In this section, we study and present several computational results on the large N limit of the Hermitian Gaussian matrix model averages \(\langle \chi _r (A)\rangle _0\) and \(\langle \chi _r (A^2)\rangle _0\) as the latter appeared as a key quantity as in (61) and the former can be an useful exercise and preparation for computing the latter. The Gaussian average of a character of A for a given representation \(\\) is given by

$$\begin \langle \chi _}(A) \rangle _0 = \frac \int dA\; e^\frac A^2} \; \chi _}(A) , \end$$

(105)

where \(Z_0\) is an Gaussian integral of A;

$$\begin Z_0 = \int dA\; e^\frac A^2}. \end$$

(106)

Similarly to (105), the Gaussian average of a character of \(A^2\) for a given representation, given by

$$\begin \langle \chi _}(A^2) \rangle _0 = \frac \int dA\; e^\frac A^2} \; \chi _}(A^2) . \end$$

(107)

For \(\^e=\|\,a\;\mathrm \}\) and \(\^o=\|\,a\;\mathrm \}\), (105) evaluates to [119]

$$\begin \langle \chi _}(A) \rangle _0 = N^} \; d_h \; \frac(h^e_i-h^o_j)} =N^} \; d_h \; \frac}(C_2)}}(C_1)} , \end$$

(108)

where \(n=\sum _i-N(N-1)/2\) is the size of the representation, and \(d_h\) is the dimension of the representation h, which can be evaluated as

$$\begin d_h=\chi _}(\mathbb 1)=\frac^k!}. \end$$

(109)

We also used

$$\begin \frac(h^e_i-h^o_j)} = \frac}(C_2)}}(C_1)} , \end$$

(110)

as presented in more detail in the appendix of [78].

Lemma 9.1

If there exists a GL(N) matrix M such that Di Francesco-Itzykson integral (105) can be computed as the character of M;

$$\begin \langle \chi _}(A) \rangle _0 =\chi _}(M), \end$$

(111)

then the integral of our interest (61) can be solved as:

$$\begin \langle \chi _}(A^2) \rangle _0 =\chi _}(M^2). \end$$

(112)

Proof

The function \(\chi _}(A^2)\) is a class function of A, therefore there exists an expansion

$$\begin \chi _}(A^2)=\sum _}c_,\}\chi _}(A). \end$$

(113)

This expansion is finite, since it is the same as the present in the identity \(\chi _}(A^2)=\chi _^2\}(A)-\chi _^2\}(A)\). Thus,

$$\begin \begin \langle \chi _}(A^2) \rangle _0&= \frac\int dA \;\chi _}(A^2) \; e^\frac A^2} \;\;\\&=\frac\int dA\;\sum _}c_,\}\chi _}(A)\;e^\frac A^2}\;\\&=\frac\sum _}c_,\}\int dA\;\chi _}(A)\;e^\frac A^2}\\&=\frac\sum _}c_,\}I^_} (\mathbb ) \\&=\sum _}c_,\}\chi _}(M)=\chi _}(M^2). \end \nonumber \\ \end$$

(114)

We note that the exchange between the summation and the integral is legal, since the summation is finite. \(\square \)

Lemma 9.1 suggests that we shall study the expression in (108) in order to see if we can find such a M as in (111). Let us present a more general normalized version of (105);

$$\begin _} (B) =\ \int dA\;\chi _}(A)e^\frac AB^AB^}\;} . \end$$

(115)

The generalized normalized Di Francesco-Itzykson integral (115) can be expressed as a product of characters

$$\begin \frac_} (B)} = \frac}(B)\chi _}(C_2)}}(C_1)} , \end$$

(116)

where we used (110) and \(Z(B) = \int dA\;e^\frac AB^AB^}\) is a normalization factor. Remark that on the right hand side of (116), the only part that depends on the choice of B is \(\chi _}(B)\).

