Chevalley groups of types $B_n$, $C_n$, $D_n$ over certain fields do not possess the $R_{\infty}$-property

TL;DR

This paper proves that Chevalley groups of types B, C, D over certain infinite transcendence degree fields do not have the $R_{ty}$-property, extending previous results known for other types.

Contribution

It demonstrates that Chevalley groups of types B, C, D over fields with infinite transcendence degree lack the $R_{ty}$-property, filling a gap in the understanding of these groups.

Findings

Chevalley groups of types B, C, D over fields with infinite transcendence degree do not possess the $R_{ty}$-property.

The result extends known properties from other types to these classical series.

The paper clarifies the behavior of automorphisms and twisted conjugacy classes in these groups.

Abstract

Let $F$ be an algebraically closed field of zero characteristic. If the transcendence degree of $F$ over $\mathbb{Q}$ is finite, then all Chevalley groups over $F$ are known to possess the $R_{\infty}$-property. If the transcendence degree of $F$ over $\mathbb{Q}$ is infinite, then Chevalley groups of type $A_n$ over $F$ do not possess the $R_{\infty}$-property. In the present paper, we consider Chevalley groups of classical series $B_n$, $C_n$, $D_n$ over $F$ in the case when the transcendence degree of $F$ over $\mathbb{Q}$ is infinite, and prove that such groups do not possess the $R_{\infty}$-property.

Equations

$$\left(\begin{array}{cc}\text{lx@intercol}\text{hfil}A\text{hfil}\text{lx@intercol}\text{vrule}\text{lx@intercol}& O_{n\times m}\\ \text{cline}1-2& \\ O_{m\times n}& \text{vrule}\text{lx@intercol}\text{hfil}B\text{hfil}\text{lx@intercol}\end{array}\right).$$

$${\mathrm{O}}_n(F)=\left\{A\in {\mathrm{GL}}_n(F)\text{leavevmode}|\text{leavevmode}A{A}^T={\mathrm{I}}_n\right\},$$

$$Y_1$$

$$Y_2$$

$$Y_3$$

$$\text{leavevmode}\text{leavevmode}⋮$$

$$Y_n$$

$$XD{X}^{-1}=\mathrm{diag}(\underset{m}{\underbrace{-1,\dots ,-1}},1,\dots ,1)$$

$$Z=\frac{1}{\sqrt{{\alpha }^2+{\beta }^2}}\left(\begin{array}{cc}\alpha & \beta \\ -\beta & \alpha \end{array}\right)$$

$$T=\underset{m/2}{\underbrace{Z\oplus \cdots \oplus Z}}\oplus {\mathrm{I}}_{n-m}.$$

$$T^{-1}{\phi }_n(T)=\mathrm{diag}(\underset{m}{\underbrace{-1,\dots ,-1}},1,\dots ,1)=XD{X}^{-1}.$$

$$T={\mathrm{I}}_1\oplus \underset{(m-1)/2}{\underbrace{Z\oplus \cdots \oplus Z}}\oplus {\mathrm{I}}_{n-m}.$$

$$T^{-1}\mathrm{diag}(-1,1,\dots ,1){\phi }_n(T)=\mathrm{diag}(\underset{m}{\underbrace{-1,\dots ,-1}},1,\dots ,1)=XD{X}^{-1}.$$

$$-1=\det (\mathrm{diag}(-1,1,\dots ,1))=\det (X^{-1})\det ({\phi }_n(X)).$$

$${\mathrm{I}}_n$$

$$=X^{-1}{\phi }_n(X)(X^{-1}{\phi }_n(X){)}^T$$

$$=X^{-1}{\phi }_n(X){\phi }_n(X{)}^T(X^{-1}{)}^T$$

$$=(LQ{)}^{-1}{\phi }_n(LQ){\phi }_n(LQ{)}^T((LQ{)}^{-1}{)}^T$$

$$=Q^{-1}{L}^{-1}{\phi }_n(L){\phi }_n(Q){\phi }_n(Q{)}^T{\phi }_n(L{)}^T(L^{-1}{)}^T(Q^{-1}{)}^T$$

