# The eigenvalue spectrum of a large real antisymmetric random matrix
with non-zero mean

Andrei Katsevicha,b, Pavel Meshcheriakova,b aMoscow Institute of Physics and
Technology, 141700 Dolgoprudny, Russia
bLandau Institute for Theoretical Physics, 142432 Chernogolovka, Russia

(September 3, 2023)

###### Abstract

We study the eigenvalue spectrum of a large real antisymmetric random matrix
$J_{ij}$. Using a fermionic approach and replica trick, we obtain a
semicircular spectrum of eigenvalues when the mean value of each matrix
element is zero, and in the case of a non-zero mean, we show that there is a
set of critical finite mean values above which eigenvalues arise that are
split off from the semicircular continuum of eigenvalues. The result converged
with numerical simulations.

##  I Introduction

For even $N$ consider a purely imaginary antisymmetric $N\times N$ matrix
$iJ_{ij}=-iJ_{ji}.$ Its eigenvalues are decomposed naturally into pairs
$\pm\lambda_{i}\in\mathbb{R}$ with $i=1,2,\dots,N/2$ and $\lambda_{i}>0$. We
define the density of eigenvalues as

$\nu(\lambda)=\frac{1}{N}\sum_{i=1}^{N/2}\delta(\lambda-\lambda_{i})+\delta(\lambda+\lambda_{i})\,.$
(1)

We would like to find $\nu(\lambda)$ in the large $N$ limit for the random
antisymmetric matrix $J_{ij}$ whose elements are taken from Gaussian ensemble
with fixed mean $J_{0}/N$ and variance $\sigma^{2}=J^{2}/N$ ($J_{0}$ and $J$
are of the order of unity), so the probability density function reads

$p(J_{ij})=\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left(-\frac{(J_{ij}-J_{0}/N)^{2}}{2\sigma^{2}}\right)\,.$
(2)

The eigenvalue probability density function for some ensembles of Gaussian
random matrices with zero mean $J_{0}=0$ is a Wigner semicircle. This result
was obtained in classical works on various matrix ensembles. See, for example,
the book Mehta (2004). Several examples include Gaussian orthogonal (GOE),
unitary (GUE), and symplectic (GSE) ensembles.
For these ensembles, it was showed that the eigenvalue probability density
function of the matrix ensemble approaches a Wigner semicircle as the size of
the matrices increases to infinity. This is a fundamental result in random
matrix theory, demonstrating the universality of the Wigner semicircle.

$\rho_{0}(\lambda)\equiv\braket{\nu(\lambda)}_{J_{ij}:J_{0}=0}=\begin{cases}\begin{aligned}
&\frac{1}{\pi
J}\sqrt{1-\frac{\lambda^{2}}{4J^{2}}}\,,&\quad&|\lambda|<2J\,;\\\
&0\,,&\quad&|\lambda|>2J\,.\end{aligned}\end{cases}$ (3)

In the case of antisymmetric matrices Wigner semicircle was obtained in Mehta
and Rosenzweig (1968). Our goal is to generalize this result to arbitrary
$J_{0}\neq 0$.

A similar problem in the case of real symmetric random matrices $\bm{M}$ with
mean $\frac{M_{0}}{N}$ was discussed in Edwards and Jones (1976); Jones et al.
(1978). Using replica trick, it was shown that if the mean value of random
matrix elements is greater than a critical one the distribution of the
eigenvalues consists of the Wigner semicircle plus a separate single state:

$\rho_{M_{0}}(\lambda)\equiv\braket{\nu(\lambda)}_{M_{ij}:M_{0}\neq
0}=\begin{cases}\begin{aligned}
&\rho_{0}(\lambda)+\frac{1}{N}\delta\left(\lambda-
M_{0}-\frac{J^{2}}{M_{0}}\right)\,,&\quad&|M_{0}|>J\,;\\\
&\rho_{0}(\lambda)\,,&\quad&|M_{0}|<J\,.\end{aligned}\end{cases}$ (4)

An algorithm for applying the replica trick:

  1. 1.

Rewrite delta-functions in $\nu(\lambda)$ as derivatives of logarithmic
functions using Sokhotski–Plemelj formula.

  2. 2.

Rewrite logarithmic function in $\nu(\lambda)$ using limit:

$\log x=\lim\limits_{n\rightarrow 0}\frac{x^{n}-1}{n}\,.$ (5)

  3. 3.

Swap the limit $n\rightarrow 0$ with the limit $N\rightarrow\infty$. Treat $n$
as an integer during calculating the Gaussian integrals.

  4. 4.

Take a limit $n\rightarrow 0$ after evaluating the integrals.

Physically, the replica trick introduces $n$ replicated copies of the original
disordered system that are coupled together. It uses the non-commutativity of
limits $n\rightarrow 0$ and $N\rightarrow\infty$. Taking the limit
$n\rightarrow 0$ at the end serves to analytically continue the results for
integer $n$ to non-integer values. The coupling between the replicated systems
enables averaging over the disorder in the original system.

This problem also arises when one considers the $\text{SYK}_{2}$ model with
$\frac{N(N-1)}{2}$ couplings Klebanov and Tarnopolsky :

$H_{SYK}=i\sum_{i<j}J_{ij}\psi_{i}\psi_{j},$ (6)

where $\psi_{i},i=1,...,N$ are Majorana fermions with anticommutation rules
$\\{\psi_{i},\psi_{j}\\}=\delta_{ij}$ and the random couplings
$J_{ij}=-J_{ji}$ are drawn from Gaussian distribution with non-zero mean
$J_{0}/N$ and variance $\sigma^{2}=J^{2}/N$:

$\braket{J_{ij}}=\frac{J_{0}}{N},\quad\braket{(J_{ij}-J_{0}/N)^{2}}=\frac{J^{2}}{N}\,.$
(7)

As in the case of a random matrix, $\text{SYK}_{2}$ is likely to have similar
properties.

