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  Matrix integral identity

+ 7 like - 0 dislike

1) How to prove that $N\times N$ matrix integral over complex matrices $Z$ $$ \int d Z d Z^\dagger e^{-Tr Z Z^\dagger} \frac{x_1\det e^Z -x_2 \det e^{AZ^\dagger}}{\det(1-x_1e^Z)\det(1-x_2e^{AZ^\dagger})} $$ does not depend on the external Hermitian matrix $A$? $x_1$ and $x_2$ are numbers. The statement is trivial for $1\times1$ case.

2)The same for

$$ \int d Z d Z^\dagger e^{-Tr Z Z^\dagger} \frac{x_1\det e^Z -x_2 \det e^{AZ^\dagger}}{\det(1-x_1e^Zg)\det(1-x_2e^{AZ^\dagger}g)} $$

where g - arbitrary $GL(N)$ matrix.

This post imported from StackExchange MathOverflow at 2014-07-29 11:48 (UCT), posted by SE-user Sasha
asked Nov 29, 2010 in Mathematics by Sasha (110 points) [ no revision ]
retagged Jul 29, 2014
I understand that $dZ$ is the Lebesgue measure on $N\times N$ complex matrices, that is $dZ=\prod_{i,j}d\Re z_{ij}d\Im z_{ij}$, but what $dZ^\dagger$ stands for ?

This post imported from StackExchange MathOverflow at 2014-07-29 11:48 (UCT), posted by SE-user Adrien Hardy
This is just a notation sometimes used in the mathphys literature to show that you integrate over $2N^2$ real variables contrary to $N^2$ for the Hermitian model.

This post imported from StackExchange MathOverflow at 2014-07-29 11:48 (UCT), posted by SE-user Sasha

How exactly is the integral defined? Isn't the denominator always going to hit 0? For instance in the 1 by 1 case, the denominator contains $1-x_1 e^z$, where $z$ is a complex number, and then at the points $z=\log (x_1)^{-1}$ the integrand diverges.

If it were true, you would just do a unitary transformation (which leaves the measure invariant) to diagonalize A, and then rescale the coordinates to get rid of the dependence on the eignevalues. But this integral is A dependent in the 1 by 1 case, which is not trivial.

Did you try to prove it for $N=1$? If it does not hold in that case I do not think it is valid for $N>1$.

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