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The renormalization group transformation in the generalized fermionic hierarchical model M. D. Missarov, D. A. Khajrullin Kazan (Volga Region) Federal University
Russian version:
Izvestiya Rossiiskoi Akademii Nauk. Seriya Matematicheskaya, 2023, Volume 87, Issue 5, DOI: https://doi.org/10.4213/im9371 
Bibliographic databases:
    § 1. IntroductionThe renormalization group method is one of the central methods of modern statistical physics and quantum field theory. The main mathematical problem in the rigorous justification of the renormalization group (RG) theory is the non-locality of the transformation of the RG. The latter means that even if the initial Hamiltonian has a simple form and depends on a finite number of coupling constants, the iterations of the RG transformation generate more and more complex interactions of spins in the Hamiltonian, which makes it impossible to mathematically analyze an infinite-dimensional dynamical system. The models of statistical physics on a hierarchical lattice, which were introduced in mathematical physics by Dyson [1], proved useful for rigorous mathematical study of the RG theory. The RG transformation in bosonic hierarchical models is local and is reduced to a non-linear integral operator in the free measure density space [2]. Later, it was shown that, for certain values of the number of points in a unit cell, the hierarchical lattice can be realized as a lattice in $p$-adic space, and $p$-models are continuous versions of hierarchical models (see [3]). Volovich [4] proposed a $p$-adic analog of the string amplitude. The results of first studies in various areas of $p$-adic mathematical physics are presented in the monograph by Vladimirov, Volovich and Zelenov [5]. More recent results are discussed in the survey [6]. The paper [7] gives an overview of research on the fermionic hierarchical model. Unlike the bosonic case, the RG transformation in fermionic hierarchical model can be calculated exactly and is represented as a birational mapping in the two-dimensional space of model coupling constants. Note that the RG in Euclidean models is studied only in lower orders of perturbation theory in the neighbourhood of a Gaussian fixed point. The unique property of the hierarchical fermionic model is that it is capable of describing the global flow of the RG in the entire plane of coupling constants [8]. The explicit description of the properties of the RG within the framework of the hierarchical fermionic model generates a number of non-trivial conjectures for hierarchical and Euclidean bosonic models [9]. A generalization of the fermionic model was proposed in [10]. A two-dimensional lattice is considered, in which the elementary cell is represented by the vertices of a square. In the standard hierarchical model, the distances between the vertices of the square are equal. In the generalized model, the distance between opposite vertices differs from that between neighbouring vertices and is actually a parameter of the new model. In the present paper, we explicitly compute the transformation of the RG in the density space of a free measure and describe some of its properties. Let $T=\{0,1,\dots\}$ and $V^s_k=\{j\in T\colon k\,{\cdot}\, 2^s\leqslant j<(k+1)\,{\cdot}\, 2^s\}$, where $k\in T$, $s\in N=\{1,2,\dots\}$. The hierarchical distance $d_2(i,j)$, $i,j\in T$, $i\ne j$, is defined as $d_2(i,j)=2^{s(i,j)}$, where
$$
\begin{equation*}
s(i,j)=\min\{ s\colon \text{is } k\in T \text{ such that } i,j\in V^s_k\}.
\end{equation*}
\notag
$$
Let $T^2=T\times T$, $k=(k_1,k_2)\in T^2$,
$$
\begin{equation*}
V^s_k=\{(j_1,j_2)\in T^2\colon k_1\cdot 2^s\leqslant j_1<(k_1+1)\cdot 2^s,\, k_2\cdot 2^s\leqslant j_2<(k_2+1)\cdot 2^s\}.