##  II The technique

In this section we give a general scheme of calculations. The presented
formulas can be applied to an arbitrary matrix $\bm{H}$, so here we do not
restrict ourselves to the case of antisymmetric ones. We can write
$\nu(\lambda)$ as

$\nu(\lambda)=\frac{1}{\pi
N}\sum_{i=1}^{N}\textrm{Im}\frac{1}{\lambda-i\epsilon-\lambda_{i}}\,,$ (8)

where we used Sokhotski–Plemelj formula

$\lim_{\epsilon\to
0^{+}}\frac{1}{x-i\epsilon}=i\pi\delta(x)+\mathcal{P}\Big{(}\frac{1}{x}\Big{)}\,.$
(9)

On the other hand we have

$\textrm{det}(\bm{I}\lambda-\bm{H})=\prod_{i=1}^{N}(\lambda-\lambda_{i})\,.$
(10)

Therefore, we obtain

$\nu(\lambda)=\frac{1}{\pi
N}\textrm{Im}\frac{\partial}{\partial\lambda}\ln\textrm{det}(\bm{I}(\lambda-i\epsilon)-\bm{H})$
(11)

and using replicas

$\nu(\lambda)=\frac{1}{\pi
N}\text{Im}\frac{\partial}{\partial\lambda}\lim_{n\to
0}\frac{1}{n}\\{\det(\bm{I}\lambda_{\epsilon}-\bm{H})^{n}-1\\}\,,$ (12)

where $\lambda_{\epsilon}\equiv\lambda-i\epsilon$. The determinant of a matrix
may be represented as an integral over the Grassmann numbers (fermionic
approach):

$\textrm{det}(\bm{I}\lambda_{\epsilon}-\bm{H})=\int\prod_{i=1}^{N}d\bar{c}_{i}dc_{i}\exp\left(-\sum\limits_{i,j}\bar{c}_{i}(I_{ij}\lambda_{\epsilon}-H_{ij})c_{j}\right)\,,$
(13)

where
$\\{\bar{c}_{i},c_{j}\\}=\\{\bar{c}_{i},\bar{c}_{j}\\}=\\{c_{i},c_{j}\\}=0$.
Thus, we obtain the basic working formula:

$\nu(\lambda)=\frac{1}{\pi
N}\textrm{Im}\frac{\partial}{\partial\lambda}\lim_{n\to
0}\frac{1}{n}\left\\{\int\prod_{i;\alpha}d\bar{c}^{\alpha}_{i}dc^{\alpha}_{i}\exp\left(-\sum\limits_{i,j;\alpha}\bar{c}^{\alpha}_{i}(I_{ij}\lambda_{\epsilon}-H_{ij})c^{\alpha}_{j}\right)-1\right\\}\,.$
(14)

##  III Spectrum of the matrix with zero mean

In this section we elaborate on the case of antisymmetric matrices with zero
mean. This is important because the case of non-zero mean will refer to the
case of zero mean. Now we consider an antisymmetric matrix in which each
element has zero mean with the probability density function:

$p(J_{ij})=\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left(-\frac{J_{ij}^{2}}{2\sigma^{2}}\right)\,.$
(15)

The average density of eigenvalues:

$\rho_{0}(\lambda)=\braket{\nu(\lambda)}_{J_{ij}:J_{0}=0}=\int\nu\left(\lambda;\\{J_{ij}:J_{0}=0\\}\right)\prod\limits_{i<j}\;p(J_{ij})\;dJ_{ij}\,.$
(16)

Using (14) and (16), we obtain

$\rho_{0}(\lambda)=\frac{1}{\pi
N}\textrm{Im}\frac{\partial}{\partial\lambda}\lim\limits_{n\rightarrow
0}\frac{1}{n}\Biggl{\\{}\int\left(\prod\limits_{i;\alpha}d\bar{c}_{i}^{\alpha}dc_{i}^{\alpha}\right)\exp\left(-\lambda_{\epsilon}\sum\limits_{i;\alpha}\bar{c}_{i}^{\alpha}c_{i}^{\alpha}\right)\times\\\
\times\left(\prod\limits_{i<j}\frac{dJ_{ij}}{2\pi\sigma^{2}}\exp\left(-\frac{J_{ij}^{2}}{2\sigma^{2}}\right)\right)\exp\left(i\sum\limits_{i,j;\alpha}\bar{c}_{i}^{\alpha}J_{ij}c^{\alpha}_{j}\right)-1\Biggl{\\}}\,.$
(17)

Let it be $P_{ij}=\sum\limits_{\alpha}\bar{c}^{\alpha}_{i}c^{\alpha}_{j}$ is
antihermitian tensor ($P_{ij}=-\bar{P}_{ji}$) and
$P^{A}_{ij}=\sum\limits_{\alpha}\frac{\bar{c}^{\alpha}_{i}c^{\alpha}_{j}-\bar{c}^{\alpha}_{j}c^{\alpha}_{i}}{2}$
is its antisymmetric part. Extract and carry out the Gaussian integrals over
$J_{ij}$:

$I_{1}=\frac{1}{(2\pi\sigma^{2})^{\frac{N(N-1)}{2}}}\int\left(\prod\limits_{i<j}dJ_{ij}\right)\exp\left(\sum\limits_{i,j}iP_{ij}J_{ij}-\sum\limits_{i<j}\frac{J_{ij}^{2}}{2\sigma^{2}}\right)=\exp\left(-\sigma^{2}\sum\limits_{i,j}(P^{A}_{ij})^{2}\right)\,.$
(18)

Substituting (18) into (17), we obtain

$\rho_{0}(\lambda)=\frac{1}{\pi
N}\textrm{Im}\frac{\partial}{\partial\lambda}\lim\limits_{n\rightarrow
0}\frac{1}{n}\Biggl{\\{}\int\left(\prod\limits_{i;\alpha}d\bar{c}_{i}^{\alpha}dc_{i}^{\alpha}\right)\exp\left(-\lambda_{\epsilon}\sum\limits_{i}P_{ii}-\sigma^{2}\sum\limits_{i,j}(P^{A}_{ij})^{2}\right)-1\Biggl{\\}}\,.$
(19)

Let us define the antihermitian tensor
$T^{\alpha\beta}=\sum\limits_{i}\bar{c}_{i}^{\alpha}c_{i}^{\beta}$
($T^{\alpha\beta}=-\bar{T}^{\beta\alpha}$) with symmetric and antisymmetric
parts
$T_{S}^{\alpha\beta}=\sum\limits_{i}\frac{\bar{c}_{i}^{\alpha}c_{i}^{\beta}+\bar{c}_{i}^{\beta}c_{i}^{\alpha}}{2}$,
$T_{A}^{\alpha\beta}=\sum\limits_{i}\frac{\bar{c}_{i}^{\alpha}c_{i}^{\beta}-\bar{c}_{i}^{\beta}c_{i}^{\alpha}}{2}$
and antisymmetric tensor
$K^{\alpha\beta}=\sum\limits_{i}c_{i}^{\alpha}c_{i}^{\beta}$
($K^{\alpha\beta}=-K^{\beta\alpha}$). We rewrite the tensor expressions
associated with $\bm{P},\bm{P}^{A}$ in terms of $\bm{T},\bm{K}$ in (19). Thus,
we obtain

$\rho_{0}(\lambda)=\frac{1}{\pi
N}\textrm{Im}\frac{\partial}{\partial\lambda}\lim\limits_{n\rightarrow
0}\frac{1}{n}\Biggl{\\{}\int\left(\prod\limits_{i;\alpha}d\bar{c}_{i}^{\alpha}dc_{i}^{\alpha}\right)\exp\left(-\lambda_{\epsilon}\sum\limits_{\alpha}T^{\alpha\alpha}-\frac{J^{2}}{2N}\sum\limits_{\alpha,\beta}\left(T^{\alpha\beta}T^{\beta\alpha}+\bar{K}^{\alpha\beta}K^{\beta\alpha}\right)\right)-1\Biggl{\\}}\,.$
(20)

