Question about Matrix Integral I am stuck with the technique details of KKN's paper . 
How to get formula (2.11)
$$Z= \int \frac{d \phi }{ \rm{ Vol(G)}} \frac{1}{\rm{Det}(\rm{ad}(\phi)+\epsilon)} $$
 from (2.10)?
$$Z=\int \frac{d \phi dX_a d\psi_a}{ \rm{ Vol(G)}}  \exp {\kappa_1 \rm{Tr}( i\phi[X_1,X_2] -\frac{1}{2}\epsilon^2(X_1^2+X_2^2)+\psi_1\psi_2) } $$
I have no idea about this calculation and I even do not know the meaning of $\rm{Det}(\rm{ad}(\phi)+\epsilon)$. Any advices are welcoming.
 A: As user BebopButUnsteady mentions in a comment, this is essentially an exercise in Gaussian integration. With the caveat that the integration variables take values in a Lie algebra representation. (Warning: We will ignore factors of 2 and $\pi$ in what follows, and sometimes use Einstein summation convention.) The three bosonic fields $X_1\equiv X$, $X_2\equiv Y$, and $\phi$ belong to the adjoint Lie algebra representation of the (real) Lie group $G$, e.g., 
$$\tag{A}\phi ~=~\sum_{a=1}^n \phi^a T_a, \qquad n~:=~\dim({\rm Lie}(G)). $$ 
Similarly for $X$ and $Y$. Here $T_a$ are the Lie algebra generators in the adjoint representation,
$$\tag{B} [T_a,T_b]= \sum_{c=1}^n f_{ab}{}^c~T_c. $$
Assume for simplicity that ${\rm Tr}(T_a T_b)=\delta_{ab}$, that we may raise and lower Lie algebra indices for free, and that $f_{abc}$ is totally antisymmetric. Define the adjoint map $A\equiv{\rm ad} \phi $ as
$$\tag{C} A(X)~:=~[\phi,X]. $$
In the basis $T_a$, we let $A\equiv{\rm ad} \phi$ be represented by an antisymmetric matrix 
$$\tag{D} A^a{}_b~=~\sum_{c=1}^n\phi^c f_{cb}{}^a,$$ 
i.e. 
$$\tag{E} A(T_b)~=~\sum_{a=1}^n T_a A^a{}_b. $$
After integrating out the fermions in eq. (2.10) of Ref. 1, the action then reads
$$S ~:=~  {\rm Tr}\left(\frac{1}{2}\epsilon(X^2+Y^2) -i\phi[X,Y]\right)
~=~ {\rm Tr}\left(\frac{1}{2}\epsilon(X^2+Y^2) +i X A(Y)\right)$$
$$\tag{F} ~=~ \frac{1}{2}\begin{pmatrix} X^a &Y^b \end{pmatrix} 
\begin{pmatrix} \epsilon{\bf 1}_{n\times n} &  iA \\
-iA &\epsilon{\bf 1}_{n\times n}  \end{pmatrix}
\begin{pmatrix}X^c \\Y^d \end{pmatrix}, \qquad\epsilon~>~0.$$
The corresponding Gaussian integral then becomes
$$\tag{G} \int_{\mathbb{R}^{2n}}\! [d^nX^a] [d^nY^b] ~e^{-S} ~\sim~ {\rm Det}\begin{pmatrix} \epsilon{\bf 1}_{n\times n} &  iA \\
-iA &\epsilon{\bf 1}_{n\times n}  \end{pmatrix} ^{-\frac{1}{2}} ~=~{\rm Det}(\epsilon{\bf 1}_{n\times n}+A) ^{-1}.$$
The right-hand side of eq. (G) is the integrand of eq. (2.11). To prove the last equality in eq. (G), use that an antisymmetric matrix $A^a{}_b$ can be brought on standard form with $2\times 2$ blocks via a real orthogonal transformation, so it is enough to prove it for $n=2$. The latter is a straightforward exercise.
References:


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*V.A. Kazakov, I.K. Kostov, and N. Nekrasov, $D$-particles, Matrix Integrals and KP hierachy, arXiv:hep-th/9810035.

