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Edit: some backgrouds added.

In quiver matrix model which is reviewed DV or CKR, the path integral reduce to the matrix integral $$Z \sim \int \prod_{i=1}^r d\Phi_i \prod_{<a,b>} dQ_{ab} e^{-\frac{1}{g_s} \rm{Tr} W(\Phi,Q)}$$ where $Q_{ab}$ are complex $N_a\times N_b$ matrces satisfying $Q_{ab}^\dagger=Q_{ba}$, and $\Phi_i$ are $N_i\times N_i$ matrices. $W$ is given by $$W(\Phi,Q) =\sum_{i,j=1}^{r} s_{ij}Q_{ij} \Phi_jQ_{ji} +\sum_{i=1}^{r} W_i{\Phi_i} $$ where the constants $s_{ij}$ are antisymmetric, $s_{ij}=-s_{ji}$, they obey $s_{ij} = 1$ if $i < j$ and the nodes in the quiver diagram are linked, and $s_{ij} = 0$ otherwise. The notation $<a,b>$ for the range of the product denotes all pairs $(a, b)$ with $1 ≤ a, b ≤ r$ s.t. $s_{ab}\neq 0$. In order to calculate the path integral, first integrate out $Q_{ab}$: $$ \int \prod_{(a,b)\in E}[dQ_{ab}dQ_{ba}] \exp \sum_{(a,b)\in E}-\frac{s_{ab}}{g_s}\rm{Tr} (Q_{ab}\Phi_b Q_{ba}-Q_{ba}\Phi_a Q_{ab})$$ where$E=\{(a,b)|1 ≤ a< b ≤ r\}$

My question is how to integrat $Q$ out to obtain something like $$ \det (\Phi_a\otimes 1-1\otimes \Phi_b)^{-1}$$. Thanks in advance.

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  • $\begingroup$ What does this mean? are $\Phi$ arbitrary? And the two matrices in the exp have different dimensions ($N_a \times N_a$ and $N_b \times N_b.$) $\endgroup$ – Igor Rivin Dec 10 '13 at 4:08
  • $\begingroup$ sorry for my mistake. There is a trace and $\Phi$ is arbitary. $\endgroup$ – Craig Thone Dec 10 '13 at 4:14
  • $\begingroup$ @CraigThone, perhaps you could use the fact that trace is a linear operator; that is $Tr(A+B) = Tr(A) + Tr(B)$ and then use some variation on Liouville's Formula? $\endgroup$ – Chris K Dec 10 '13 at 4:46
  • $\begingroup$ The answer is something like this: $\det [{\Phi_a\otimes I_{N_b\times N_b}-I_{N_a\times N_a}\otimes \Phi_b}]^{-1}$. But I do not know how to derive it. $\endgroup$ – Craig Thone Dec 10 '13 at 5:22
  • $\begingroup$ @Chris K, from the form of answer, it seems that the two terms are calculated together. $\endgroup$ – Craig Thone Dec 10 '13 at 8:18

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