For convenience, we introduce

$I_{2}=\int\left(\prod\limits_{i;\alpha}d\bar{c}_{i}^{\alpha}dc_{i}^{\alpha}\right)\exp\left(-\lambda_{\epsilon}\sum\limits_{\alpha}T^{\alpha\alpha}-\frac{J^{2}}{2N}\sum_{\alpha,\beta}\left(T^{\alpha\beta}T^{\beta\alpha}+\bar{K}^{\alpha\beta}K^{\beta\alpha}\right)\right)\,.$
(21)

The next step is to apply the Hubbard-Stratonovich transformation to the
tensors $\bm{T},\bm{K}$, which allows us to replace quadratic expressions with
tensors with linear ones by introducing additional integration variables. For
$\bm{T}$ we have

$\exp\left(-\frac{J^{2}}{2N}\sum\limits_{\alpha,\beta}T^{\alpha\beta}T^{\beta\alpha}\right)=\exp\left(-\frac{J^{2}}{2N}\text{tr}\;\bm{T}^{2}\right)=\exp\left(-\frac{J^{2}}{2N}\text{tr}\;\bm{T}_{S}^{2}-\frac{J^{2}}{2N}\text{tr}\;\bm{T}_{A}^{2}\right)=\\\
=2^{-\frac{n}{2}}\left(\frac{N\lambda^{2}}{\pi
J^{2}}\right)^{\frac{n^{2}}{2}}\int[d\bm{R}_{S}][d\bm{R}_{A}]\exp\left(-\frac{N\lambda^{2}}{2J^{2}}\text{\;tr\;}\bm{R}_{S}^{2}+\frac{N\lambda^{2}}{2J^{2}}\text{\;tr\;}\bm{R}_{A}^{2}-i\lambda\text{\;tr\;}\bm{T}_{S}\bm{R}_{S}+\lambda\text{\;tr\;}\bm{T}_{A}\bm{R}_{A}\right)\,,$
(22)

where $\bm{R}_{S}$ is a real symmetric $\bm{R}_{S}^{T}=\bm{R}_{S}$ and
$\bm{R}_{A}$ is a real antisymmetric $\bm{R}_{A}^{T}=-\bm{R}_{A}$ matrices.

$[d\bm{R}_{S}]=\prod\limits_{\alpha\leq\beta}dR_{S}^{\alpha\beta},\quad[d\bm{R}_{A}]=\prod\limits_{\alpha<\beta}dR_{A}^{\alpha\beta}\,.$
(23)

For the tensor $\bm{K}$ we have:

$\exp\left(-\frac{J^{2}}{2N}\sum\limits_{\alpha,\beta}\bar{K}^{\alpha\beta}K^{\beta\alpha}\right)=\exp\left(-\frac{J^{2}}{2N}\text{tr\;}\bar{\bm{K}}\bm{K}\right)=\\\
=2^{n(n-1)}\left(\frac{N\lambda^{2}}{\pi
J^{2}}\right)^{\frac{n(n-1)}{2}}\int[d\bm{M}]\exp\left(\frac{2N\lambda^{2}}{J^{2}}\text{\;tr\;}\bar{\bm{M}}\bm{M}+\lambda\text{\;tr\;}\bm{K}\bm{M}+\lambda\text{\;tr\;}\bar{\bm{K}}\bar{\bm{M}}\right)\,,$
(24)

where $\bm{M}$ is an antisymmetric complex matrix and
$[d\bm{M}]=\prod\limits_{i<j}d^{2}M_{ij}$. Substitute (22) and (24) in (21)
and obtain

$I_{2}=2^{n(n-\frac{3}{2})}\left(\frac{N\lambda^{2}}{\pi
J^{2}}\right)^{n(n-\frac{1}{2})}\int\left(\prod\limits_{i;\alpha}d\bar{c}_{i}^{\alpha}dc_{i}^{\alpha}\right)[d\bm{R}_{S}][d\bm{R}_{A}][d\bm{M}]\exp\left(-\lambda_{\epsilon}\text{tr}\;\bm{T}-\frac{N\lambda^{2}}{2J^{2}}\text{tr\;}\bm{R}_{S}^{2}+\frac{N\lambda^{2}}{2J^{2}}\text{tr\;}\bm{R}_{A}^{2}-\right.\\\
\left.-i\lambda\text{\;tr\;}\bm{T}_{S}\bm{R}_{S}+\lambda\text{\;tr\;}\bm{T}_{A}\bm{R}_{A}+\frac{2N\lambda^{2}}{J^{2}}\text{tr}\;\bar{\bm{M}}\bm{M}+\lambda\text{\;tr\;}\bm{K}\bm{M}+\lambda\text{\;tr\;}\bar{\bm{K}}\bar{\bm{M}}\right)\,.$
(25)

Let us introduce the vector of $2n$ parameters
$\bm{\theta}_{i}=\\{c^{1}_{i},...,c^{n}_{i},\bar{c}^{1}_{i},...,\bar{c}^{n}_{i}\\}$
with the measure
$d\bm{\theta}_{i}=\prod\limits_{\alpha}d\bar{c}_{i}^{\alpha}dc_{i}^{\alpha}$
and we represent the linear tensor part in a more compact form

$-\lambda_{\epsilon}\text{tr}\;\bm{T}-i\lambda\text{\;tr\;}\bm{T}_{S}\bm{R}_{S}+\lambda\text{\;tr\;}\bm{T}_{A}\bm{R}_{A}+\lambda\;\text{tr}\;\bm{K}\bm{M}+\lambda\text{\;tr\;}\bar{\bm{K}}\bar{\bm{M}}=-\sum_{i}\frac{1}{2}\bm{\theta}^{T}_{i}\left(\lambda_{\epsilon}\bm{\Omega}+i\lambda\bm{\sigma}_{1}\bm{L}\right)\bm{\theta}_{i}\,,$
(26)

where $\bm{R}=\frac{\bm{R}_{S}+i\bm{R}_{A}}{2}$ is the Hermitian matrix,
$\bm{\sigma}_{1}$ is the Pauli matrix and also

$\bm{\Omega}=\begin{pmatrix}\bm{0}&-\bm{I}_{n}\\\ \
\bm{I}_{n}&\bm{0}\end{pmatrix},\quad\bm{L}=2\begin{pmatrix}\bm{R}^{T}&-i\bar{\bm{M}}\\\
-i\bm{M}&-\bm{R}\end{pmatrix}\,.$ (27)

It should be noted that the matrix $\bm{L}$ has the following properties

$\bm{L}=\bm{L}^{\dagger},\quad\bm{\sigma}_{1}\bm{L}\bm{\sigma}_{1}=-\bm{L}^{T}\,.$
(28)

In terms of the Cartan classification of Hamiltonians, we work with class $D$
with $P^{2}=1$ symmetry (see appendix B). This class was considered, for
example, in Ivanov (2002) using a supersymmetric approach.