If we set \(B = \mathbb \) in (116), we indeed recover (108). By setting \(B=C_1\) in (116), we find that

$$\begin I^_} (C_1) = \int dA\;\chi _}(A) \; e^\frac AC_1^AC_1^}\; = Z(C_1) \; \chi _}(C_2). \end$$

(117)

Since \((C_2)^2=C_1\) in the large N limit, by summing over all representations as in (113) on both sides of (117) we obtain

$$\begin I^_} (C_1) \; \sim \; \chi _}(C_1), \end$$

(118)

or performing change of variables,

$$\begin I^_} (C_1) \; \sim \; \int dA\;\chi _}((C_1A)^2)e^\frac A^2}\; \sim \; \chi _}(C_1), \end$$

(119)

which can also be written as

$$\begin \langle \chi _}(C_1 A)^2 \rangle _0 \sim \; \chi _}(C_1), \end$$

(120)

where \(\sim \) denotes equal up to a constant in N, and \(\langle \rangle _0\) denotes the Gaussian average over A as defined in (14). This result is nearly the expression we aim at computing in (61), except for the extra \(C_1\) in the left hand side.

On this note, let us now investigate further what the condition in (108) tells us about the matrix M in (111).

Proposition 9.2

Given a representation r of \(\textrm(N)\) defined through the set of shifted weights \(_i\), \(i=1,...N,\), consider \(\chi _r\) the character in that representation r. If M is a GL(N) matrix that satisfies

$$\begin \int dA\;\chi _}(A)\; e^\frac A^2} = \chi _}(M), \end$$

(121)

then, for a positive integer q,

$$\begin \frac\textrm(M^) = q^q. \end$$

(122)

Proof

From (50) we can deduce that

$$\begin \textrm(M^p) =\sum _^N\frac\}}(M)}}(M)} \end$$

(123)

by using a representation \(\\) where \(\chi _}(M)\ne 0\) and the modified representations \(\\}\) where the shifted weights are related to the ones of \(\\) by \(h_i^=h_i+p\,\delta _\), for \(i=1,...,N\).

Using (109) in (108) we can deduce that, for some constant \(c_N\),

$$\begin \chi _}(M)=c_N\,N^} \Delta (\^e)\Delta (\^o)\prod _i (h_i^e-1)!!h_i^o!!. \end$$

(124)

We assume N and p even. For i even, we set \(h_i=p (N-i)/2+1\), therefore \(h_i\) is always odd for ieven. For i odd, we set \(h_i=p /+2\), therefore \(h_i\) is always even for i odd. Consequently, for i even, \(h_i^=p /+1+p\,\delta _\), making \(h_i^\) odd and, for i odd, \(h_i^=p (N-i-1)/2+2+p\,\delta _\), making \(h_i^\) even. Notice that \(\^e=\^o\}\), thus the Vandermonde determinants \(\Delta (\^e)\) and \(\Delta (\^o)\) are

$$\begin \Delta (\^o) & = \Delta (\^e)=\prod _(_i-_j)=\prod _\left( p \frac+1-p \frac-1\right) \nonumber \\ & =\prod _p \frac, \end$$

(125)

where the indices i and j are always even. Because of this, by setting \(i=2\,m\) and \(j=2n\) we find

$$\begin \Delta (\^o)= \Delta (\^e) =p^\frac\left( \frac-1\right) }\prod _(n-m)=p^\frac\left( \frac-1\right) }\prod _^-1}m!. \end$$

(126)

We then get that

$$\begin \Delta (\^o)\Delta (\^e)=p^\left( \frac-1\right) }\prod _^-1}(m!)^2, \end$$

(127)

which will be necessary for the normalization in (123). The terms involving double factorials in (124) are:

$$\begin \prod _i (h_i^e-1)!!=\prod _i h_i^o!!= \prod _^}\left( p\left( \frac-m\right) +1\right) !!. \end$$

(128)

The size of the representation is

$$\begin n=\sum _ih_i-\frac=\frac\frac\left( p(N-2)+6\right) -\frac. \end$$

(129)

