead of $903$).

$\bullet$ Continuing with $n=5$, we have $$a_{2k}=\frac{1}{2\pi^2}\int_0^{2\pi}dx\int_0^{2\pi}dy\,(1-\cos x)(1-\cos y)(\cos x-\cos y)^2(1+2\cos x+2\cos y)^{2k}$$ which evaluates to $$a_2=1,\;\;a_4=3,\;\;a_6=15,\;\;a_8=105,\;\;a_{10}=945,\;\;a_{12}=10263.$$ The formula $[*]$ for $a_{2k}$ holds for $k\leq 5$ (for $k=6$ it gives $10395$ instead of $10263$).

I am now tempted to $$\text{conjecture}\quad\quad a_{2k}=\pi^{-1/2}2^{k}\Gamma(k+1/2)\;\;\text{for}\;\;k\leq n\quad\quad[**]$$ thus doubling the range of validity of $[\ast]$. Is it true?

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Thanks. This is the formula I spotted. However, I want a formula for that holds for large $k,$ (in some sense the $n$ I am choosing is small and fixed) for example if $n=3.$ The above formula doesn't give me enough information. – J. E. Pascoe Sep 4 '14 at 21:19
For $n=3$ you could try using Weyl's integral formula which reduces to the computation of an integral over the maximal torus of $O(3)$. This is an $S^1$ and there is a greater hope you can say something concrete. – Liviu Nicolaescu Sep 4 '14 at 21:35
This is an amazing amount of data. I wonder if it's in any of the triangles in the OEIS (Like Pascal's triangle or Stirling numbers.) – J. E. Pascoe Sep 5 '14 at 16:22
I searched a bit, but only the $n=2$ series (A001700) and $n=3$ series (A099251) seem to be on OEIS. – Carlo Beenakker Sep 5 '14 at 18:36
@CarloBeenakker: I've now written an explicit formula for $a_{2k}$ in my post; would be amazing if it simplifies down to your conjecture! – Suvrit Sep 5 '14 at 21:37

Here is a general comment. Let $G$ be a compact group and let $V$ be a (finite-dimensional, continuous, complex) representation of $G$. This data determines a locally finite directed graph, the representation graph $\Gamma$ of $G$ and $V$, as follows. The vertices of $\Gamma$ are the irreducible representations of $G$, and the number of edges from an irreducible representation $U_1$ to an irreducible representation $U_2$ is the multiplicity of $U_2$ in $U_1 \otimes V$.

Let $\chi_V$ denote the character of $G$.

Claim: The moment

$$\int_G \chi_V(g)^k \, dg$$

is the number of closed walks of length $k$ from the trivial representation to itself in $\Gamma$.

The idea of the proof is to look at the relationship between the decomposition of $V^{\otimes k}$ into irreducibles and the decomposition of $V^{\otimes (k+1)}$ into irreducibles. The application is that if $G$ is sufficiently nice then we can try to figure out what $\Gamma$ looks like and work from there.

Let's do $G = \text{O}(3)$ ($V$ the complexification of the standard representation) this way to demonstrate how the method works. First observe that we can immediately reduce to $G = \text{SO}(3)$, since the integral over $\text{O}(3)$ splits up into an integral over $\text{SO}(3)$ and over the other coset, and since the latter is $-1$ times the former (here we are using the fact that $3$ is odd) we get that

$$\int_{\text{O}(3)} \chi_V(g)^{2k+1} \, dg = 0$$


$$\int_{\text{O}(3)} \chi_V(g)^{2k} \, dg = \int_{\text{SO}(3)} \chi_V(g)^{2k} \, dg.$$

Recall that the irreducible representations of $\text{SO}(3)$ are precisely the odd-dimensional irreducible representations $V_1, V_3, V_5 \dots$ of its Lie algebra $\mathfrak{so}(3) \cong \mathfrak{su}(2)$, with $V = V_3$.

Claim: For $n \ge 3$, we have $V_n \otimes V_3 \cong V_{n-2} \oplus V_n \oplus V_{n+2}$.

This should be familiar from the representation theory of $\mathfrak{su}(2)$, and it gets us a very explicit description of the representation graph in this case: the vertices are labeled by the odd positive integers $1, 3, 5 \dots$ and for every vertex there are three edges, one from each vertex to its left neighbor, to itself, and to its right neighbor. (The special case $V_1$ is easy to handle since $V_1 \otimes V_3 \cong V_3$.)

As a corollary, closed walks from $1$ to itself of length $k$ are very close to being counted by the Motzkin numbers $M_k$ (A001006), which would be the correct answer if the representation graph had an additional edge from $1$ to itself. Some computations reveal that the actual sequence we get is A005043, which I've never seen before but which OEIS says are called the Riordan numbers. The sequence begins

$$1, 0, 1, 1, 3, 6, 15, 36, 91 \dots$$

so to get the original sequence of moments we wanted, we start with this sequence and replace every odd term with $0$, giving

$$1, 0, 1, 0, 3, 0, 15, 0, 91 \dots$$

This sequence appears in the OEIS with the zeroes removed as A099251.