$$L^{-1}{\phi }_n(L)=L^T{\phi }_n(L^T{)}^{-1}.$$

$$A=X^{-1}{\phi }_n(X)=(LQ{)}^{-1}{\phi }_n(L)\phi (Q)=Q^{-1}D{\phi }_n(Q).$$

$$J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right),$$

$${\mathrm{Sp}}_{2n}(F)=\left\{A\in {\mathrm{GL}}_{2n}(F)\text{leavevmode}|\text{leavevmode}A{\Omega }_n{A}^T={\Omega }_n\right\}.$$

$$⟨U,V⟩=U{\Omega }_n{V}^T=u_{1}J{v}_1^T+\cdots +a_{n}J{v}_n^T.$$

$$⟨U+V,W⟩$$

$$⟨xU,V⟩$$

$$⟨U,U⟩=\mathrm{d}(U)J$$

$$Y_1$$

$$Y_2$$

$$Y_3$$

$$\text{leavevmode}\text{leavevmode}⋮$$

$$Y_n$$

$$f_1({\mathrm{I}}_{2n})=\mathrm{d}\left(({\mathrm{I}}_2,O_{2\times 2},\dots ,O_{2\times 2})\right)=\det ({\mathrm{I}}_2)=1,$$

$$f_2({\mathrm{I}}_{2n})=\left((O_{2\times 2},{\mathrm{I}}_2,O_{2\times 2},\dots ,O_{2\times 2})\right)=\det ({\mathrm{I}}_2)=1,$$

$$⟨Y_2,Y_1⟩$$

$$\stackrel{(\text{ref}lin1)}{=}⟨X_2,Y_1⟩+\mathrm{d}(Y_1{)}^{-1}⟨X_2,Y_1⟩J⟨Y_1,Y_1⟩$$

$$\stackrel{(\text{ref}inv1)}{=}⟨X_2,Y_1⟩+\mathrm{d}(Y_1{)}^{-1}⟨X_2,Y_1⟩J\mathrm{d}(Y_1)J$$

$$=⟨X_2,Y_1⟩-⟨X_2,Y_1⟩=0.$$

Full text

Keywords: twisted conjugacy classes, Reidemeister number, Chevalley groups.

1 Introduction

Let $G$ be a group and $\varphi$ be an automorphism of $G$. Elements $x$, $y$ from $G$ are said to be $\varphi$-conjugated if there exists an element $z\in G$ such that $x=zy\varphi(z)^{-1}$. The relation of $\varphi$-conjugation is an equivalence relation and it divides $G$ into $\varphi$-conjugacy classes. The number $R(\varphi)$ of these classes is called the Reidemeister number of $\varphi$.

Twisted conjugacy classes appear naturally in Nielsen-Reidemeister fixed point theory. Let $X$ be a finite polyhedron and $f:X\to X$ be a homeomorphism of $X$. Two fixed points $x,y$ of $f$ are said to belong to the same fixed point class of $f$ if there exists a path $c:[0,1]\to X$ with $c(0)=x$ and $c(1)=y$ such that $c\simeq f\circ c$, where $\simeq$ denotes the homotopy with fixed endpoints. The relation of being in the same fixed point class is an equivalence relation on the set of fixed points of $f$. The number $R(f)$ of fixed point classes of $f$ is called the Reidemeister number of the homeomorphism $f$. The Reidemeister number $R(f)$ is the homotopic invariant of $f$ and it plays a crucial role in the Nielsen-Reidemeister fixed point theory [8]. Denote by $G=\pi_{1}(X)$ the fundamental group of $X$, and by $\varphi$ the automorphism of $G$ induced by $f$. In this notation the number $R(f)$ of fixed point classes of $f$ is equal to the number $R(\varphi)$ of $\varphi$-conjugacy classes in $G$ (see [8, Chapter III, Lemma 1.2]). Thus, the topological problem of finding $R(f)$ reduces to the purely algebraic problem of finding $R(\varphi)$.