This way the character of M in the representation \(\\) is

$$\begin \chi _}(M)=c_N\;N^}\;p^\left( \frac-1\right) }\prod _^-1}(m!)^2\prod _^}\left( p\left( \frac-m\right) +1\right) !!^2.\qquad \end$$

(130)

Now, for the representations \(\}\}\), set \(\}\}^e=\\}|\,a\;\mathrm \}\) and \(\}\}^o=\\}|\,a\;\mathrm \}\). For k odd, \(\}\}^e=\\}^e\), then

$$\begin \Delta (\}\}^e)=\Delta (\\}^e)=p^\frac\left( \frac-1\right) }\prod _^-1}m!. \end$$

(131)

For k even, \(\}\}^o=\\}^o\), then

$$\begin \Delta (\}\}^o)=\Delta (\\}^o)=p^\frac\left( \frac-1\right) }\prod _^-1}m!. \end$$

(132)

Another consequence for k even is

$$\begin \Delta (\}\}^e) & =\prod _ i<j\\ i,j\;\textrm \end}(}_i-}_j)\nonumber \\ & =\prod _ i<j\\ i,j\;\textrm \end}\left( p \frac+2+p\,\delta _-p \frac-2-p\,\delta _\right) \nonumber \\ & =\prod _ i<j\\ i,j\;\textrm \end}\left( p \frac+p(\delta _-\delta _)\right) . \end$$

(133)

Setting \(i=2\,m\), \(j=2n\), and \(k=2r\) we find

$$\begin \Delta (\}\}^e)=p^\frac\left( \frac-1\right) }\prod _\left( n-m+(\delta _-\delta _)\right) . \end$$

(134)

Let us emphasize that \(n>1\) since \(n>m\ge 1\). For \(k=2\),

$$\begin \Delta (\}\}^e)=p^\frac\left( \frac-1\right) }\prod _\left( n-m+\delta _\right) =p^\frac\left( \frac-1\right) }\frac\prod _^-1}i!.\qquad \end$$

(135)

For \(k> 2\), k even, since the factor for \(m=r-1\) and \(n=r\) in the product is zero,

$$\begin \Delta (\}\}^e)=0. \end$$

(136)

It is also true that \(\}\}^o=\}\}^e\}\), then \(\Delta (\}\}^o)=\Delta (\}\}^e)\), therefore

$$\begin \Delta (\}\}^o)=p^\frac\left( \frac-1\right) }\frac\prod _^-1}i!. \end$$

(137)

For k odd with \(k>2\), since \(\Delta (\}\}^o)=\Delta (\}\}^e)\),

$$\begin \Delta (\}\}^o)=0. \end$$

(138)

Using (131), (132), (135), (136), (137) and (138) we get

$$\begin \Delta (\}\}^e)\Delta (\}\}^o)= p^\left( \frac-1\right) }\frac\prod _^-1} (m!)^2 & \text k=1,2 \\ 0 & \text k>2 \end\right. }. \end$$

(139)

The consequence of (139) in the sum in (123) is that only the terms for \(k=1\) and \(k=2\) are nonzero. For \(k=1\) or \(k=2\), the double factorial terms in (124) is

$$\begin \prod _i (h_i^e-1)!! h_i^o!! =\frac +1)!!}-1\right) +1\right) !!}\prod _^} \left( p \left( \frac-m\right) +1\right) !!^2. \end$$

(140)

The size of the representation is

$$\begin n^=\sum _ih^_i-\frac=p+\frac\frac\left( p(N-2)+6\right) -\frac. \nonumber \\ \end$$

(141)

This way the character of M in the representation \(\}\}\) is, for \(k=1\) or \(k=2\),

$$\begin \chi _}\}}(M) & =c_N\,N^}}\;p^\left( \frac-1\right) }\frac\prod _^-1}(m!)^2\frac +1)!!}-1\right) +1\right) !!}\nonumber \\ & \quad \prod _^}\left( p\left( \frac-m\right) +1\right) !!^2, \end$$

