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This article is cited in 2 scientific papers (total in 2 papers) Proper cyclic symmetries of multidimensional continued fractions I. A. Tlyustangelovab a Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia b Moscow Center of Fundamental and Applied Mathematics, Moscow, Russia
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     § 1. IntroductionThe classical concept of a continued fraction associated with a real number admits several generalizations, one of which is based on Klein’s geometric interpretation of a continued fraction (see [1]). Namely, let $l_1,\dots,l_n$ be one-dimensional subspaces of the space $\mathbb{R}^n$ spanning the whole of $\mathbb{R}^n$. Then the hyperplanes spanned by all possible systems of $n-1$ subspaces in this set partition $\mathbb{R}^n$ into $2^n$ simplicial cones. The set of these cones is denoted by A simplicial cone with vertex at the origin $\mathbf{0}$ is called irrational if the linear hull of each facet of this cone contains no integer points other than the origin $\mathbf{0}$. Definition 1. Let $C \in \mathcal{C}(l_1, \dots, l_n)$ be an irrational cone. The convex hull $\mathcal{K}(C)=\operatorname{conv}(C \cap \mathbb{Z}^{n}\setminus\{\mathbf{0}\} )$ and its boundary $\partial(\mathcal{K}(C))$ are called the Klein polyhedron and Klein sail corresponding to the cone $C$, respectively. The union of all $2^n$ sails is called an $(n-1)$-dimensional continued fraction. Lagrange’s classical theorem on continued fractions (see [2] and [3]) asserts that a number $\alpha$ is a quadratic irrationality if and only if the continued fraction of $\alpha$ is eventually periodic. A geometric interpretation of Lagrange’s theorem is based on the fact that a vector $(1, \alpha)$ is an eigenvector of some operator with distinct real eigenvalues in $\operatorname{SL}_{2}(\mathbb{Z})$ if and only if $\alpha$ is a quadratic irrationality (see, for example, [4]). This result extends naturally to arbitrary $n$. Recall that an operator in $\operatorname{GL}_{n}(\mathbb{Z})$ is called hyperbolic if it has only real eigenvalues and its characteristic polynomial is irreducible over $\mathbb{Q}$. Definition 2. Let $l_1,\dots,l_n$ be the eigensubspaces of some hyperbolic operator $A\in\operatorname{GL}_n(\mathbb{Z})$. Then the $(n-1)$-dimensional continued fraction $\operatorname{CF}(l_1,\dots,l_n)$ is said to be algebraic. We also say that this continued fraction is associated with $A$ (and write $\operatorname{CF}(A)=\operatorname{CF}(l_1,\dots,l_n)$). The set of all $(n-1)$-dimensional algebraic continued fractions is denoted by $\mathfrak{A}_{n-1}$. The extension referred to above is as follows (for details, see, for example, [5]). Proposition 1. Numbers $1,\alpha_1,\dots,\alpha_{n-1}$ form a basis for some totally real extension $K$ of the field $\mathbb{Q}$ if and only if $(1,\alpha_1,\dots,\alpha_{n-1})$ is an eigenvector of some hyperbolic operator $A\in\operatorname{SL}_n(\mathbb{Z})$. Moreover, the vectors $(1,\sigma_i(\alpha_1),\dots,\sigma_i(\alpha_{n-1}))$, $i=1,\dots,n$, where $\sigma_1(=\operatorname{id}),\sigma_2,\dots,\sigma_n$ are all the possible embeddings of $K$ into $\mathbb{R}$, form an eigenbasis for $A$. The symmetry group of an algebraic continued fraction $\operatorname{CF}(A)=\operatorname{CF}(l_1,\dots,l_n)$ is defined as the set
$$
\begin{equation*}
\mathrm{Sym}_{\mathbb{Z}}(\operatorname{CF}(A)) =\bigl\{ G\in\operatorname{GL}_n(\mathbb{Z}) \mid G(\operatorname{CF}(A))=\operatorname{CF}(A) \bigr\}.
\end{equation*}
\notag
$$
By continuity, it is clear that for each $G\in\mathrm{Sym}_{\mathbb{Z}}(\operatorname{CF}(A))$ there is a unique permutation $\sigma_G$ such that Conversely, if for $G\in\operatorname{GL}_n(\mathbb{Z})$ there exists a permutation $\sigma_{G}$ satisfying (1.1), then $G\in\mathrm{Sym}_{\mathbb{Z}}(\operatorname{CF}(A))$. Definition 3. An operator $G\in\mathrm{Sym}_{\mathbb{Z}}(\operatorname{CF}(A))$ such that $\sigma_G=\operatorname{id}$ is called a Dirichlet symmetry of the continued fraction $\operatorname{CF}(A)\in\mathfrak{A}_{n-1}$. Definition 4. An operator $G\in\mathrm{Sym}_{\mathbb{Z}}(\operatorname{CF}(A))$ that is not a Dirichlet symmetry is called a palindromic symmetry of $\operatorname{CF}(A)$. If the set of palindromic symmetries of a continued fraction is nonempty, then this continued fraction is said to be palindromic. Definition 5. A symmetry $G\in\mathrm{Sym}_{\mathbb{Z}}(\operatorname{CF}(A))$ is called cyclic if $\sigma_G$ is a cyclic permutation. It is clear that all cyclic symmetries of a continued fraction $\operatorname{CF}(A)$ are palindromic symmetries of it. Definition 6. A palindromic symmetry $G\in\mathrm{Sym}_{\mathbb{Z}}(\operatorname{CF}(A))$ is called proper if $G$ has a fixed point on some sail of $\operatorname{CF}(A)$. An improper palindromic symmetry $G\in\mathrm{Sym}_{\mathbb{Z}}(\operatorname{CF}(A))$ is a palindromic symmetry that is not proper. Let $\partial(\mathcal{K}(C))$ be one of the $2^n$ sails of $\operatorname{CF}(A)$. By Dirichlet’s theorem on algebraic units, all Dirichlet symmetries of the continued fraction $\operatorname{CF}(A)$ form a group, and this group has a subgroup whose action on the sail $\partial(\mathcal{K}(C))$ has a compact fundamental domain (see, for example, [5] and [6]). Thus, one can speak about the period region of a sail. In this paper we are concerned with proper cyclic symmetries of $\operatorname{CF}(A)$. For $n=2$, that is, for one-dimensional continued fractions, palindromicity is directly related to the symmetry of the periods of regular continued fractions for quadratic irrationalities. Given a quadratic irrationality, a criterion for the period of its continued fraction to be symmetric dates back to Galois [7], Legendre [8], Perron [9] and Kraitchik [10]. A geometric proof of this criterion was given in [4], where both proper and improper symmetries appeared. However, for $n=3$ each palindromic continued fraction features a proper cyclic symmetry (see [5]). In [4] and [5] the following criteria for the existence of proper cyclic symmetries of algebraic continued fractions were proved. Proposition 2. Let $\operatorname{CF}(l_1,l_2)\in\mathfrak{A}_1$, and let the subspace $l_1$ be spanned by a vector $(1, \alpha)$. Then $\operatorname{CF}(l_1, l_2)$ features a proper cyclic symmetry if and only if there exists an algebraic number $\omega$ of degree $2$ such that at least one of the following conditions is satisfied: (a) $(1, \alpha) \sim(1,\omega)\colon \operatorname{Tr}(\omega)=\omega+\omega'=0$; (b) $(1, \alpha) \sim(1,\omega)\colon \operatorname{Tr}(\omega)=\omega+\omega'=1$; where $\omega'$ is the conjugate of $\omega$. Proposition 3. Let $\operatorname{CF}(l_1,l_2,l_3)\in\mathfrak{A}_2$, and let the subspace $l_1$ be spanned by a vector $(1, \alpha, \beta)$. Then $\operatorname{CF}(l_1, l_2, l_3)$ features a proper cyclic symmetry if and only if there exists an algebraic number $\omega$ of degree $3$ such that at least one of the following conditions is satisfied: (a) $(1, \alpha, \beta)\sim(1, \omega, \omega')\colon \operatorname{Tr}(\omega)=\omega+\omega'+\omega''=0$; (b) $(1, \alpha, \beta)\sim(1, \omega, \omega')\colon \operatorname{Tr}(\omega)=\omega+\omega'+\omega''=1$; where $\omega'$ and $\omega''$ are the conjugates of $\omega$. If either (a) or (b) holds, then the cubic extension $\mathbb{Q}(\alpha, \beta)$ is normal. In these results, given two vectors $\mathbf v_1, \mathbf v_2\in \mathbb{R}^n$, the relation $\mathbf v_1\sim\mathbf v_2$ means that there exist an operator $X\in\operatorname{GL}_n(\mathbb{Z})$ and a nonzero number $\mu\in\mathbb{R}$ such that $X\mathbf v_1=\mu\mathbf v_2$. § 2. Main resultsThe first result in this paper is the proof that palindromic continued fractions exist in all dimensions. Theorem 1. For each integer $n > 1$ there exists an $(n-1)$-dimensional continued fraction $\operatorname{CF}(A)$ featuring a proper cyclic palindromic symmetry. The second result of this paper generalizes Propositions 2 and 3 to $n=4$. Theorem 2. Let $\operatorname{CF}(l_1,l_2,l_3,l_4)\in\mathfrak{A}_3$, and let the subspace $l_1$ be generated by a vector $(1, \alpha, \beta, \gamma)$. Also let $K=\mathbb{Q}(\alpha, \beta, \gamma)$. Then $\operatorname{CF}(l_1, l_2, l_3, l_4)$ features a proper cyclic symmetry if and only if $K$ is cyclic Galois extension of degree $4$ whose Galois group is generated by an embedding $\sigma$, and there exists an algebraic number $\omega \in K$ of degree $4$ such that at least one of the following conditions is satisfied: Theorems 1 and 2 are proved in § 3 and § 4, respectively, Remark. In dimensions $n=2, 3$ all the palindromic symmetries of a continued fraction $\operatorname{CF}(A)$ are cyclic symmetries of it, and so in Propositions 2 and 3 we can replace the term ‘cyclic’ by ‘palindromic’. But this is not so in dimension $n=4$. However, a complete classification of palindromic symmetries is too bulky for this paper. We address this question in a separate study. § 3. Existence of palindromic symmetries for finite totally real cyclic Galois extensionsHere and in what follows we denote the norm $\operatorname{N}_{\mathbb{Q}(\alpha)/\mathbb{Q}}(\alpha)$ and trace $\operatorname{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}(\alpha)$ of the algebraic number $\alpha$ by $\operatorname{N}(\alpha)$ and $\operatorname{Tr}(\alpha)$, respectively. Given a continued fraction $\operatorname{CF}(l_1, \dots, l_n)=\operatorname{CF}(A)\in\mathfrak{A}_{n-1}$, we assume that the subspace $l_1$ is spanned by a vector $\mathbf l_1=(1,\alpha_1, \dots, \alpha_{n-1})$ (this assumption is correct in view of Proposition 1). Now Proposition 1 implies that $1,\alpha_1, \dots, \alpha_{n-1}$ form a basis for the field ${K=\mathbb{Q}(\alpha_1, \dots, \alpha_{n-1})}$ over $\mathbb{Q}$ and each $l_i$ is spanned by a vector $\mathbf l_i=(1,\sigma_i(\alpha_1), \dots, \sigma_i(\alpha_{n-1}))$, where $\sigma_1(=\operatorname{id}),\sigma_2, \dots, \sigma_n$ are all the embeddings of $K$ into $\mathbb{R}$. Consider the matrix
$$
\begin{equation}
\begin{pmatrix} 0 & \mu_{1} & 0 & \dots & 0 & 0 \\ 0 & 0 & \mu_{2} & \dots & 0 & 0 \\ 0 & 0 & 0 & \dots & 0 & 0 \\ \dots & \dots & \dots & \dots &\dots &\dots \\ 0 & 0 & 0 & \dots & 0 & \mu_{n-1} \\ \mu_{n} & 0 & 0 & \dots & 0 & 0 \end{pmatrix}.
\end{equation}
\tag{3.1}
$$
Lemma 1. Let $G$ be a cyclic symmetry of $\operatorname{CF}(l_1, \dots, l_n) \in\mathfrak{A}_{n-1}$, and let the matrix of the operator $G$ have the form (3.1) in the basis $\mathbf{l}_{1}, \dots, \mathbf{l}_{n}$. Then $G$ is a proper cyclic symmetry of the continued fraction $\operatorname{CF}(l_1, \dots, l_n)$ if and only if $\mu_{1}\mu_{2}\dotsb\mu_{n}= 1$. Proof. Let $G$ be a proper cyclic symmetry of $\operatorname{CF}(l_1, \dots, l_n)$. Then there exist numbers $\varepsilon_{1}, \varepsilon_{2}, \dots, \varepsilon_{n}\in \{-1, 1\}$ such that and Consequently, $\mu_{1}\mu_{2}\dotsb\mu_{n} > 0$, and so $\mu_{1}\mu_{2}\dotsb\mu_{n} =1$.
If $\mu_{1}\mu_{2}\dotsb\mu_{n}=1$, then the operator $G$ has an eigendirection corresponding to eigenvalue $1$ and lying inside some cone $C \in \mathcal{C}(l_1, \dots, l_n)$. This completes the proof. Denote $\mathfrak{A}_{n-1}'$ denote the set of all $(n-1)$-dimensional algebraic continued fractions for which the field $K$ in Proposition 1 is a totally real cyclic Galois extension. Let $\sigma$ be a generator of the Galois group $\operatorname{Gal}(K/\mathbb{Q})$. We also assume that the lines $l_1, \dots, l_n$ are labelled so that if $(\mathbf l_1,\mathbf l_2, \dots, \mathbf l_{n-1}, \mathbf l_{n} )$ is the matrix with columns $\mathbf l_1,\mathbf l_2, \dots, \mathbf l_{n-1}, \mathbf l_{n}$, then
$$
\begin{equation*}
(\mathbf l_1,\mathbf l_2, \dots, \mathbf l_{n-1}, \mathbf l_{n} )= \begin{pmatrix} 1 & 1 & \dots & 1 & 1 \\ \alpha_1 & \sigma(\alpha_1) & \dots & \sigma^{n-2}(\alpha_1) & \sigma^{n-1}(\alpha_1) \\ \alpha_2 & \sigma(\alpha_2) & \dots & \sigma^{n-2}(\alpha_2) & \sigma^{n-1}(\alpha_2) \\ \dots & \dots & \dots & \dots & \dots \\ \alpha_{n-1} & \sigma(\alpha_{n-1}) & \dots & \sigma^{n-2}(\alpha_{n-1}) & \sigma^{n-1}(\alpha_{n-1}) \end{pmatrix}.
\end{equation*}
\notag
$$
Consider the class of $(n-1)$-dimensional algebraic continued fractions
$$
\begin{equation*}
\mathbf{CF}=\biggl\{ \operatorname{CF}(l_1, \dots, l_n)\in\mathfrak{A}_{n-1}' \Bigm|\alpha_{j}=\prod_{k=0}^{j-1}\sigma^{k}(\alpha_1), \,\operatorname{N}(\alpha_1)=1\biggr\}.
\end{equation*}
\notag
$$
We set
$$
\begin{equation*}
H= \begin{pmatrix} 0 & 1 & 0 & \dots & 0 & 0 \\ 0 & 0 & 1 & \dots & 0 & 0 \\ 0 & 0 & 0 & \dots & 0 & 0 \\ \dots & \dots & \dots & \dots &\dots &\dots \\ 0 & 0 & 0 & \dots & 0 & 1 \\ 1 & 0 & 0 & \dots & 0 & 0 \end{pmatrix}.
