1
$\begingroup$

Though it is a mathematical problem, maybe more physicists know it well.

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.

$\endgroup$
3
  • $\begingroup$ Cross-posted from math.stackexchange.com/q/600896/11127 and mathoverflow.net/q/151344/13917 $\endgroup$
    – Qmechanic
    Dec 11, 2013 at 0:46
  • $\begingroup$ Dear Craig Thone. In general, it is frown upon to cross-post simultaneously, because it may waste potential answerer's time. As a minimum OP should mention the cross-posts (on all sites!). The preferred procedure is to not cross-post, and if the post hasn't received an acceptable answer after, say, a couple of days, then OP could flag for migration. $\endgroup$
    – Qmechanic
    Dec 11, 2013 at 0:53
  • $\begingroup$ @Qmechanic, I am very sorry for this. What should I do? Delete some of them? or... $\endgroup$
    – thone
    Dec 11, 2013 at 2:58

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.