The Reidemeister number is either a positive integer or infinity and we do not distinguish different infinite cardinal numbers denoting all of them by the symbol $\infty$. If $R(\varphi)=\infty$ for all automorphisms $\varphi$ of $G$, then $G$ is said to possess the $R_{\infty}$-property. The problem of determining groups which possess the $R_{\infty}$-property was formulated by A. Fel’shtyn and R. Hill [4]. The study of this problem has been quite an active research topic in recent years. We refer to the paper [5] for an overview of the families of groups which have been studied in this context until 2016. More recent results can be found in [3, 9, 15, 17, 18, 19]. For the immediate consequences of the $R_{\infty}$-property for topological fixed point theory see [7]. Some aspects of the $R_{\infty}$-property can be found in [6].

The author studied conditions which imply the $R_{\infty}$-property for different linear groups over rings [11, 13, 15] and fields [5, 12, 14, 15]. In particular, it was proved that if $F$ is a field of zero characteristic which has either finite transcendence degree over $\mathbb{Q}$, or periodic group of automorphisms, then every Chevalley group (of normal type) over $F$ possesses the $R_{\infty}$-property [12]. In the present paper we consider Chevalley groups of classical series $B_{n},C_{n},D_{n}$ over algebraically closed fields of zero characteristic which have infinite transcendence degree over $\mathbb{Q}$, and prove the following main result.

Theorem 7. *Let $F$ be an algebraically closed field of zero characteristic which has infinite transcendence degree over $\mathbb{Q}$. Then Chevalley groups of types $B_{n},C_{n},D_{n}$ over $F$ do not possess the $R_{\infty}$-property.

In [14, Theorem 7] a similar result was proved for Chevalley groups of type $A_{n}$, therefore we do not consider the case of root systems $A_{n}$ in the present paper. In order to prove Theorem 7 we use classical matrix representations of adjoint Chevalley groups of types $B_{n}$, $C_{n}$, $D_{n}$. Chevalley groups of types $E_{6}$, $E_{7}$, $E_{8}$, $F_{4}$, $G_{2}$ are not isomorphic to certain classical linear groups and we cannot use the same technique.

Theorem 7 together with the results [5, Theorem 3.2] and [14, Theorem 7] imply the following result which gives necessary and sufficient condition of the $R_{\infty}$-property for Chevalley groups of classical series $A_{n},B_{n},C_{n},D_{n}$ over algebraically closed fields.

Theorem 8. *Let $F$ be an algebraically closed field of zero characteristic. Then Chevalley groups of types $A_{n},B_{n},C_{n},D_{n}$ over $F$ possess the $R_{\infty}$-property if and only if the transcendence degree of $F$ over $\mathbb{Q}$ is finite.

The paper is organized as follows. In Section 2 we give necessary preliminaries from field theory and recall the construction from [14] of one specific automorphism of the field $\overline{\mathbb{Q}(s_{1},s_{2},\dots)}$, where $s_{1},s_{2},\dots$ is a countable set of variables. In Sections 3, 4 we prove that orthogonal and symplectic linear groups over $\overline{\mathbb{Q}(s_{1},s_{2},\dots)}$ do not possess the $R_{\infty}$-property. In Section 5 we prove that orthogonal and symplectic linear groups over an arbitrary algebraically closed field $F$ of zero characteristic with infinite transcendence degree over $\mathbb{Q}$ do not possess the $R_{\infty}$-property. In Section 6 we consider classical Chevalley groups over an arbitrary algebraically closed field of zero characteristic with infinite transcendence degree over $\mathbb{Q}$ and prove Theorem 7 and Theorem 8.