\end{equation*}
\notag
$$
For any $k=(k_1,k_2)\in T^2$, $l=(l_1,l_2)\in T^2$, $k\ne l$, we define $s(k,l)=\max(s(k_1,l_1)$, $s(k_2,l_2))$. The hierarchical distance on $T^2$ is defined as $d_2(k,l)=2^{s(k,l)}$. Consider the 4-component fermionic field
$$
\begin{equation*}
\psi^*(i)=\bigl(\overline\psi_1(i), \psi_1(i),\overline\psi_2(i),\psi_2(i)\bigr),\qquad i\in T^2,
\end{equation*}
\notag
$$
where the components are generators of a Grassman algebra. Let us redenote $V^N_0$ by $\Lambda_N$. Let $\Gamma_N$ be the Grassmann subalgebra generated by $4\,{\cdot}\,4^N$ generators $\overline\psi_1(i)$, $\psi_1(i)$, $\overline\psi_2(i)$, $\psi_2(i)$, $i \in \Lambda_N$. The action of the RG transformation $R(\alpha,\delta)$ on $\psi^*$ is defined by
$$
\begin{equation}
{\psi^*}'(i)\equiv \bigl(r(\alpha)\psi^*\bigr)(i)=2^{-\alpha/2} \sum_{j\in V^1_i}\psi^*(j),
\end{equation}
\tag{1.1}
$$
where $\alpha\in R^1$ is the RG parameter. The Gaussian fermionic field with zero mean and binary correlation function
$$
\begin{equation*}
\langle \psi_{n}(k)\overline\psi_{m}(l)\rangle=\delta_{n,m}b(k,l),\qquad n,m=1,2,\quad k,l\in T^2,
\end{equation*}
\notag
$$
is defined on the whole lattice as a quasistate (expectation value) $\langle\,{\cdot}\, \rangle$ on the algebra of all monomials such that $\langle F\rangle$ for an even degree monomial $F$ is calculated according to the Wick rules and $\langle F\rangle=0$ for any odd degree monomial, $\delta_{n,m}$ is the Kronecker delta. Consider the following functions on $T^2$:
$$
\begin{equation*}
\begin{gathered} \, \begin{alignedat}{2} d(k,l;\lambda) &=d_2(k,l) &\quad &\text{if }\ s(k_1,l_1)\ne s(k_2,l_2), \\ d(k,l;\lambda) &=\lambda d_2(k,l) &\quad &\text{if }\ s(k_1,l_1)=s(k_2,l_2), \end{alignedat} \\ f(k,l;\lambda;\alpha)=d^{\alpha}(k,l;\lambda)\quad \text{if }\ k\ne l,\qquad f(k,k;\lambda;\alpha)=\frac{2+\lambda^{\alpha}}{4(1-2^{-(2+\alpha)})}, \end{gathered}
\end{equation*}
\notag
$$
where $\lambda>0$ is a real parameter. Let $b(k,l;\lambda;\alpha)=f(k,l;\lambda;\alpha-4)$. It was shown in [10] that a zero-mean Gaussian fermionic field with binary correlation function
$$
\begin{equation*}
\langle \psi_{n}(k)\overline\psi_{m}(l)\rangle=\delta_{n,m}b(k,l;\lambda;\alpha),\qquad n,m=1,2,\quad k,l\in T^2,
\end{equation*}
\notag
$$
is invariant under the transformation of the RG with parameter $\alpha$ (see (1.1)). Let us denote the corresponding Gaussian quasi-state as $\rho_0(\lambda;\alpha)$. To construct non-Gaussian states, we will use the Gibbs description of the field. Consider the restriction of the Gaussian field $\psi^*$ to the volume $\Lambda_N$. Let $B_N(\lambda;\alpha)=(b(k,l;\lambda;\alpha))_{k,l\in\Lambda_N}$ be the correlation matrix of this restriction. We also set $H_N(\lambda,\mu;\alpha)=(h_N(k,l;\lambda,\mu;\alpha))_{k,l\in\Lambda_N}$, where
$$
\begin{equation}
h_N(k,l;\lambda,\mu;\alpha) =g(\lambda,\mu;\alpha)h_0(k,l;\mu;\alpha)-C(N;\lambda,\mu;\alpha),
\end{equation}
\tag{1.2}
$$
$g(\lambda,\mu;\alpha)$ and $C(N;\lambda,\mu;\alpha)$ are some normalizing functions [10]. It was proved in [10] that if $\alpha>2$, $\alpha\ne 4$, then, for all $\lambda$ satisfying
$$
\begin{equation}
\biggl(\frac{3\cdot 2^{2-\alpha}}{1-2^{2-\alpha}}\biggr)^{1/(\alpha-4)}<\lambda< \biggl(\frac{4(1-2^{2-\alpha})}{3}\biggr)^{1/(\alpha-4)},
\end{equation}
\tag{1.4}
$$
the matrix $H_N(\lambda,\mu(\lambda);\alpha)$ is the inverse of the matrix $B_N(\lambda;\alpha)$ if
$$
\begin{equation}
\mu(\lambda)=\biggl(\frac{4(1-2^{2-\alpha})-3\lambda^{\alpha-4}} {\lambda^{\alpha-4}(1-2^{2-\alpha}) -3\cdot 2^{2-\alpha}}\biggr)^{-1/\alpha}.
\end{equation}
\tag{1.5}
$$
Let $\lambda$, $\mu$ satisfy (1.4), (1.5). For brevity, we write
$$
\begin{equation*}
\begin{gathered} \, a=g(\lambda,\mu(\lambda);\alpha),\qquad C(N)=C(N;\lambda,\mu(\lambda);\alpha), \\ h_{0,N}(k,l)=af(k,l;\mu(\lambda);-\alpha)-C(N). \end{gathered}
\end{equation*}
\notag
$$
Let $H_{0,N}(\psi^*;\lambda;\alpha)$ be of the form
$$
\begin{equation*}
H_{0,N}(\psi^*;\lambda;\alpha) =\sum_{k,l\in\Lambda_N}h_{0,N}(k,l)\bigl(\overline\psi_1(k)\psi_1(l)+\overline\psi_2(k)\psi_2(l) \bigr).