(142)

and zero for \(k\ge 3\). Hence, inserting (130) and (142) in (123), we find that

$$\begin \textrm(M^p) =N^}\frac+1\right) !!}-1\right) +1\right) !!}. \end$$

(143)

We evaluate a large N limit by keeping only the largest order in N,

$$\begin \textrm(M^p) = N^}\prod _^\left( p\left( \frac-1\right) +1+2j\right) = N\left( \frac\right) ^}\left( 1+(N^)\right) . \nonumber \\ \end$$

(144)

Setting the integer \(q=p/2\), we get the trace property

$$\begin \frac\textrm(M^) = q^q. \end$$

(145)

\(\square \)

In principle, this information should be enough to find M up to matrix conjugation, similarly to what is done in (30) for the matrix \(C_m\). We leave this possibility of computation of M to future studies.

Let us now evaluate \(\langle \chi _r(A^2)\rangle _0\) for a finite N. In particular, in Theorem 9.3, the integral over matrices has been evaluated and replaced with a summation over integers.

Theorem 9.3

Let A be a random variable for a \(N \times N\) Hermitian matrix under the Gaussian measure. Given a representation r of \(\textrm(N)\) defined through the set of shifted weights \( h_i \), \(i=1,...N,\) and considering \(\chi _r\) the character in that representation r the following holds true:

$$\begin \langle \chi _r(A^2)\rangle _0 =\frac}}^k!}\frac(2h_i)!}}\underset}\sum _ k+l=2h_i\\ u+v=2h_j \end}\frac}\frac. \nonumber \\ \end$$

(146)

Proof

For a Hermitian matrix A of size N whose eigenvalues are denoted by x, we write X as a diagonal matrix whose diagonal elements are the eigenvalues x’s. For a given representation r, we are interested in:

(147)

where

$$\begin }}}_N=\frac\frac (U(N))}=N^}(2\pi )^}\left( \prod _^k!\right) ^. \end$$

(148)

See [120, 121], and [122] for the change of variables from A to X. We use de Bruijn’s formula [123],

(149)

so we can reduce the integration over the N variables on the left hand side to a Pfaffian of an integral over two variables on the right hand side. \(d\mu (x)\) sets the measure on x. Here, it is a Gaussian measure \(d\mu (x)=e^x^2}\). Since [124]

$$\begin \prod _\frac = \underset} \Bigg (\frac \Bigg ) , \end$$

(150)

let us first rewrite (147),

(151)

An important thing to notice is that when \(x_i+x_j=0\), the Pfaffian has a divergence that is controlled by the zero in the determinant. However, when we use the de Bruijn’s formula the determinant is removed. Therefore, it is best to deal with this divergence already here. We introduce a damping with a small constant \(\epsilon \) to prevent a divergence from appearing. We take care of this by multiplying the elements in the Pfaffian by the term \(\frac\). We regularize (151) by defining, for \(\epsilon >0\),

(152)

We expand the Pfaffian in (152), a polynomial, into its monomials. We do the same for the Pfaffian in (151). By comparing the absolute values of these integrands, term by term, we see that the ones from (152) are bounded by the ones from (151). Therefore, the dominated convergence theorem [125] tells us that

$$\begin \langle \chi _r(A^2)\rangle _0=\lim _\langle \chi _r(A^2)\rangle _\epsilon . \end$$

(153)

Now we can use the de Bruijn’s formula in (149) and we obtain the damped average

$$\begin \langle \chi _r(A^2)\rangle _\epsilon = }}}_N \; \underset}\int _}^2} dxdy\; \fracx^y^ \; e^(x^2+y^2)}. \end$$

(154)

We remark that at this point, if we take the \(\epsilon \rightarrow 0\) limit, the integral in (154) turns into a principal value integral. Let us define

$$\begin T_=\int _}^2} dxdy\; \fracx^y^ e^(x^2+y^2)}. \end$$

(155)

Then, according to (154),

$$\begin \langle \chi _r(A^2)\rangle _\epsilon = }}}_N \underset}\;T_. \end$$

(156)

