## Sunday, October 25, 2015

### Invariance of the Gaussian orthogonal ensemble

On page 17 in his book , Mehta proves the following result about the ensemble of symmetric $n \times n$ matrices $H$
1. If the ensemble is invariant under every transformation $H \mapsto R H R^T$ with $R$ an orthogonal matrix
2. and if all components $H_{ij}, i \le j$ are independent
then the probability measure has the form $$\label{eq:20151025e} \prod_{ i \le j} dH_{ij} \ \exp \left( -a\ \mathrm{tr}(H^2) + b\ \mathrm{tr} H + c \right)$$ with $a, b$ and $c$ constants.

I prove here the converse, namely, the probability measure \eqref{eq:20151025e} is invariant under transformations $H \mapsto R H R^T$.
I first prove that the exponent is invariant and secondly that the absolute value of the Jacobian of the transformation is $1$.

1. The exponent is clearly invariant because a short calculation gives \begin{equation*} \mathrm{tr}(RHR^T) = \mathrm{tr}(R^TRH) = \mathrm{tr}(H) \end{equation*} and similarly \begin{equation*} \mathrm{tr}(RHR^T)^2 = \mathrm{tr}(RHR^TRHR^T) = \mathrm{tr}(R^T RHR^TR H) = \mathrm{tr}(H^2) \end{equation*} where I used the cyclicity of the trace and $R^T R = 1$.
2. The Jacobian of a linear transformation is equal to the determinant of this linear transformation. I therefore need to know the determinant of the linear transformation $H \mapsto R H R^T$ on symmetric matrices $H$. A bit of work shows that this determinant is equal to $\left( \mathrm{det} R \right) ^{n+1}$. Because $R$ is orthogonal, it follows that $\mathrm{det} R = \pm 1$ and thus $\left( \mathrm{det} R \right) ^{n+1} = \pm 1$.