\end{equation*}
\notag
$$
Consider the restriction of the Gaussian state $\rho_0(\lambda;\alpha)$ to the volume $\Gamma_N$. According to Theorem 1 in [10], for any $F(\psi^*)\in\Gamma_N$, we have
$$
\begin{equation*}
\begin{gathered} \, \rho_0(\lambda;\alpha)(F(\psi^*))= Z^{-1}_N(\lambda;\alpha)\int F(\psi^*) \exp \{-H_{0,N}(\psi^*;\lambda;\alpha)\}\,d\psi^*, \\ d\psi^*=\prod_{i\in\Lambda_N}\,d\psi_1(i) \,d\overline\psi_1(i)\,d\psi_2(i)\,d\overline\psi_2(i), \end{gathered}
\end{equation*}
\notag
$$
where the integration is carried out according to the rules of superanalysis,
$$
\begin{equation*}
Z_N(\lambda;\alpha)=\int\exp\{-H_{0,N}(\psi^*;\lambda;\alpha)\} \,d\psi^*.
\end{equation*}
\notag
$$
Consider the local potential (self-action) for a 4-component fermionic field
$$
\begin{equation*}
L(\psi^*;r,g)=r\bigl(\overline\psi_1\psi_1+\overline\psi_2\psi_2\bigr)+ g\overline\psi_1\psi_1\overline\psi_2\psi_2.
\end{equation*}
\notag
$$
We set $u(\psi^*(i))=\exp\{-L(\psi^*(i);r,g)\}$, and denote by $\rho_N(\lambda;\alpha;u)$ the state on algebra $\Gamma_N$ defined by
$$
\begin{equation*}
\begin{gathered} \, \rho_N(\lambda;\alpha;u)(F)=Z^{-1}_N(\lambda;\alpha;u) \int F\exp\{-H_{0,N} (\psi^*;\lambda;\alpha)\} \prod_{i\in \Lambda_N}u(\psi^*(i))\,d\psi^*, \\ Z_N(\lambda;\alpha;u)=\int \exp\{-H_{0,N}(\psi^*;\lambda;\alpha)\} \prod_{i\in \Lambda_N}u(\psi^*(i))\,d\psi^*. \end{gathered}
\end{equation*}
\notag
$$
If $\rho$ is a state on $\Gamma_N$, then the renormalized state of $\rho'$ is defined on $\Gamma_{N-1}$ by
$$
\begin{equation*}
\rho'\bigl(F({\psi^*}')\bigr)=\rho\bigl(F(r(\alpha)\psi^*)\bigr).
\end{equation*}
\notag
$$
Consider the quadratic form
$$
\begin{equation}
\begin{gathered} \, a\sum_{i,j\in V^1_k} f(i,j;\mu(\lambda);-\alpha)\overline\eta^*(i)\eta^*(j)=Q_k(\eta^*), \qquad k\in\Lambda_{N-1}; \nonumber \\ f(i,j;\mu;-\alpha)= \begin{cases} \dfrac{2+\mu^{-\alpha}}{4(1-2^{\alpha-2})} &\text{if } i= j, \\ 2^{-\alpha} &\text{if } d_2(i,j)=1 \text{ and } s(i_1,j_1)\ne s(i_2,j_2), \\ (2\mu)^{-\alpha} &\text{if } d_2(i,j)=1 \text{ and } s(i_1,j_1)= s(i_2,j_2). \end{cases} \end{gathered}
\end{equation}
\tag{1.6}
$$
According to Theorem 2 in [10],
$$
\begin{equation*}
\rho'_N(\lambda;\alpha;u)=\rho_{N-1}(\lambda;\alpha;{u}'),
\end{equation*}
\notag
$$
where
$$
\begin{equation}
u'({\psi^*}')=\int\delta\biggl(\sum_{i\in V^1_0}\eta^*(i)\biggr)\exp\{-Q_0(\eta^*)\} \prod_{i\in V^1_0}u\bigl(2^{\alpha/2-2}{\psi^*}'+\eta^*(i)\bigr)\, d\eta^*(i).
\end{equation}
\tag{1.7}
$$
We denote by $R(\alpha)u$ the transformation in the density space defined by the right-hand side of (1.7). This formula implies that the RG transformation $R(\alpha)$ is local and independent of $N$. § 2. Calculation of the renormalization group transformationIn this section, instead of regular densities of the form we will use Grassmann-valued densities of the general form
$$
\begin{equation*}
u(\psi^*)=c_0+c_1(\overline\psi_1\psi_1+\overline\psi_2\psi_2) +c_2\overline\psi_1\psi_1\overline\psi_2\psi_2.
\end{equation*}
\notag
$$
The triples $c=(c_0,c_1,c_2)$, which define the densities $u(\psi^*)$, will be considered as points in the two-dimensional real projective space, because two sets that differ by a non-zero factor represent the same Gibbs state. The coefficients of the quadratic form $Q_0$ are as follows:
$$
\begin{equation*}
s_1=a\cdot\frac{2+\mu(\lambda)^{-\alpha}}{4(1-2^{\alpha-2})},\qquad s_2=a\cdot 2^{-\alpha}, \qquad s_3=a\cdot (2\mu)^{-\alpha}.