The next step is to perform integration over the Grassmannian variables in
(25). To do this, we use the formula for the Berezin integral:

$\int
d\bm{\theta}_{i}\exp{\left(-\frac{1}{2}\bm{\theta}^{T}_{i}\left(\lambda_{\epsilon}\bm{\Omega}+i\lambda\bm{\sigma}_{1}\bm{L}\right)\bm{\theta}_{i}\right)}=\text{Pf}(\lambda_{\epsilon}\bm{\Omega}+i\lambda\bm{\sigma}_{1}\bm{L})\,.$
(29)

We obtain

$I_{2}=c_{n}\int[d\bm{R}][d\bm{M}]\exp\left(-\frac{2N\lambda^{2}}{J^{2}}\text{tr}\left(\bm{R}^{2}-\bar{\bm{M}}\bm{M}\right)+\frac{N}{2}\log\det\left(\lambda_{\epsilon}\bm{\Omega}+i\lambda\bm{\sigma}_{1}\bm{L}\right)\right),$
(30)

where $c_{n}=2^{n(n-\frac{3}{2})}\left(\frac{N\lambda^{2}}{\pi
J^{2}}\right)^{n(n-\frac{1}{2})}$. We keep in mind that

$\text{tr}\;\bm{L}^{2}=8\text{tr}\;(\bm{R}^{2}-\bar{\bm{M}}\bm{M}),\quad\bm{\Omega}^{2}=-\bm{I},\quad\bm{\Omega}\bm{\sigma}_{1}=\begin{pmatrix}-\bm{I}_{n}&0\\\
0&\bm{I}_{n}\end{pmatrix}=-\bm{\sigma}_{3}\,.$ (31)

Therefore, we can rewrite (30) through one matrix $\bm{L}$ over which the
integration is performed

$I_{2}=c_{n}\int[d\bm{L}]\exp\left(-N\text{tr}\left(\frac{\lambda^{2}}{4J^{2}}\bm{L}^{2}-\frac{1}{2}\log\left(\lambda_{\epsilon}\bm{I}+i\lambda\bm{\sigma}_{3}\bm{L}\right)\right)\right)\,.$
(32)

By varying the action, the equation of the saddle-point manifold is

$\frac{\lambda^{2}}{2J^{2}}\bm{L}_{0}-\frac{1}{2}i\lambda\bm{\sigma}_{3}(\lambda_{\epsilon}\bm{I}+i\lambda\bm{\sigma}_{3}\bm{L}_{0})^{-1}=0\,,$
(33)

and it is solved by

$\bm{L}_{0}=\text{diag}(x_{1},...,x_{n},x_{n+1},...,x_{2n})\,,$ (34)

where $x_{i}$ are saddle points

$\begin{cases}x^{\pm}_{i}=\frac{i\lambda_{\epsilon}\pm\sqrt{4J^{2}-\lambda_{\epsilon}^{2}}}{2\lambda},\qquad
i=\\{1,...,n\\}\,;\\\
x^{\pm}_{j}=-\frac{i\lambda_{\epsilon}\pm\sqrt{4J^{2}-\lambda_{\epsilon}^{2}}}{2\lambda},\quad
j=\\{n+1,...,2n\\}\,.\end{cases}$ (35)

Let us introduce $g_{\pm}(x)$ as

$g_{\pm}(x)=\frac{\lambda^{2}}{4J^{2}}x^{2}-\frac{1}{2}\log\left(\lambda_{\epsilon}\pm
ix\lambda\right)\,.$ (36)

By deforming the integration contour onto the saddle-point manifold, the
integral (32) reduces to

$I_{2}\simeq
c_{n}\prod\limits_{i=1}^{n}\left(\sum_{x_{i}\in\\{x_{i}^{+},x_{i}^{-}\\}}\int_{-\infty}^{\infty}dx_{i}e^{-Ng_{+}(x_{i})}\right)\prod\limits_{j=n+1}^{2n}\left(\sum_{x_{j}\in\\{x_{j}^{+},x_{j}^{-}\\}}\int_{-\infty}^{\infty}dx_{j}e^{-Ng_{-}(x_{j})}\right)\,.$
(37)

Considering $g_{+}(x)=g_{-}(-x)\equiv g(x)$, we get

$I_{2}\simeq
c_{n}\prod\limits_{i=1}^{n}\left(\sum_{x_{i}\in\\{x_{i}^{+},x_{i}^{-}\\}}\int_{-\infty}^{\infty}dx_{i}e^{-Ng_{+}(x_{i})}\right)^{2}\,.$
(38)

Without loss of generality, let it be $\lambda>0$. We know that
$\rho_{0}(\lambda)=\rho_{0}(-\lambda)$ due to the matrix antisymmetry.

Figure 1: Here $x^{\pm}$ are the saddle points of $g(x)$. We choose only
$x^{-}$. Red lines means branch cut.

We investigate 2 cases:

  * •

$\lambda<2J$.

Value of $g(x)$ in $x_{i}^{\pm}$:

$g(x^{\pm}_{i})=\frac{1}{4}+\frac{\lambda_{\epsilon}^{2}}{8J^{2}}\left(-1\pm
i\sqrt{\frac{4J^{2}}{\lambda_{\epsilon}^{2}}-1}\right)-\frac{1}{2}\log\left(\frac{\lambda_{\epsilon}}{2}\left(1\pm
i\sqrt{\frac{4J^{2}}{\lambda_{\epsilon}^{2}}-1}\right)\right)\,.$ (39)

In the limit $\epsilon\rightarrow 0$:

$g^{\prime\prime}(x^{\pm}_{i})=\frac{\lambda^{4}}{4J^{4}}\left(\frac{4J^{2}}{\lambda^{2}}-1\pm
i\sqrt{\frac{4J^{2}}{\lambda^{2}}-1}\right)\,,$ (40)
$|g^{\prime\prime}(x^{\pm}_{i})|=\frac{\lambda^{2}}{2J^{3}}\sqrt{4J^{2}-\lambda^{2}},\quad\arg
g^{\prime\prime}(x^{\pm}_{i})=\pm\arctan\left(\frac{1}{\sqrt{\frac{4J^{2}}{\lambda^{2}}-1}}\right)\,.$
(41)

For all $i$ there are two variants: $x_{i}=x^{-}_{i}$ or $x_{i}=x^{+}_{i}$.
Therefore, we have a $n$-dimensional complex saddle-point manifold. We choose
all $x_{i}=x^{-}_{i}\equiv x^{-}$ to $\rho_{0}(\lambda)>0$ because
contributions of all other variants are suppressed (see fig. 1). Similar
reasoning was made in Kamenev and Mézard (1999) for GUE, which consists of
Hermitian matrices.