\end{equation*}
\notag
$$
Lemma 2. Let $\operatorname{CF}(l_1, \dots, l_n)\in\mathfrak{A}_{n-1}$. Then the following two assertions are equivalent: (a) $\operatorname{CF}(l_{1}, \dots, l_{n})$ lies in the class $\mathbf{CF}$; (b) $H$ is a proper palindromic symmetry of $\operatorname{CF}(l_1, \dots, l_n)$ and Proof. By Lemma 1 the operator $H\in\operatorname{GL}_n(\mathbb{Z})$ is a proper palindromic symmetry of $\operatorname{CF}(A)$ and $\sigma_{H}=(1, 2, \dots, n-1, n)$ if and only if there exist real numbers $\mu_1,\mu_2, \dots,\mu_n$ such that $\mu_1\mu_2 \dotsb \mu_n=1$ and
Let $\operatorname{CF}(l_{1}, \dots, l_{n}) \in \mathbf{CF}$. Then
$$
\begin{equation*}
\begin{aligned} \, &(\mathbf l_{1},\mathbf l_{2}, \dots, \mathbf l_{n-1}, \mathbf l_{n}) \\ &= { \begin{pmatrix} 1& 1 & \dots & 1 & 1 \\ \alpha_1 & \sigma(\alpha_1) & \dots& \sigma^{n-2}(\alpha_1) & \sigma^{n-1}(\alpha_1) \\ \alpha_1\sigma(\alpha_1) & \sigma(\alpha_1)\sigma^{2}(\alpha_1) & \dots& \sigma^{n-2}(\alpha_1)\sigma^{n-1}(\alpha_1) & \sigma^{n-1}(\alpha_1)\alpha_1 \\ \dots & \dots& \dots & \dots & \dots \\ \displaystyle\prod_{k=1}^{j-1}\sigma^{k-1}(\alpha_1) &\displaystyle \prod_{k=2}^{(j-1)+1}\sigma^{k-1}(\alpha_1) & \dots &\displaystyle \prod_{k=n-1}^{(j-1)+ n-2}\sigma^{k-1}(\alpha_1) &\displaystyle \prod_{k=n}^{(j-1)+n- 1}\sigma^{k-1}(\alpha_1) \\ \dots & \dots & \dots & \dots & \dots \end{pmatrix}. } \end{aligned}
\end{equation*}
\notag
$$
In other words, we have
$$
\begin{equation*}
(\mathbf l_{1},\mathbf l_{2}, \dots, \mathbf l_{n-1}, \mathbf l_{n})=(a_{ji}),
\end{equation*}
\notag
$$
where $a_{ji}=\prod_{k=i}^{(j-1)+i-1}\sigma^{k-1}(\alpha_1)$ for $j=2, \dots, n$ and $a_{1i}=1$ for any $i=1, \dots, n$. Consequently,
$$
\begin{equation*}
H(\mathbf l_{1},\mathbf l_{2}, \dots, \mathbf l_{n-1}, \mathbf l_{n}) =\bigl(\alpha_1\mathbf{l}_2, \sigma(\alpha_1)\mathbf{l}_3, \dots, \sigma^{n-2}(\alpha_1)\mathbf{l}_{n}, \sigma^{n-1}(\alpha_1)\mathbf{l}_1\bigr).
\end{equation*}
\notag
$$
Hence $H$ is a proper palindromic symmetry of the continued fraction $\operatorname{CF}(l_1, \dots, l_n)$ and $\sigma_{H}=(1, 2, \dots, n-1, n)$.
Conversely, assume that $H$ is a proper palindromic symmetry of the continued fraction $\operatorname{CF}(l_1, \dots, l_n)$ and $\sigma_{H}=(1, 2, \dots, n-1, n)$. In view of (3.2) we have
$$
\begin{equation*}
H\mathbf{l}_1=\begin{pmatrix} \alpha_1 \\ \alpha_2 \\ \dots \\ \alpha_{n-1} \\ 1 \end{pmatrix}=\mu_2\begin{pmatrix} 1 \\ \sigma_{2}(\alpha_1) \\ \dots \\ \sigma_{2}(\alpha_{n-2}) \\ \sigma_{2}(\alpha_{n-1}) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\begin{gathered} \, \mu_2=\alpha_1, \qquad\alpha_2=\alpha_1\sigma_{2}(\alpha_1), \qquad\alpha_3=\alpha_1\sigma_{2}(\alpha_2)=\alpha_1\sigma_{2}(\alpha_1)\sigma^{2}_{2}(\alpha_1), \qquad \dots, \\ \alpha_{n-1}=\alpha_1\sigma_{2}(\alpha_1)\cdots\sigma^{n-2}_{2}(\alpha_1), \qquad 1= \alpha_1\sigma_{2}(\alpha_1)\cdots\sigma^{n-1}_{2}(\alpha_1). \end{gathered}
\end{equation*}
\notag
$$
For any $i=2, \dots, n-1$, from (3.2) we obtain
$$
\begin{equation*}
H\mathbf{l}_i=\begin{pmatrix} \sigma_{i}(\alpha_1) \\ \sigma_{i}(\alpha_2) \\ \dots \\ \sigma_{i}(\alpha_{n-1}) \\ 1 \end{pmatrix}=\mu_{i+1}\begin{pmatrix} 1 \\ \sigma_{i+1}(\alpha_1) \\ \dots \\ \sigma_{i+1}(\alpha_{n-2}) \\ \sigma_{i+1}(\alpha_{n-1}) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us $\mu_{i+1}=\sigma_{i}(\alpha_1)$ and
$$
\begin{equation*}
\begin{gathered} \, \sigma_{i+1}(\alpha_1)= \frac{\sigma_{i}(\alpha_2)}{\sigma_{i}(\alpha_1)}=\sigma_{i}(\sigma_{2}(\alpha_1)), \\ \sigma_{i+1}(\alpha_2)= \frac{\sigma_{i}(\alpha_3)}{\sigma_{i}(\alpha_1)}=\sigma_{i}(\sigma_{2}(\alpha_2)), \\ \dots \\ \sigma_{i+1}(\alpha_{n-1})= \frac{1}{\sigma_{i}(\alpha_1)}=\sigma_{i}(\sigma_{2}(\alpha_{n-1})). \end{gathered}
\end{equation*}
\notag
$$
Using induction we conclude that
$$
\begin{equation*}
\sigma_{i+1}(\alpha_1)= \sigma_{2}^{i}(\alpha_1), \qquad\sigma_{i+1}(\alpha_2)= \sigma_{2}^{i}(\alpha_2), \qquad \dots,\qquad \sigma_{i+1}(\alpha_{n-1})= \sigma_{2}^{i}(\alpha_{n-1}).
\end{equation*}
\notag
$$
Hence $\operatorname{CF}(l_1, \dots, l_n) \in \mathbf{CF}$, since the numbers $1,\alpha_1, \dots, \alpha_{n-1}$ form a basis for the field $K$.
Lemma 2 is proved. Proof of Theorem 1. First we show that there exists a finite totally real cyclic Galois extension of degree $n$ of the field $\mathbb{Q}$. By Dirichlet’s theorem on primes in arithmetic progressions there exists a prime number $p$ such that $p\equiv 1\pmod {2n}$. Let $\zeta_{p}$ be a $p$th root of unity and let $E= \mathbb{Q}(\zeta_{p})$.
Note that $\operatorname{Gal}(E/\mathbb{Q})= \mathbb{Z}/(p-1)\mathbb{Z}$. Consider the field $K_0=\mathbb{Q}(\zeta_{p}+\zeta^{-1}_{p})$. This field is a finite totally real extension of $\mathbb{Q}$ of degree $(p-1)/2$ (see [11]). We also have $[E:K_0]=2$, since $x^2-(\zeta_{p}+\zeta^{-1}_{p})x+1$ is a minimal polynomial for $\zeta_{p}$ over $K_0$. The group $\operatorname{Gal}(E/\mathbb{Q})$ is cyclic, hence all of its subgroups are normal; moreover, all the quotient groups of $\operatorname{Gal}(E/\mathbb{Q})$ by subgroups of $\operatorname{Gal}(E/\mathbb{Q})$ are cyclic. Therefore, by the main theorem of Galois theory $K_0$ is a cyclic Galois extension of $\mathbb{Q}$. Since $n$ divides $(p-1)/2$, the cyclic group $\operatorname{Gal}(K_0/\mathbb{Q})$ contains a subgroup $F$ of index $n$. Let $K=K_{0}^{F}$. Another application of the main theorem of Galois theory shows that $K$ is a cyclic Galois extension of $\mathbb{Q}$ and $[K:\mathbb{Q}]=[\operatorname{Gal}(K_{0}/\mathbb{Q}):F]=n$. Moreover, $K \subset K_{0}$ is a totally real extension of $\mathbb{Q}$. Let $K$ be a finite totally real cyclic Galois extension of degree $n$ of $\mathbb{Q}$, and let $\sigma$ be a generator of the Galois group of this extension. By the normal basis theorem there exist numbers
$$
\begin{equation*}
\omega,\ \ \sigma(\omega), \ \ \dots, \ \ \sigma^{n-1}(\omega)
\end{equation*}
\notag
$$
that form a basis for the extension $K$. Then $1,{\sigma(\omega)}/{\omega}, \dots,{\sigma^{n-1}(\omega)}/{\omega}$ also form a basis for $K$. Now an appeal to Proposition 1 shows that the vector $(1, {\sigma(\omega)}/{\omega}, \dots, {\sigma^{n-1}(\omega)}/{\omega})$ is an eigenvector of some hyperbolic operator $A\in\operatorname{SL}_n(\mathbb{Z})$. Note that, for any $j=2, \dots, n-1$,
$$
\begin{equation*}
\frac{\sigma(\omega)}{\omega}\sigma\biggl(\frac{\sigma(\omega)}{\omega}\biggr) \sigma^{2}\biggl(\frac{\sigma(\omega)}{\omega}\biggr)\cdots\sigma^{j-1} \biggl(\frac{\sigma(\omega)}{\omega}\biggr) =\frac{\sigma^{j}(\omega)}{\omega},
\end{equation*}
\notag
$$
and, in addition, $\operatorname{N}(\sigma(\omega)/\omega)=1$. So we have $\operatorname{CF}(A) \in \mathbf{CF}$. Now it suffices to invoke Lemma 2 to complete the proof.
Theorem 1 is proved. § 4. Palindromic symmetries in the case $n=4$In what follows we assume that $n=4$, that is, we deal with three-dimensional continued fractions. Recall that the set of all three-dimensional algebraic continued fractions is denoted by $\mathfrak{A}_3$. Next, as in § 3, we assume that, given a continued fraction $\operatorname{CF}(l_1,l_2,l_3, l_4)\in\mathfrak{A}_3$, the subspace $l_i$, $i=1,2,3,4$, is spanned by a vector $\mathbf l_i$ whose first coordinate is $1$. Let $G$ be a cyclic symmetry of $\operatorname{CF}(l_1,l_2,l_3, l_4)\in\mathfrak{A}_3$. Relabelling the subspaces $l_1$, $l_2$, $l_3$ and $l_4$ if necessary, we can take real numbers $\mu_{1}$, $\mu_{2}$, $\mu_{3}$ and $\mu_{4}$ such that in the basis $\mathbf{l}_{1}$, $\mathbf{l}_{2}$, $\mathbf{l}_{3}$, $\mathbf{l}_{4}$ the matrix of the operator $G$ has the form
$$
\begin{equation}
\begin{pmatrix} 0 & 0 & 0 & \mu_{1} \\ \mu_{2} & 0 & 0 & 0 \\ 0 & \mu_{3} & 0 & 0 \\ 0 & 0 & \mu_{4} & 0 \end{pmatrix}.
\end{equation}
\tag{4.1}
$$
For $n=4$ Lemma 1 (for the relabelled subspaces $l_1$, $l_2$, $l_3$ and $l_4$) has the following form. Corollary. Let $G$ be a cyclic symmetry of a continued fraction $\operatorname{CF}(l_1,l_2,l_3, l_4)\in\mathfrak{A}_3$ and let the matrix of the operator $G$ in the basis $\mathbf{l}_{1}$, $\mathbf{l}_{2}$, $\mathbf{l}_{3}$, $\mathbf{l}_{4}$ have the form (4.1). Then $G$ is a proper cyclic symmetry of $\operatorname{CF}(l_1,l_2, l_3, l_4)$ if and only if $\mu_{1}\mu_{2}\mu_{3}\mu_{4}=1$. Lemma 3. Let the operator $G$ be a proper cyclic symmetry of a continued fraction $\operatorname{CF}(l_1,l_2,l_3, l_4)\in\mathfrak{A}_3$. Then $G$ has eigenvalues $1$, $-1$, $i$ and $-i$. Further, the eigensubspaces $l_{+}$ (corresponding to eigenvalue $1$), $l_{-}$ (corresponding to eigenvalue $-1$) and $L$ (corresponding to eigenvalues $i$ and $-i$) are rational. In particular, the subspace $L$ does not contain one-dimensional eigensubspaces of $G$ and, moreover, $G^2({\mathbf v})=-\mathbf v$ for any $\mathbf v \in L$. Proof. Relabelling the subspaces $l_1$, $l_2$, $l_3$ and $l_4$ if necessary and taking the above corollary into account, we can assume that there exist real numbers $\mu_{1}$, $\mu_{2}$, $\mu_{3}$ and $\mu_{4}$ such that $\mu_{1}\mu_{2}\mu_{3}\mu_{4}=1$ and the matrix of the operator $G$ has the form (4.1) in the basis $\mathbf{l}_{1}$, $\mathbf{l}_{2}$, $\mathbf{l}_{3}$, $\mathbf{l}_{4}$. Since $\chi_{G}(x)=x^4-\mu_{1}\mu_{2}\mu_{3}\mu_{4}$, the eigenvalues of $G$ are $1$, $-1$, $i$ and $-i$. Hence $G$ has precisely two rational one-dimensional eigensubspaces $l_{+}$ and $l_{-}$ and one two-dimensional invariant subspace $L$, which does not contain one-dimensional eigensubspaces of $G$. We claim that the subspace $L$ is rational.
Since $l_{-}+l_{+}+L=\mathbb{R}^{4}$, for any vector $\mathbf v \in \mathbb{R}^{4}$ there exist unique vectors $\mathbf{p}(\mathbf v, l_{-})\in l_{-},\mathbf{p}(\mathbf v, l_{+}) \in l_{+}$ and $\mathbf{p}(\mathbf v, L) \in L$ such that
$$
\begin{equation*}
\mathbf v=\mathbf{p}(\mathbf v, l_{-})+\mathbf{p}(\mathbf v, l_{+})+\mathbf{p}(\mathbf v, L).
\end{equation*}
\notag
$$
Note that $\mathbf{p}(G^{2}(\mathbf{v}), L)=\mathbf{p}(-\mathbf v, L)$, $\mathbf{p}(G^{2}(\mathbf{v}), l_{-})= \mathbf{p}(\mathbf v, l_{-})$ and $\mathbf{p}(G^{2}(\mathbf{v}), l_{+})=\mathbf{p}(\mathbf v, l_{+})$ for any vector $\mathbf v \in \mathbb{R}^{4}$. So, for each $\mathbf{z} \in \mathbb{Z}^{4} \setminus (l_{+} +l_{-})$ the nonzero integer vectors $\mathbf{z}-G^{2}(\mathbf{z})$ and $G(\mathbf{z})-G^{3}(\mathbf{z})$ lie in the two-dimensional subspace $L$. These two integer vectors are noncollinear, because $G(\mathbf{z})-G^{3}(\mathbf{z})= G(\mathbf{z}-G^{2}(\mathbf{z}))$ and $L$ does not contain one-dimensional eigensubspaces of $G$. Thus we have shown that the subspace $L$ is rational.