\end{equation*}
\notag
$$
Let $t_1=2(s_1-s_2)$, $t_2=s_1-s_3$, $t_3=s_1-2s_2+s_3$, $\delta=t_1/t_2$, $\gamma=2^{\alpha - 2}$. Theorem 1. Let $u'(\psi^{*\prime})=R(\alpha)u$. Then Proof. Let the vertices $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$ of the square $V_0^1$ be denoted by $1$, $2$, $3$, $4$, respectively. We also write $\eta^*(1) = \eta^*(0,0)$, $\eta^*(2) = \eta^*(1,0)$, $\eta^*(3)=\eta^*(1,1)$, $\eta^*(4)=\eta^*(0,1)$ for $\eta^*(0,0)$, $\eta^*(1,0)$, $\eta^*(1,1)$, $\eta^*(0,1)$, respectively.
By integrating the Grassmann $\delta$-function
$$
\begin{equation*}
\delta\bigl(\eta^*(1)+\eta^*(2)+\eta^*(3)+\eta^*(4)\bigr),
\end{equation*}
\notag
$$
we will eliminate the 4-component variable $\eta^*(4)$, for which we substitue
$$
\begin{equation}
\eta^*(4)= - \bigl(\eta^*(1)+\eta^*(2)+\eta^*(3)\bigr).
\end{equation}
\tag{2.5}
$$
Now the quadratic form $Q_0$ in (1.7) assumes the form where
$$
\begin{equation*}
\begin{aligned} \, A_k &=t_1\overline\eta_k(1)\eta_k(1)+t_2\overline\eta_k(1)\eta_k(2) +t_3\overline\eta_k(1)\eta_k(3)+2t_2\overline\eta_k(2)\eta_k(2) +t_2\overline\eta_k(2)\eta_k(1) \\ &\qquad +t_2\overline\eta_k(2)\eta_k(3)+t_1\overline\eta_k(3)\eta_k(3)+ t_2\overline\eta_k(3)\eta_k(2)+t_3\overline\eta_k(3)\eta_k(1), \qquad k=1,2. \end{aligned}
\end{equation*}
\notag
$$
Setting $2^{\alpha/2-2}\psi^{*\prime}=z^*$, consider the change of variables In what follows, we set
$$
\begin{equation*}
a_k(0,0)=\overline z_k z_k,\ \ a_k(0,i)=\overline z_k\eta_k(i),\ \ a_k(i,j)=\overline\eta_k(i)\eta_k(j), \quad i,j=1,2,3,\ \ k=1,2.
\end{equation*}
\notag
$$
Now, in the new variables, the forms $A_k$, $k=1,2$, can be written as where
$$
\begin{equation*}
\begin{aligned} \, A_k^1 &=t_1a_k(1,1)-2t_1\bigl(a_k(0,1)+a_k(1,0)\bigr)+ t_2\bigl(a_k(1,2)+a_k(2,1)\bigr) \\ &\qquad+t_3\bigl(a_k(1,3)+a_k(3,1)\bigr), \\ A_k^2 &=\bigl(4(t_1+t_2)\bigr)a_k(0,0)-4t_2\bigl(a_k(0,2)+a_k(2,0)\bigr) -2t_1\bigl(a_k(0,3)+a_k(3,0)\bigr) \\ &\qquad +t_2\bigl(a_k(2,3)+a_k(3,2)\bigr)-2t_1\bigl(a(0,3)+a(3,0)\bigr)+2t_2a(2,2)+t_1a(3,3). \end{aligned}
\end{equation*}
\notag
$$
Note that in the form $A_k^1$ all the terms depend on the variables $\overline\zeta_k(1)$ or $\zeta_k(1)$, while the form $A_k^2$ does not involve such terms. We also set Next, we note that
$$
\begin{equation*}
a_k(i_1,i_2) a_k(i_1,i_3)= a_k(i_2,i_1) a_k(i_3,i_1)=0
\end{equation*}
\notag
$$
for all $i_1$, $i_2$, $i_3$; we also have for all $i_1$, $i_2$, $i_3$, $i_4$. Using these relations, we obtain
$$
\begin{equation*}
\begin{aligned} \, L_k^1 &=1+a_k(1,1)\bigl[-t_1-t_2^2a_k(2,2)-t_3^2a_k(3,3)-t_2t_3\bigl(a_k(2,3)+a_k(3,2)\bigr) \\ &\qquad-4t_1^2a_k(0,0)+ 2t_1t_2\bigl(a_k(0,2)+a_k(2,0)\bigr)+2t_1t_3\bigl(a_k(0,3)+a_k(3,0)\bigr)\bigr] \\ &\qquad -t_2\bigl(a_k(1,2)+a_k(2,1)\bigr)-t_3\bigl(a_k(1,3)+a_k(3,1)\bigr)+2t_1\bigr(a_k(0,1) +a_k(1,0)\bigr), \\ L_k^2 &= 1 - 4(t_1+t_2)a_k(0,0) - 2t_2a_k(2,2) - t_1a_k(3,3)+4t_2\bigl(a_k(0,2)+a_k(2,0)\bigr) \\ &\qquad- t_2\bigl(a_k(2,3)+a_k(3,2)\bigr)+2t_1\bigl(a_k(0,3)+a_k(3,0)\bigr) \\ &\qquad +8t_2(t_1-t_2)a_k(0,0)a_k(2,2) +4t_2(t_2-t_1)a_k(0,0)\bigl(a_k(2,3)+a_k(3,2)\bigr) \\ &\qquad +4t_2(t_2-t_1)a_k(2,2)\bigl(a_k(0,3) +a_k(3,0)\bigr)+ (2t_1t_2-t_1^2)a_k(2,2)a_k(3,3) \\ &\qquad - 4t_2^2(t_1+t_2)a_k(0,0)a_k(2,2)a_k(3,3)+ 4t_1t_2a_k(0,0)a_k(3,3). \end{aligned}
\end{equation*}
\notag
$$
After changing variables (2.5), (2.6), the product of the densities assumes the form
$$
\begin{equation*}
u\bigl(\zeta^*(1)\bigr)u\bigl(\zeta^*(2)\bigr)u\bigl(\zeta^*(3)\bigr)u\bigl(4z^* - \zeta^*(1) - \zeta^*(2) - \zeta^*(3)\bigr).
