On the Correlation Functions of the Characteristic Polynomials of the Sparse Hermitian Random Matrices

Abstract

We consider asymptotics of the correlation functions of characteristic polynomials corresponding to random weighted $G(n, \frac{p}{n})$ Erdős–Rényi graphs with Gaussian weights in the case of finite p and also when $p \rightarrow \infty$. It is shown that for finite p the second correlation function demonstrates a kind of transition: when $p < 2$ it factorizes in the limit $n \rightarrow \infty$, while for $p > 2$ there appears an interval $(-\lambda _*(p), \lambda _*(p))$ such that for $\lambda _0 \in (-\lambda _*(p), \lambda _*(p))$ the second correlation function behaves like that for Gaussian unitary ensemble (GUE), while for $\lambda _0$ outside the interval the second correlation function is still factorized. For $p \rightarrow \infty$ there is also a threshold in the behavior of the second correlation function near $\lambda _0 = \pm 2$: for $p \ll n^{2/3}$ the second correlation function factorizes, whereas for $p \gg n^{2/3}$ it behaves like that for GUE. For any rate of $p \rightarrow \infty$ the asymptotics of correlation functions of any even order for $\lambda _0 \in (-2, 2)$ coincide with that for GUE.

Appendix

1.1 Grassmann Variables

Consider the set of formal variables $\{\psi _j,\, \overline{\psi }_j\}_{j=1}^n$ which satisfy the anticommutation relations

$$\begin{aligned} \psi _j \psi _k + \psi _k \psi _j = \overline{\psi }_j \psi _k + \psi _k \overline{\psi }_j = \overline{\psi }_j \overline{\psi }_k + \overline{\psi }_k \overline{\psi }_j = 0. \end{aligned}$$

This set generates a graded algebra $\mathcal {A}$, which is called the Grassmann algebra. Taking into account that $\psi _j^2 = \overline{\psi }_j^2 = 0$, we have that all elements of $\mathcal {A}$ are polynomials of $\{\psi _j,\, \overline{\psi }_j\}_{j=1}^n$. We can also define functions of Grassmann variables. Let $\chi$ be an element of $\mathcal {A}$ and f be any analytical function. By $f(\chi )$ we mean the element of $\mathcal {A}$ obtained by substituting $\chi - z_0$ in the Taylor series of f near $z_0$, where $z_0$ is a free term of $\chi$. Since $\chi - z_0$ is a polynomial of $\{\psi _j,\, \overline{\psi }_j\}_{j=1}^n$ with zero free term, there exists $l \in \mathbb {N}$ such that $(\chi - z_0)^l = 0$, and hence the series terminates after a finite number of terms.

The integral over the Grassmann variables is a linear functional, defined on the basis by the relations

$$\begin{aligned} \int d\psi _j=\int d\overline{\psi }_k = 0, \qquad \int \psi _j d\psi _j=\int \overline{\psi }_k d\overline{\psi }_k = 1. \end{aligned}$$

A multiple integral is defined to be the repeated integral. Moreover “differentials” $\{d\psi _j,\, d\overline{\psi }_j\}_{j=1}^n$ anticommute with each other and with $\{\psi _j,\, \overline{\psi }_j\}_{j=1}^n$. Hence for a function f

$$\begin{aligned} f(\psi _1, \ldots , \psi _n) = a_0 + \sum \limits _{j=1}^n a_j \psi _j + \ldots + a_{1, \ldots , n} \prod \limits _{j=1}^n \psi _j \end{aligned}$$

we have by definition

$$\begin{aligned} \int f(\psi _1, \ldots , \psi _n)d\psi _n \ldots d\psi _1 = a_{1, \ldots , n}. \end{aligned}$$

The use of Grassmann variables for computing averages of determinants rests on the following identity, valid for any $n \times n$ matrix A:

$$\begin{aligned} \int \exp \bigg \{ -\sum \limits _{j,k=1}^n \overline{\psi }_jA_{jk}\psi _k \bigg \} \prod \limits _{j=1}^n d\overline{\psi }_j d\psi _j = \det {A}. \end{aligned}$$