2 One automorphism of $\overline{\mathbb{Q}(s_{1},s_{2},\dots)}$

Let $L$ be a subfield of a field $F$, and $X$ be a subset of $F$. The minimal subfield of $F$ which contains $L$ and $X$ is denoted by $L(X)$. Elements $x_{1},\dots,x_{k}$ of $F$ are called algebraically independent over $L$ if there is no polynomial $f(t_{1},\dots,t_{k})\neq 0$ with coefficients from $L$ such that $f(x_{1},\dots,x_{k})=0$. An infinite set $X$ of elements from $F$ is called algebraically independent over $L$ if every finite subset of $X$ is algebraically independent over $L$. A maximal set of algebraically independent over $L$ elements from $F$ is called a transcendence basis of $F$ over $L$. The cardinality of a transcendence basis of $F$ over $L$ does not depend on this basis and is called the transcendence degree of $F$ over $L$. If $X$ is a transcendence basis of $F$ over $L$, then the subfield $L(X)$ of $F$ is isomorphic to the field of rational functions over the set of variables $X$ with coefficients from $L$. A field $F$ is said to be algebraically closed if for every polynomial $f(t)$ of non-zero degree with coefficients from $F$ there exists an element $x$ in $F$ such that $f(x)=0$. The minimal algebraically closed field which contains $F$ is called the algebraic closure of $F$ and is denoted by $\overline{F}$. For every field there exists a unique (up to isomorphism) algebraic closure. If $F$ is an algebraically closed field with the prime subfield $L$ (i. e. $L=F_{p}$ or $L=\mathbb{Q}$) and the trascendence basis $X$ of $F$ over $L$, then $F=\overline{L(X)}$, therefore every algebraically closed field is completely determined by its characteristic and the trascendence degree over the prime subfield.

For $n\geq 2$ denote by ${\rm M}_{n}(F)$ the set of all $n\times n$ matrices over $F$. If $\varphi$ is an automorphism of $F$, then $\varphi$ induces a map $\varphi_{n}:{\rm M}_{n}(F)\to{\rm M}_{n}(F)$ which maps a matrix $A=(a_{i,j})\in{\rm M}_{n}(F)$ to $\varphi_{n}(A)=(\varphi(a_{i,j}))$. The map $\varphi_{n}$ induces an automorphism of classical linear groups (general linear group ${\rm GL}_{n}(F)$, special linear group ${\rm SL}_{n}(F)$, orthogonal group ${\rm O}_{n}(F)$, $\dots$) which we denote by the same symbol $\varphi_{n}$. Note that if an $n\times n$ matrix $A$ has rational coefficients, then $\varphi_{n}(A)=A$.

Let $F$ be an algebraically closed field of zero characteristic with countable transcendence degree over $\mathbb{Q}$. This field is isomorphic to $\overline{\mathbb{Q}(S)}$, where $S=\{s_{1},s_{2},\dots\}$ is a countable set of variables. In [14, Theorem 6] it is proved that there exists an automorphism $\varphi$ of $F=\overline{\mathbb{Q}(S)}$ which induces an automorphism $\varphi_{n}$ of ${\rm GL}_{n}(F)$ with $R(\varphi_{n})=1$. The following theorem gives some details about this automorphism $\varphi$ which follow from the proof of [14, Theorem 6].

Theorem** 1****.**

Let $S$ be a countable set of variables, $F=\overline{\mathbb{Q}(S)}$, and $n\geq 1$ be a positive integer. Then there exists an automorphism $\varphi$ of $F$ (depending on $n$) such that for every matrix $A\in{\rm GL}_{n}(F)$ there exists an $n\times n$ matrix $X$ such that the entries of $X$ are algebraically independent over $\mathbb{Q}$ and $\varphi_{n}(X)=XA$.

Since entries of the matrix $X$ from Theorem 1 are algebraically independent over $\mathbb{Q}$, and ${\rm det}(X)$ is the polynomial over the entries of $X$ with integer coefficients, we have ${\rm det}(X)\neq 0$ and $X\in{\rm GL}_{n}(F)$. So, from Theorem 1 follows that for each matrix $A\in{\rm GL}_{n}(F)$ there exists a matrix $X\in{\rm GL}_{n}(F)$ such that $A=X^{-1}\varphi_{n}(X)$, i. e. all matrices from ${\rm GL}_{n}(F)$ are $\varphi_{n}$-conjugated to the identity matrix and $R(\varphi_{n})=1$.