This proves Lemma 3. Lemma 4. Let $G$ be a proper cyclic symmetry of a continued fraction $\operatorname{CF}(l_1,l_2, l_3, l_4)\in\mathfrak{A}_3$. Then there exist $\mathbf{z}_1,\mathbf{z}_2,\mathbf{z}_3,\mathbf{z}_4 \in \mathbb{Z}^4$ such that and at least one of the following seven assertions holds: 1) the vectors $\mathbf{z}_{1}$, $\mathbf{z}_{2}$, $\mathbf{z}_{3}$ and $\frac{1}{4}(\mathbf{z}_{1}+\mathbf{z}_{2}+\mathbf{z}_{3}+\mathbf{z}_{4})$ form a basis of the lattice $\mathbb{Z}^4$; 2) the vectors $\mathbf{z}_{1}$, $\mathbf{z}_{2}$, $\mathbf{z}_{3}$ and $\mathbf{z}_{4}$ form a basis of $\mathbb{Z}^4$; 3) the vectors $\mathbf{z}_{1}$, $\frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{2})$, $\frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{3})$ and $\frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{4})$ form a basis of $\mathbb{Z}^4$; 4) the vectors $\mathbf{z}_{1}$, $\mathbf{z}_{2}$, $\frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{3})$ and $\frac{1}{4}(\mathbf{z}_{1}+\mathbf{z}_{2}+\mathbf{z}_{3}+\mathbf{z}_{4})$ form a basis of $\mathbb{Z}^4$; 5) the vectors $\mathbf{z}_{1}$, $\mathbf{z}_{2}$, $\frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{3})$ and $\frac{1}{2}(\mathbf{z}_{2}+\mathbf{z}_{4})$ form a basis of $\mathbb{Z}^4$; 6) the vectors $\mathbf{z}_{1}$, $\mathbf{z}_{2}$, $\mathbf{z}_{3}$ and $\frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{3}+\mathbf{z}_{4}-\mathbf{z}_{2})$ form a basis of $\mathbb{Z}^4$; 7) the vectors $\mathbf{z}_{1}$, $\mathbf{z}_{2}$, $\mathbf{z}_{3}$ and $\frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{2}) +\frac{1}{4}(\mathbf{z}_{1}+\mathbf{z}_{4}-\mathbf{z}_{3}- \mathbf{z}_{2})$ form a basis of $\mathbb{Z}^4$. Proof. A plane will be said to be rational if the set of integer points contained in this plane is an (affine) lattice of rank equal to the dimension of this plane.
Consider the subspaces $l_{+}$, $l_{-}$ and $L$ from Lemma 3 and set $S= l_{-}+L$. Let $S_1$ be a rational hyperplane parallel to $S$ and distinct from $S$ which is closest to $S$ (we choose one of the two such planes). Then $G(S_1)=S_1$. Let $\mathbf{p}$ be the intersection point of the hyperplanes $S_1$ and $l_{+}$. Next, let $l$ and $\pi$ be, respectively, the straight line and the plane passing through the point $\mathbf{p}$ parallel to $l_{-}$ and $L$. By Lemma 3 we have $G(\mathbf{p})=\mathbf{p}$ and $G(\mathbf{v}-\mathbf{p})=\mathbf{p}-\mathbf{v}$ for any vector $\mathbf v \in l$ and $G^{2}(\mathbf{v}-\mathbf{p})=\mathbf{p}-\mathbf{v}$ for any $\mathbf v \in \pi$. The subspace $L$ does not contain one-dimensional eigensubspaces of $G$. Hence, for any point $\mathbf{v} \in \pi \setminus l$ the quadrilateral $\mathrm{conv}(\mathbf{v}, G(\mathbf{v}), G^{2}(\mathbf{v}), G^{3}(\mathbf{v}))$ is a parallelogram whose diagonals intersect at the point $\mathbf{p}=\frac{1}{2}(\mathbf{v}+G^{2}(\mathbf{v}))=\frac{1}{2}(G(\mathbf{v})+G^{3}(\mathbf{v}))$. Let $Q$ be a rational plane closest to $\pi$ that lies in the hyperplane $S_1$ and is parallel to $\pi$ and distinct from $\pi$. We have $G(\mathbf{v}-\mathbf{p})=\mathbf{p}-\mathbf{v}$ for any vector $\mathbf v \in l$. Hence $R=G(Q)$ and $Q$ are the rational planes closest to $\pi$ that are equidistant from $p$, parallel to $\pi$, distinct from $\pi$, and lie in the hyperplane $S_1$ on different sides of $\pi$, We set $\mathbf{p}^{Q}=Q \cap l$ and $\mathbf{p}^{R}=R \cap l$. Let us construct the points $\mathbf{z}_{1}$, $\mathbf{z}_{2}$, $\mathbf{z}_{3}$ and $\mathbf{z}_{4}$ via the following iterative procedure. Consider an arbitrary integer point $\mathbf{v}_{1,1} \in Q \setminus l$. Set $\mathbf{v}_{1,2}=G(\mathbf{v}_{1,1})$, $\mathbf{v}_{1,3}=G^{2}(\mathbf{v}_{1,1})$ and $\mathbf{v}_{1,4}=G^{3}(\mathbf{v}_{1,1})$. Let $\mathbf{v}^{\pi}_{1,1}$, $\mathbf{v}^{\pi}_{1,2}$, $\mathbf{v}^{\pi}_{1,3}$ and $\mathbf{v}^{\pi}_{1,4}$ be the projections of the points $\mathbf{v}_{1,1}$, $\mathbf{v}_{1,2}$, $\mathbf{v}_{1,3}$ and $\mathbf{v}_{1,4}$, respectively, onto the plane $\pi$ parallel to $l$. We also denote by $\mathbf{v}^{R}_{1,1}$ and $\mathbf{v}^{R}_{1,3}$ the projections of the points $\mathbf{v}_{1,1}$ and $\mathbf{v}_{1,3}$, respectively, onto the plane $R$ parallel to $l$, and we denote by $\mathbf{v}^{Q}_{1,2}$ and $\mathbf{v}^{Q}_{1,4}$ the projections of the points $\mathbf{v}_{1,2}$ and $\mathbf{v}_{1,4}$ onto the plane $Q$ parallel to $l$. By the above, the set $\Delta^{\pi}_{1}=\mathrm{conv}(\mathbf{v}^{\pi}_{1,1}, \mathbf{v}^{\pi}_{1,2}, \mathbf{v}^{\pi}_{1,3}, \mathbf{v}^{\pi}_{1,4})$ is a parallelogram whose diagonals intersect at the point $\mathbf{p}=\frac{1}{4}(\mathbf{v}_{1,1} + \mathbf{v}_{1,2} + \mathbf{v}_{1,3} + \mathbf{v}_{1,4})$. So $\Delta^{Q}_{1}=\mathrm{conv}(\mathbf{v}_{1,1}, \mathbf{v}^{Q}_{1,2}, \mathbf{v}_{1,3}, \mathbf{v}^{Q}_{1,4})$ and $\Delta^{R}_{1}= \mathrm{conv}(\mathbf{v}^{R}_{1,1}, \mathbf{v}_{1,2}, \mathbf{v}^{R}_{1,3}, \mathbf{v}_{1,4})$ are also parallelograms. Note that $\mathbf{p}^{Q}=\frac{1}{2}(\mathbf{v}_{1,1}+\mathbf{v}_{1,3})$ and $\mathbf{p}^{R}=\frac{1}{2}(\mathbf{v}_{1,2}+\mathbf{v}_{1,4})$. Assume that we have constructed the parallelograms $\Delta^{\pi}_{j}$, $\Delta^{Q}_{j}$ and $\Delta^{R}_{j}$. If the planes $Q$ and $R$ contain an integer point distinct from $\mathbf{p}^{Q}$, $\mathbf{p}^{R}$ and the vertices of $\Delta^{Q}_{j}$ and $\Delta^{R}_{j}$ that lies in one of these parallelograms (we assume without loss of generality that this point lies inside $\Delta^{Q}_{j}$), then we denote it by $\mathbf{v}_{j+1,1}$. We set $\mathbf{v}_{j+1,2} = G(\mathbf{v}_{j+1,1})$, $\mathbf{v}_{j+1,3} = G^{2}(\mathbf{v}_{j+1,1})$ and $\mathbf{v}_{j+1,4} = G^{3}(\mathbf{v}_{j+1,1})$. Let $\mathbf{v}^{\pi}_{j+1,1}$, $\mathbf{v}^{\pi}_{j+1,2}$, $\mathbf{v}^{\pi}_{j+1,3}$ and $\mathbf{v}^{\pi}_{j+1,4}$ be the projections of the points $\mathbf{v}_{j+1,1}$, $\mathbf{v}_{j+1,2}$, $\mathbf{v}_{j+1,3}$ and $\mathbf{v}_{j+1,4}$, respectively, onto the plane $\pi$ parallel to $l$. We also denote by $\mathbf{v}^{R}_{j+1,1}$ and $\mathbf{v}^{R}_{j+1,3}$ the projections of the points $\mathbf{v}_{j+1,1}$ and $\mathbf{v}_{j+1,3}$, respectively, onto the plane $R$ parallel to $l$, and by $\mathbf{v}^{Q}_{j+1,2}$ and $\mathbf{v}^{Q}_{j+1,4}$ the projections of the points $\mathbf{v}_{j+1,2}$ and $\mathbf{v}_{j+1,4}$ onto the plane $Q$ parallel to $l$. Consider the parallelograms
$$
\begin{equation*}
\begin{aligned} \, \Delta^{\pi}_{j+1} &=\mathrm{conv}(\mathbf{v}^{\pi}_{j+1,1}, \mathbf{v}^{\pi}_{j+1,2}, \mathbf{v}^{\pi}_{j+1,3}, \mathbf{v}^{\pi}_{j+1,4}), \\ \Delta^{Q}_{j+1} &=\mathrm{conv}(\mathbf{v}_{j+1,1}, \mathbf{v}^{Q}_{j+1,2}, \mathbf{v}_{j+1,3}, \mathbf{v}^{Q}_{j+1,4}) \end{aligned}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\Delta^{R}_{j+1} =\mathrm{conv}(\mathbf{v}^{R}_{j+1,1}, \mathbf{v}_{j+1,2}, \mathbf{v}^{R}_{j+1,3}, \mathbf{v}_{j+1,4}).
\end{equation*}
\notag
$$
Note that $\mathbf{p}=\frac{1}{4}(\mathbf{v}_{j+1,1}+\mathbf{v}_{j+1,2} +\mathbf{v}_{j+1,3}+\mathbf{v}_{j+1,4})$, $\mathbf{p}^{Q}=\frac{1}{2}(\mathbf{v}_{j+1,1}+\mathbf{v}_{j+1,3})$ and $\mathbf{p}^{R}= \frac{1}{2}(\mathbf{v}_{j+1,2}+\mathbf{v}_{j+1,4})$.
The sequence of triples $(\Delta^{\pi}_{j}, \Delta^{Q}_{j}, \Delta^{R}_{j})$ is finite. Let $(\Delta^{\pi}_{k}, \Delta^{Q}_{k}, \Delta^{R}_{k})$ be the last element of this sequence. Set
$$
\begin{equation*}
\begin{gathered} \, \mathbf{z}_{1}= \mathbf{v}_{k,1}, \qquad \mathbf{z}_{2}=\mathbf{v}_{k,2}, \qquad \mathbf{z}_{3}=\mathbf{v}_{k,3}, \qquad \mathbf{z}_{4}= \mathbf{v}_{k,4}, \\ \mathbf{z}^{\pi}_{1}=\mathbf{v}^{\pi}_{k,1}, \qquad \mathbf{z}^{\pi}_{2}=\mathbf{v}^{\pi}_{k,2}, \qquad \mathbf{z}^{\pi}_{3}=\mathbf{v}^{\pi}_{k,3}, \qquad \mathbf{z}^{\pi}_{4}=\mathbf{v}^{\pi}_{k,4}, \\ \mathbf{z}^{R}_{1}=\mathbf{v}^{R}_{k,1}, \qquad \mathbf{z}^{R}_{3}=\mathbf{v}^{R}_{k,3}, \qquad \mathbf{z}^{Q}_{2}=\mathbf{v}^{Q}_{k,2}, \qquad \mathbf{z}^{Q}_{4}=\mathbf{v}^{Q}_{k,4}. \end{gathered}
\end{equation*}
\notag
$$
We claim that $(\Delta^{\pi}_{k} \cup \Delta^{Q}_{k} \cup \Delta^{R}_{k}) \cap \mathbb{Z}^{4}$ coincides with one of the sets
$$
\begin{equation*}
\begin{gathered} \, \bigl\{\mathbf{p}, \, \mathbf{z}_{1}, \, \mathbf{z}^{Q}_{2}, \, \mathbf{z}_{3}, \, \mathbf{z}^{Q}_{4}, \, \mathbf{z}^{R}_{1}, \, \mathbf{z}_{2}, \, \mathbf{z}^{R}_{3}, \, \mathbf{z}_{4} \bigr\},\qquad \bigl\{\mathbf{z}_{1}, \, \mathbf{z}_{3}, \, \mathbf{z}_{2}, \, \mathbf{z}_{4}\bigr\}, \\ \biggl\{\mathbf{z}_{1}, \, \mathbf{z}^{Q}_{2}, \, \mathbf{z}_{3}, \, \mathbf{z}^{Q}_{4}, \, \mathbf{z}^{R}_{1}, \, \mathbf{z}_{2}, \, \mathbf{z}^{R}_{3}, \, \mathbf{z}_{4}, \, \mathbf{p}^{Q}, \, \mathbf{p}^{R}, \, \frac{\mathbf{z}^{\pi}_{1}+\mathbf{z}^{\pi}_{2}}{2}, \, \frac{\mathbf{z}^{\pi}_{2}+\mathbf{z}^{\pi}_{3}}{2}, \, \frac{\mathbf{z}^{\pi}_{3}+\mathbf{z}^{\pi}_{4}}{2}, \, \frac{\mathbf{z}^{\pi}_{4}+\mathbf{z}^{\pi}_{1}}{2}\biggr\}, \\ \bigl\{\mathbf{p}, \, \mathbf{z}_{1}, \, \mathbf{z}^{Q}_{2}, \, \mathbf{z}_{3}, \, \mathbf{z}^{Q}_{4}, \, \mathbf{z}^{R}_{1}, \, \mathbf{z}_{2}, \, \mathbf{z}^{R}_{3}, \, \mathbf{z}_{4}, \, \mathbf{p}^{Q}, \, \mathbf{p}^{R}, \, \mathbf{z}^{\pi}_{1}, \, \mathbf{z}^{\pi}_{2}, \, \mathbf{z}^{\pi}_{3}, \, \mathbf{z}^{\pi}_{4}\bigr\}, \\ \bigl\{\mathbf{z}_{1}, \, \mathbf{z}^{Q}_{2}, \, \mathbf{z}_{3}, \, \mathbf{z}^{Q}_{4}, \, \mathbf{z}^{R}_{1}, \, \mathbf{z}_{2}, \, \mathbf{z}^{R}_{3}, \, \mathbf{z}_{4}, \, \mathbf{p}^{Q}, \, \mathbf{p}^{R}\bigr\}, \qquad \bigl\{\mathbf{z}_{1}, \, \mathbf{z}^{Q}_{2}, \, \mathbf{z}_{3}, \, \mathbf{z}^{Q}_{4}, \, \mathbf{z}^{R}_{1}, \, \mathbf{z}_{2}, \, \mathbf{z}^{R}_{3}, \, \mathbf{z}_{4}\bigr\}, \\ \bigl\{\mathbf{z}_{1}, \, \mathbf{z}^{Q}_{2}, \, \mathbf{z}_{3}, \, \mathbf{z}^{Q}_{4}, \, \mathbf{z}^{R}_{1}, \, \mathbf{z}_{2}, \, \mathbf{z}^{R}_{3}, \, \mathbf{z}_{4}, \, \mathbf{z}^{\pi}_{1}, \, \mathbf{z}^{\pi}_{2}, \, \mathbf{z}^{\pi}_{3}, \, \mathbf{z}^{\pi}_{4}\bigr\}. \end{gathered}
\end{equation*}
\notag
$$
To begin with, we show that
$$
\begin{equation*}
(\Delta^{\pi}_{k}\cap \mathbb{Z}^{4}) \subset \biggl\{\mathbf{p}, \mathbf{z}^{\pi}_{1}, \, \mathbf{z}^{\pi}_{2}, \, \mathbf{z}^{\pi}_{3}, \, \mathbf{z}^{\pi}_{4}, \, \frac{\mathbf{z}^{\pi}_{1}+\mathbf{z}^{\pi}_{2}}{2}, \, \frac{\mathbf{z}^{\pi}_{2}+\mathbf{z}^{\pi}_{3}}{2}, \, \frac{\mathbf{z}^{\pi}_{3}+\mathbf{z}^{\pi}_{4}}{2}, \, \frac{\mathbf{z}^{\pi}_{4}+\mathbf{z}^{\pi}_{1}}{2}\biggr\}.