Since $N\gg 1$, we use the complex steepest descent method:

$\rho_{0}(\lambda)\simeq\frac{1}{\pi
N}\text{Im}\frac{\partial}{\partial\lambda}\log\left(\sqrt{\frac{2\pi}{N|g^{\prime\prime}(x^{-})|}}e^{-Ng(x^{-})-i\frac{\arg
g^{\prime\prime}(x^{-})}{2}}\right)^{2}\,.$ (42)

As the matrix size $N$ grows to infinity, $|g^{\prime\prime}(x^{-}_{i})|$ and
$\arg g^{\prime\prime}(x^{-}_{i})$ remain approximately constant, scaling as
$\mathcal{O}(1)$ for each $i$. We derive

$\rho_{0}(\lambda)\simeq\frac{1}{\pi}\text{Im}\frac{\partial}{\partial\lambda}(-2g(x^{-}))=\frac{1}{\pi
J}\sqrt{1-\frac{\lambda^{2}}{4J^{2}}}\,.$ (43)

  * •

$\lambda>2J$.
Both saddle points are purely imaginaries:

$x^{\pm}_{i}=\frac{i}{2}\left(1\pm\sqrt{1-\frac{4J^{2}}{\lambda^{2}}}\right)\,.$
(44)

We have $g(x^{\pm}_{i})\in\mathbb{R}$, consequently $\rho_{0}(\lambda)=0$.

Combining the results of both cases and remembering that $\lambda$ can be
negative, we get

$\rho_{0}(\lambda)=\begin{cases}\begin{aligned} &\frac{1}{\pi
J}\sqrt{1-\frac{\lambda^{2}}{4J^{2}}},&\quad&|\lambda|<2J\,;\\\
&0,&\quad&|\lambda|>2J\,.\end{aligned}\end{cases}$ (45)

This is the Wigner semicircle (see fig. 2). We reproduce this well known
result for the $D$ class using a replica approach with fermions.

Figure 2: Wigner semicircle. The dotted line corresponds $\rho_{0}(\lambda).$

##  IV Spectrum of the antisymmetric matrix with non-zero mean

In this section we consider antisymmetric matrices with elements having non-
zero mean $\frac{J_{0}}{N}$ with the probability density function:

$p(J_{ij})=\frac{1}{\sqrt{2\pi\sigma^{2}}}e^{-\frac{(J_{ij}-J_{0}/N)^{2}}{2\sigma^{2}}}\,.$
(46)

The average density of eigenvalues:

$\rho_{J_{0}}(\lambda)=\braket{\nu(\lambda)}_{J_{ij}:J_{0}\neq
0}=\int\nu\left(\lambda;\\{J_{ij}:J_{0}\neq
0\\}\right)\prod\limits_{i<j}\;p(J_{ij})\;dJ_{ij}\,.$ (47)

Similarly to (17) we have

$\rho_{J_{0}}(\lambda)=\frac{1}{\pi
N}\textrm{Im}\frac{\partial}{\partial\lambda}\lim\limits_{n\rightarrow
0}\frac{1}{n}\Biggl{\\{}\int\left(\prod\limits_{i;\alpha}d\bar{c}_{i}^{\alpha}dc_{i}^{\alpha}\right)\exp\left(-\sum\limits_{i,j;\alpha}\bar{c}_{i}^{\alpha}(\lambda_{\epsilon}I_{ij}-iJ_{ij})c^{\alpha}_{j}\right)\times\\\
\times\left(\prod\limits_{i<j}\frac{dJ_{ij}}{\sqrt{2\pi\sigma^{2}}}\exp\left(-\frac{(J_{ij}-\frac{J_{0}}{N})^{2}}{2\sigma^{2}}\right)\right)-1\Biggl{\\}}\,.$
(48)

Now we are reducing the problem with a non-zero mean to a zero-mean case. For
this purpose, add and subtract term $A_{ij}\frac{J_{0}}{N}$ to $J_{ij}$:

$J_{ij}=\underbrace{J_{ij}-A_{ij}\frac{J_{0}}{N}}_{\text{zero-mean
matrix}}+A_{ij}\frac{J_{0}}{N}\,,$ (49)

where $A_{ij}$ is a constant antisymmetric matrix (see appendix A). Using the
result of (20), we get

$\rho_{J_{0}}(\lambda)=\frac{1}{\pi
N}\textrm{Im}\frac{\partial}{\partial\lambda}\lim_{n\to
0}\frac{1}{n}\Biggl{\\{}\int\left(\prod\limits_{i;\alpha}d\bar{c}_{i}^{\alpha}dc_{i}^{\alpha}\right)\exp\bigg{(}-\sum\limits_{i,j;\alpha}\bar{c}^{\alpha}_{i}\left(\bm{I}\lambda_{\epsilon}-i\bm{A}\frac{J_{0}}{N}\right)_{ij}c^{\alpha}_{j}-\\\
-\frac{J^{2}}{2N}\sum_{\alpha,\beta}\left(T^{\alpha\beta}T^{\beta\alpha}+\bar{K}^{\alpha\beta}K^{\beta\alpha}\right)\bigg{)}-1\Biggl{\\}}\,.$
(50)

As in the case of the zero mean for the integral in (50), we perform the
Hubbard-Stratonovich transformation and obtain an analogous formula to (25):

$I_{2}=c_{n}\int\left(\prod\limits_{i;\alpha}d\bar{c}_{i}^{\alpha}dc_{i}^{\alpha}\right)[d\bm{R}_{S}][d\bm{R}_{A}][d\bm{M}]\exp\Biggl{(}-\sum\limits_{i,j;\alpha}\bar{c}^{\alpha}_{i}\left(\lambda_{\epsilon}\bm{I}-i\bm{A}\frac{J_{0}}{N}\right)_{ij}c^{\alpha}_{j}-\frac{N\lambda^{2}}{2J^{2}}\text{tr\;}\bm{R}_{S}^{2}+\frac{N\lambda^{2}}{2J^{2}}\text{tr\;}\bm{R}_{A}^{2}-\\\
-i\lambda\text{\;tr\;}\bm{T}_{S}\bm{R}_{S}+\lambda\text{\;tr\;}\bm{T}_{A}\bm{R}_{A}+\frac{2N\lambda^{2}}{J^{2}}\text{tr}\;\bar{\bm{M}}\bm{M}+\lambda\text{\;tr\;}\bm{K}\bm{M}+\lambda\text{\;tr\;}\bar{\bm{K}}\bar{\bm{M}}\Biggl{)}\,.$
(51)

Performing calculations for the linear part in (51) similarly to (26), we come
to