We can simplify the N dependence by changing variables \(x,y\rightarrow N^}x,N^}y\),

$$\begin T_=\frac}\int _}^2} dxdy\; \fracx^y^ e^(x^2+y^2)}. \end$$

(157)

Introducing sources \(\alpha \) and \(\beta \) through the terms \(\alpha x\) and \(\beta y\) in the exponential, we can turn the factors \(x^\) and \(y^\) into derivatives:

$$\begin T_=\frac}\left. \frac}}\frac}}\int _}^2} dxdy\; \frac e^(x^2+y^2)+\alpha x+\beta y}\right| _. \nonumber \\ \end$$

(158)

Let us also define

$$\begin \textrm(\alpha ,\beta )=\int _}^2} dxdy\; \frac e^(x^2+y^2)+\alpha x+\beta y}, \end$$

(159)

thus

$$\begin T_ = \frac}\left. \frac}}\frac}} \textrm(\alpha ,\beta )\right| _. \end$$

(160)

By changing integration variables to \(u=(x+y)/\sqrt\) and \(v=(x-y)/\sqrt\), and also defining \(a=(\alpha +\beta )/\sqrt\) and \(b=(\alpha -\beta )/\sqrt\), it becomes

$$\begin \textrm(\alpha ,\beta )=\int _}^2} dudv\; \frac e^(u^2+v^2)+au+bv}. \end$$

(161)

Here we notice that the integration over u and over v are independent, hence we can separate them,

$$\begin \textrm(\alpha ,\beta )=\int _}} du\;u\;e^u^2+au} \int _}} dv\; \frac e^v^2+bv} = \textrm^(a) \textrm^_(b), \nonumber \\ \end$$

(162)

where we define the integrals

$$\begin \textrm^(a) = \int _}} du\;u\;e^u^2+au}\qquad \textrm \qquad \textrm^_\epsilon (b)=\int _}} dv\; \frac e^v^2+bv}.\nonumber \\ \end$$

(163)

The first integral is easily evaluated as

$$\begin \textrm^(a)=\sqrt\,a\;e^a^2}. \end$$

(164)

The second integral we solve by introducing another integral,

$$\begin \textrm^_\epsilon (b)=\int _0^b d}}\int _}} dv\; \frac e^v^2+}}}v}. \end$$

(165)

At this point we can evaluate the \(\epsilon \rightarrow 0\) limit due to the dominated convergence theorem. Thus, by defining

$$\begin \textrm^_0(b)=\lim _ \textrm^_\epsilon (b), \end$$

(166)

we find that

$$\begin \textrm^_0(b)=\int _0^b d}}\int _}} dv\; e^v^2+}}}v}. \end$$

(167)

The integral over v in (167) is a simple Gaussian, and we can evaluate it to find

$$\begin \textrm^_0(b) =\sqrt \int _0^bd}}}\; e^}}}^2}. \end$$

(168)

This function is, up to normalization conventions, the imaginary error function. Joining (164) and (168) in (162), we deduce that

$$\begin \textrm(\alpha ,\beta )= 2\pi \,a\;e^a^2}\int _0^bd}\; e^}}}^2}. \end$$

(169)

Going back through (160) and (156), we obtain

$$\begin \langle \chi _r(A^2)\rangle _0 = \left. }_N\underset}\frac}\frac}}\frac}}\,a \, e^a^2}\int _0^b d}\; e^}^2}\right| _,\nonumber \\ \end$$

(170)

and by using that

$$\begin a \, e^a^2}=\sum _^\infty \frac}\qquad \textrm\qquad \int _0^b d}\; e^}}}^2}=\sum _^\infty \frac} \end$$

(171)

and that \(a=(\alpha +\beta )/\sqrt\) and \(b=(\alpha -\beta )/\sqrt\), we can evaluate the derivatives in (170) to find that

$$\begin \langle \chi _r(A^2)\rangle _0 =}_N\underset}\frac}\frac}\sum _ k+l=2h_i\\ u+v=2h_j\\ k+u\mathrm \end}(-1)^u\frac,\nonumber \\ \end$$