\end{equation*}
\notag
$$
Assume that this is not so. Without loss of generality we can assume that there exists an integer point $\mathbf{w}$ lying in the parallelogram $\mathrm{conv}\bigl(\mathbf{z}^{\pi}_{1}, (\mathbf{z}^{\pi}_{1}+ \mathbf{z}^{\pi}_{2})/2, \mathbf{p}, (\mathbf{z}^{\pi}_{4}+ \mathbf{z}^{\pi}_{1})/2\bigr)$ and distinct from the vertices of this parallelogram. If $\mathbf{w}$ is the point of intersection of the diagonals of $\mathrm{conv}\bigl(\mathbf{z}^{\pi}_{1}, (\mathbf{z}^{\pi}_{1}+\mathbf{z}^{\pi}_{2})/2, \mathbf{p}, (\mathbf{z}^{\pi}_{4}+\mathbf{z}^{\pi}_{1})/2\bigr)$, then the integer point $\mathbf{z}_{1}+ (G(\mathbf{w})-\mathbf{w})$, which is distinct from $\mathbf{p}^{Q}$ and the vertices of $\Delta^{Q}_{k}$, lies in the parallelogram $\Delta^{Q}_{k}$. If $\mathbf{w}$ is not the point of intersection of the diagonals of $\mathrm{conv}\bigl(\mathbf{z}^{\pi}_{1}, (\mathbf{z}^{\pi}_{1}+\mathbf{z}^{\pi}_{2})/2, \mathbf{p}, (\mathbf{z}^{\pi}_{4}+ \mathbf{z}^{\pi}_{1})/2\bigr)$, then the integer point $\mathbf{z}_{1}+(G^{2}(\mathbf{w})-\mathbf{w})$, which is distinct from $\mathbf{p}^{Q}$ and the vertices of $\Delta^{Q}_{k}$, also lies in $\Delta^{Q}_{k}$. In either case we obtain a contradiction with the construction of the triplet of parallelograms $(\Delta^{\pi}_{k}, \Delta^{Q}_{k}, \Delta^{R}_{k})$.
A. Assume that the plane $\pi$ is rational. Then there are four fundamentally different cases to consider. A.1. Assume that $\mathbf{p} \in \mathbb{Z}^{4}$. In this case each of the points $\mathbf{z}^{R}_{1}$, $\mathbf{z}^{R}_{3}$, $\mathbf{z}^{Q}_{2}$ and $\mathbf{z}^{Q}_{4}$ belongs to the lattice $\mathbb{Z}^{4}$. None of the points $(\mathbf{z}^{\pi}_{1}+\mathbf{z}^{\pi}_{2})/2$, $(\mathbf{z}^{\pi}_{2}+\mathbf{z}^{\pi}_{3})/2$ and ${(\mathbf{z}^{\pi}_{3} + \mathbf{z}^{\pi}_{4})/2}$, $(\mathbf{z}^{\pi}_{4} + \mathbf{z}^{\pi}_{1})/2$ lies in $\mathbb{Z}^{4}$, because otherwise the midpoints of some sides of the parallelograms $\Delta^{Q}_{k}$ and $\Delta^{R}_{k}$ lie in $\mathbb{Z}^{4}$. We have the following two cases to consider: A.1.1 (corresponding to assertion 1)). Let $\mathbf{p}^{Q} \notin \mathbb{Z}^{4}$. Then $\mathbf{p}^{R} \notin \mathbb{Z}^{4}$. We have $\mathbf{p} \in \mathbb{Z}^{4}$, hence none of the points $\mathbf{z}^{\pi}_{1}$, $\mathbf{z}^{\pi}_{2}$, $\mathbf{z}^{\pi}_{3}$ and $\mathbf{z}^{\pi}_{4}$ lies in $\mathbb{Z}^{4}$. Consequently,
$$
\begin{equation*}
(\Delta^{\pi}_{k} \cup \Delta^{Q}_{k} \cup \Delta^{R}_{k}) \cap \mathbb{Z}^{4}= \bigl\{\mathbf{p}, \, \mathbf{z}_{1}, \, \mathbf{z}^{Q}_{2}, \, \mathbf{z}_{3}, \, \mathbf{z}^{Q}_{4}, \, \mathbf{z}^{R}_{1}, \, \mathbf{z}_{2}, \, \mathbf{z}^{R}_{3}, \, \mathbf{z}_{4} \bigr\},
\end{equation*}
\notag
$$
and the vectors $\mathbf{z}_{1}$, $\mathbf{z}_{2}$, $\mathbf{z}_{3}$, $\mathbf{p}= \frac{1}{4}(\mathbf{z}_{1}+\mathbf{z}_{2}+\mathbf{z}_{3}+\mathbf{z}_{4})$ form a basis of $\mathbb{Z}^{4}$. Hence assertion 1) holds (see Figure 1).
A.1.2 (corresponding to assertion 4)). Let $\mathbf{p}^{Q} \in \mathbb{Z}^{4}$. Then $\mathbf{p}^{R} \in \mathbb{Z}^{4}$. We have $\mathbf{p} \in \mathbb{Z}^{4}$, hence the points $\mathbf{z}^{\pi}_{1}$, $\mathbf{z}^{\pi}_{2}$, $\mathbf{z}^{\pi}_{3}$ and $\mathbf{z}^{\pi}_{4}$ belong to $\mathbb{Z}^{4}$. Consequently,
$$
\begin{equation*}
(\Delta^{\pi}_{k} \cup \Delta^{Q}_{k} \cup \Delta^{R}_{k}) \cap \mathbb{Z}^{4}=\bigl\{\mathbf{p}, \, \mathbf{z}_{1}, \, \mathbf{z}^{Q}_{2}, \, \mathbf{z}_{3}, \, \mathbf{z}^{Q}_{4}, \, \mathbf{z}^{R}_{1}, \, \mathbf{z}_{2}, \, \mathbf{z}^{R}_{3}, \, \mathbf{z}_{4}, \, \mathbf{p}^{Q}, \mathbf{p}^{R}, \, \mathbf{z}^{\pi}_{1}, \, \mathbf{z}^{\pi}_{2}, \, \mathbf{z}^{\pi}_{3}, \, \mathbf{z}^{\pi}_{4} \bigr\}
\end{equation*}
\notag
$$
and the vectors $\mathbf{z}_{1}$, $\mathbf{z}_{2}$, $\mathbf{p}_{Q}=\frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{3})$ and $\mathbf{p}=\frac{1}{4}(\mathbf{z}_{1}+\mathbf{z}_{2}+\mathbf{z}_{3}+\mathbf{z}_{4})$ form a basis of $\mathbb{Z}^{4}$. Hence assertion 4) holds; see Figure 2. A.2. Assume that $\mathbf{p} \notin \mathbb{Z}^{4}$. Consider the following two cases. A.2.1 (corresponding to assertion 3)). The points $(\mathbf{z}^{\pi}_{1}+ \mathbf{z}^{\pi}_{2})/2$, $(\mathbf{z}^{\pi}_{2}+\mathbf{z}^{\pi}_{3})/2$, $(\mathbf{z}^{\pi}_{3}+\mathbf{z}^{\pi}_{4})/2$ and $(\mathbf{z}^{\pi}_{4}+\mathbf{z}^{\pi}_{1})/2$ belong to $\mathbb{Z}^{4}$. In this case all point $\mathbf{p}^{Q}$, $\mathbf{p}^{R}$, $ \mathbf{z}^{R}_{1}$, $ \mathbf{z}^{R}_{3}$, $\mathbf{z}^{Q}_{2}$ and $\mathbf{z}^{Q}_{4}$ belong to $\mathbb{Z}^{4}$, but none of the points $\mathbf{z}^{\pi}_{1}$, $\mathbf{z}^{\pi}_{2}$, $\mathbf{z}^{\pi}_{3}$ and $\mathbf{z}^{\pi}_{4}$ belongs to $\mathbb{Z}^{4}$. Consequently,
$$
\begin{equation*}
\begin{aligned} \, &(\Delta^{\pi}_{k} \cup \Delta^{Q}_{k} \cup \Delta^{R}_{k}) \cap \mathbb{Z}^{4} =\biggl\{\mathbf{z}_{1}, \, \mathbf{z}^{Q}_{2}, \, \mathbf{z}_{3}, \, \mathbf{z}^{Q}_{4}, \\ &\qquad\qquad \mathbf{z}^{R}_{1}, \, \mathbf{z}_{2}, \, \mathbf{z}^{R}_{3}, \, \mathbf{z}_{4}, \, \mathbf{p}^{Q}, \, \mathbf{p}^{R}, \, \frac{\mathbf{z}^{\pi}_{1}+\mathbf{z}^{\pi}_{2}}{2}, \, \frac{\mathbf{z}^{\pi}_{2}+\mathbf{z}^{\pi}_{3}}{2}, \, \frac{\mathbf{z}^{\pi}_{3}+\mathbf{z}^{\pi}_{4}}{2}, \, \frac{\mathbf{z}^{\pi}_{4}+\mathbf{z}^{\pi}_{1}}{2}\biggr\}, \end{aligned}
\end{equation*}
\notag
$$
and the vectors
$$
\begin{equation*}
\mathbf{z}_{1}, \quad\frac{\mathbf{z}^{\pi}_{1}+\mathbf{z}^{\pi}_{2}}{2}= \frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{2}), \quad\mathbf{p}^{Q}=\frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{3})\quad\text{and} \quad\frac{\mathbf{z}^{\pi}_{4}+\mathbf{z}^{\pi}_{1}}{2}=\frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{4})
\end{equation*}
\notag
$$
form a basis of $\mathbb{Z}^{4}$. Hence assertion 3) holds (see Figure 3). A.2.2 (corresponding to assertion 7)). None of the points $(\mathbf{z}^{\pi}_{1}+ \mathbf{z}^{\pi}_{2})/2$, $(\mathbf{z}^{\pi}_{2}+\mathbf{z}^{\pi}_{3})/2$, $(\mathbf{z}^{\pi}_{3}+\mathbf{z}^{\pi}_{4})/2$ and $(\mathbf{z}^{\pi}_{4}+\mathbf{z}^{\pi}_{1})/2$ belongs to $\mathbb{Z}^{4}$. Assume that none of $\mathbf{z}^{R}_{1}$, $\mathbf{z}^{R}_{3}$, $\mathbf{z}^{Q}_{2}$ and $\mathbf{z}^{Q}_{4}$ lies in $\mathbb{Z}^{4}$. Let $\mathbf{r}$ be some vector connecting two integer points lying on the planes $\pi$ and $Q$, respectively (such a vector exists because by the plane $\pi$ is rational by assumption). Consider the set $\{\mathbf{z}_{3}- \mathbf{z}_{1}$, $\mathbf{z}_{4}-\mathbf{z}_{2}$, $\mathbf{z}_{1}$, $\mathbf{r}\}$, which is a basis of $\mathbb{Z}^{4}$. In this basis the first two coordinates of the point $\mathbf{z}_{1}$ (of $\mathbf{z}_3$, $\mathbf{z}^{Q}_{4}$ and $\mathbf{z}^{Q}_{2}$) have the form $(0, 0)$ (the form $(1, 0)$, $(1/2, 1/2)$ and $(1/2, -1/2)$, respectively). Let $(z_1, z_2)$ be the first two coordinates of $\mathbf{z}_{4}$. Then the first two coordinates of $\mathbf{z}_2$ (of $\mathbf{z}^{R}_{1}$ and $\mathbf{z}^{R}_{3}$) have the form $(z_1, z_{2}-1)$ (the form $(z_{1}-1/2, z_{2}-1/2)$ and $(z_{1}+1/2, z_{2}- 1/2)$, respectively). It follows that the first two coordinates of $(\mathbf{z}^{\pi}_{1}+\mathbf{z}^{\pi}_{2})/2$ (of $(\mathbf{z}^{\pi}_{2}+\mathbf{z}^{\pi}_{3})/2$, $(\mathbf{z}^{\pi}_{3}+\mathbf{z}^{\pi}_{4})/2$ and $(\mathbf{z}^{\pi}_{4}+\mathbf{z}^{\pi}_{1})/2$) have the form $(z_{1}/2, (z_{2}-1)/2)$ ($((z_{1}+1)/2, (z_{2}-1)/2)$, $((z_{1}+1)/2, z_{2}/2)$ and $(z_{1}/2, z_{2}/2)$, respectively). Now using parity arguments we arrive at a contradiction with the assumption. Thus, the points $\mathbf{z}^{R}_{1}$, $\mathbf{z}^{R}_{3}$, $\mathbf{z}^{Q}_{2}$ and $\mathbf{z}^{Q}_{4}$ belong to $\mathbb{Z}^{4}$. Hence $\mathbf{z}^{\pi}_{1}$, $\mathbf{z}^{\pi}_{2}$, $\mathbf{z}^{\pi}_{3}$ and $\mathbf{z}^{\pi}_{4}$ belong to $\mathbb{Z}^{4}$ too and, since $\mathbf{p} \notin \mathbb{Z}^{4}$, we have $\mathbf{p}^{Q} \notin \mathbb{Z}^4$ and $\mathbf{p}^{R} \notin \mathbb{Z}^4$. Consequently,
$$
\begin{equation*}
(\Delta^{\pi}_{k} \cup \Delta^{Q}_{k} \cup \Delta^{R}_{k}) \cap \mathbb{Z}^{4}=\bigl\{\mathbf{z}_{1}, \, \mathbf{z}^{Q}_{2}, \, \mathbf{z}_{3}, \, \mathbf{z}^{Q}_{4}, \, \mathbf{z}^{R}_{1}, \, \mathbf{z}_{2}, \, \mathbf{z}^{R}_{3}, \, \mathbf{z}_{4}, \, \mathbf{z}^{\pi}_{1}, \, \mathbf{z}^{\pi}_{2}, \, \mathbf{z}^{\pi}_{3}, \, \mathbf{z}^{\pi}_{4}\bigr\}
\end{equation*}
\notag
$$
and the vectors $\mathbf{z}_{1}$, $\mathbf{z}_{2}$, $\mathbf{z}_{3}$ and $\mathbf{z}^{\pi}_{1}= \frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{2})+\frac{1}{4}(\mathbf{z}_{1}+\mathbf{z}_{4}-\mathbf{z}_{3}-\mathbf{z}_{2})$ form a basis of $\mathbb{Z}^{4}$. Hence assertion 7) holds (see Figure 4). B. Assume that $\pi$ is not a rational plane. In this case there are three fundamentally different cases to consider. B.1 (corresponding to assertion 5)). Let $\mathbf{p}^{Q} \in \mathbb{Z}^{4}$. Then $\mathbf{p}^{R} \in \mathbb{Z}^{4}$. Hence
$$
\begin{equation*}
(\Delta^{\pi}_{k} \cup \Delta^{Q}_{k} \cup \Delta^{R}_{k}) \cap \mathbb{Z}^{4}= \bigl\{\mathbf{z}_{1}, \, \mathbf{z}^{Q}_{2}, \, \mathbf{z}_{3}, \, \mathbf{z}^{Q}_{4}, \, \mathbf{z}^{R}_{1}, \, \mathbf{z}_{2}, \, \mathbf{z}^{R}_{3}, \, \mathbf{z}_{4}, \, \mathbf{p}^{Q}, \, \mathbf{p}^{R}\bigr\}
\end{equation*}
\notag
$$
and the vectors $\mathbf{z}_{1}$, $\mathbf{z}_{2}$, $\mathbf{p}^{Q}=\frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{3})$ and $\mathbf{p}^{R}=\frac{1}{2}(\mathbf{z}_{2}+\mathbf{z}_{4})$ form a basis of $\mathbb{Z}^{4}$. Hence assertion 5) holds (see Figure 5). B.2. Assume that $\mathbf{p}^{Q} \notin \mathbb{Z}^{4}$. Then $\mathbf{p}^{R} \notin \mathbb{Z}^{4}$. Under this assumption we have the following two cases to consider. B.2.1 (corresponding to assertion 6)). Assume that $\mathbf{z}^{Q}_{2} \in \mathbb{Z}^{4}$, $\mathbf{z}^{Q}_{4} \in \mathbb{Z}^{4}$, $\mathbf{z}^{R}_{1} \in \mathbb{Z}^{4}$ and $\mathbf{z}^{R}_{2} \in \mathbb{Z}^{4}$. Then
$$
\begin{equation*}
(\Delta^{\pi}_{k} \cup \Delta^{Q}_{k} \cup \Delta^{R}_{k}) \cap \mathbb{Z}^{4}= \bigl\{\mathbf{z}_{1}, \, \mathbf{z}^{Q}_{2}, \, \mathbf{z}_{3}, \, \mathbf{z}^{Q}_{4}, \, \mathbf{z}^{R}_{1}, \, \mathbf{z}_{2}, \, \mathbf{z}^{R}_{3}, \, \mathbf{z}_{4}\bigr\}
\end{equation*}
\notag
$$
and the vectors $\mathbf{z}_{1}$, $\mathbf{z}_{2}$, $\mathbf{z}_{3}$ and $\mathbf{z}^{Q}_{4}= \frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{3}+\mathbf{z}_{4}-\mathbf{z}_{2})$ form a basis of $\mathbb{Z}^{4}$. Hence assertion 6) holds (see Figure 6). B.2.2 (corresponding to assertion 2)). Assume that $\mathbf{z}^{Q}_{2} \notin \mathbb{Z}^{4}$, $\mathbf{z}^{Q}_{4} \notin \mathbb{Z}^{4}$, $\mathbf{z}^{R}_{1} \notin \mathbb{Z}^{4}$ and $\mathbf{z}^{R}_{2} \notin \mathbb{Z}^{4}$. Then
$$
\begin{equation*}
(\Delta^{\pi}_{k} \cup \Delta^{Q}_{k} \cup \Delta^{R}_{k}) \cap \mathbb{Z}^{4}=\bigl\{\mathbf{z}_{1}, \, \mathbf{z}_{3}, \, \mathbf{z}_{2}, \, \mathbf{z}_{4}\bigr\}
\end{equation*}
\notag
$$
and the vectors $\mathbf{z}_{1}$, $\mathbf{z}_{2}$, $\mathbf{z}_{3}$ and $\mathbf{z}_{4}$ form a basis of $\mathbb{Z}^{4}$. Hence assertion 2) holds (see Figure 7). Lemma 4 is proved. Given a continued fraction $\operatorname{CF}(l_1,l_2,l_3,l_4)=\operatorname{CF}(A)\in\mathfrak{A}_3$, we assume that the subspace $l_1$ is spanned by a vector $\mathbf l_1=(1,\alpha,\beta,\gamma)$. Now it follows from Proposition 1 that the numbers $1$, $\alpha$, $\beta$ and $\gamma$ form a basis for the field $K=\mathbb{Q}(\alpha,\beta, \gamma)$ over $\mathbb{Q}$, and each $l_i$ is spanned by a vector $\mathbf l_i=(1,\sigma_i(\alpha),\sigma_i(\beta), \sigma_i(\gamma))$, where $\sigma_1(=\operatorname{id})$, $\sigma_2$, $\sigma_3$ and $\sigma_4$ are all the embeddings of $K$ into $\mathbb{R}$. In other words, if $(\mathbf l_1,\mathbf l_2,\mathbf l_3,\mathbf l_4)$ denotes the matrix with columns $\mathbf l_1$, $\mathbf l_2$, $\mathbf l_3$ and $\mathbf l_4$, then we have
$$
\begin{equation*}
(\mathbf l_1,\mathbf l_2,\mathbf l_3,\mathbf l_4)=\begin{pmatrix} 1 & 1 & 1 & 1 \\ \alpha & \sigma_2(\alpha) & \sigma_3(\alpha) & \sigma_4(\alpha) \\ \beta & \sigma_2(\beta) & \sigma_3(\beta) & \sigma_4(\beta) \\ \gamma & \sigma_2(\gamma) & \sigma_3(\gamma) & \sigma_4(\gamma) \end{pmatrix}.