(6.1)

The one more identity is the Hubbard-Stratonovich transformation

$$\begin{aligned} e^{y^2} = \frac{a}{\sqrt{\pi }}\int e^{2axy - a^2x^2} dx,\nonumber \\ e^{yt} = \frac{a^2}{\pi }\int e^{ay(u + iv) + at(u - iv) - a^2u^2 - a^2v^2} du dv. \end{aligned}$$

(6.2)

which valid for any complex numbers y, t and any positive number a. The identities (6.2) also hold when y, t are arbitrary even Grassmann variables (i.e. sums of the products of even number of Grassmann variables). For even Grassmann variables the formulas can be proved by Taylor-expanding $e^{2axy}$ and $e^{ay(u + iv) + at(u - iv)}$ into the series and integrating each term.

The properties explained so far suffice to obtain the integral representation for $F_{2m}$, $m = 1$, whereas the general case $m > 1$ requires some additional preliminaries, pertaining to antisymmetric tensor products. Further details about antisymmetric tensor products may be found in [30, Chapter 8.4].

1.1.1 Grassmann Variables and the Exterior Product

The exterior product of vectors is well-known, as well as the exterior product of alternating multilinear forms (see [30]). However, to prove Proposition 1 we need the exterior product of alternating operators. Define it as following. Let A be a linear operator on $\Lambda ^q \mathbb {C}^n$ and B be a linear operator on $\Lambda ^r \mathbb {C}^n$. Then the exterior product $A \wedge B$ is the restriction of the linear operator ${\text {Alt}} \circ (A \otimes B)$ on the $\Lambda ^{q + r} \mathbb {C}^n$. Here ${\text {Alt}}$ is the operator of the alternation, i.e.,

$$\begin{aligned} {\text {Alt}}(t) = \frac{1}{k!}\sum \limits _{\pi \in S_k} {\text {sgn}} \pi f_\pi (t), \quad t \in \Lambda ^k V, \end{aligned}$$

where $S_k$ is the group of permutations of length k; ${\text {sgn}} \pi$ is the sign of permutation $\pi$; $f_\pi$ is the canonical automorphism of $V^{\otimes k}$, which carries $v_1 \otimes \ldots \otimes v_k$ to $v_{\pi (1)} \otimes \ldots \otimes v_{\pi (k)}$, $v_j \in V$; V is some finite-dimensional linear space. Note, that for $A \in {{\mathrm{End}}}V$ the exterior product $A \wedge A$ coincides with the well-known second exterior power of linear operator A.

Fix some basis $\{ e_j \}_{j = 1}^n$ of $\mathbb {C}^n$. Let $A \in {{\mathrm{End}}}\Lambda ^k \mathbb {C}^n$ and $\alpha , \beta \in I_{n,k}$, where $I_{n, k}$ is defined in (2.2). By $A_{\alpha \beta }$ we denote the corresponding entry of the matrix of A in the basis $\{e_{\alpha _1} \wedge \ldots \wedge e_{\alpha _k},\, \alpha \in I_{n,k}\}$.

To obtain the integral representation for $F_{2m}$ we use the lemma:

Lemma 6

Let A and B be linear operators on $\Lambda ^q \mathbb {C}^n$ and $\Lambda ^r \mathbb {C}^n$ respectively. Then

$$\begin{aligned} \sum \limits _{\alpha , \beta \in I_{n, q}} A_{\alpha \beta } \prod \limits _{j = 1}^q \overline{\psi }_{\alpha _j}\psi _{\beta _j} \cdot \sum \limits _{\gamma , \delta \in I_{n, r}} B_{\gamma \delta } \prod \limits _{j = 1}^r \overline{\psi }_{\gamma _j}\psi _{\delta _j} \\ = \left( {\begin{array}{c}q + r\\ q\end{array}}\right) \sum \limits _{\alpha , \beta \in I_{n, q + r}}(A \wedge B)_{\alpha \beta } \prod \limits _{j = 1}^{q + r} \overline{\psi }_{\alpha _j}\psi _{\beta _j}. \end{aligned}$$