\end{equation*}
\notag
$$
Consider separately the product of the densities depending on $\zeta^*(1)$ and independent of $\zeta^*(1)$. Let
$$
\begin{equation*}
\begin{aligned} \, I_1 &=u\bigl(\zeta^*(1)\bigr)u\bigl(4z^* - \zeta^*(1) - \zeta^*(2) - \zeta^*(3)\bigr), \\ I_2 &=u\bigl(\zeta^*(2)\bigr)u\bigl(\zeta^*(3)\bigr). \end{aligned}
\end{equation*}
\notag
$$
Let $c_0 \ne 0$, $b_1=c_1/c_0$, $b_2=c_2/c_0$. Then
$$
\begin{equation*}
\begin{aligned} \, u(\zeta^*(1)) &= c_0\bigl[\bigl(1+b_1\overline\zeta_1(1)\zeta_1(1)\bigr)\bigl(1+b_1\overline\zeta_2(1)\zeta_2(1) \bigr){+}\,(b_2 \,{-}\, b_1^2)\overline\zeta_1(1)\zeta_1(1)\overline\zeta_2(1)\zeta_2(1)\bigr] \\ &= c_0\bigl[\bigl(1+b_1a_1(1,1)\bigr)\bigl(1+b_1a_2(1,1)\bigr)+(b_2-b_1^2)a_1(1,1)a_2(1,1) \bigr]. \end{aligned}
\end{equation*}
\notag
$$
A similar analysis shows that
$$
\begin{equation*}
u\bigl(\zeta^*(1)\bigr)u\bigl(4z - \zeta^*(1) - \zeta^*(2) - \zeta^*(3)\bigr)= c_0[(1+b_1T_1)(1+b_2T_2)+(b_2-b_1^2)T_1T_2],
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\begin{aligned} \, T_k &=\bigl(4\overline z^* - \overline\zeta_k(1) - \overline\zeta_k(2) - \overline\zeta_k(3)\bigr)\bigl(4z_k - \zeta_k(1) - \zeta_k(2) - \zeta_k(3)\bigr)=M_k^1+M_k^2, \\ M_k^1 &=a_k(1,1)+a_k(1,2)+a_k(1,3)+a_k(2,1)+a_k(3,1) - 4(a_k(0,1)+a_k(1,0)), \\ M_k^2 &=16a_k(0,0) - 4\bigl(a_k(0,2)+a_k(0,3)+a_k(2,0)+a_k(3,0)\bigr) +a_k(2,2)+a_k(2,3) \\ &\qquad+a_k(3,2) + a_k(3,3)+a_k(3,2)+a_k(3,3) . \end{aligned}
\end{equation*}
\notag
$$
Here, $M_k^1$ involves the contains depending on $\zeta_k(1)$ or $\overline\zeta_k(1)$, and $M_k^2$ contains all other terms, $k=1,2$. So, we have
$$
\begin{equation*}
\begin{aligned} \, I^1 &=c_0^2\bigl[ \bigl(1+b_1a_1(1,1)\bigr)\bigl(1+b_1a_1(1,1)\bigr)+(b_2 - b_1^2)a_1(1,1)a_2(1,1) \bigr] \\ &\qquad\times [(1+b_1T_1)(1+b_1T_2)+(b_2-b_1^2)T_1T_2] \\ &=c_0^2 (P_1^1P_2^1+dP_1^2P_2^2+ dP_1^3P_2^3+d^2P_1^4P_2^4), \end{aligned}
\end{equation*}
\notag
$$
where $d=b_2 - b_1^2$ and
$$
\begin{equation*}
\begin{aligned} \, P_k^1 &=\bigl(1+b_1a_k(1,1)\bigr)(1+b_1T_k)=1+b_1a_k(1,1)+b_1M_k^1+b_1M_k^2 +b_1^2a_k(1,1)M_k^2, \\ P_k^2 &=\bigl(a+b_1a_k(1,1)\bigr)T_k=1+b_1a_k(1,1)+b_1M_k^1+b_1M_k^2+b_1^2a_k(1,1)M_k^2, \\ P_k^3 &=a_k(1,1)(1+b_1T_k)=a_k(1,1)+b_1a_k(1,1)M_k^2, \\ P_k^4 &=a_k(1,1)T_k=a_k(1,1)M_k^2. \end{aligned}
\end{equation*}
\notag
$$
Similarly, one can expand the product
$$
\begin{equation*}
I^2=u\bigl(\zeta^*(2)\bigr)u\bigl(\zeta^*(3)\bigr)=c_0^2(G_1^1G_2^1+dG_1^2G_2^2+dG_1^3G_2^3+ d^2G_1^4G_2^4),