\end{equation*}
\notag
$$
Recall that $\mathfrak{A}_{3}'$ is the set of three-dimensional algebraic continued fractions for which the field $K$ in Proposition 1 is a totally real cyclic Galois extension. Let $\sigma$ be a generator of the Galois group $\operatorname{Gal}(K/\mathbb{Q})$. Recall also that the straight lines $l_1$, $l_2$, $l_3$ and $l_4$ are labelled so that if $(\mathbf l_1,\mathbf l_2, \mathbf l_3, \mathbf l_4 )$ is the matrix with columns $\mathbf l_1$, $\mathbf l_2$, $\mathbf l_3$ and $\mathbf l_4$, then
$$
\begin{equation*}
(\mathbf l_1,\mathbf l_2, \mathbf l_3, \mathbf l_4 )= \begin{pmatrix} 1 & 1 & 1 & 1 \\ \alpha & \sigma(\alpha) & \sigma^{2}(\alpha) & \sigma^{3}(\alpha) \\ \beta & \sigma(\beta) & \sigma^{2}(\beta) & \sigma^{3}(\beta) \\ \gamma & \sigma(\gamma) & \sigma^{2}(\gamma) & \sigma^{3}(\gamma) \end{pmatrix}.
\end{equation*}
\notag
$$
Consider the following classes of three-dimensional algebraic continued fractions:
$$
\begin{equation*}
\begin{aligned} \, \mathbf{CF}_1 &=\bigl\{ \operatorname{CF}(l_1,l_2,l_3,l_4)\in\mathfrak{A}_3' \mid \beta= \sigma(\alpha),\ \gamma=\sigma^{2}(\alpha), \ \operatorname{Tr}(\alpha)=0 \bigr\}, \\ \mathbf{CF}_2 &=\bigl\{ \operatorname{CF}(l_1,l_2,l_3,l_4)\in\mathfrak{A}_3' \mid \beta= \sigma(\alpha) ,\ \gamma=\sigma^{2}(\alpha), \ \operatorname{Tr}(\alpha)=1 \bigr\}, \\ \mathbf{CF}_3 &=\bigl\{ \operatorname{CF}(l_1,l_2,l_3,l_4)\in\mathfrak{A}_3' \mid \beta= \sigma(\alpha) ,\ \gamma=\sigma^{2}(\alpha), \ \operatorname{Tr}(\alpha)=2 \bigr\}, \\ \mathbf{CF}_4 &=\biggl\{ \operatorname{CF}(l_1,l_2,l_3,l_4)\in\mathfrak{A}_3' \Bigm| \beta= \sigma(\alpha),\ \gamma=\frac{\alpha+\sigma^{2}(\alpha)}{2}, \ \operatorname{Tr}(\alpha)=0 \biggr\}, \\ \mathbf{CF}_5 &=\biggl\{ \operatorname{CF}(l_1,l_2,l_3,l_4)\in\mathfrak{A}_3' \Bigm| \beta= \sigma(\alpha) ,\ \gamma=\frac{\alpha+\sigma^{2}(\alpha)}{2}, \ \operatorname{Tr}(\alpha)=2 \biggr\}, \\ \mathbf{CF}_6 &=\biggl\{ \operatorname{CF}(l_1,l_2,l_3,l_4)\in\mathfrak{A}_3' \Bigm| \beta= \sigma(\alpha) ,\ \gamma=\frac{\alpha+\sigma^{2}(\alpha)+1}{2}, \ \operatorname{Tr}(\alpha)=0 \biggr\}, \\ \mathbf{CF}_7 &=\biggl\{ \operatorname{CF}(l_1,l_2,l_3,l_4)\in\mathfrak{A}_3' \Bigm| \beta= \sigma(\alpha) ,\ \gamma=\frac{\alpha+\sigma^{2}(\alpha)+1}{2}, \ \operatorname{Tr}(\alpha)=2 \biggr\}. \end{aligned}
\end{equation*}
\notag
$$
We claim that each continued fraction in each class $\mathbf{CF}_{i}$, $i=1,\dots, 7$, is palindromic. Set
$$
\begin{equation*}
\begin{gathered} \, G_{1}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & -1 & -1 & -1 \end{pmatrix}, \qquad G_{2}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & -1 & -1 & -1 \end{pmatrix}, \\ G_{3}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 2 & -1 & -1 & -1 \end{pmatrix}, \qquad G_{4}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 2 \\ 0 &0 & 0 & -1 \end{pmatrix}, \\ G_{5}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 2 \\ 1 & 0 & 0 & -1 \end{pmatrix}, \qquad G_{6}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -1 & -1 & 0 & 2 \\ 1 & 0 & 0 & -1 \end{pmatrix}, \\ G_{7}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -1 & -1 & 0 & 2 \\ 2 & 0 & 0 & -1 \end{pmatrix}. \end{gathered}
\end{equation*}
\notag
$$
Lemma 5. Let $\operatorname{CF}(l_1,l_2,l_3,l_4)\in\mathfrak{A}_3$ and $i\in\{1,2,3,4,5,6,7\}$. Then the continued fraction $\operatorname{CF}(l_1,l_2,l_3,l_4)$ lies in the class $\mathbf{CF}_{i}$ if and only if $G_{i}$ is a proper cyclic symmetry of it. Proof. By the above corollary the operator $G\in\operatorname{GL}_4(\mathbb{Z})$ is a proper cyclic symmetry of $\operatorname{CF}(l_1,l_2,l_3,l_4)$ if and only if, up to a permutation of indices, there exist real numbers $\mu_1$, $\mu_2$, $\mu_3$ and $\mu_4$ such that $G(\mathbf l_1,\mathbf l_2,\mathbf l_3,\mathbf l_4)=(\mu_2\mathbf l_2,\mu_3\mathbf l_3,\mu_4\mathbf l_4,\mu_1\mathbf l_1)$ and $\mu_{1}\mu_{2}\mu_{3}\mu_{4}=1$.
Let $\operatorname{CF}(l_1, l_2, l_3, l_4) \in \mathbf{CF}_{1}$. Then
$$
\begin{equation*}
\begin{gathered} \, G_{1}\mathbf{l}_1= \begin{pmatrix} 1 \\ \sigma(\alpha) \\ \sigma^{2}(\alpha) \\ -\alpha - \sigma(\alpha) - \sigma^{2}(\alpha) \end{pmatrix}, \qquad G_{1}\mathbf{l}_2= \begin{pmatrix} 1 \\ \sigma^{2}(\alpha) \\ \sigma^{3}(\alpha) \\ -\sigma(\alpha) - \sigma^{2}(\alpha) - \sigma^{3}(\alpha) \end{pmatrix}, \\ G_{1}\mathbf{l}_3= \begin{pmatrix} 1 \\ \sigma^{3}(\alpha) \\ \alpha \\ -\sigma^{2}(\alpha) - \sigma^{3}(\alpha) - \alpha \end{pmatrix}, \qquad G_{1}\mathbf{l}_4= \begin{pmatrix} 1 \\ \alpha \\ \sigma(\alpha) \\ -\sigma^{3}(\alpha) - \alpha - \sigma(\alpha) \end{pmatrix}, \end{gathered}
\end{equation*}
\notag
$$
that is, $G_{1}(\mathbf{l}_1, \mathbf{l}_2, \mathbf{l}_3, \mathbf{l}_4) = (\mathbf{l}_2, \mathbf{l}_3, \mathbf{l}_4, \mathbf{l}_1)$. Therefore, $G_{1}$ is a proper cyclic symmetry of $\operatorname{CF}(l_1,l_2,l_3,l_4)$. Conversely, assume that $G_{1}$ is a proper cyclic symmetry of the continued fraction $\operatorname{CF}(l_1,l_2,l_3,l_4)$. Then there exists $\mu_2$ such that, up to a permutation of indices,
$$
\begin{equation*}
G_{1}\mathbf{l}_1=\begin{pmatrix} 1 \\ \beta \\ \gamma \\ -\alpha-\beta-\gamma \end{pmatrix}=\mu_2\begin{pmatrix} 1 \\ \sigma_{2}(\alpha) \\ \sigma_{2}(\beta) \\ \sigma_{2}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\mu_2=1, \qquad \beta=\sigma_{2}(\alpha), \qquad\gamma=\sigma_{2}(\beta)\quad\text{and} \quad- \alpha-\beta- \gamma=\sigma_{2}(\gamma).
\end{equation*}
\notag
$$
There exists $\mu_3$ such that
$$
\begin{equation*}
G_{1}\mathbf{l}_2=\begin{pmatrix} 1 \\ \gamma \\ -\alpha-\beta-\gamma \\ \alpha \end{pmatrix}=\mu_3\begin{pmatrix} 1 \\ \sigma_{3}(\alpha) \\ \sigma_{3}(\beta) \\ \sigma_{3}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\begin{gathered} \, \mu_3=1, \qquad\sigma_{3}(\alpha)=\gamma=\sigma_{2}(\beta)=\sigma^{2}_{2}(\alpha), \\ \sigma_{3}(\beta)=- \alpha-\beta-\gamma=\sigma_{2}(\gamma)=\sigma^{2}_{2}(\beta) \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\sigma_{3}(\gamma)=\alpha=-\beta-\gamma-(- \alpha-\beta-\gamma)=\sigma_{2}(- \alpha-\beta- \gamma)=\sigma^{2}_{2}(\gamma).
\end{equation*}
\notag
$$
There exists $\mu_4$ such that
$$
\begin{equation*}
G_{1}\mathbf{l}_3=\begin{pmatrix} 1 \\ -\alpha-\beta-\gamma \\ \alpha \\ \beta \end{pmatrix}=\mu_4\begin{pmatrix} 1 \\ \sigma_{4}(\alpha) \\ \sigma_{4}(\beta) \\ \sigma_{4}(\gamma) \end{pmatrix}.
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
\begin{gathered} \, \mu_4=1, \qquad\sigma_{4}(\alpha)=- \alpha-\beta-\gamma=\sigma_{2}(\gamma)= \sigma^{3}_{2}(\alpha), \\ \sigma_{4}(\beta)=\alpha=-\beta-\gamma-(- \alpha-\beta- \gamma)=\sigma_{2}(-\alpha-\beta-\gamma)= \sigma^{3}_{2}(\beta), \\ \sigma_{4}(\gamma)= \beta =\sigma_{2}(\alpha)=\sigma^{3}_{2}(\gamma)\quad\text{and} \quad\operatorname{Tr}(\alpha)=0. \end{gathered}
\end{equation*}
\notag
$$
Consequently, $\operatorname{CF}(l_1, l_2, l_3, l_4) \in \mathbf{CF}_{1}$ since the numbers $1$, $\alpha$, $\beta$ and $\gamma$ form a basis for the field $K=\mathbb{Q}(\alpha,\beta, \gamma)$.