For general compact connected Lie groups the representation graph should look a bit like the intersection of a Weyl chamber with the weight lattice, or something like that. I think the combinatorics gets much hairier as soon as the rank is higher than $1$ though.

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hmm, for $a_{2k}$ in $O_3$ I got indeed $a_2=1$, $a_4=3$, but then $a_6=1/\pi+29/2$ instead of 15... I might be in error.... – Carlo Beenakker Sep 5 '14 at 10:21
@Carlo: the moments must all be non-negative integers, since as mentioned in the OP they count the multiplicity of the trivial representation in tensor powers of the standard representation of $\text{O}(n)$. – Qiaochu Yuan Sep 5 '14 at 10:24
my silly mistake, fooled by Mathematica, thanks for correcting me… – Carlo Beenakker Sep 5 '14 at 12:23
For a more general discussion of the connections between characters of a compact connected Lie group and random walks on lattice points in the fundamental domain, see {\par} [H93] David Handelman, Representation rings as invariants for compact groups and ratio limit theorems for them, J Pure Appl Algebra 66 (1990) 165--184, including asymptotics, especially towards the end. (Typically, no one looks beyond the first result in that article, which unfortunately had been done before.) Also, Iterated multiplication of characters of compact connected Lie groups, J Algebra 173 (1995) 67--96. – David Handelman Sep 6 '14 at 20:33

This answer is a follow-up to the other answers (particularly to Carlo Beenakker's answer from September 4).

First off, Carlo's conjecture is indeed true. That is:

Theorem. If $n \geq k$ then $$a_{2k}=\pi^{-1/2}2^{k}\Gamma(k+1/2).\quad [*]$$

Idea of proof: Recall that the Brauer algebra describes the set of invariants of $2k$-fold tensor copies of orthogonal matrices, which the following operator projects onto: $$ \int_{O_n}X^{\otimes 2k} dX. \quad [\dagger] $$ If we let $\{e_i\}$ be the standard basis of $\mathbb{R}^n$, then after you unwrap everything, you find that $[\dagger]$ is the orthogonal projection onto the span of the $\pi^{-1/2}2^{k}\Gamma(k+1/2)$ different vectors of the form $$ W_\pi\sum_{i_1,i_2,\ldots,i_k=1}^n e_{i_1} \otimes e_{i_1} \otimes e_{i_2} \otimes e_{i_2} \otimes \cdots \otimes e_{i_k} \otimes e_{i_k}, \quad [**] $$ where $W_\pi$ is a unitary operator that permutes the $2k$ different tensor factors according to an arbitrary permutation $\pi$ (this is essentially Theorem 3.1 in [1]). The reason that there are exactly $\pi^{-1/2}2^{k}\Gamma(k+1/2)$ vectors of the form $[**]$ is that this is the number of perfect matchings of $2k$ objects (which in turn is equal to the dimension of the Brauer algebra, as expected).

So far, this shows that $a_{2k} \leq \pi^{-1/2}2^{k}\Gamma(k+1/2)$ always (regardless of $n$ and $k$). To show that the other inequality holds when $n \geq k$, it suffices to show that the vectors $[**]$ are linearly independent. I will just refer to Theorem 3.4 of [1] for this claim, but it's not difficult to prove directly using standard linear algebra tools.

This still leaves the question of what $a_{2k}$ equals when $n < k$. Here are some random observations and conjectures based on the numerical evidence found by Carlo and that I've dug up elsewhere:

Conjecture. If $n = k-1$ then $a_{2k} = \pi^{-1/2}2^{k}\Gamma(k+1/2) - \frac{(2k)!}{k!(k+1)!}$. This is the number of perfect matchings of $2k$ objects that have at least one "crossing" (in the same sense that the Catalan numbers give the number of perfect matchings of $2k$ objects with no "crossings").

Conjecture. If $n = k-2$ then $a_{2k}$ seems to be the number of perfect matchings of $2k$ objects with at least three crossings (there is a known formula for this quantity, but it's messy so I won't write it here).

