# For a matrix model with $O(n)$ as a symmetry group is $\langle \det A \rangle = \det \langle A \rangle$

Suppose that I have a model of random $n \times n$ symmetric, positive definite matrices where the probability distribution is generated by the partition function:

$Z = \int dA \; e^{-V(A)}$

where $V(A)$ is left invariant under orthogonal transformations $A \rightarrow O^T \, A \, O$ and $dA$ is the Haar measure

$dA = \prod_i \, dA_{ii} \, \prod_{i<j} dA_{ij}$.

Because this problem can be reduced to finding the probability distribution for the eigenvalues, it seems intuitive to me that one may be able to show:

$\langle \det A \rangle = \det \langle A \rangle$.

Clearly this wouldn't be true for any random matrix theory as the determinant is a highly non-linear operation, but with the large $O(n)$ symmetry it at least seems possible.

Does anyone know if this is discussed anywhere? Or can someone see how to prove it or show that it is not true?

Thanks

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This cannot be generally true. As you wrote, the problem reduces to a weighted integration over the eigenvalues; we don't have to determine the weighting because $V$ can be chosen arbitrarily to make up for it, so we have
$$Z=\int\mathrm d^n\lambda\,\mathrm e^{-V'(\lambda)}$$
with $\lambda$ the vector of eigenvalues and $V'$ an arbitrary function. If $n$ and $V'$ are even, then $\langle A\rangle=0$ and thus $\det\langle A\rangle=0$, but generally not $\langle\det A\rangle=0$.