The question I have is quite specific. So in the hope that this post might help others in the future, my problem boils down to solving the following integral:

$$I_\tau=\int \prod_{i, j=1}^{N} d J_{i j} \exp \left\{-\frac{N}{2(1-\tau^2)}\left( \sum_{i, j, k} J_{k i} A_{i j} J_{k j}-\tau\sum_{ij}J_{ij}J_{ji}\right)\right\}$$ Assuming we can compute the determinant of $A$, is there a general identity to compute $I_\tau$?

Simplifying their argument, in the following paper [1], the authors compute the following Gaussian integral over all $N\times N$ real asymmetric matrices $J_{ij}$:

$$I=\int \prod_{i, j=1}^{N}d J_{i j} \exp \left\{-\frac{N}{2} \sum_{i, j, k} J_{k i} A_{i j} J_{k j}\right\}$$

Where $A_{i j}=\frac{z_{i}^{*} z_{j}}{N}+\frac{z_{j}^{*} z_{i}}{N}+\delta_{i j}$ and $z_i\in \mathbb{C}$.

If I define the vector $\mathbf{x}$ such that: $$\mathbf{x}:=\left(\begin{array}{c} J_{11} \\ J_{12} \\ \vdots \\ J_{1 n} \\ J_{21} \\ J_{22} \\ \vdots \\ J_{n n} \end{array}\right) \in \mathbb{R}^{n^{2}}$$ $$\implies x_{N(i-1))+j}=J_{ij}$$ Then we can represent $I$ as the following integral: $$I=\int \left(\prod_{i, j=1}^{N} \mathrm{d}x_{N(i-1))+j} \right)\exp \left(-\frac{N}{2} \mathbf{x}^T\Sigma \mathbf{x}\right)$$ Where $\Sigma$ simply is: $$\Sigma=\left(\begin{array}{cccc} \mathbf{A} & 0 & \cdots & 0 \\ 0 & \mathbf{A} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathbf{A} \end{array}\right)$$

Thus $I\propto \frac{1}{\sqrt{|\Sigma|}}=|A|^\frac{-N}{2}$.

Defining $r\equiv\frac{1}{N}\sum_i |z_i|^2$, the authors observe that to order $1/\sqrt{N}$, the matrix $A$ has all but two of its eigenvalues equal to one. The two exceptions are the eigenvectors $z_i$ and $z^*_i$ , both with eigenvalue $(1+r)$. This observation allows them to compute the determinant of $\Sigma$ and thus the Gaussian integral $I$.

Now, considering the integral $I_\tau$ using the same method: $$I_\tau=\int \left(\prod_{i j} \mathrm{d} J_{i j}\right)\exp \left\{-\frac{N}{2(1-\tau^2)}\left( \sum_{i, j, k} J_{k i} A_{i j} J_{k j}-\tau\sum_{ij}J_{ij}J_{ji}\right)\right\}$$ $$I_\tau=\int\left(\prod_{i j} \mathrm{d} J_{i j}\right) \exp \left\{-\frac{N}{2(1-\tau^2)} \sum_{i j, k, l} J_{i j}\left(A_{j l} \delta_{i k}-\tau\delta_{i l} \delta_{j k}\right) J_{k l}\right\}$$ $$\implies I_\tau=\int\left(\prod_{i, j=1}^{N} \mathrm{d} x_{N(i-1))+j}\right) \exp \left(-\frac{N}{2} \mathbf{x}^{T} \Sigma \mathbf{x}\right)$$ Where this time: $$\Sigma_{n(i-1)+j,n(k-1)+l}=A_{j l} \delta_{i k}-\tau\delta_{i l} \delta_{j k} \quad \forall i, j, k, l \in \mathbb{N} \cap[1, N]$$

For large $N$, how can we compute the determinant of $\Sigma$ this time? It seems to be a non-trivial question.

Any remark or comment is always appreciated, thank you.

[1]: Rajan, K., & Abbott, L. F. (2006). Eigenvalue spectra of random matrices for neural networks. Physical review letters, 97(18), 188104.

Link: https://www.researchgate.net/profile/Kanaka_Rajan/publication/6643146_Eigenvalue_Spectra_of_Random_Matrices_for_Neural_Networks/links/0912f50f7347f4863b000000.pdf


1 Answer 1


Only the symmetric part of $A$ contributes to the integrand, so we may assume $A$ is symmetric and diagonalize it as $A=O\Lambda O^T$ with $O$ orthogonal and $\Lambda={\rm diag}\,(\lambda_1,\lambda_2,\ldots\lambda_N)$. The orthogonal matrix may be removed by a change of variables, so without loss of generality we can take a diagonal $A$. So we seek the integral $$I=\int \prod_{i, j=1}^{N} d J_{i j} \exp \left\{-\frac{N}{2(1-\tau^2)}\left( \sum_{i, j} \lambda_j J_{ij}^2-\tau\sum_{ij}J_{ij}J_{ji}\right)\right\}.$$ I now proceed similarly to here. Decompose the sum over $i,j$ as $$\sum_{ij}\lambda_j J_{ij}^2-\tau\sum_{ij}J_{ij}J_{ji}=$$ $$\qquad\qquad=\sum_{i}(\lambda_i-\tau)J_{ii}^2+\sum_{i<j}\left(\lambda_j J_{ij}^2+\lambda_i J_{ji}^2-2\tau J_{ij}J_{ji}\right)$$ $$\qquad\qquad\equiv\sum_{i} A_i+\sum_{i<j} B_{ij}.$$ Then perform the Gaussian integrals separately for each term in the sum, $$I=\left(\prod_{i=1}^N\int e^{-\beta A_i}dJ_{ii}\right)\left(\prod_{i<j=1}^{N}\int\int e^{-\beta B_{ij}}dJ_{ij}dJ_{ji}\right)$$ $$\qquad\qquad=(\pi/\beta)^{N^2/2}\left(\prod_{i=1}^N(\lambda_i-\tau)^{-1/2}\right)\left(\prod_{i<j=1}^N(\lambda_i\lambda_j-\tau^2)^{-1/2}\right),$$ where I have defined $\beta=\frac{1}{2}N(1-\tau^2)^{-1}$, and assumed that $\beta>0$, $\lambda_i>\tau$ for all $i$.

So knowledge of only the determinant of $A$ is not sufficient to evaluate the integral, you need to know the individual eigenvalues -- not just their product. In particular, if $\tau$ happens to be close to one particular eigenvalue of $A$, then that eigenvalue will dominate the integral.


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