Proof

Let $S_{q, r}$ be the set of such $\pi \in S_{q + r}$ that satisfy inequalities $\pi (1) < \ldots < \pi (q)$ and $\pi (q + 1) < \ldots < \pi (q + r)$. Then

$$\begin{aligned}&\sum \limits _{\alpha , \beta \in I_{n, q}} A_{\alpha \beta } \prod \limits _{j = 1}^q \overline{\psi }_{\alpha _j}\psi _{\beta _j} \cdot \sum \limits _{\gamma , \delta \in I_{n, r}} B_{\gamma \delta } \prod \limits _{j = 1}^r \overline{\psi }_{\gamma _j}\psi _{\delta _j} \\&\quad = \sum \limits _{\alpha , \beta \in I_{n, q + r}}\sum \limits _{\pi , \sigma \in S_{q, r}} {\text {sgn}} \pi {\text {sgn}} \sigma A_{\alpha _{\pi }' \beta _{\sigma }'} B_{\alpha _{\pi }'' \beta _{\sigma }''} \prod \limits _{j = 1}^{q + r} \overline{\psi }_{\alpha _j}\psi _{\beta _j}, \end{aligned}$$

where

$$\begin{aligned} \alpha _{\pi }&= (\alpha _{\pi (1)}, \ldots , \alpha _{\pi (q + r)}), \\ \alpha '&= (\alpha _{1}, \ldots , \alpha _{q}) \in I_{n,q}, \\ \alpha ''&= (\alpha _{q + 1}, \ldots , \alpha _{q + r}) \in I_{n,r}. \end{aligned}$$

On the other hand,

$$\begin{aligned} (A \otimes B)(e_{\beta '} \wedge e_{\beta ''})= & {} (A \otimes B)\bigg (\frac{1}{(q + r)!} \sum \limits _{\sigma \in S_{q + r}} {\text {sgn}} \sigma f_\sigma (e_{\beta '} \otimes e_{\beta ''})\bigg ) \\= & {} \frac{q!r!}{(q + r)!} (A \otimes B)\bigg ( \sum \limits _{\sigma \in S_{q,r}} {\text {sgn}} \sigma f_\sigma (e_{\beta '} \otimes e_{\beta ''})\bigg ) \\= & {} \frac{q!r!}{(q + r)!} \sum \limits _{\begin{array}{c} \alpha \in I_{n,q} \\ \gamma \in I_{n,r} \end{array}} \bigg ( \sum \limits _{\sigma \in S_{q, r}} {\text {sgn}} \sigma A_{\alpha \beta _\sigma '} B_{\gamma \beta _\sigma ''}\bigg ) e_{\alpha } \otimes e_{\gamma }, \end{aligned}$$

where $e_\alpha = e_{\alpha _1} \wedge \ldots \wedge e_{\alpha _q}$, $\alpha \in I_{n, q}$.