The other answers gave formulas for when $n = 2$ or $n = 3$. It turns out that there is also already a known formula for the $n = 5$ case (see OEIS A095922: the even-indexed terms are the values of $a_{2k}$ that we want). That OEIS entry cites the book [2], but I haven't been able to track down a copy of it. Numerical evidence also suggests the following:

Conjecture. If $n = 4$ then $$ a_{2k} = \frac{1}{2}\frac{(2k)!}{k!(k+1)!}\left(\frac{(2k)!}{k!(k+1)!}+1\right).$$


  1. G. Lehrer and R. Zhang, The second fundamental theorem of invariant theory for the orthogonal group, Ann. of Math. 176:2031-2054 (2012).
  2. A. Mihailovs, A Combinatorial Approach to Representations of Lie Groups and Algebras, Springer-Verlag New York (2004).
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Below I'll mention a classical result (due to Alan T. James (1964), * ). For the moments to be nonzero, $k$ must be even, so following Carlo, I'll also write $a_{2k}$ to denote the following quantity: \begin{equation*} a_{2k}(X) := E_X[ \mbox{tr}(AX)^{2k}] = \sum_{\tau \in P_k} \frac{2^k \cdot k!\, d_\tau}{(2k)!C_\tau(I_n)}C_\tau(AA^T) \end{equation*} where $n\times n$ orthogonal $X$ is uniformly distributed, $C_\tau(\cdot)$ denotes the (normalized) Zonal symmetric polynomial indexed by partition $\tau$ ($P_k$ is the set of partitions of integer $k$), and it is known that \begin{equation*} d_\tau = \frac{\chi_{[2\tau]}(1)2^kk!}{(2k)!} = \frac{2^k k!\prod_{i< j}^{\ell(\tau)} (2\tau_i -2\tau_j -i + j)} {\prod_{i=1}^{\ell(\tau)} (2\tau_i + \ell(\tau) - i)!}, \end{equation*} where $\chi_{[2\tau]}(1)$ is the dimension of the representation of $[2\tau]=[2\tau_1,\ldots,2\tau_{\ell(\tau)}]$ of the symmetric group on $2k$ symbols, and $\ell(\tau)$ is the length (i.e., number of parts) of the partition $\tau$.

This quantity was obtained by James (1964) using representation theory (as is now evident from Qiaochu's more general answer).

As noted, in the OP, $A=I_n$, so actually the desired moment is (for $k\le n$) \begin{equation*} a_{2k}(I_n) = \frac{2^k k!}{(2k)!}\sum_{\tau \in P_k} d_\tau, \end{equation*} which should hopefully simplify further, as conjectured by Carlo.

The reference (that on page 40 includes the abovecited result) from which I took the above material: Zonal Polynomials by Akimichi Takemura (see e.g., google books link) --- though note that this book uses $n$ where I've written $k$ (to be consistent with the OP).

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I don't really know what you mean by $A'$ above, nor do I know enough about the values of zonal polynomials. However since I think I am calculating $E_X[\text{tr}(X)^k]$ in your language, is it the case that we will have some cancellation in this formula (the $\lambda$ might cancel the zonal polynomial factors out, but I don't know what you mean by scaling)? (Also in the denominator do you mean $(2k)!$ or $2(k!)$? I guess there's also a similar question about the numerator.) – J. E. Pascoe Sep 4 '14 at 21:51
$A'$ just denotes transpose of the matrix $A$. In your case $A=I$, so this does not matter anyways. Yes, that is $(2k)!$, but the numerator does indeed have $2^k \cdot k!$ --- scaled version means: there is a complicated looking term with which you have to multiply $C_\kappa(I_n)$, but I did not have time to spell out... --- the link that I mentioned includes all the details--- and you guessed it, since $A=I$ in your case, once you know the scaling the Zonal polynomials that appear will cancel... – Suvrit Sep 4 '14 at 21:56
apologies for being slow in understanding, but your final result for $a_{2k}$ no longer seems to depend on $n$, while the correct result does depend on $n$ for $k>n$... – Carlo Beenakker Sep 5 '14 at 17:25
@CarloBeenakker: sorry! it seems that the result only holds for $k \le n$; am trying to check what happens for $k > n$. – Suvrit Sep 5 '14 at 17:47
intriguing --- where does the condition $k\leq n$ arise in your derivation? – Carlo Beenakker Sep 6 '14 at 0:38

I am writing more than a year after the question was posted, only to spell out some more details regarding the calculation implicit in the solution presented by Suvrit, and to clarify the dependence on $n$ in that solution.

Write $[{\rm Tr}(AX)]^{2k}=p_{1^{2k}}(AX)$, using power sum symmetric functions. Next, expand into Schur functions, $p_{1^{2k}}(AX)=\sum_{\lambda\vdash 2k}d_\lambda s_\lambda(AX)$. Notice that in this expansion we must have $\ell(\lambda)\le n$, otherwise the Schur function is identically zero.

Now use the definition of zonal polynomials (average of irreducible characters of $U(n)$ over the subgroup $O(n)$), $$ \int_{O(n)}s_\lambda(AX)dX=\frac{Z_\tau(AA^T)}{Z_\tau(1^n)},$$ which holds if $\lambda=2\tau$ (the integral vanishes otherwise) to get $$ \int_{O(N)}[{\rm Tr}(X)]^{2k}dX=\sum_{\tau\vdash k}d_{2\tau}, $$ with the condition $\ell(\tau)\le n$ (which is the only dependence of the answer on $n$).

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