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
\begin{alignedat}{2} G_k^1 &=\bigl(1+b_1a_k(2,2)\bigr)\bigl(1+c_1a_k(3,3)\bigr), &\qquad G_k^2 &=\bigl(1+b_1a_k(2,2)\bigr)a_k(3,3), \\ G_k^3 &=a_k(2,2)\bigl(1+b_1a_k(3,3)\bigr), &\qquad G_k^4 &=a_k(2,2)a_k(3,3). \end{alignedat}
\end{equation*}
\notag
$$
Let
$$
\begin{equation*}
R_k^i=\int L_k^1 P_k^i \,d\zeta_k(1)\, d\overline\zeta_k(1),\qquad k=1,2,\quad i=1,2,3,4.
\end{equation*}
\notag
$$
The expressions $R_k^i$ depend only on the variables $\zeta_k(2)$, $\overline\zeta_k(2)$, $\zeta_k(3)$, $\overline\zeta_k(3)$. Hence
$$
\begin{equation*}
\begin{aligned} \, &\int L_1^1L_2^1I^1\,d\zeta_1(1)\, d\overline\zeta_1(1)\, d\zeta_2(1)\, d\overline\zeta_2(1) \\ &\qquad= c_0^2 (R_1^1R_2^1+dR_1^2R_2^2+dR_1^3R_2^3+d^2R_1^4R_2^4). \end{aligned}
\end{equation*}
\notag
$$
We set
$$
\begin{equation*}
\int R_k^{m} L_k^2 G_k^n \,d\zeta_k(2)\, d\overline\zeta_k(2)\, d\zeta_k(3)\, d\overline\zeta_k(3)=F_k^{n,m}.
\end{equation*}
\notag
$$
For the complete integral, we have
$$
\begin{equation*}
R(\alpha)u =\int L_1^1L_2^1L_1^2L_2^2I^1I^2\prod_{i=1}^3 d\zeta_1(i)\, d\overline\zeta_1(i)\, d\zeta_2(i)\, d\overline\zeta_2(i) =c_0^4\sum_{\substack{0\leqslant i \leqslant4 \\ 0\leqslant j \leqslant4}} d_id_{j}F_1^{m,n}F_2^{m,n},
\end{equation*}
\notag
$$
where Since $t_3=t_1 - t_2$, we have, for the integrals $F_k^{m,n}$,
$$
\begin{equation*}
\begin{aligned} \, F_k^{1,1} &=a_k(0,0)(16c_1^4 - 16c_1^3\delta t_2 - 16c_1^3t_2+32c_1^2\delta t_2^2 - 16c_1^2t_2^2 - 16c_1\delta t_2^3+16c_1t_2^3) \\ &\qquad+4c_1^3 - 4c_1^2\delta t_2 - 4c_1^2t_2+8c_1\delta t_2^2 - 4c_1t_2^2 - 4\delta t_2^3+ 4t_2^3, \\ F_k^{1,2} &=F_k^{1,3}=F_k^{2,1}=F_k^{3,1}=a_k(0,0)(16c_1^3 - 12c_1^2\delta t_2 - 12c_1^2t_2+ 16c_1\delta t_2^2 - 8c_1t_2^2 \\ &\qquad - 4\delta t_2^3+4t_2^3)+ 3c_1^2 - 2c_1\delta t_2 - 2c_1t_2+2\delta t_2^2 - t_2^2, \\ F_k^{1,4} &=F_k^{2,2}=F_k^{3,3}=F_k^{4,1}=a_k(0,0)(16c_1^2 - 8c_1\delta t_2 - 8c_1t_2+ 4\delta t_2^2)+2c_1 - \delta t_2, \\ F_k^{2,3} &=F_k^{3,2}=F_k^{4,4}=a_k(0,0)(16c_1^2 - 8c_1\delta t_2 - 8c_1t_2+8\delta t_2^2 - 8t_2^2)+2c_1 - 2t_2, \\ F_k^{2,4}&=F_k^{3,4}=F_k^{4,2}=F_k^{4,3}=a_k(0,0)(16c_1 - 4\delta t_2 - 4t_2)+1. \end{aligned}
\end{equation*}
\notag
$$
Since the set of coefficients ($c_0$, $c_1$, $c_2$) defining the general density is determined up to a non-zero factor, the RG transformation can be written as
$$
\begin{equation*}
R(\alpha)u=\frac{1}{16} \cdot c_0^4\sum_{\substack{0\leqslant i \leqslant4 \\ 0\leqslant j \leqslant4}} d_id_{j}F_1^{m,n}F_2^{m,n}.