Hence,

$$\begin{aligned} {\text {Alt}}((A \otimes B)(e_{\beta '} \wedge e_{\beta ''}))= & {} \frac{q!r!}{(q + r)!} \sum \limits _{\begin{array}{c} \alpha \in I_{n,q} \\ \gamma \in I_{n,r} \end{array}} \bigg ( \sum \limits _{\sigma \in S_{q, r}} {\text {sgn}} \sigma A_{\alpha \beta _\sigma '} B_{\gamma \beta _\sigma ''}\bigg ) e_{\alpha } \wedge e_{\gamma } \\= & {} \frac{q!r!}{(q + r)!} \sum \limits _{\alpha \in I_{n,q + r}} \sum \limits _{\pi , \sigma \in S_{q, r}} {\text {sgn}} \sigma A_{\alpha _\pi ' \beta _\sigma '} B_{\alpha _\pi '' \beta _\sigma ''} e_{\alpha _\pi '} \wedge e_{\alpha _\pi ''} \\= & {} \frac{q!r!}{(q + r)!} \sum \limits _{\alpha \in I_{n,q + r}} \sum \limits _{\pi , \sigma \in S_{q, r}} {\text {sgn}} \pi {\text {sgn}} \sigma A_{\alpha _\pi ' \beta _\sigma '} B_{\alpha _\pi '' \beta _\sigma ''} e_{\alpha '} \wedge e_{\alpha ''}. \end{aligned}$$

Thus,

$$\begin{aligned} (A \wedge B)_{\alpha \beta } = \frac{q!r!}{(q + r)!} \sum \limits _{\pi , \sigma \in S_{q, r}} {\text {sgn}} \pi {\text {sgn}} \sigma A_{\alpha _\pi ' \beta _\sigma '} B_{\alpha _\pi '' \beta _\sigma ''}, \end{aligned}$$

which completes the proof of the lemma. $\square$

We also need some properties of the exterior product of the operators.

Proposition 2

Let $A_j \in {\text {End}} \Lambda ^{q_j} \mathbb {C}^n$, $j = \overline{1, k}$, and $B \in {\text {End}} \mathbb {C}^n$. Then

• (i) $A_1 \wedge A_2 = A_2 \wedge A_1$;

• (ii) $(A_1 \wedge A_2) \wedge A_3 = A_1 \wedge (A_2 \wedge A_3)$;

• (iii) $\bigwedge \limits _{j = 1}^k A_j = \Big ( {\text {Alt}} \circ \bigotimes \limits _{j = 1}^k A_j \Big )\Big |_{\Lambda ^q \mathbb {C}^n}$;

• (iv) $\bigwedge \limits _{j = 1}^k A_jB^{\wedge q_j} = \Big ( \bigwedge \limits _{j = 1}^k A_j \Big )B^{\wedge q}$ and $\bigwedge \limits _{j = 1}^k B^{\wedge q_j}A_j = B^{\wedge q}\Big ( \bigwedge \limits _{j = 1}^k A_j \Big )$;

where $q = \sum \limits _{j = 1}^k q_j$, $B^{\wedge q} = \underbrace{B \wedge \ldots \wedge B}_{q \text { times}}$.

Proof

Assertions (i) and (ii) follow from Lemma 1 and from Grassmann variables multiplication’s anticommutativity and associativity.

(iii) Consider the case $k = 3$.

$$\begin{aligned}&A_1 \wedge A_2 \wedge A_3 = {\text {Alt}} \circ ((I \circ A_1) \otimes ({\text {Alt}} \circ (A_2 \otimes A_3))) \\&\quad = {\text {Alt}} \circ (I \otimes {\text {Alt}}) \circ (A_1 \otimes A_2 \otimes A_3) = {\text {Alt}} \circ (A_1 \otimes A_2 \otimes A_3), \end{aligned}$$

where I is the identity operator.

The general case follows from the induction.

(iv) By definition, we have

$$\begin{aligned} \bigwedge \limits _{j = 1}^k B^{\wedge q_j}A_j = {\text {Alt}} \circ \bigotimes \limits _{j = 1}^k B^{\wedge q_j} A_j. \end{aligned}$$