$-\sum\limits_{i,j;\alpha}\bar{c}^{\alpha}_{i}\left(\lambda_{\epsilon}\bm{I}-i\bm{A}\frac{J_{0}}{N}\right)_{ij}c^{\alpha}_{j}-i\lambda\text{\;tr\;}\bm{T}_{S}\bm{R}_{S}+\lambda\text{\;tr\;}\bm{T}_{A}\bm{R}_{A}+\frac{2N\lambda^{2}}{J^{2}}\text{tr}\;\bar{\bm{M}}\bm{M}+\lambda\text{\;tr\;}\bm{K}\bm{M}+\lambda\text{\;tr\;}\bar{\bm{K}}\bar{\bm{M}}=\\\
=\sum\limits_{i,j}\bm{A}^{ij}\otimes\frac{1}{2}\bm{\theta}^{T}_{i}\left(\frac{iJ_{0}}{N}\bm{\sigma}_{1}\right)\bm{\theta}_{j}-\sum_{i,j}\bm{I}^{ij}\otimes\frac{1}{2}\bm{\theta}^{T}_{i}\left(\lambda_{\epsilon}\bm{\Omega}+i\lambda\bm{\sigma}_{1}\bm{L}\right)\bm{\theta}_{j}\,,$
(52)

where $\bm{A}^{ij}=\bm{A}$ and $\bm{I}^{ij}=\bm{I}$ for all $i$ and $j$. After
integration by all $\bm{\theta}_{i}$, we get a formula similar to (30):

$I_{3}=c_{n}\int[d\bm{R}][d\bm{M}]\exp\left(-\frac{2N\lambda^{2}}{J^{2}}\text{tr}\left(\bm{R}^{2}-\bar{\bm{M}}\bm{M}\right)+\frac{N}{2}\log\text{det}\left(\bm{A}\otimes\frac{iJ_{0}}{N}\bm{\sigma}_{1}-\bm{I}\otimes\left(\lambda_{\epsilon}\bm{\Omega}+i\lambda\bm{\sigma}_{1}\bm{L}\right)\right)\right)\,.$
(53)

Later we diagonalize $\bm{A}$:

$\text{det}\left(\bm{A}\otimes\frac{iJ_{0}}{N}\bm{\sigma}_{1}-\bm{I}\otimes\left(\lambda_{\epsilon}\bm{\Omega}+i\lambda\bm{\sigma}_{1}\bm{L}\right)\right)=\text{det}\left(\bm{D}\otimes\frac{iJ_{0}}{N}\bm{\sigma}_{1}-\bm{I}\otimes\left(\lambda_{\epsilon}\bm{\Omega}+i\lambda\bm{\sigma}_{1}\bm{L}\right)\right)\,,$
(54)

where $\bm{D}=\text{diag}(i\lambda_{1},...,i\lambda_{N})$ (see appendix A).
After substitution of $\bm{D}$, we obtain

$I_{3}=c_{n}\int[d\bm{L}]\prod_{i=1}^{N}\text{det}^{\frac{1}{2}}\left(\bm{I}+\frac{J_{0}\lambda_{i}}{N}\bm{\sigma}_{3}(\lambda_{\epsilon}\bm{I}+i\lambda\bm{\sigma}_{3}\bm{L})^{-1}\right)\exp\left(-N\left\\{\text{tr}\left(\frac{\lambda^{2}}{4J^{2}}\bm{L}^{2}-\frac{1}{2}\log\left(\lambda_{\epsilon}\bm{I}+i\lambda\bm{\sigma}_{3}\bm{L}\right)\right)\right\\}\right)\,.$
(55)

Let us use again the method of steepest descent. It should be noted that
prefactor

$\prod\limits_{i=1}^{N}\text{det}^{\frac{1}{2}}\left(\bm{I}+\frac{J_{0}\lambda_{i}}{N}\bm{\sigma}_{3}(\lambda_{\epsilon}\bm{I}+i\lambda\bm{\sigma}_{3}\bm{L})^{-1}\right)$
(56)

does not influence the saddle-point equations.

$\rho_{J_{0}}(\lambda)=\frac{1}{\pi
N}\textrm{Im}\frac{\partial}{\partial\lambda}\left(\sum_{k=1}^{N}\log\left(1-\frac{J_{0}\lambda_{k}}{N(\lambda+i\lambda
x^{-}-i\epsilon)}\right)-Ng(x^{-})\right)\,.$ (57)

Substituting $x^{-}$, we obtain

$\rho_{J_{0}}(\lambda)=\rho_{0}(\lambda)+\frac{1}{\pi
N}\textrm{Im}\frac{\partial}{\partial\lambda}\sum_{k=1}^{N}\log\left(1-\frac{J_{0}\lambda_{k}}{2J^{2}N}(\lambda+i\sqrt{4J^{2}-\lambda^{2}})-iu_{k}\epsilon\right),$
(58)

where $u_{k}(\lambda)=\frac{J_{0}\lambda_{k}}{N\lambda^{2}(1+ix^{-})^{2}}$.
Without loss of generality, let $\lambda>0$. We know that
$\rho_{0}(\lambda)=\rho_{0}(-\lambda)$ due to the matrix antisymmetry. We
consider 2 cases:

  * •

$\lambda>2J$.

$\rho_{J_{0}}(\lambda)=\rho_{0}(\lambda)+\frac{1}{\pi
N}\textrm{Im}\frac{\partial}{\partial\lambda}\sum_{k=1}^{N}\log\left(1-\frac{J_{0}\lambda_{k}}{2J^{2}N}(\lambda-\sqrt{\lambda^{2}-4J^{2}})-iu_{k}\epsilon\right).$
(59)

We define a temporary function
$f(\lambda)=1-\frac{J_{0}\lambda_{k}}{2J^{2}N}(\lambda-\sqrt{\lambda^{2}-4J^{2}})$.
Expand $f(\lambda)$ in the Taylor series near its zero $f(\mu_{k})=0$, where
$\mu_{k}=\lambda^{*}_{k}+\frac{J^{2}}{\lambda^{*}_{k}}$ and
$\lambda^{*}_{k}=\frac{J_{0}}{N}\lambda_{k}$:

$f(\lambda)=f^{\prime}(\mu_{k})(\lambda-\mu_{k})+\mathcal{O}((\lambda-\mu_{k})^{2}),\quad
f^{\prime}(\mu_{k})=\frac{NJ_{0}\lambda_{k}}{J_{0}^{2}\lambda_{k}^{2}-J^{2}N^{2}}\,,$
(60)
$\frac{\partial}{\partial\lambda}\log(f^{\prime}(\mu_{k})(\lambda-\mu_{k})+\mathcal{O}((\lambda-\mu_{k})^{2})-iu_{k}\epsilon)=\frac{1-\frac{iu^{\prime}_{k}(\lambda)\epsilon}{f^{\prime}(\mu_{k})}+\mathcal{O}(\lambda-\mu_{k})}{\lambda-\mu_{k}-\frac{iu_{k}\epsilon}{f^{\prime}(\mu_{k})}+\mathcal{O}((\lambda-\mu_{k})^{2})}\,.$
(61)