(172)

which becomes (146) by using (148), Pfaffian properties and that \((-1)^k=-(-1)^u\) for \(k+u\) odd and \((-1)^k=(-1)^u\) for \(k+u\) even. \(\square \)

We wish to remark here about the quantity \( \langle \chi _r(A^2)\rangle _0 \) which we computed by taking the limit of \(\epsilon \rightarrow 0\) in the expression \( \langle \chi _r(A^2)\rangle _\epsilon \) given in (156). For a given (in other words, finite) N, the expression we obtained in (172) is valid and therefore Theorem 9.3. However, if we pay attention the expression (157), we notice that N comes with \(\epsilon \). Then, one notices that once we send N to infinity, this procedure becomes sensitive to the ratio in which \(N \rightarrow \infty \) and \(\epsilon \rightarrow 0\) are sent.

One naturally wonders if the expression obtained in Theorem 9.3 may become simpler in large N limit. Let us explore this possibility in Proposition 9.4.

Proposition 9.4

Let A be a random variable for a \(N \times N\) Hermitian matrix under the Gaussian measure. Given a representation r of \(\textrm(N)\) defined through the set of normalized shifted weights \(}}}_i = h_i/N\), \(i=1,...N,\). Consider \(\chi _r\) the character in that representation r. Defining

(173)

which satisfies \( \langle \chi _r(A^2)\rangle _0=\lim _\langle \chi _r(A^2)\rangle _\epsilon \), the following holds true:

$$\begin \lim _ \frac}}}_N\prod _k 2e^}_k}(2 }}}_k)^}}}_k} \underset\left[ \frac}}}_i- }}}_j)( }}}_i+ }_j+\epsilon ^2/2)}}}}_i- }}}_j)^2+\epsilon ^2( }}}_i+ }}}_j)+\epsilon ^4/4}\right] } = 1, \end$$

(174)

where \(}_N=N^}(2\pi )^}(\prod _^k!)^\).

Proof

We apply the saddle point method to compute (154). Let us first rescale integers \(h_i\) to \(}}}_i = h_i/N\);

$$\begin \langle \chi _r(A^2)\rangle _\epsilon = }}}_N\; \underset}\int _}^2} dxdy\; \fracx^}}}_i}y^}_j} \; e^(x^2+y^2)} \nonumber \\ \end$$

(175)

and prepare in a form proper to use the saddle point approximation,

$$\begin \langle \chi _r(A^2)\rangle _\epsilon = }}}_N \; \textrm \int _}^2} dxdy\; \frac \; e^x^2+\fracy^2-2 }}}_i\textrm|x|-2 }}}_j\textrm|y|)}. \nonumber \\ \end$$

(176)

Laplace’s method of integration [126] can be expressed as

$$\begin \int _}^d} dX \, g(X) \, e^ = \sum _ \left( \frac\right) ^ \frac}\big (H(f)(X_0)\big )}} (1+}}(N^)) ,\nonumber \\ \end$$

(177)

where X is a set of d real variables, f is a twice-differentiable complex valued function of X, H(f) is the Hessian matrix of f, the points \(X_0\) are local maxima of f, and g is a complex valued function of X nonzero at \(X_0\). Comparing (176) and (177) we identify

$$\begin f(X) = \fracx^2+\fracy^2-2 }}}_i\textrm|x|-2 }}}_j\textrm|y| \quad \textrm \quad g(X) = \frac.\nonumber \\ \end$$

(178)

Computing the saddle point equations \(x-2 }}}_i x^=0\) and \(y-2 }}}_j y^=0\), we find four saddle points:

$$\begin x=\pm \sqrt}}}_i}\quad \textrm\quad y=\pm \sqrt}}}_j}. \end$$

(179)