3 Orthogonal groups over $\overline{\mathbb{Q}(s_{1},s_{2},\dots)}$

We use classical notation. Symbols $I_{n}$ and $O_{n\times m}$ denote the identity $n\times n$ matrix and the $n\times m$ matrix with zero entries, respectively. If $A$ an $n\times n$ matrix and $B$ an $m\times m$ matrix over a field $F$, then the symbol $A\oplus B$ denotes the direct sum of the matrices $A$ and $B$, i. e. the block-diagonal $(m+n)\times(m+n)$ matrix

[TABLE]

It is clear that if $\varphi$ is an automorphism of $F$, then $\varphi_{n+m}(A\oplus B)=\varphi_{n}(A)\oplus\varphi_{m}(B)$. We denote by ${\rm GL}_{n}(F)$, ${\rm SL}_{n}(F)$, respectively, the group of $n\times n$ invertible matrices over $F$, and the group of $n\times n$ matrices over $F$ with determinant $1$. The orthogonal group ${\rm O}_{n}(F)$ over $F$ is the group

[TABLE]

where T denotes transpose. Denote by ${\rm SO}_{n}(F)={\rm O}_{n}(F)\cap{\rm SL}_{n}(F)$ the special orthogonal group, and by $\Omega_{n}(F)$ the derived subgroup of ${\rm O}_{n}(F)$. Symbols ${\rm PO}_{n}(F)$, ${\rm PSO}_{n}(F)$, ${\rm P}\Omega_{n}(F)$ denote, respectively, the quotients of the groups ${\rm O}_{n}(F)$, ${\rm SO}_{n}(F)$, $\Omega_{n}(F)$ by its centers. If $n\geq 2$, and $F$ is an algebraically closed field, then from [1, Theorem 5.17] follows that $\Omega_{n}(F)={\rm SO}_{n}(F)$. The group of lower triangular matrices over $F$ is denoted by ${\rm T}_{n}(F)=\{(a_{i,j})\in{\rm GL}_{n}(F)\leavevmode\nobreak\ |\leavevmode\nobreak\ a_{i,i}\neq 0,\leavevmode\nobreak\ \text{for}\leavevmode\nobreak\ i=1,\dots,n,\leavevmode\nobreak\ a_{i,j}=0\leavevmode\nobreak\ \text{for}\leavevmode\nobreak\ i<j\}$.

It is well known that every matrix $X$ from ${\rm GL}_{n}(\mathbb{R})$ can be presented as a product $X=LQ$ for some $L\in{\rm T}_{n}(\mathbb{R})$, $Q\in{\rm O}_{n}(\mathbb{R})$ (see, for example, [16, Section II, Lecture 7]). The following lemma says that some invertible matrices over an algebraically closed field of zero characteristic can be also presented in such form.

Lemma** 1****.**

Let $F$ be an algebraically closed field of zero characteristic, and $X$ be an $n\times n$ matrix such that entries of $X$ are algebraically independent over $\mathbb{Q}$. Then there exist matrices $L\in{\rm T}_{n}(F)$, $Q\in{\rm O}_{n}(F)$ such that $X=LQ$.

Proof. For $U=(u_{1},\dots,u_{n}),V=(v_{1},\dots,v_{n})\in F^{n}$ denote by $\langle U,V\rangle$ the bilinear form $\langle U,V\rangle=u_{1}v_{1}+u_{2}v_{2}+\dots+u_{n}v_{n}$ on $F^{n}\times F^{n}$. Let $X_{1},\dots,X_{n}$ be the rows of the matrix $X$. Denote by $Y_{1},\dots,Y_{n}$ the following vectors from $F^{n}$.