Let $\operatorname{CF}(l_1, l_2, l_3, l_4) \in \mathbf{CF}_{2}$. Then
$$
\begin{equation*}
\begin{gathered} \, G_{2}\mathbf{l}_1= \begin{pmatrix} 1 \\ \sigma(\alpha) \\ \sigma^{2}(\alpha) \\ 1 -\alpha - \sigma(\alpha) - \sigma^{2}(\alpha) \end{pmatrix}, \qquad G_{2}\mathbf{l}_2= \begin{pmatrix} 1 \\ \sigma^{2}(\alpha) \\ \sigma^{3}(\alpha) \\ 1 -\sigma(\alpha) - \sigma^{2}(\alpha) - \sigma^{3}(\alpha) \end{pmatrix}, \\ G_{2}\mathbf{l}_3= \begin{pmatrix} 1 \\ \sigma^{3}(\alpha) \\ \alpha \\ 1 -\sigma^{2}(\alpha) - \sigma^{3}(\alpha) - \alpha \end{pmatrix}, \qquad G_{2}\mathbf{l}_4= \begin{pmatrix} 1 \\ \alpha \\ \sigma(\alpha) \\ 1 -\sigma^{3}(\alpha) - \alpha - \sigma(\alpha) \end{pmatrix}, \end{gathered}
\end{equation*}
\notag
$$
that is, $G_{2}(\mathbf{l}_1, \mathbf{l}_2, \mathbf{l}_3, \mathbf{l}_4) = (\mathbf{l}_2, \mathbf{l}_3, \mathbf{l}_4, \mathbf{l}_1)$. Therefore, $G_{2}$ is a proper cyclic symmetry of $\operatorname{CF}(l_1,l_2,l_3,l_4)$. Conversely, assume that $G_{2}$ is a proper cyclic symmetry of $\operatorname{CF}(l_1,l_2,l_3,l_4)$. Then there exists $\mu_2$ such that, up to a permutation of indices,
$$
\begin{equation*}
G_{2}\mathbf{l}_1= \begin{pmatrix} 1 \\ \beta \\ \gamma \\ 1-\alpha-\beta-\gamma \end{pmatrix}=\mu_2\begin{pmatrix} 1 \\ \sigma_{2}(\alpha) \\ \sigma_{2}(\beta) \\ \sigma_{2}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\mu_2=1, \qquad \beta=\sigma_{2}(\alpha), \qquad\gamma=\sigma_{2}(\beta)\quad\text{and} \quad 1-\alpha-\beta- \gamma=\sigma_{2}(\gamma).
\end{equation*}
\notag
$$
There exists $\mu_3$ such that
$$
\begin{equation*}
G_{2}\mathbf{l}_2= \begin{pmatrix} 1 \\ \gamma \\ 1-\alpha-\beta-\gamma \\ \alpha \end{pmatrix}=\mu_3\begin{pmatrix} 1 \\ \sigma_{3}(\alpha) \\ \sigma_{3}(\beta) \\ \sigma_{3}(\gamma) \end{pmatrix}.
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
\begin{gathered} \, \mu_3=1, \qquad\sigma_{3}(\alpha)=\gamma=\sigma_{2}(\beta)=\sigma^{2}_{2}(\alpha), \\ \sigma_{3}(\beta)=1-\alpha-\beta-\gamma=\sigma_{2}(\gamma)=\sigma^{2}_{2}(\beta) \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\sigma_{3}(\gamma)=\alpha=1 -\beta-\gamma-(1-\alpha-\beta-\gamma)=\sigma_{2}(1-\alpha-\beta- \gamma)=\sigma^{2}_{2}(\gamma).
\end{equation*}
\notag
$$
There exists $\mu_4$ such that
$$
\begin{equation*}
G_{2}\mathbf{l}_3=\begin{pmatrix} 1 \\ 1-\alpha-\beta-\gamma \\ \alpha \\ \beta \end{pmatrix}=\mu_4\begin{pmatrix} 1 \\ \sigma_{4}(\alpha) \\ \sigma_{4}(\beta) \\ \sigma_{4}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\begin{gathered} \, \mu_4=1, \qquad \sigma_{4}(\alpha)=1-\alpha-\beta-\gamma=\sigma_{2}(\gamma)= \sigma^{3}_{2}(\alpha), \\ \sigma_{4}(\beta)=\alpha=1-\beta-\gamma-(1-\alpha-\beta- \gamma)=\sigma_{2}(1-\alpha-\beta-\gamma)= \sigma^{3}_{2}(\beta), \\ \sigma_{4}(\gamma)= \beta =\sigma_{2}(\alpha)=\sigma^{3}_{2}(\gamma)\quad\text{and} \quad \operatorname{Tr}(\alpha)=1. \end{gathered}
\end{equation*}
\notag
$$
Consequently, $\operatorname{CF}(l_1, l_2, l_3, l_4) \in \mathbf{CF}_{2}$ since the numbers $1$, $\alpha$, $\beta$ and $\gamma$ form a basis for the field $K=\mathbb{Q}(\alpha,\beta, \gamma)$.
Let $\operatorname{CF}(l_1, l_2, l_3, l_4) \in \mathbf{CF}_{3}$. Then
$$
\begin{equation*}
\begin{gathered} \, G_{3}\mathbf{l}_1= \begin{pmatrix} 1 \\ \sigma(\alpha) \\ \sigma^{2}(\alpha) \\ 2 -\alpha - \sigma(\alpha) - \sigma^{2}(\alpha) \end{pmatrix}, \qquad G_{3}\mathbf{l}_2= \begin{pmatrix} 1 \\ \sigma^{2}(\alpha) \\ \sigma^{3}(\alpha) \\ 2 -\sigma(\alpha) - \sigma^{2}(\alpha) - \sigma^{3}(\alpha) \end{pmatrix}, \\ G_{3}\mathbf{l}_3= \begin{pmatrix} 1 \\ \sigma^{3}(\alpha) \\ \alpha \\ 2 -\sigma^{2}(\alpha) - \sigma^{3}(\alpha) - \alpha \end{pmatrix}, \qquad G_{3}\mathbf{l}_4= \begin{pmatrix} 1 \\ \alpha \\ \sigma(\alpha) \\ 2 -\sigma^{3}(\alpha) - \alpha - \sigma(\alpha) \end{pmatrix}, \end{gathered}
\end{equation*}
\notag
$$
that is, $G_{3}(\mathbf{l}_1, \mathbf{l}_2, \mathbf{l}_3, \mathbf{l}_4) = (\mathbf{l}_2, \mathbf{l}_3, \mathbf{l}_4, \mathbf{l}_1)$. Therefore, $G_{3}$ is a proper cyclic symmetry of the continued fraction $\operatorname{CF}(l_1,l_2,l_3,l_4)$. Conversely, assume that $G_{3}$ is a proper cyclic symmetry of $\operatorname{CF}(l_1,l_2,l_3,l_4)$. Then there exists $\mu_2$ such that, up to a permutation of indices,
$$
\begin{equation*}
G_{3}\mathbf{l}_1= \begin{pmatrix} 1 \\ \beta \\ \gamma \\ 2-\alpha-\beta-\gamma \end{pmatrix}=\mu_2\begin{pmatrix} 1 \\ \sigma_{2}(\alpha) \\ \sigma_{2}(\beta) \\ \sigma_{2}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\mu_2=1, \qquad \beta=\sigma_{2}(\alpha), \qquad \gamma=\sigma_{2}(\beta)\quad\text{and} \quad 2-\alpha-\beta- \gamma=\sigma_{2}(\gamma).
\end{equation*}
\notag
$$
There exists $\mu_3$ such that
$$
\begin{equation*}
G_{3}\mathbf{l}_2= \begin{pmatrix} 1 \\ \gamma \\ 2-\alpha-\beta-\gamma \\ \alpha \end{pmatrix}=\mu_3\begin{pmatrix} 1 \\ \sigma_{3}(\alpha) \\ \sigma_{3}(\beta) \\ \sigma_{3}(\gamma) \end{pmatrix}.
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
\begin{gathered} \, \mu_3=1, \qquad \sigma_{3}(\alpha)=\gamma=\sigma_{2}(\beta)=\sigma^{2}_{2}(\alpha), \\ \sigma_{3}(\beta)=2-\alpha-\beta-\gamma=\sigma_{2}(\gamma)=\sigma^{2}_{2}(\beta) \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\sigma_{3}(\gamma)=\alpha=2-\beta-\gamma-(2 -\alpha-\beta -\gamma)=\sigma_{2}(2-\alpha-\beta- \gamma)=\sigma^{2}_{2}(\gamma).
\end{equation*}
\notag
$$
There exists $\mu_4$ such that
$$
\begin{equation*}
G_{3}\mathbf{l}_3=\begin{pmatrix} 1 \\ 2-\alpha-\beta-\gamma \\ \alpha \\ \beta \end{pmatrix}=\mu_4\begin{pmatrix} 1 \\ \sigma_{4}(\alpha) \\ \sigma_{4}(\beta) \\ \sigma_{4}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\begin{gathered} \, \mu_4=1, \qquad \sigma_{4}(\alpha)=2-\alpha-\beta-\gamma=\sigma_{2}(\gamma)= \sigma^{3}_{2}(\alpha), \\ \sigma_{4}(\beta)=\alpha=2-\beta-\gamma-(2 -\alpha-\beta -\gamma)=\sigma_{2}(2-\alpha-\beta-\gamma)=\sigma^{3}_{2}(\beta), \\ \sigma_{4}(\gamma)= \beta=\sigma_{2}(\alpha)=\sigma^{3}_{2}(\gamma)\quad\text{and} \quad \operatorname{Tr}(\alpha)=2. \end{gathered}
\end{equation*}
\notag
$$
Consequently, $\operatorname{CF}(l_1, l_2, l_3, l_4) \in \mathbf{CF}_{3}$ since the numbers $1$, $\alpha$, $\beta$ and $\gamma$ form a basis for $K=\mathbb{Q}(\alpha,\beta, \gamma)$.
Let $\operatorname{CF}(l_1, l_2, l_3, l_4) \in \mathbf{CF}_{4}$. Then
$$
\begin{equation*}
\begin{gathered} \, G_{4}\mathbf{l}_1= \begin{pmatrix} 1 \\ \sigma(\alpha) \\ \sigma^{2}(\alpha) \\ -\dfrac{\alpha + \sigma^{2}(\alpha)}{2} \end{pmatrix}, \qquad G_{4}\mathbf{l}_2= \begin{pmatrix} 1 \\ \sigma^{2}(\alpha) \\ \sigma^{3}(\alpha) \\ -\dfrac{\sigma(\alpha) + \sigma^{3}(\alpha)}{2} \end{pmatrix}, \\ G_{4}\mathbf{l}_3= \begin{pmatrix} 1 \\ \sigma^{3}(\alpha) \\ \alpha \\ -\dfrac{\sigma^{2}(\alpha) + \alpha}{2} \end{pmatrix}, \qquad G_{4}\mathbf{l}_4= \begin{pmatrix} 1 \\ \alpha \\ \sigma(\alpha) \\ -\dfrac{\sigma^{3}(\alpha) + \sigma(\alpha)}{2} \end{pmatrix}, \end{gathered}
\end{equation*}
\notag
$$
that is, $G_{4}(\mathbf{l}_1, \mathbf{l}_2, \mathbf{l}_3, \mathbf{l}_4) = (\mathbf{l}_2, \mathbf{l}_3, \mathbf{l}_4, \mathbf{l}_1)$. Therefore, $G_{4}$ is a proper cyclic symmetry of $\operatorname{CF}(l_1,l_2,l_3,l_4)$. Conversely, assume that $G_{4}$ is a proper cyclic symmetry of $\operatorname{CF}(l_1,l_2,l_3,l_4)$. Then there exists $\mu_2$ such that, up to a permutation of indices,
$$
\begin{equation*}
G_{4}\mathbf{l}_1= \begin{pmatrix} 1 \\ \beta \\ -\alpha+2\gamma \\ -\gamma \end{pmatrix}=\mu_2\begin{pmatrix} 1 \\ \sigma_{2}(\alpha) \\ \sigma_{2}(\beta) \\ \sigma_{2}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\mu_2=1, \qquad \beta=\sigma_{2}(\alpha), \qquad - \alpha+2\gamma=\sigma_{2}(\beta)\quad\text{and} \quad - \gamma=\sigma_{2}(\gamma).
\end{equation*}
\notag
$$
There exists $\mu_3$ such that
$$
\begin{equation*}
G_{4}\mathbf{l}_2=\begin{pmatrix} 1 \\ -\alpha+2\gamma \\ -\beta-2\gamma \\ \gamma \end{pmatrix}=\mu_3\begin{pmatrix} 1 \\ \sigma_{3}(\alpha) \\ \sigma_{3}(\beta) \\ \sigma_{3}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\begin{gathered} \, \mu_3=1, \qquad \frac{\alpha+\sigma_{3}(\alpha)}{2}=\gamma, \sigma_{3}(\alpha)=- \alpha+2\gamma=\sigma_{2}(\beta)=\sigma^{2}_{2}(\alpha), \\ \sigma_{3}(\beta)=- \beta-2\gamma= \sigma_{2}(-\alpha+2\gamma)=\sigma^{2}_{2}(\beta)\quad\text{and} \quad \sigma_{3}(\gamma)=\gamma= \sigma_{2}(-\gamma)=\sigma^{2}_{2}(\gamma). \end{gathered}
\end{equation*}
\notag
$$
There exists $\mu_4$ such that
$$
\begin{equation*}
G_{4}\mathbf{l}_3=\begin{pmatrix} 1 \\ -\beta- 2\gamma \\ \alpha \\ -\gamma \end{pmatrix}=\mu_4\begin{pmatrix} 1 \\ \sigma_{4}(\alpha) \\ \sigma_{4}(\beta) \\ \sigma_{4}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\begin{gathered} \, \mu_4=1, \qquad \sigma_{4}(\alpha)=- \beta- 2\gamma=\sigma_{2}(-\alpha)+\sigma_{2}(2\gamma)=\sigma_{2}(-\alpha+2\gamma)=\sigma^{3}_{2}(\alpha), \\ \sigma_{4}(\beta)= \alpha=2\gamma-\sigma_{2}(\beta) =\sigma_{2}(-2\gamma-\beta)= \sigma^{3}_{2}(\beta), \\ \sigma_{4}(\gamma)=-\gamma=\sigma_{2}(\gamma) = \sigma^{3}_{2}(\gamma)\quad\text{and} \quad \operatorname{Tr}(\alpha)=0. \end{gathered}
\end{equation*}
\notag
$$
Consequently, $\operatorname{CF}(l_1, l_2, l_3, l_4) \in \mathbf{CF}_{4}$ since the numbers $1$, $\alpha$, $\beta$ and $\gamma$ form a basis for the field $K=\mathbb{Q}(\alpha,\beta, \gamma)$.
Let $\operatorname{CF}(l_1, l_2, l_3, l_4) \in \mathbf{CF}_{5}$. Then
$$
\begin{equation*}
\begin{gathered} \, G_{5}\mathbf{l}_1= \begin{pmatrix} 1 \\ \sigma(\alpha) \\ \sigma^{2}(\alpha) \\ \dfrac{2 - \alpha - \sigma^{2}(\alpha)}{2} \end{pmatrix}, \qquad G_{5}\mathbf{l}_2= \begin{pmatrix} 1 \\ \sigma^{2}(\alpha) \\ \sigma^{3}(\alpha) \\ \dfrac{2 - \sigma(\alpha) - \sigma^{3}(\alpha)}{2} \end{pmatrix}, \\ G_{5}\mathbf{l}_3= \begin{pmatrix} 1 \\ \sigma^{3}(\alpha) \\ \alpha \\ \dfrac{2 - \sigma^{2}(\alpha) - \alpha}{2} \end{pmatrix}, \qquad G_{5}\mathbf{l}_4= \begin{pmatrix} 1 \\ \alpha \\ \sigma(\alpha) \\ \dfrac{2 - \sigma^{3}(\alpha) - \sigma(\alpha)}{2} \end{pmatrix}, \end{gathered}
\end{equation*}
\notag
$$
that is, $G_{5}(\mathbf{l}_1, \mathbf{l}_2, \mathbf{l}_3, \mathbf{l}_4) = (\mathbf{l}_2, \mathbf{l}_3, \mathbf{l}_4, \mathbf{l}_1)$. Therefore, $G_{5}$ is a proper cyclic symmetry of $\operatorname{CF}(l_1,l_2,l_3,l_4)$. Conversely, assume that $G_{5}$ is a proper cyclic symmetry of $\operatorname{CF}(l_1,l_2,l_3,l_4)$. Then there exists $\mu_2$ such that, up to a permutation of indices,
$$
\begin{equation*}
G_{5}\mathbf{l}_1= \begin{pmatrix} 1 \\ \beta \\ -\alpha+2\gamma \\ 1-\gamma \end{pmatrix}=\mu_2\begin{pmatrix} 1 \\ \sigma_{2}(\alpha) \\ \sigma_{2}(\beta) \\ \sigma_{2}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\mu_2=1, \qquad \beta=\sigma_{2}(\alpha), \qquad - \alpha+2\gamma=\sigma_{2}(\beta)\quad\text{and} \quad 1- \gamma=\sigma_{2}(\gamma).