\end{equation*}
\notag
$$
Returning to the coordinates ($c_0$, $c_1$, $c_2$) and since $d=c_2/c_0 - (c_1/c_0)^2$, and
$$
\begin{equation*}
a_k(0,0)=\overline z_kz_k=\bigl(2^{\alpha/2 - 2} \overline\psi_k^{\,\prime}\bigr)(2^{\alpha/2 - 2} \psi_k')= 2^{\alpha - 4}\overline\psi_k^{\,\prime}\psi_k',
\end{equation*}
\notag
$$
we have
$$
\begin{equation*}
u'(\psi^{*\prime})= R(\alpha)u= c_0'+c_1'\bigl(\overline{\psi}^{\,\prime}_1\psi'_1 +\overline{\psi}^{\,\prime}_2\psi'_2\bigr) +c_2'\overline{\psi}^{\,\prime}_1\psi'_1\overline{\psi}^{\,\prime}_2\psi'_2,
\end{equation*}
\notag
$$
where $c_0'$, $c_1'$, $c_2'$ are given by (2.2), (2.3), (2.4), proving the theorem. § 3. Properties of the transformation of the renormalization groupConsider the operator Note that formulas (2.2)–(2.4), which define the mapping, involve the coefficients $t_2,\delta$, which, in turn, depend on $\alpha$. For further purposes, we denote the mapping of the renomalization group $R(\alpha)$ in the $c$-space by
$$
\begin{equation*}
R(\alpha;\delta, t_2)(c)=\bigl(f_0(c; \delta, t_2), f_1(c; \delta, t_2), f_2(c; \delta, t_2)\bigr).
\end{equation*}
\notag
$$
It is easy to see that the mapping of the transformation $R(\alpha;\delta, t_2)$ in the projective space is conjugate to the transformation $R(\alpha;\delta, 1)$. Indeed, from (2.2), (2.3), (2.4) we have
$$
\begin{equation*}
\begin{gathered} \, f_0(c;\delta, t_2)=t_2^6f_0(T(t_2)c; \delta, 1), \qquad f_1(c;\delta,t_2)=t_2^7f_1\bigl(T(t_2)c;\delta,1\bigr), \\ f_2(c, \delta, t_2)=t_2^8f_2\bigl(T(t_2)c; \delta, 1\bigr). \end{gathered}
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
R(\alpha;\delta,t_2)c=t_2^6T^{-1}(t_2)R(\alpha;\delta,1)T(t_2)c.
\end{equation*}
\notag
$$
Since $t_2 \ne 0$, we obtain the conjugacy condition
$$
\begin{equation*}
R(\alpha;t_2,\delta)c=T^{-1}(t_2)R(\alpha;\delta,1)T(t_2)c.
\end{equation*}
\notag
$$
So, in what follows, we will consider the mapping $R(\alpha;\delta, 1)$, which we denote by $R(\alpha;\delta)$. Note that if $\lambda=1$, then $\mu=1$, and in this case $s_2=s_3$, and so $\delta= t_2/t_1=2$, which corresponds to the usual hierarchical lattice. The case $\lambda=1$ corresponds to the standard hierarchical lattice. From the expressions for $f_0$, $f_1$, $f_2$ we can find a common factor $(c_0-2c_1+c_2)^2$, and the transformation $R(\alpha;2)$ coincides with the RG transformation in the fermionic model on the usual hierarchical lattice [10]. Next, consider the representation of our model in the $(r,g)$-coordinates. Let
$$
\begin{equation*}
u(\psi^*)=\exp\bigl\{-r\bigl(\overline\psi_1\psi_1+\overline\psi_2\psi_2\bigr)+ g\overline\psi_1\psi_1\overline\psi_2\psi_2\bigr\}.
\end{equation*}
\notag
$$
In the projective representation, this density is given by the triple of coefficients The transformation of the RG in the plane of coupling constants $(r,g)$ will also be denoted by $R(\alpha; \delta)$: Lemma 1. The RG transformation $R(\alpha;\delta)$ in the $(r,g)$-plane has the form The proof is by a direct calculation with the use of the inverse transformation
$$
\begin{equation*}
r=- \frac{c_1}{c_0},\qquad g=\frac{c_1^2 - c_2c_0}{c_0^2}.
