In the course of a physics problem (arXiv:1206.6687), I stumbled on a curious correspondence between the eigenvalue distributions of the matrix product $U\bar{U}$, with $U$ a random unitary matrix and $\bar{U}$ its complex conjugate, on the one hand, and the random orthogonal matrix $O$ with determinant $-1$, on the other hand. The random matrices $U$ and $O$ are uniformly distributed with respect to the Haar measure on the unitary group $U(N)$ and the orthogonal group $O(N+1)$.

One eigenvalue of $O$ is fixed at $-1$ (to ensure that det $O=-1$). The other $N$ eigenvalues have a distribution $p_O$ which was known since Girko (1985). We calculated the distribution $p_{U\bar{U}}$ of the eigenvalues of $U\bar{U}$ (which we did not find in the literature --- has anyone seen it before?). We discovered to our surprise that $p_{U\bar{U}}=p_O$. This holds for both $N$ even and odd (in the latter case both $U\bar{U}$ and $O$ have an eigenvalue fixed at $+1$).

Question: Is there a more direct route to arrive at this identity between the two eigenvalue distributions, without going through a separate calculation of each one? (You can find two such separate calculations in the Appendix of arXiv:1206.6687, but this seems a rather unsatisfactory way of understanding the correspondence.)

Some intuition for what is going on: for both $U\bar{U}$ and $O$ the eigenvalues different from $\pm 1$ come in complex conjugate pairs $e^{\pm i\theta}$. The matrix $O$ has an unpaired eigenvalue at $-1$, which repels $\theta$ from $\pi$. The matrix $U\bar{U}$ cannot have an unpaired eigenvalue at $-1$ by construction and somehow this leads to a repulsion of $\theta$ from $\pi$ which is mathematically equivalent to what happens for the matrix $O$.

For example, when $N=2$ the eigenvalue $e^{i\theta}$ has the same distribution $P(\theta)=(2\pi)^{-1}(1+\cos\theta)$ for both $U\bar{U}$ and $O$. For $N=3$ the distribution is $P(\theta)=\pi^{-1}(1-\cos^2\theta)$, in addition to an eigenvalue fixed at $+1$, again the same for $U\bar{U}$ and $O$. The correspondence continues for larger $N$, when factors $(\cos\theta_k-\cos\theta_l)^2$ appear in both distributions.

  • $\begingroup$ well, $U\bar{U}$ is not itself orthogonal (its inverse is not equal to its transpose); I can construct an orthogonal matrix $O$ with the same eigenvalues as $U\bar{U}$, but this orthogonal matrix is not uniformly distributed, so that does not seem to help much. $\endgroup$ Aug 27, 2012 at 10:31
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    $\begingroup$ argh...i made stupid calculation error! $\endgroup$
    – Suvrit
    Aug 27, 2012 at 12:00
  • $\begingroup$ There is an identity for the eigenvalue distributions in this paper: web.williams.edu/go/math/sjmiller/public_html/ntandrmt/handouts/… (see page 6) Maybe this would help. It is also a known fact that the orthogonal group is similar to a standard normal distribution: www-stat.stanford.edu/~cgates/PERSI/papers/random_matrices.pdf (see page 56). Thus it is possible to solve equality of these two distributions to obtain the answer to your question. $\endgroup$
    – user22393
    Aug 30, 2012 at 13:24
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    $\begingroup$ indeed, the eigenvalue distributions of the classical compact groups are related, notably $p_O$ for $N$ even is also the eigenvalue distribution of the compact symplectic group; the matrices of the form $U\bar{U}$ do not form a group, so this does not seem to help much; note also that the correspondence between $p_{U\bar{U}}$ and $p_O$ is not an asymptotic large-$N$ result (like the correspondence to a normal distribution): it holds exactly for any finite $N$. $\endgroup$ Aug 30, 2012 at 14:11
  • $\begingroup$ Is the observation that $U\bar{U}$ is a unitary symmetric matrix (and every symmetric unitary matrix $S$ can be written as such a product) of value to you here (I say that because such matrices seem to be used when studying orthogonal ensembles...) $\endgroup$
    – Suvrit
    Sep 5, 2012 at 21:12


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