We are interested in singular terms because of the factor $\frac{1}{N}$, which
suppresses any regular term. Using the Sokhotski–Plemelj formula, we get

$\rho_{J_{0}}(\lambda)=\rho_{0}(\lambda)+\frac{1}{N}\sum\limits_{k}\delta(\lambda-\mu_{k}),\quad
k\in\left\\{j:f(\mu_{j})=0\right\\}$ (62)

Value $\mu_{k}$ can be a zero of the function $f(\lambda)$ only in several
cases, depending on $J_{0}$, $J$, $N$. Let us investigate it. Notice that

$\sqrt{\lambda^{2}-4J^{2}}-\lambda>\begin{cases}\begin{aligned}
&2J\,,&\quad&\lambda<-2J\,;\\\
&-2J\,,&\quad&\lambda>2J\,.\end{aligned}\end{cases}$ (63)

Let us simplify the value of the $f(\lambda)$. This function can vanish in
several cases:

$\begin{cases}\begin{aligned}
&2J^{2}N<2JJ_{0}\lambda_{k},&\quad&\lambda>2J,\;\lambda_{k}>0\,;\\\
&2J^{2}N>-2JJ_{0}\lambda_{k},&\quad&\lambda<-2J,\;\lambda_{k}<0\,.\end{aligned}\end{cases}$
(64)

We will remain from (62) $\delta$-functions with $k$ which satisfy
$\frac{J_{0}}{N}\lambda_{k}>J$ and $-\frac{J_{0}}{N}\lambda_{k}<J$, which is
trivial.

$\rho_{J_{0}}(\lambda)=\rho_{0}(\lambda)+\frac{1}{N}\sum_{k}\delta(\lambda-\mu_{k}),\quad
k\in\left\\{j:\lambda^{*}_{j}>J\right\\}\,.$ (65)

  * •

$\lambda<2J$.
In this case we only have a regular term.

$\text{Im}\frac{\partial}{\partial\lambda}\sum_{k=1}^{N}\log\left(1-\frac{J_{0}\lambda_{k}(\lambda+i\sqrt{4J^{2}-\lambda^{2}})}{2J^{2}N}\right)=\sum\limits_{k=1}^{N}\frac{J_{0}\lambda_{k}}{\sqrt{4J^{2}-\lambda^{2}}}\frac{\lambda
N/2-J_{0}\lambda_{k}}{J^{2}_{0}\lambda^{2}_{k}-J_{0}\lambda_{k}\lambda
N+J^{2}N^{2}}\,.$ (66)

In the limit $N\gg 1$ we get

$\rho_{J_{0}}(\lambda)=\rho_{0}(\lambda)\,.$ (67)

Considering the spectrum is even and combining the results of both cases, we
get

$\rho_{J_{0}}(\lambda)=\rho_{0}(\lambda)+\frac{1}{N}\sum\limits_{k\in\left\\{j:\lambda^{*}_{j}>J\right\\}}\delta(\lambda-\mu_{k})+\delta\left(\lambda+\mu_{k}\right)\,.$
(68)

This result corresponds with numerical simulation from Wolfram Mathematica
13.0. We can see this in fig. 3.

Figure 3: Wigner semicircle and emerging delta peaks. The dotted line
corresponds $\rho_{0}(\lambda).$

Thus there are a set of critical values
$J^{crit}_{0,k}=\frac{JN}{\lambda_{k}},\;k\in\\{1,...,N/2\\}.$ When the value
of $J_{0}$ passes the critical value $J^{crit}_{0,k},$ a two new delta peaks
$\delta(\lambda\pm\mu_{k})$ appear. In order to make the form of the answer
record similar to the answer in the case of a symmetric matrix, let us
separately identify the smallest among the critical values
$J^{crit}_{0}=J^{crit}_{0,1}=JN\tan{\frac{\pi}{2N}}\underset{N\rightarrow\infty}{=}\frac{\pi}{2}J$
below which no effect occurs. We found that the value of $J_{0}$ can
significantly affect the properties of the spectrum. A phase transition
occurs: depending on $J_{0}$, we can have a separation of the largest
eigenvalues. We rewrite (68) in a more comprehensible way:

$\rho_{J_{0}}(\lambda)\equiv\braket{\nu(\lambda)}_{J_{ij}:J_{0}\neq
0}=\begin{cases}\begin{aligned}
&\rho_{0}(\lambda),&\quad&J_{0}<J^{crit}_{0,1}\,;\\\
&\rho_{0}(\lambda)+\frac{1}{N}(\delta(\lambda-\mu_{1})+\delta(\lambda+\mu_{1})),&\quad&J^{crit}_{0,1}<J_{0}<J^{crit}_{0,2}\,;\\\
&\ldots\\\
&\rho_{0}(\lambda)+\frac{1}{N}\sum_{k=1}^{N/2}\delta(\lambda-\mu_{k})+\delta\left(\lambda+\mu_{k}\right),&\quad&J^{crit}_{0,\frac{N}{2}}<J_{0}\,.\end{aligned}\end{cases}$
(69)

Let us notice interesting fact. We present the antisymmetric matrix with a
non-zero mean as the sum of the non-zero mean matrix (with the Wigner
semicircle spectrum) and the constant matrix (with a spectrum of
$\lambda_{k}$). And the spectrum of the sum of these two matrices is similar
to the superposition of two spectra but there are several conditions on the
appearance of the single states (69) and delta-peaks are shifted further away
from the semicircle in comparison with constant matrix ($\mu_{k}$ instead of
$\lambda_{k}$). Let us call this effect ”spectrum interference”. The same
situation was in the case of symmetric matrices in Edwards and Jones (1976);
Jones et al. (1978).

##  V Acknowledgements

The authors are grateful to acknowledge helpful discussions with A. Litvinov
and A. Kamenev. Special thanks to G. Tarnopolsky, who inspired us to solve
this problem and helped during the writing of this article. This work has been
supported by the Russian Science Foundation under the grant 22-22-00991.