\end{equation*}
\notag
$$
There exists $\mu_3$ such that
$$
\begin{equation*}
G_{5}\mathbf{l}_2=\begin{pmatrix} 1 \\ -\alpha+2\gamma \\ 2-\beta-2\gamma \\ \gamma \end{pmatrix}=\mu_3\begin{pmatrix} 1 \\ \sigma_{3}(\alpha) \\ \sigma_{3}(\beta) \\ \sigma_{3}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\begin{gathered} \, \mu_3=1, \qquad \frac{\alpha+\sigma_{3}(\alpha)}{2}=\gamma, \sigma_{3}(\alpha)=- \alpha+2\gamma=\sigma_{2}(\beta)=\sigma^{2}_{2}(\alpha), \\ \sigma_{3}(\beta)=2-\beta-2\gamma= \sigma_{2}(-\alpha+2\gamma)=\sigma^{2}_{2}(\beta)\quad\!\!\text{and}\!\! \quad \sigma_{3}(\gamma)=\gamma=\sigma_{2}(1- \gamma)=\sigma^{2}_{2}(\gamma). \end{gathered}
\end{equation*}
\notag
$$
There exists $\mu_4$ such that
$$
\begin{equation*}
G_{5}\mathbf{l}_3=\begin{pmatrix} 1 \\ 2-\beta- 2\gamma \\ \alpha \\ 1 -\gamma \end{pmatrix}=\mu_4\begin{pmatrix} 1 \\ \sigma_{4}(\alpha) \\ \sigma_{4}(\beta) \\ \sigma_{4}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\begin{gathered} \, \mu_4=1, \qquad \sigma_{4}(\alpha)=2-\beta- 2\gamma=\sigma_{2}(-\alpha)+\sigma_{2}(2\gamma)=\sigma_{2}(-\alpha+2\gamma)=\sigma^{3}_{2}(\alpha), \\ \sigma_{4}(\beta)= \alpha=2\gamma-\sigma_{2}(\beta) =\sigma_{2}(2-2\gamma-\beta)= \sigma^{3}_{2}(\beta), \\ \sigma_{4}(\gamma)=1 -\gamma=\sigma_{2}(\gamma) = \sigma^{3}_{2}(\gamma)\quad\text{and} \quad \operatorname{Tr}(\alpha)=2. \end{gathered}
\end{equation*}
\notag
$$
Consequently, $\operatorname{CF}(l_1, l_2, l_3, l_4) \in \mathbf{CF}_{5}$ since the numbers $1$, $\alpha$, $\beta$ and $\gamma$ form a basis for $K=\mathbb{Q}(\alpha,\beta, \gamma)$.
Let $\operatorname{CF}(l_1, l_2, l_3, l_4) \in \mathbf{CF}_{6}$. Then
$$
\begin{equation*}
\begin{gathered} \, G_{6}\mathbf{l}_1= \begin{pmatrix} 1 \\ \sigma(\alpha) \\ \sigma^{2}(\alpha) \\ \dfrac{1 - \alpha - \sigma^{2}(\alpha)}{2} \end{pmatrix}, \qquad G_{6}\mathbf{l}_2= \begin{pmatrix} 1 \\ \sigma^{2}(\alpha) \\ \sigma^{3}(\alpha) \\ \dfrac{1 - \sigma(\alpha) - \sigma^{3}(\alpha)}{2} \end{pmatrix}, \\ G_{6}\mathbf{l}_3= \begin{pmatrix} 1 \\ \sigma^{3}(\alpha) \\ \alpha \\ \dfrac{1 - \sigma^{2}(\alpha) - \alpha}{2} \end{pmatrix}, \qquad G_{6}\mathbf{l}_4= \begin{pmatrix} 1 \\ \alpha \\ \sigma(\alpha) \\ \dfrac{1 - \sigma^{3}(\alpha) - \sigma(\alpha)}{2} \end{pmatrix}, \end{gathered}
\end{equation*}
\notag
$$
that is, $G_{6}(\mathbf{l}_1, \mathbf{l}_2, \mathbf{l}_3, \mathbf{l}_4) = (\mathbf{l}_2, \mathbf{l}_3, \mathbf{l}_4, \mathbf{l}_1)$. Therefore, $G_{6}$ is a proper cyclic symmetry of the continued fraction $\operatorname{CF}(l_1,l_2,l_3,l_4)$. Conversely, assume that $G_{6}$ is a proper cyclic symmetry of $\operatorname{CF}(l_1,l_2,l_3,l_4)$. Then there exists $\mu_2$ such that, up to a permutation of indices,
$$
\begin{equation*}
G_{6}\mathbf{l}_1= \begin{pmatrix} 1 \\ \beta \\ -1-\alpha+2\gamma \\ 1-\gamma \end{pmatrix}=\mu_2\begin{pmatrix} 1 \\ \sigma_{2}(\alpha) \\ \sigma_{2}(\beta) \\ \sigma_{2}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\mu_2=1, \qquad \beta=\sigma_{2}(\alpha), \qquad -1-\alpha+2\gamma =\sigma_{2}(\beta)\quad\text{and} \quad 1- \gamma=\sigma_{2}(\gamma).
\end{equation*}
\notag
$$
There exists $\mu_3$ such that
$$
\begin{equation*}
G_{6}\mathbf{l}_2= \begin{pmatrix} 1 \\ -1-\alpha+2\gamma \\ 1-\beta-2\gamma \\ \gamma \end{pmatrix}=\mu_3\begin{pmatrix} 1 \\ \sigma_{3}(\alpha) \\ \sigma_{3}(\beta) \\ \sigma_{3}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\begin{gathered} \, \mu_3=1, \qquad \frac{\alpha+\sigma_{3}(\alpha)+1}{2}=\gamma, \sigma_{3}(\alpha)=-1- \alpha+2\gamma=\sigma_{2}(\beta)=\sigma^{2}_{2}(\alpha), \\ \sigma_{3}(\beta)=1-\beta-2\gamma=-1 -\beta+(2-2\gamma)=\sigma_{2}(-1 -\alpha+2\gamma)=\sigma^{2}_{2}(\beta) \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\sigma_{3}(\gamma)=\gamma=\sigma_{2}(1-\gamma)=\sigma^{3}_{2}(\gamma).
\end{equation*}
\notag
$$
There exists $\mu_4$ such that
$$
\begin{equation*}
G_{6}\mathbf{l}_3= \begin{pmatrix} 1 \\ 1-\beta- 2\gamma \\ \alpha \\ 1-\gamma \end{pmatrix}=\mu_4\begin{pmatrix} 1 \\ \sigma_{4}(\alpha) \\ \sigma_{4}(\beta) \\ \sigma_{4}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\begin{gathered} \, \mu_4=1, \qquad \sigma_{4}(\alpha)=1-\beta- 2\gamma=-1 -\beta+(2-2\gamma)= \sigma_{2}(-1 -\alpha+2\gamma)=\sigma^{3}_{2}(\alpha), \\ \sigma_{4}(\beta)=\alpha=1 -(-1 -\alpha+2\gamma)-(2- 2\gamma)=\sigma_{2}(1 -\beta-2\gamma)=\sigma^{3}_{2}(\beta), \\ \sigma_{4}(\gamma)=1- \gamma=\sigma_{2}(\gamma)=\sigma^{3}_{2}(\gamma)\quad\text{and} \quad \operatorname{Tr}(\alpha)=0. \end{gathered}
\end{equation*}
\notag
$$
Consequently, $\operatorname{CF}(l_1, l_2, l_3, l_4) \in \mathbf{CF}_{6}$ since the numbers $1$, $\alpha$, $\beta$ and $\gamma$ form a basis for the field $K=\mathbb{Q}(\alpha,\beta, \gamma)$.
Let $\operatorname{CF}(l_1, l_2, l_3, l_4) \in \mathbf{CF}_{7}$. Then
$$
\begin{equation*}
\begin{gathered} \, G_{7}\mathbf{l}_1= \begin{pmatrix} 1 \\ \sigma(\alpha) \\ \sigma^{2}(\alpha) \\ \dfrac{3 - \alpha - \sigma^{2}(\alpha)}{2} \end{pmatrix}, \qquad G_{7}\mathbf{l}_2= \begin{pmatrix} 1 \\ \sigma^{2}(\alpha) \\ \sigma^{3}(\alpha) \\ \dfrac{3 - \sigma(\alpha) - \sigma^{3}(\alpha)}{2} \end{pmatrix}, \\ G_{7}\mathbf{l}_3= \begin{pmatrix} 1 \\ \sigma^{3}(\alpha) \\ \alpha \\ \dfrac{3 - \sigma^{2}(\alpha) - \alpha}{2} \end{pmatrix}, \qquad G_{7}\mathbf{l}_4= \begin{pmatrix} 1 \\ \alpha \\ \sigma(\alpha) \\ \dfrac{3 - \sigma^{3}(\alpha) - \sigma(\alpha)}{2} \end{pmatrix}, \end{gathered}
\end{equation*}
\notag
$$
that is, $G_{7}(\mathbf{l}_1, \mathbf{l}_2, \mathbf{l}_3, \mathbf{l}_4) = (\mathbf{l}_2, \mathbf{l}_3, \mathbf{l}_4, \mathbf{l}_1)$. Therefore, $G_{7}$ is a proper cyclic symmetry of $\operatorname{CF}(l_1,l_2,l_3,l_4)$. Conversely, assume that $G_{7}$ is a proper cyclic symmetry of $\operatorname{CF}(l_1,l_2,l_3,l_4)$. Then there exists $\mu_2$ such that, up to a permutation of indices,
$$
\begin{equation*}
G_{7}\mathbf{l}_1= \begin{pmatrix} 1 \\ \beta \\ -1-\alpha+2\gamma \\ 2-\gamma \end{pmatrix}=\mu_2\begin{pmatrix} 1 \\ \sigma_{2}(\alpha) \\ \sigma_{2}(\beta) \\ \sigma_{2}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\mu_2=1, \qquad \beta=\sigma_{2}(\alpha), \qquad-1-\alpha+2\gamma=\sigma_{2}(\beta)\quad\text{and} \quad 2- \gamma=\sigma_{2}(\gamma).
\end{equation*}
\notag
$$
There exists $\mu_3$ such that
$$
\begin{equation*}
G_{7}\mathbf{l}_2= \begin{pmatrix} 1 \\ -1-\alpha+2\gamma \\ 3-\beta-2\gamma \\ \gamma \end{pmatrix}=\mu_3\begin{pmatrix} 1 \\ \sigma_{3}(\alpha) \\ \sigma_{3}(\beta) \\ \sigma_{3}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\begin{gathered} \, \mu_3=1, \qquad \frac{\alpha+\sigma_{3}(\alpha)+1}{2}=\gamma, \qquad \sigma_{3}(\alpha)=-1- \alpha+2\gamma=\sigma_{2}(\beta)=\sigma^{2}_{2}(\alpha), \\ \sigma_{3}(\beta)=3-\beta-2\gamma=-1 -\beta+(4-2\gamma)=\sigma_{2}(-1 -\alpha+2\gamma)=\sigma^{2}_{2}(\beta) \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\sigma_{3}(\gamma)=\gamma=\sigma_{2}(2-\gamma)=\sigma^{3}_{2}(\gamma).
\end{equation*}
\notag
$$
There exists $\mu_4$ such that
$$
\begin{equation*}
G_{7}\mathbf{l}_3= \begin{pmatrix} 1 \\ 3-\beta- 2\gamma \\ \alpha \\ 2- \gamma \end{pmatrix}=\mu_4\begin{pmatrix} 1 \\ \sigma_{4}(\alpha) \\ \sigma_{4}(\beta) \\ \sigma_{4}(\gamma) \end{pmatrix},
\end{equation*}
\notag
$$
which gives us
$$
\begin{equation*}
\begin{gathered} \, \mu_4=1, \qquad \sigma_{4}(\alpha)=3-\beta- 2\gamma=-1 -\beta+(4-2\gamma)= \sigma_{2}(-1 -\alpha+2\gamma)=\sigma^{3}_{2}(\alpha), \\ \sigma_{4}(\beta)=\alpha=3 -(-\alpha+2\gamma-1)-(4- 2\gamma)=\sigma_{2}(3 -\beta-2\gamma)=\sigma^{3}_{2}(\beta), \\ \sigma_{4}(\gamma)=2- \gamma=\sigma_{2}(\gamma)=\sigma^{3}_{2}(\gamma)\quad\text{and} \quad \operatorname{Tr}(\alpha)=2. \end{gathered}
\end{equation*}
\notag
$$
Consequently, $\operatorname{CF}(l_1, l_2, l_3, l_4) \in \mathbf{CF}_{7}$ since the numbers $1$, $\alpha$, $\beta$ and $\gamma$ form a basis for the field $K=\mathbb{Q}(\alpha,\beta, \gamma)$.
Lemma 5 is proved. For $i=1,\dots,7$ we let $\overline{\mathbf{CF}}_i$ denote the image of $\mathbf{CF}_i$ under the action of $\operatorname{GL}_4(\mathbb{Z})$:
$$
\begin{equation*}
\overline{\mathbf{CF}}_i= \bigl\{ \operatorname{CF}(l_1,l_2,l_3,l_4)\in\mathfrak{A}_3' \mid \exists\, X\in\operatorname{GL}_4(\mathbb{Z}):X(\operatorname{CF}(l_1,l_2,l_3,l_4))\in\mathbf{CF}_i \bigr\}.
\end{equation*}
\notag
$$
Lemma 6. A continued fraction $\operatorname{CF}(l_1,l_2,l_3,l_4)\in\mathfrak{A}_3$ satisfies condition i) in Theorem 2 if and only if it lies in the class $\overline{\mathbf{CF}}_i$, where $i\in\{1,2,3,4,5,6,7\}$. Proof. For $X\in\operatorname{GL}_4(\mathbb{Z})$ an operator $A\in\operatorname{GL}_4(\mathbb{Z})$ is hyperbolic if and only if so is $XAX^{-1}$. Moreover, the eigensubspaces of a hyperbolic operator can uniquely be recovered from any of its eigenvectors. It now remains to use the definition of the equivalence $\sim$ from § 1.
Lemma 6 is proved. In view of Lemma 6 we can reformulate Theorem 2 as follows. A continued fraction $\operatorname{CF}(l_1,l_2,l_3,l_4)\in\mathfrak{A}_3$ features a proper cyclic symmetry $G$ if and only if $\operatorname{CF}(l_1,l_2,l_3,l_4)$ lies in one of the classes $\overline{\mathbf{CF}}_i$, where $i\in\{1,2,3,4,5,6,7\}$. Proof of Theorem 2. If $\operatorname{CF}(l_1,l_2,l_3,l_4)$ lies in some $\overline{\mathbf{CF}}_i$, then by Lemma 5 it has a proper cyclic symmetry $G$, because the action of operators in $\operatorname{GL}_4(\mathbb{Z})$ preserves the property of an algebraic continued fraction to have a proper cyclic symmetry.
Conversely, assume that a continued fraction $\operatorname{CF}(l_1,l_2,l_3,l_4)\in\mathfrak{A}_3$ has a proper cyclic symmetry $G$ with fixed point on some sail $\partial(\mathcal{K}(C)) \in \operatorname{CF}(l_1, l_2, l_3, l_4)$. Consider the points $\mathbf z_1$, $\mathbf z_2$, $\mathbf z_3$ and $\mathbf z_4$ from Lemma 4. Let $\mathbf{e}_1$, $\mathbf{e}_2$, $\mathbf{e}_3$, $\mathbf{e}_4$ be the standard basis for $\mathbb{R}^4$. At least one of assertions 1)–7) of Lemma 4 holds for $\mathbf z_1$, $\mathbf z_2$, $\mathbf z_3$ and $\mathbf z_4$. Let assertion 1) of Lemma 4 hold. Consider the operator $X_{1} \in \operatorname{GL}_4(\mathbb{Z})$ such that
$$
\begin{equation*}
X_{1}\biggl(\mathbf{z}_1, \mathbf{z}_2, \mathbf{z}_3, \frac{1}{4}(\mathbf{z}_{1}+\mathbf{z}_{2}+\mathbf{z}_{3}+\mathbf{z}_{4})\biggr) =(\mathbf{e}_1-\mathbf{e}_2, \mathbf{e}_1+\mathbf{e}_4, \mathbf{e}_1+\mathbf{e}_3-\mathbf{e}_4, \mathbf{e}_1).