Therefore, we see that the term associated with f is the same for any \(X_0\) and is

$$\begin e^=(2 }}}_i)^}}}_i}(2 }}}_j)^}}}_j}e^}}}_i+ }}}_j)}, \end$$

(180)

and therefore, for the Hessian,

$$\begin H(f)(X_0)= \begin 2&0\\ 0&2 \end\; \quad \mathrm} \quad \sqrt\big (H(f)(X_0)\big )}=2. \end$$

(181)

Additionally,

$$\begin \sum _g(X_0) & =2\frac}}}_i-2 }}}_j}}}}_i}+\sqrt}}}_j})^2+\epsilon ^2}+2\frac}_i-2 }}}_j}}}}_i}-\sqrt}_j})^2+\epsilon ^2} \nonumber \\ & =4\frac}}}_i- }}}_j)( }}}_i+ }}}_j+\epsilon ^2/2)}}}}_i- }_j)^2+\epsilon ^2( }}}_i+ }}}_j)+\epsilon ^4/4} . \end$$

(182)

Joining everything, we obtain the saddle point approximate for the regularized character of \(A^2\) as

$$\begin \langle \chi _r(A^2)\rangle _\epsilon & = }}}_N \underset\left[ \frace^}}}_i+ }}}_j)}(2 }}}_i)^}}}_i}(2 }}}_j)^}_j}4\frac}}}_i- }}}_j)( }}}_i+ }_j+\epsilon ^2/2)}}}}_i-}}}_j)^2+\epsilon ^2(}_i+}}}_j)+\epsilon ^4/4}\right] \nonumber \\ & \quad \times (1+}}(N^)). \end$$

(183)

Using some Pfaffian properties we can simplify the expression (183) and get

$$\begin \langle \chi _r(A^2)\rangle _\epsilon= & }_N2^}\prod _k e^}}}_k}(2 }}}_k)^}}}_k} \underset\left[ \frac}}}_i- }}}_j)( }}}_i+ }}}_j+\epsilon ^2/2)}}_i- }}}_j)^2+\epsilon ^2( }}}_i+ }_j)+\epsilon ^4/4}\right] \nonumber \\ & (1+}}(N^)). \end$$

(184)

Applying the limit \(N\rightarrow \infty \), we find

$$\begin \lim _ \frac}}}_N2^}\prod _k e^}}}_k}(2 }}}_k)^}}}_k} \underset\left[ \frac}}}_i- }}}_j)( }}}_i+ }}}_j+\epsilon ^2/2)}}}}_i- }_j)^2+\epsilon ^2( }}}_i+ }_j)+\epsilon ^4/4}\right] } = 1. \end$$

(185)

\(\square \)

One may wish to apply \(\epsilon \rightarrow 0\) limit to (185) and if the limits \(\epsilon \rightarrow 0\) and \(N \rightarrow \infty \) commute, then, (185) can be manipulated to say that in the large N limit, \(\) is equal to

$$\begin }_N2^}\prod _k e^}}}_k}(2 }}}_k)^}}}_k} \underset\left[ \frac}}}_i+ }}}_j}}}}_i- }}}_j}\right] } . \end$$

(186)

We computed \(\langle \chi _r(A^2)\rangle _0\) using (146) and (186) for \(1 \le N \le 30\) (we should note here that if we use integral form for \(\langle \chi _r(A^2)\rangle _0\), then even \(N=6\) the computation becomes slow) using Mathematica for trivial, defining, and determinant representations. However, for the above values of N that we tested, the values of the expression (186) do not converge to the values computed using the expression achieved in (146).

Now, we present first a new way of computing \(\langle \chi _R(A)\rangle _0\) below in Proposition 9.5, whose similar technique is used to compute \(\langle \chi _R(A^2)\rangle _0\) in Theorem 9.6.