Consider $({\text {Alt}} \circ B^{\otimes q_j})(v_1 \otimes \ldots \otimes v_{q_j})$, $v_l \in \mathbb {C}^n$

$$\begin{aligned} ({\text {Alt}} \circ B^{\otimes q_j})(v_1 \otimes \ldots \otimes v_{q_j})= & {} \frac{1}{q_j!} \sum \limits _{\pi \in S_{q_j}} {\text {sgn}} \pi f_\pi (Bv_1 \otimes \ldots \otimes Bv_{q_j}) \\= & {} \frac{1}{q_j!} \sum \limits _{\pi \in S_{q_j}} {\text {sgn}} \pi Bv_{\pi (1)} \otimes \ldots \otimes Bv_{\pi (q_j)} \\= & {} \frac{1}{q_j!} \sum \limits _{\pi \in S_{q_j}} {\text {sgn}} \pi B^{\otimes q_j}(f_\pi (v_1 \otimes \ldots \otimes v_{q_j}))\\= & {} (B^{\otimes q_j} \circ {\text {Alt}})(v_1 \otimes \ldots \otimes v_{q_j}). \end{aligned}$$

Therefore, ${\text {Alt}} \circ B^{\otimes q_j} = B^{\otimes q_j} \circ {\text {Alt}}$, in particular, $B^{\wedge q_j} = \left. B^{\otimes q_j} \right| _{\Lambda ^{q_j} \mathbb {C}^n}$. Thus,

$$\begin{aligned} \bigwedge \limits _{j = 1}^k B^{\wedge q_j}A_j = {\text {Alt}} \circ \bigotimes \limits _{j = 1}^k B^{\otimes q_j} A_j&= {\text {Alt}} \circ B^{\otimes q} \circ \bigg (\bigotimes \limits _{j = 1}^k A_j\bigg ) \\&= B^{\otimes q} \circ {\text {Alt}} \circ \bigg (\bigotimes \limits _{j = 1}^k A_j\bigg ) = B^{\wedge q}\bigg ( \bigwedge \limits _{j = 1}^k A_j \bigg ). \end{aligned}$$

The proof of the second formula is similar. $\square$

1.2 The Harish-Chandra/Itsykson–Zuber formula

For computing the integral over the unitary group, the following Harish-Chandra/Itsykson–Zuber formula is used

Proposition 3

Let A be a normal $n \times n$ matrix with distinct eigenvalues $\{a_j\}_{j = 1}^n$ and $B = {{\mathrm{diag}}}\{b_j\}_{j = 1}^n$, $b_j \in \mathbb {R}$. Then

$$\begin{aligned} \int \limits _{U_n} \exp \{z{{\mathrm{tr}}}AU^*BU\}dU_n(U) = \bigg (\prod \limits _{j = 1}^{n - 1} j!\bigg ) \frac{\det \{\exp (za_jb_k)\}_{j,k = 1}^n}{z^{(n^2 - n)/2}\Delta (A)\Delta (B)}, \end{aligned}$$

where z is come constant, $\Delta (A) = \Delta (\{a_j\}_{j = 1}^n)$, $\Delta$ is defined in (2.1). Moreover, for any symmetric domain $\Omega$ and any symmetric function f(B) of $\{b_j\}_{j = 1}^n$

$$\begin{aligned}&\int \limits _{U_n}\int \limits _{\Omega } \exp \{z{{\mathrm{tr}}}AU^*BU\}\Delta ^2(B)f(B)dU_n(U)dB \nonumber \\&\quad = \bigg (\prod \limits _{j = 1}^n j!\bigg ) \frac{z^{-(n^2 - n)/2}}{\Delta (A)}\int \limits _{\Omega } \exp \bigg \{z\sum \limits _{j = 1}^n a_jb_j\bigg \}\Delta (B)f(B) dB, \end{aligned}$$

(6.3)

where $dB = \prod \nolimits _{j = 1}^{n} db_j$.

For the proof see, e.g., [16, Appendix 5].

Remark 2

Notice, that (6.3) is also valid if A has equal eigenvalues.

Indeed, if $a_{j_1} = a_{j_2}$, $j_1 \ne j_2$, the integrand in r.h.s. of (6.3) changes sign when $b_{j_1}$ and $b_{j_2}$ are swapped. Thus, the ratio of the integral and $a_{j_1} - a_{j_2}$ is well defined.