\end{equation*}
\notag
$$
Then
$$
\begin{equation*}
X_{1}(\mathbf{z}_4)=X_{1}\biggl(4\cdot\frac{1}{4}(\mathbf{z}_{1} +\mathbf{z}_{2}+\mathbf{z}_{3}+\mathbf{z}_{4})- \mathbf{z}_1-\mathbf{z}_2-\mathbf{z}_3\biggr) =\mathbf{e}_1+\mathbf{e}_2-\mathbf{e}_3
\end{equation*}
\notag
$$
and since by Lemma 4
$$
\begin{equation*}
\begin{aligned} \, &X_{1}GX_{1}^{-1}(\mathbf{e}_1-\mathbf{e}_2, \mathbf{e}_1+\mathbf{e}_4, \mathbf{e}_1+\mathbf{e}_3-\mathbf{e}_4, \mathbf{e}_1+\mathbf{e}_2-\mathbf{e}_3) \\ &\qquad=(\mathbf{e}_1+\mathbf{e}_4, \mathbf{e}_1+\mathbf{e}_3-\mathbf{e}_4, \mathbf{e}_1+\mathbf{e}_2-\mathbf{e}_3, \mathbf{e}_1- \mathbf{e}_2). \end{aligned}
\end{equation*}
\notag
$$
Consequently, $X_{1}(\operatorname{CF}(l_1,l_2,l_3,l_4)) \in \mathbf{CF}_{1}$, that is, $\operatorname{CF}(l_1,l_2,l_3,l_4)\in\overline{\mathbf{CF}}_1$.
Let assertion 2) of Lemma 4 hold. Consider the operator $X_{2} \in \operatorname{GL}_4(\mathbb{Z})$ such that
$$
\begin{equation*}
X_{2}(\mathbf{z}_1, \mathbf{z}_2, \mathbf{z}_3, \mathbf{z}_{4})=(\mathbf{e}_1, \mathbf{e}_1+\mathbf{e}_4, \mathbf{e}_1+\mathbf{e}_3, \mathbf{e}_1+\mathbf{e}_2).
\end{equation*}
\notag
$$
We have $X_{2}GX_{2}^{-1}=G_{2}$, since by Lemma 4
$$
\begin{equation*}
X_{2}GX_{2}^{-1}(\mathbf{e}_1, \mathbf{e}_1+\mathbf{e}_4, \mathbf{e}_1+\mathbf{e}_3, \mathbf{e}_1+\mathbf{e}_2) =( \mathbf{e}_1+\mathbf{e}_4, \mathbf{e}_1+\mathbf{e}_3, \mathbf{e}_1+\mathbf{e}_2, \mathbf{e}_1).
\end{equation*}
\notag
$$
Consequently, $X_{2}(\operatorname{CF}(l_1,l_2,l_3,l_4)) \in \mathbf{CF}_{2}$, that is, $\operatorname{CF}(l_1,l_2,l_3,l_4)\in\overline{\mathbf{CF}}_2$.
Let assertion 3) of Lemma 4 hold. Consider the operator $X_{3} \in \operatorname{GL}_4(\mathbb{Z})$ such that
$$
\begin{equation*}
X_{3}\biggl(\mathbf{z}_1, \frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{2}), \frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{3}), \frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{4})\biggr) =(\mathbf{e}_1, \mathbf{e}_1+\mathbf{e}_4, \mathbf{e}_1+\mathbf{e}_3, \mathbf{e}_1+\mathbf{e}_2 ).
\end{equation*}
\notag
$$
Then
$$
\begin{equation*}
\begin{gathered} \, X_{3}(\mathbf{z}_2)=X_{3}\biggl(2\cdot\frac{1}{2}(\mathbf{z}_1 +\mathbf{z}_2)-\mathbf{z}_1\biggr)=\mathbf{e}_1+ 2\mathbf{e}_4, \\ X_{3}(\mathbf{z}_3)=X_{3}\biggl(2\cdot\frac{1}{2}(\mathbf{z}_1+\mathbf{z}_3)-\mathbf{z}_1\biggr) = \mathbf{e}_1+2\mathbf{e}_3, \\ X_{3}(\mathbf{z}_4)= X_{3}\biggl(2\cdot\frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{4})-\mathbf{z}_1\biggr) =\mathbf{e}_1+2\mathbf{e}_2 \end{gathered}
\end{equation*}
\notag
$$
and since by Lemma 4
$$
\begin{equation*}
X_{3}GX_{3}^{-1}(\mathbf{e}_1, \mathbf{e}_1+2\mathbf{e}_4, \mathbf{e}_1+2\mathbf{e}_3, \mathbf{e}_1+2\mathbf{e}_2) =(\mathbf{e}_1+2\mathbf{e}_4, \mathbf{e}_1+2\mathbf{e}_3, \mathbf{e}_1+2\mathbf{e}_2, \mathbf{e}_1).
\end{equation*}
\notag
$$
Consequently, $X_{3}(\operatorname{CF}(l_1,l_2,l_3,l_4)) \in \mathbf{CF}_{3}$, that is, $\operatorname{CF}(l_1,l_2,l_3,l_4)\in\overline{\mathbf{CF}}_3$.
Let assertion 4) of Lemma 4 hold. Consider the operator $X_{4} \in \operatorname{GL}_4(\mathbb{Z})$ such that
$$
\begin{equation*}
\begin{aligned} \, & X_{4}\biggl(\mathbf{z}_1, \mathbf{z}_2, \frac{1}{2}(\mathbf{z}_1+\mathbf{z}_3), \frac{1}{4}(\mathbf{z}_{1}+\mathbf{z}_{2}+\mathbf{z}_{3}+\mathbf{z}_{4})\biggr) \\ &\qquad=(\mathbf{e}_1-\mathbf{e}_3+\mathbf{e}_4, \mathbf{e}_1 -\mathbf{e}_2+2\mathbf{e}_3-\mathbf{e}_4, \mathbf{e}_1+\mathbf{e}_2-\mathbf{e}_3+\mathbf{e}_4, \mathbf{e}_1). \end{aligned}
\end{equation*}
\notag
$$
Then
$$
\begin{equation*}
\begin{gathered} \, X_{4}(\mathbf{z}_3)=X_{4}\biggl(2\cdot\frac{1}{2}(\mathbf{z}_1+\mathbf{z}_3)- \mathbf{z}_1\biggr) =\mathbf{e}_1+2\mathbf{e}_2-\mathbf{e}_3+\mathbf{e}_4, \\ X_{4}(\mathbf{z}_4)= X_{4}\biggl(4\cdot\frac{1}{4}(\mathbf{z}_{1}+\mathbf{z}_{2}+\mathbf{z}_{3} +\mathbf{z}_{4})-\mathbf{z}_1-\mathbf{z}_2- \mathbf{z}_3\biggr) =\mathbf{e}_1-\mathbf{e}_2-\mathbf{e}_4 \end{gathered}
\end{equation*}
\notag
$$
and since by Lemma 4
$$
\begin{equation*}
\begin{aligned} \, &X_{4}GX_{4}^{-1}(\mathbf{e}_1-\mathbf{e}_3+\mathbf{e}_4, \mathbf{e}_1 -\mathbf{e}_2+2\mathbf{e}_3-\mathbf{e}_4, \mathbf{e}_1+2\mathbf{e}_2-\mathbf{e}_3+\mathbf{e}_4, \mathbf{e}_1-\mathbf{e}_2-\mathbf{e}_4) \\ &\qquad=(\mathbf{e}_1 -\mathbf{e}_2+2\mathbf{e}_3-\mathbf{e}_4, \mathbf{e}_1+2\mathbf{e}_2-\mathbf{e}_3+\mathbf{e}_4, \mathbf{e}_1- \mathbf{e}_2-\mathbf{e}_4, \mathbf{e}_1-\mathbf{e}_3+\mathbf{e}_4). \end{aligned}
\end{equation*}
\notag
$$
Consequently, $X_{4}(\operatorname{CF}(l_1,l_2,l_3,l_4)) \in \mathbf{CF}_{4}$, that is, $\operatorname{CF}(l_1,l_2,l_3,l_4)\in\overline{\mathbf{CF}}_4$.
Let assertion 5) of Lemma 4 hold. Consider the operator $X_{5} \in \operatorname{GL}_4(\mathbb{Z})$ such that
$$
\begin{equation*}
X_{5}\biggl(\mathbf{z}_1, \mathbf{z}_2, \frac{1}{2}(\mathbf{z}_1+\mathbf{z}_3), \frac{1}{2}(\mathbf{z}_2+\mathbf{z}_4)\biggr)=(\mathbf{e}_1, \mathbf{e}_1+\mathbf{e}_4, \mathbf{e}_1+\mathbf{e}_3, \mathbf{e}_1+\mathbf{e}_2).
\end{equation*}
\notag
$$
Then
$$
\begin{equation*}
\begin{gathered} \, X_{5}(\mathbf{z}_3)=X_{5}\biggl(2\cdot\frac{1}{2}(\mathbf{z}_1+\mathbf{z}_3)-\mathbf{z}_1\biggr) =\mathbf{e}_1+ 2\mathbf{e}_3, \\ X_{5}(\mathbf{z}_4)=X_{5}\biggl(2\cdot\frac{1}{2}(\mathbf{z}_2+\mathbf{z}_4)-\mathbf{z}_2\biggr) = \mathbf{e}_1+2\mathbf{e}_2+\mathbf{e}_4 \end{gathered}
\end{equation*}
\notag
$$
and since by Lemma 4
$$
\begin{equation*}
\begin{aligned} \, &X_{5}GX_{5}^{-1}(\mathbf{e}_1, \mathbf{e}_1+\mathbf{e}_4, \mathbf{e}_1+2\mathbf{e}_3, \mathbf{e}_1+2\mathbf{e}_2+\mathbf{e}_4) \\ &\qquad=( \mathbf{e}_1+\mathbf{e}_4, \mathbf{e}_1+2\mathbf{e}_3, \mathbf{e}_1+2\mathbf{e}_2+\mathbf{e}_4, \mathbf{e}_1). \end{aligned}
\end{equation*}
\notag
$$
Consequently, $X_{5}(\operatorname{CF}(l_1,l_2,l_3,l_4)) \in \mathbf{CF}_{5}$, that is, $\operatorname{CF}(l_1,l_2,l_3,l_4)\in\overline{\mathbf{CF}}_5$.
Let assertion 6) of Lemma 4 hold. Consider the operator $X_{6} \in \operatorname{GL}_4(\mathbb{Z})$ such that
$$
\begin{equation*}
X_{6}\biggl(\mathbf{z}_1, \mathbf{z}_{2}, \mathbf{z}_{3}, \frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{3}+\mathbf{z}_{4}- \mathbf{z}_{2})\biggr) =(\mathbf{e}_1+\mathbf{e}_2+\mathbf{e}_4, \mathbf{e}_1, \mathbf{e}_1-\mathbf{e}_3+\mathbf{e}_4, \mathbf{e}_1+\mathbf{e}_4).
\end{equation*}
\notag
$$
Then
$$
\begin{equation*}
X_{6}(\mathbf{z}_4)=X_{6}\biggl(2\cdot\frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{3}+\mathbf{z}_{4} -\mathbf{z}_{2})- \mathbf{z}_1+\mathbf{z}_2-\mathbf{z}_3\biggr) =\mathbf{e}_1-\mathbf{e}_2+\mathbf{e}_3
\end{equation*}
\notag
$$
and since by Lemma 4
$$
\begin{equation*}
\begin{aligned} \, &X_{6}GX_{6}^{-1}(\mathbf{e}_1+\mathbf{e}_2+\mathbf{e}_4, \mathbf{e}_1, \mathbf{e}_1-\mathbf{e}_3+\mathbf{e}_4, \mathbf{e}_1-\mathbf{e}_2+\mathbf{e}_3) \\ &\qquad=(\mathbf{e}_1, \mathbf{e}_1-\mathbf{e}_3+\mathbf{e}_4, \mathbf{e}_1-\mathbf{e}_2+\mathbf{e}_3, \mathbf{e}_1+\mathbf{e}_2+\mathbf{e}_4). \end{aligned}
\end{equation*}
\notag
$$
Consequently, $X_{6}(\operatorname{CF}(l_1,l_2,l_3,l_4)) \in \mathbf{CF}_{6}$, that is, $\operatorname{CF}(l_1,l_2,l_3,l_4)\in\overline{\mathbf{CF}}_6$.
Let assertion 7) of Lemma 4 hold. Consider the operator $X_{7} \in \operatorname{GL}_4(\mathbb{Z})$ such that
$$
\begin{equation*}
\begin{aligned} \, &X_{7}\biggl(\mathbf{z}_1, \mathbf{z}_{2}, \mathbf{z}_{3}, \frac{1}{2}(\mathbf{z}_{1}+\mathbf{z}_{2})+\frac{1}{4}(\mathbf{z}_{1}+\mathbf{z}_{4}-\mathbf{z}_{3}- \mathbf{z}_{2})\biggr) \\ &\qquad=(\mathbf{e}_1+\mathbf{e}_2-\mathbf{e}_3+2\mathbf{e}_4, \mathbf{e}_1-\mathbf{e}_2+2\mathbf{e}_3, \mathbf{e}_1+2\mathbf{e}_2+2\mathbf{e}_4, \mathbf{e}_1+\mathbf{e}_4). \end{aligned}
\end{equation*}
\notag
$$
Then
$$
\begin{equation*}
X_{7}(\mathbf{z}_4)= X_{7}\biggl(4\cdot\frac{3\mathbf{z}_1+\mathbf{z}_2-\mathbf{z}_3+\mathbf{z}_4}{4} -3\mathbf{z}_1-\mathbf{z}_2+\mathbf{z}_3\biggr) = \mathbf{e}_1+\mathbf{e}_3
\end{equation*}
\notag
$$
and since by Lemma 4
$$
\begin{equation*}
\begin{aligned} \, &X_{7}GX_{7}^{-1}(\mathbf{e}_1+\mathbf{e}_2-\mathbf{e}_3+2\mathbf{e}_4, \mathbf{e}_1-\mathbf{e}_2+2\mathbf{e}_3, \mathbf{e}_1+2\mathbf{e}_2+2\mathbf{e}_4, \mathbf{e}_1+\mathbf{e}_3) \\ &\qquad=(\mathbf{e}_1-\mathbf{e}_2+2\mathbf{e}_3, \mathbf{e}_1+2\mathbf{e}_2+2\mathbf{e}_4, \mathbf{e}_1+\mathbf{e}_3, \mathbf{e}_1+\mathbf{e}_2-\mathbf{e}_3+2\mathbf{e}_4). \end{aligned}
\end{equation*}
\notag
$$
Consequently, $X_{7}(\operatorname{CF}(l_1,l_2,l_3,l_4)) \in \mathbf{CF}_{7}$, that is, $\operatorname{CF}(l_1,l_2,l_3,l_4)\in\overline{\mathbf{CF}}_7$.
Theorem 2 is proved. AcknowledgementsThe author is grateful to O. N. German for his constant attention to this work and for numerous extremely helpful discussions. Being a winner of the “Junior Leader” contest arranged by the “BASIS” Theoretical Physics and Mathematics Advancement Foundation, the author would like to thank the sponsors and jury of the contest. 
Citation in format AMSBIB
\Bibitem{Tly22} |
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