[TABLE]

The element $Y_{1}$ is uniquely defined in (1). In order the element $Y_{2}$ be well defined we need to check that $\langle Y_{1},Y_{1}\rangle\neq 0$. If $\langle Y_{1},Y_{1}\rangle\neq 0$, then in order $Y_{3}$ be well defined we need to check that $\langle Y_{2},Y_{2}\rangle\neq 0$. So, in order formulas (1) be well defined, we need subsequently check that $\langle Y_{k},Y_{k}\rangle\neq 0$ for all $k=1,\dots,n-1$.

Since the entries of $X$ are algebraically independent over $\mathbb{Q}$, we can think about these entries as about variables, and we can think about $\langle Y_{1},Y_{1}\rangle$ as about rational function from $\mathbb{Q}^{n\times n}$ to $\mathbb{Q}$ (elements in entries of $Y_{1}$ are rational functions in terms of entries of the matrix $X$, and $\langle Y_{1},Y_{1}\rangle$ is a rational function over entries of $Y_{1}$). Denote this function by $f_{1}(X)=\langle Y_{1},Y_{1}\rangle$. Since $f_{1}({\rm I}_{n})=\langle(1,0,\dots,0),(1,0,\dots,0)\rangle=1$ , the function $f_{1}(X)$ is not constantly equal to zero, therefore $\langle Y_{1},Y_{1}\rangle\neq 0$ as an element from $F$, and $Y_{2}$ is well defined. In a similar way, we can think about $\langle Y_{2},Y_{2}\rangle$ as about rational function from $\mathbb{Q}^{n\times n}$ to $\mathbb{Q}$. Denote this function by $f_{2}(X)=\langle Y_{2},Y_{2}\rangle$. Since $f_{2}({\rm I}_{n})=\langle(0,1,0,\dots,0),(0,1,0,\dots,0)\rangle=1$, the function $f_{2}(X)$ is not constantly equal to zero, therefore $\langle Y_{2},Y_{2}\rangle\neq 0$ as an element from $F$, and $Y_{3}$ is well defined. Using the same considerations, we prove that $\langle Y_{k},Y_{k}\rangle\neq 0$ for all $k=1,\dots,n-1$ and therefore formulas (1) uniquely define $Y_{1},\dots,Y_{n}$ from $X_{1},\dots,X_{n}$.

For $1\leq i\neq j\leq n$ it is easy to check that $\langle Y_{i},Y_{j}\rangle=0$ (formulas (1) are the same as in the classical Gramm-Schmidt orthogonalization process. See, for example, [16, Section II, Lecture 8]). So, denoting by $Q_{k}=Y_{k}/\sqrt{\langle Y_{k},Y_{k}\rangle}$ (we can take square roots since $F$ is algebraically closed) we see that $\langle Q_{i},Q_{j}\rangle=0$ if $i\neq j$ and $\langle Q_{i},Q_{i}\rangle=1$ for $1\leq i,j\leq n$. It means that the matrix $Q$ with the first line $Q_{1}$, the second line $Q_{2}$, $\dots$, the $n$-th line $Q_{n}$, is orthogonal. From formulas (1) follows that $Q=TX$ for $T\in{\rm T}_{n}(F)$. Denoting by $L=T^{-1}$, we conclude the proof of the lemma. $\square$

Lemma** 2****.**

Let $S$ be a countable set of variables, $F=\overline{\mathbb{Q}(S)}$, $n\geq 1$ be an integer, and $\varphi$ be an automorphism of $F$ introduced in Theorem 1. Then every diagonal matrix $D\in{\rm O}_{n}(F)$ is $\varphi_{n}$-conjugated in ${\rm O}_{n}(F)$ either to ${\rm I}_{n}$, or to ${\rm diag}(-1,\underbrace{1,\dots,1}_{n-1})$.