\end{equation*}
\notag
$$
Corollary. The domains $\{(r,g)\colon g >0\}$ (the upper half-plane), $\{(r,g)\colon g < 0\}$ (the lower half-plane) and $\{(r,g)\colon g=0\}$ are invariant domains of the mapping $R(\alpha; \delta)$. The action of $R(\alpha,\delta)$ on the line $\{(r,g):g=0\}$ is linear: Consider the singular points The points $A_0$ and $A_1$ are fixed points of $R(\alpha,\delta)$ for all $\alpha$ and $\delta$. The point $A_2$ is a singular point of the mapping $R(\alpha;\delta)$ for all $\alpha$ and $\delta$: In the $(r,g)$-coordinates, the point $A_0$ is $(0,0)$, and so we say that $A_0$ is a Gaussian fixed point. The point $A_1$ does not lie in the regular plane $\{(r,g)\}$, and to it there corresponds the density $u(\psi^*)=\overline\psi_1\psi\overline\psi_2\psi_2$ (the Grassmann delta function).Theorem 2. The fixed point $A_0$ is repulsive for $\alpha>3$, is a saddle point for $2<\alpha<3$, and is an attracting point for $\alpha<2$. The fixed point $A_1$ is attracting for $\alpha>2$, a saddle point for $1< \alpha< 2$, and a repulsive point for $\alpha<1$. Proof. We set near the point $A_0=(1,0,0)$. In these coordinates, $A_0=(0,0)$. From (2.2)–(2.4) we have
$$
\begin{equation*}
\begin{aligned} \, \! x' &=\frac{c_1'(c_0,c_1,c_2)} {c_0' (c_0,c_1,c_2)}=\frac{c_1' (1,x,y)} {c_0'(1,x,y)} = 2^{\alpha-2} \frac{x(\delta - 1)^2+y(-\delta^2/2+3\delta/4 - 1)+ P_1(x,y)}{(\delta - 1)^2+P_2(x,y)}, \\ y' &=2^{2(\alpha - 2)} \frac{y((\delta -1)^2/4)+P_3(x,y)}{(\delta -1)^2+ P_2(x,y)}. \end{aligned}
\end{equation*}
\notag
$$
The degree of any monomial in the polynomials $P_1(x,y)$ and $P_3(x,y)$ exceeds $1$, and the degree of any monomial in $P_2(x,y)$ is positive. Therefore, the differential of the mapping $R(\alpha;\delta)$ at the point $x=0$, $y=0$ has the form
The eigenvalues of the differential $D_0$ are $\gamma_1=2^{\alpha - 2}$, $\gamma_2= 2^{2\alpha - 6}$. So if $\alpha>3$, then $\gamma_1>1$, $\gamma_2>1$ and the fixed point $A_0$ is repulsive. For $2<\alpha<3$, $A_0$ is a saddle point, and for $\alpha<2$, $A_0$ is an attracting point. The value $\alpha=3$ is a bifurcation value, and, therefore, a new (non-Gaussian) branch of fixed points exists near the point $A_0$. In a neighbourhood of the point $A_1=(0,0,1)$, we introduce the local coordinates $u=c_1/c_2$, $v=c_0/c_2$.
$$
\begin{equation*}
\begin{aligned} \, u' &=\frac{c_1'(c_0,c_1,c_2)}{c_2'(c_0,c_1,c_2)}=\frac{c_1'(v,u,1)} {c_2'(u,v,1)} =2^{-(\alpha-2)} \frac{u+v(-\delta/4 - 1/4)+P_4(u,v)}{1+P_5(u,v)}, \\ v' &=2^{-2(\alpha - 2)} \frac{v/4+P_6(u,v)}{1+P_5(u,v)}. \end{aligned}
\end{equation*}
\notag
$$
The degree of any monomial in the polynomials polynomials $P_4(u,v)$ and $P_6(u,v)$ exceeds 1, and the degree of any monomial in the polynomial $P_5(u,v)$ is positive. Therefore, the differential of the mapping $R(\alpha,\delta)$ at the point $u=0$, $v=0$ has the form
$$
\begin{equation*}
D_1 =\begin{pmatrix} 2^{-(\alpha-2)} & 2^{\alpha-2} \biggl(-\dfrac{\delta}4 - \dfrac14\biggr) \\ 0 & 2^{-2\alpha+2} \end{pmatrix}.
\end{equation*}
\notag
$$
The eigenvalues of the differential $D_1$ are $\lambda_1=2^{-(\alpha - 2)}$, $\lambda_2= 2^{-2\alpha+2}$. Therefore, $A_1$ is attracting for $\alpha>2$, is a saddle fixed point for $ 1<\alpha<2$, and is a repulsive point for $\alpha<1$. The value $\alpha =1$ is a bifurcation value, and, therefore, there is a new branch of fixed points near the point $A_1$. Theorem 2 is proved. 
Citation in format AMSBIB
\Bibitem{MisKha23} |
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