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How can one compute each of the following matrices, explicitly:

$$\int_{O(n)} e^{g}dg$$ or $$\int_{O(n)} g^{n}dg \;\;\;\;n\in \mathbb{N} \;\;n>1$$ What is the explicite entries of the resulting matrices, for $n=2$?

Moreover, for $n,m\in \mathbb{Z}$ define the linear operator $T_{n,m}$ on $M_{n}(\mathbb{R})$ as follow: $$T_{n,m}(A)=\int_{O(n)}g^{n}Ag^{m}$$ Under what suficcient and necessary condition, $T_{n,m}$ is conjugate to $T_{n',m'}$? Is there an explicit formulation for $T_{n,m}$?

The integration is based on the Haar measear defined on orthogonal group $O(n)$.

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    $\begingroup$ since the first two integrals are proportional to the unit matrix, you can take the trace to find the coefficient of proportionality and then use theorem 4 from statweb.stanford.edu/~cgates/PERSI/papers/random_matrices.pdf $\endgroup$ Commented Jan 25, 2015 at 16:34
  • $\begingroup$ @CarloBeenakker why these are scalar matrices? $\endgroup$ Commented Jan 25, 2015 at 16:49
  • $\begingroup$ @CarloBeenakker thanks very much for the link paper. $\endgroup$ Commented Jan 25, 2015 at 16:56
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    $\begingroup$ The first two integrals are invariant under automorphisms of $O(n)$, so in particular under conjugation, so must be multiples of the identity by Schur's lemma. $\endgroup$
    – Ben McKay
    Commented Jan 25, 2015 at 18:38
  • $\begingroup$ The key phrase is "higher Frobenius-Schur indicators" --- these can be computed from ac.els-cdn.com/S0021869307003420/… , I believe. Unfortunately if I remember right Diaconis et al. need $n$ to be large in terms of the power, but the two combined should likely give you what you need (sorry for not helping much! --- I'm in a bit of a rush now). $\endgroup$
    – alpoge
    Commented Jan 30, 2015 at 11:32

1 Answer 1

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let me work out the comments a bit further, starting from the identity (equation 8.2 from Diaconis and Evans, correcting an earlier paper by Diaconis and Shahshahani)

$$\int_{{\rm O}(n)}{\rm tr}\,g^p\,dg=\begin{cases} 0&{\rm if}\;\;p\;\;\text{is an odd integer}\\ 1&{\rm if}\;\;p\;\;\text{is an even integer} \end{cases} $$

so the second integral of the OP evaluates to $$\int_{{\rm O}(n)}g^p\,dg=\begin{cases} 0&{\rm if}\;\;p\;\;\text{is an odd integer}\\ n^{-1}\mathbb{1}&{\rm if}\;\;p\;\;\text{is an even integer} \end{cases} $$

Taylor expansion of the exponent in the first integral of the OP and term-by-term integration gives $$ \int_{{\rm O}(n)}e^g\,dg=n^{-1}\mathbb{1}\sum_{p=0}^\infty \frac{1}{(2p)!}=\frac{\cosh 1}{n}\mathbb{1} $$

now for the third integral of the OP, we need the fourth-order tensor $$\int_{{\rm O}(n)}(g^p)_{ij}(g^q)_{kl}\,dg=a_{pq}(n)\delta_{ij}\delta_{kl}+b_{pq}(n)\delta_{ik}\delta_{jl}+c_{pq}(n)\delta_{il}\delta_{jk}$$ so that the required integral takes the form $$\int_{{\rm O}(n)}g^p Ag^q\,dg=a_{pq}(n)A+b_{pq}(n)A^{\rm t}+c_{pq}(n)\mathbb{1}\,{\rm tr}\,A $$ by taking traces I find, using again equation 8.2 from Diaconis and Evans, that $$na_{pq}(n)+nb_{pq}(n)+n^2 c_{pq}(n)=\int_{{\rm O}(n)}{\rm tr}\,g^{p+q}\,dg=\begin{cases} 0&{\rm if}\;\;p+q\;\;\text{is an odd integer}\\ 1&{\rm if}\;\;p+q\;\;\text{is an even integer} \end{cases} $$ $$n^2a_{pq}(n)+nb_{pq}(n)+nc_{pq}(n)=\int_{{\rm O}(n)}({\rm tr}\,g^p)({\rm tr}\,g^q)\,dg=\begin{cases} {\rm min}\,(p,2n)&{\rm if}\;\;p=q\;\;{\rm odd}\\ {\rm min}\,(p,2n)+1&{\rm if}\;\;p=q\;\;{\rm even}\\ 1&{\rm if}\;\;p\neq q\;\;\text{both even}\\ 0&{\rm otherwise} \end{cases} $$ $$na_{pq}(n)+n^2 b_{pq}(n)+nc_{pq}(n)=\int_{{\rm O}(n)}{\rm tr}\,g^{|p-q|}\,dg=\begin{cases} 0&{\rm if}\;\;p+q\;\;\text{is an odd integer}\\ 1&{\rm if}\;\;p+q\;\;\text{is an even integer} \end{cases} $$ Three equations with three unknowns solves the third integral.

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  • $\begingroup$ Why do the complex eigenvalues average out to 0? $\endgroup$
    – Will Sawin
    Commented Jan 26, 2015 at 22:59
  • $\begingroup$ @WillSawin --- good point, this is not obvious, so I have removed the "simple argument", which was apparently too simple; for the record, it follows after this comment. Thanks for correcting me. $\endgroup$ Commented Jan 27, 2015 at 9:10
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    $\begingroup$ deleted "simplistic argument" for the first integral in my answer: the result for $p$ odd is obvious because contributions $\pm g$ to the integral cancel; the result for $p$ even follows without calculation from the fact that only eigenvalues $\pm 1$ of the orthogonal matrix $g$ contribute to the average of the trace (the complex eigenvalues average out to zero); for $n$ odd every $g$ has one such eigenvalue, so the average is one, for $n$ even and ${\rm det}\,g=+1$ there are no such eigenvalues but for ${\rm det}\,g=-1$ there are two, so the average is again one. $\endgroup$ Commented Jan 27, 2015 at 9:17
  • $\begingroup$ @CarloBeenakker thank you very much for your very interesting answer. $\endgroup$ Commented Jan 28, 2015 at 6:00

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