Proof. If $n=1$, then the statement is obvious. So, we can assume that $n\geq 2$. Since $D$ is orthogonal diagonal matrix, $D={\rm diag}(\varepsilon_{1},\dots,\varepsilon_{n})$ for $\varepsilon_{1},\dots,\varepsilon_{n}\in\{\pm 1\}$. Denote by $m$ the number of $-1$’s on the diagonal of $D$. Then there exists a permutation matrix $X$ such that

[TABLE]

By Theorem 1 there exists a matrix $Y=(y_{i,j})$ such that the entries of $Y$ are algebraically independent over $\mathbb{Q}$ and $\varphi_{n}(Y)=-Y$. Since $n\geq 2$, there exist two algebraically independent over $\mathbb{Q}$ elements $\alpha=y_{1,1}$, $\beta=y_{1,2}$ such that $\varphi(\alpha)=-\alpha$, $\varphi(\beta)=-\beta$. Since $\alpha,\beta$ are algebraically independent over $\mathbb{Q}$, we have $\alpha^{2}+\beta^{2}\neq 0$. Denote by $Z$ the following $2\times 2$ matrix.

[TABLE]

If $m$ is even, then denote by

[TABLE]

Using direct calculations, it is clear that $T\in{\rm O}_{n}(F)$. From the formulas for matrix $Z$ follows that

[TABLE]

Since $X$ is permutation matrix, $X\in{\rm O}_{n}(F)$ and $\varphi_{n}(X)=X$ (since entries of $X$ are integers). Therefore $D=(TX)^{-1}\varphi_{n}(TX)$, where $TX\in{\rm O}_{n}(F)$, i. e. $D$ is $\varphi_{n}$-conjugated in ${\rm O}_{n}(F)$ to ${\rm I}_{n}$.

If $m$ is odd, then denote by

[TABLE]

Using direct calculations, it is clear that $T\in{\rm O}_{n}(F)$. From the formulas for matrix $Z$ follows that

[TABLE]

Similar to the case when $m$ is even we see that $D=(TX)^{-1}{\rm diag}(-1,1,\dots,1)\varphi_{n}(TX)$, where $TX\in{\rm O}_{n}(F)$, i. e. $D$ is $\varphi_{n}$-conjugated in ${\rm O}_{n}(F)$ to ${\rm diag}(-1,1,\dots,1)$. $\square$

Theorem** 2****.**

Let $S$ be a countable set of variables, $F=\overline{\mathbb{Q}(S)}$, $n\geq 1$ be an integer, and $\varphi$ be an automorphism of $F$ introduced in Theorem 1. Then $\varphi$ induces an automorphism $\varphi_{n}$ of ${\rm O}_{n}(F)$ with $R(\varphi_{n})=2$.

Proof. At first, note that ${\rm I}_{n}$ and ${\rm diag}(-1,1,\dots,1)$ cannot be $\varphi_{n}$-conjugated. Indeed, if ${\rm I}_{n}$ and ${\rm diag}(-1,1,\dots,1)$ are $\varphi_{n}$-conjugated, then there exists a matrix $X\in{\rm O}_{n}(F)$ such that ${\rm diag}(-1,1,\dots,1)=X^{-1}\varphi_{n}(X)$. Therefore

[TABLE]

Since $X\in{\rm O}_{n}(F)$, ${\rm det}(X)=\varepsilon$, where $\varepsilon\in\{\pm 1\}$. Since $\varphi_{n}$ acts as $\varphi$ on entries of $X$, and ${\rm det}(X)$ is a polynomial with integer coefficients over entries of $X$, we have ${\rm det}(\varphi_{n}(X))=\varphi({\rm det}(X))=\varphi(\varepsilon)=\varepsilon$. From equality (2) follows that $-1=1$ and we have contradiction, i. e. ${\rm I}_{n}$ and ${\rm diag}(-1,1,\dots,1)$ are not $\varphi_{n}$-conjugated and $R(\varphi_{n})\geq 2$. Let us prove that $R(\varphi_{n})=2$, i. e. that every matrix $A\in{\rm O}_{n}(F)$ is $\varphi_{n}$-conjugated in ${\rm O}_{n}(F)$ either to ${\rm I}_{n}$ or to ${\rm diag}(-1,1,\dots,1)$.