##  Appendix A Spectrum of the constant antisymmetric matrix

Let us consider a constant antisymmetric matrix $A_{ij}$ of size $N$:

$\bm{A}=\left(\begin{array}[]{ccccc}0&1&\dots&1&1\\\ -1&0&\dots&1&1\\\
\vdots&\vdots&\ddots&\vdots&\vdots\\\ -1&-1&\dots&0&1\\\ -1&-1&\dots&-1&0\\\
\end{array}\right)\,.$ (70)

We obtain the expression of the characteristic polynomial using linear
transformations for the determinant

$\det(\bm{A}-\lambda\bm{I})=\underbrace{\left|\begin{array}[]{cccc}-\lambda&1&\ldots&1\\\
-1&-\lambda&\ldots&1\\\ \vdots&\vdots&\ddots&\vdots\\\
-1&-1&\ldots&-\lambda\end{array}\right|}_{D_{N}(\lambda)}=-2\lambda
D_{N-1}(\lambda)-(\lambda^{2}-1)D_{N-2}(\lambda)\,.$ (71)

The characteristic polynomial

$D_{N}(\lambda)=-2\lambda D_{N-1}(\lambda)-(\lambda^{2}-1)D_{N-2}(\lambda)\,,$
(72)

has boundary conditions $D_{1}=-\lambda,\;D_{2}=\lambda^{2}+1$. Solving the
recurrent equation, we obtain

$D_{N}(\lambda)=\frac{1}{2}(-\lambda+1)^{N}+\frac{1}{2}(-\lambda-1)^{N}\,.$
(73)

Therefore, the eigenvalues have the following form

$i\lambda_{k}=i\cot{\left(\pi\frac{2k-1}{2N}\right)},\;k\in\\{1,2,..,N\\}\,.$
(74)

The set $\lambda_{k}$ is ordered in descending order.

##  Appendix B Cartan classes

Cartan classification is based on the fact that all random matrices can be
divided into 10 symmetrical classes (see table 1). The set of 10 classes
corresponds to a special set of manifolds known as Riemannian symmetric
spaces, which have exceptional symmetry properties. For example, you can see
Schnyder et al. (2008).

Class | T | P | S | Structure
---|---|---|---|---
A ”Unitary” | $0$ | $0$ | $0$ | $\mathfrak{u}(N)$
D | $0$ | $1$ | $0$ | $\mathfrak{so}(2N)$
C | $0$ | $-1$ | $0$ | $\mathfrak{sp}(2N)$
AI ”Orthogonal” | $1$ | $0$ | $0$ | $\mathfrak{u}(N)/\mathfrak{so}(N)$
BDI | $-1$ | $0$ | $0$ | $\mathfrak{so}(2N)/\mathfrak{so}(N)\times\mathfrak{so}(N)$
CI | $1$ | $-1$ | $1$ | $\mathfrak{sp}(2N)/\mathfrak{u}(N)$
AII ”Symplectic” | $-1$ | $0$ | $0$ | $\mathfrak{u}(2N)/\mathfrak{sp}(2N)$
DIII | $-1$ | $1$ | $1$ | $\mathfrak{so}(2N)/\mathfrak{u}(N)$
CII | $-1$ | $-1$ | $1$ | $\mathfrak{sp}(4N)/\mathfrak{sp}(2N)\times\mathfrak{sp}(2N)$
AIII | $0$ | $0$ | $1$ | $\mathfrak{u}(2N)/\mathfrak{u}(N)\times\mathfrak{u}(N)$
Table 1: The Cartan Classification of Topological Insulators.

Consider the model of spinless non-interacting fermions on a lattice with a
Hamiltonian:

$H=\sum\limits_{i,j}t_{ij}c_{i}^{\dagger}c_{i}+\frac{1}{2}\Delta_{ij}c_{i}^{\dagger}c_{j}^{\dagger}+\frac{1}{2}\Delta^{\dagger}_{ij}c_{i}c_{j}\,,$
(75)

where matrix $t_{ij}$ is Hermitian $\bm{t}^{\dagger}=\bm{t}$.

It encodes how electrons “tunnel” from site to site on the lattice. The
anomalous terms are proportional to $\Delta$ and $\Delta^{\dagger}$, where
$\Delta_{ij}$ is the superconducting gap parameter. A normal metal or
insulator has $\Delta=\Delta^{\dagger}=0$.
It is convenient to rewrite Hamiltonian in a different way. Introduce the
Majorana spinor:

$\chi\equiv\begin{pmatrix}c\\\
(c^{\dagger})^{T}\end{pmatrix}\Leftrightarrow\chi_{i,\sigma}=\begin{cases}c_{i}\,,\quad\sigma=1\,;\\\
c^{\dagger}_{i}\,,\quad\sigma=2\,,\end{cases}$ (76)

where $i\in\\{1,...,N\\}$ is a site index and $\sigma\in\\{1,2\\}$ is a
”particle-hole” index. The Hermitian conjugate quantity $\chi^{\dagger}$ is
independent of $\chi$

$\chi^{\dagger}=\begin{pmatrix}c^{\dagger}&c^{T}\end{pmatrix}=\chi^{T}\bm{\sigma}_{1}\,.$
(77)

Pauli $2N\times 2N$ matrices:

$\bm{\sigma}_{1}=\begin{pmatrix}0&\bm{I}\\\
\bm{I}&0\end{pmatrix},\quad\bm{\sigma}_{2}=\begin{pmatrix}0&-i\bm{I}\\\
i\bm{I}&0\end{pmatrix},\quad\bm{\sigma}_{3}=\begin{pmatrix}\bm{I}&0\\\
0&-\bm{I}\end{pmatrix}\,.$ (78)

We can rewrite the Hamiltonian (75) as

$H=\frac{1}{2}\chi^{T}\bm{M}_{P}\bm{h}\chi,\quad\bm{M}_{P}=\bm{\sigma}_{1},\quad\bm{h}=\begin{pmatrix}\bm{t}&\bm{\Delta}\\\
\bm{\Delta}^{\dagger}&-\bm{t}^{T}\end{pmatrix},$ (79)

where the Hermitian matrix $\bm{h}=\bm{h}^{\dagger}$ is the effective single-
particle Hamiltonian. In general case $\bm{M}_{P}$ is not necessarily
$\bm{\sigma}_{1}$.

Now let us define symmetry conditions on $\bm{h}$. We do not consider spatial
symmetries (translations and rotations) because we apply this theory to random
matrices.

  * •

Particle-hole symmetry $P$

$\bm{M}_{P}\bm{h}^{T}\bm{M}_{P}=-\bm{h},\quad\bm{M}_{P}=\begin{cases}+\bm{M}^{T}_{P}\,,\quad
P^{2}=1\,;\\\ -\bm{M}^{T}_{P}\,,\quad P^{2}=-1.\end{cases}$ (80)

For example, $\bm{M}_{P}=\bm{\sigma}_{1}$ – symmetric matrix and it
corresponds to $P^{2}=1$.

  * •

Time-reversal invariance $T$

$\bm{M}_{T}\bm{h}^{*}\bm{M}_{T}=\bm{h},\quad\bm{M}_{T}=\begin{cases}+\bm{M}^{T}_{T}\,,\quad
T^{2}=1\,;\\\ -\bm{M}^{T}_{T}\,,\quad T^{2}=-1\,.\end{cases}$ (81)

  * •

Chiral (sublattice) symmetry $S$

$-\bm{M}_{S}\bm{h}\bm{M}_{S}=\bm{h},\quad\bm{M}_{S}=\bm{M}^{\dagger}_{S}=\bm{M}^{-1}_{S}\,.$
(82)

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