# Matrix integral identity

+ 7 like - 0 dislike
192 views

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
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 $N>1$.

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.