Proposition 9.5

Let A be an \(N \times N\) Hermitian matrix under the Gaussian measure. Given a representation R of \(\textrm(N)\) defined through the set of shifted weights \(h_i\), \(i=1,...N,\) and considering \(\chi _R\) the character in the representation R, the following holds true:

$$\begin \langle \chi _R(A)\rangle _0 & = \frac)\chi _R(C_2)}\nonumber \\ & = (-1)^\frac\left( \frac-1\right) }N^}\frac^i!}\frac(h_i^o-h_j^e)}, \end$$

(187)

where the numbers h are separated in a set of \(\lceil N/2\rceil \) even numbers \(h^e\) and \(\lfloor N/2 \rfloor \) odd numbers \(h^o\). If such a separation is not possible, then the average is 0.

Proof

For \(A\in \textrm(N)\), using character orthogonality for the symmetric group \(S_n\), we can write the character of A as

$$\begin \chi _R(A)= & \sum _ \delta _\chi _r(A) = \sum _ \Big ( \sum _\frac}_R(\sigma )\chi _r(\sigma ) \Big ) \chi _r(A) , \end$$

(188)

where the bar on \(}_R(\sigma )\) denotes complex conjugate. Interchanging the sums and using Schur-Weil duality,

$$\begin \sum _ \chi _r(\sigma )\chi _r(A)=\textrm(\sigma A^) , \end$$

(189)

we obtain

$$\begin \chi _R(A) =\sum _\frac}_R(\sigma )\textrm(\sigma A^) . \end$$

(190)

Now, we take the average of the above quantity. Wick’s probability theorem tells us that

$$\begin \langle A^ \rangle _0 = N^}\sum _}]}\gamma , \end$$

(191)

where \([2^\frac]\) is the conjugacy class of permutations in \(S_n\) with n/2 2-cycles. Therefore we obtain

(192)

where we used (189) in the last equality, with \(A =}\). Again using orthogonality relations, rewrite

$$\begin \sum _\frac}_R(\sigma )\chi _r(\sigma \gamma )=\delta _\frac\chi _R(\gamma ) , \end$$

(193)

where we denote the dimension of the \(S_n\) representation \(s_R=\chi _R(\textrm)\), where \(\textrm\) is the identity permutation, and

$$\begin \langle \chi _R(A) \rangle _0= & \sum _}]}\sum _N^}\delta _\frac\chi _R(\gamma ) \; d_r = N^}\frac\sum _}]}\chi _R(\gamma ).\nonumber \\ \end$$

(194)

In order to perform \(\sum _}]}\), we use the following trick. Let us first return to the relation (190) for some matrix M,

$$\begin \chi _R(M) =\sum _\frac}_R(\sigma )\textrm(\sigma M^). \end$$

(195)

We observe that (195) sums over all elements in \(S_n\), whereas (194) sums over a subset of \(S_n\). We aim to extract (194) from (195). We choose M such that the summation in \(S_n\) is restricted to elements of \([2^}]\), achieved by \(\textrm(\sigma M^)= 0\) for \(\sigma \) not in \([2^\frac]\) and constant for when \(\sigma \) is in \([2^\frac]\). Using the notation \([\prod _k k^]\) for the cycle \([\sigma ]\) which \(\sigma \) belongs to, we wish to find M such that

$$\begin \textrm(\sigma M^)= a \;\delta _}]}=a\, \delta _}\prod _ \delta _ . \end$$

(196)

for some constant a (that might depend on N or n but not on \(\sigma \)). But also, we can write

$$\begin \textrm(\sigma M^)=\prod _k \textrm(M^k)^ , \end$$

(197)

leading us to the identification, for \(k\ne 2\),

$$\begin \textrm(M^k)^= \delta _ . \end$$

(198)

Hence, the possibility of a nonzero \(c_k\) with a general permutation \(\sigma \) tells us that

$$\begin \textrm(M^k)=0\; \end$$

(199)

for \(k\ne 2\). Then, recalling

$$\begin \textrm(C_2^k)=N\delta _ , \end$$

(200)

we conclude \(M=C_2\) with \(a=N^\frac\) is a possible solution. Then, setting \(M=C_2\) in (195) with (197), we find that

$$\begin \chi _R(C_2) =\sum _\frac}_R(\sigma )\prod _k \textrm(C_2^k)^. \end$$

(201)

Finally, using (