Let $A$ be a matrix from ${\rm O}_{n}(F)$. By Theorem 1 there exists a matrix $X\in{\rm GL}_{n}(F)$ such that the entries of $X$ are algebraically independent over $\mathbb{Q}$ and $A=X^{-1}\varphi_{n}(X)$. From Lemma 1 follows that $X$ can be presented as a product $X=LQ$, where $L$ is a lower triangular matrix over $F$, and $Q$ is an orthogonal matrix over $F$. Since $A$ is orthogonal matrix, we have

[TABLE]

Since $Q$ is orthogonal, from equality (3) follows that ${\rm I}_{n}=L^{-1}\varphi_{n}(L)\varphi_{n}(L)^{T}(L^{-1})^{T}$ or

[TABLE]

Since $L$ is a lower triangular matrix, the matrix on the left in equality (4) is lower triangular, and the matrix on the right in equality (4) is upper triangular. It is possible only in the case when $L^{-1}\varphi_{n}(L)=D$ is diagonal. Therefore we have the following equality.

[TABLE]

Since $A,Q$ are orthogonal, from equality (5) follows that $D$ is diagonal orthogonal matrix. From Lemma 2 follows that $D$ is $\varphi_{n}$-conjugated in ${\rm O}_{n}(F)$ either to ${\rm I}_{n}$ or to ${\rm diag}(-1,1,\dots,1)$. Therefore $A$ is $\varphi_{n}$-conjugated in ${\rm O}_{n}(F)$ either to ${\rm I}_{n}$ or to ${\rm diag}(-1,1,\dots,1)$. $\square$

4 Symplectic groups over $\overline{\mathbb{Q}(s_{1},s_{2},\dots)}$

Denote by $J$, $\Omega_{n}$ the following $2\times 2$ and $2n\times 2n$ matrices

[TABLE]

The symplectic group ${\rm Sp}_{2n}(F)$ over a field $F$ is the group

[TABLE]

It is easy to see that ${\rm Sp}_{2}(F)={\rm SL}_{2}(F)$. Denote by ${\rm PSp}_{2n}(F)$ the quotient of ${\rm Sp}_{2n}(F)$ by its center.

The set ${\rm M}_{2}(F)$ of all $2\times 2$ matrices over $F$ forms a (non commutative) ring and we can think about a matrix $A\in{\rm M}_{2n}(F)$ as about a matrix from ${\rm M}_{n}({\rm M}_{2}(F))$ over ${\rm M}_{2}(F)$ dividing $A$ into $2\times 2$ blocks. We say that a matrix $A\in{\rm M}_{2n}(F)={\rm M}_{n}({\rm M}_{2}(F))$ is a lower block triangular with $2\times 2$ blocks matrix if it belongs to the set $\{(a_{i,j})\in{\rm M}_{n}({\rm M}_{2}(F))\leavevmode\nobreak\ |\leavevmode\nobreak\ a_{i,j}=O_{2\times 2}\leavevmode\nobreak\ \text{for}\leavevmode\nobreak\ i<j\}$.

Lemma** 3****.**

Let $F$ be an algebraically closed field of zero characteristic, and $X$ be a $2n\times 2n$ matrix such that the entries of $X$ are algebraically independent over $\mathbb{Q}$. Then there exist a matrix $Q\in{\rm Sp}_{2n}(F)$ and a lower block triangular with $2\times 2$ blocks matrix $L$ such that $X=LQ$.

Proof. For $U=(u_{1},\dots,u_{n}),V=(v_{1},\dots,v_{n})\in{\rm M}_{2}(F)^{n}$ denote by $\langle U,V\rangle$ the following $2\times 2$ matrix with entries from $F$

[TABLE]

We can think about $\langle U,V\rangle$ as about function from ${\rm M}_{2}(F)^{n}\times{\rm M}_{2}(F)^{n}\to{\rm M}_{2}(F)$. Using direct calculations it is easy to check that the equalities

[TABLE]

hold for arbitrary $U,V,W\in{\rm M}_{2}(F)^{n}$, $x\in{\rm M}_{2}(F)$. For $U=(u_{1},\dots,u_{n})\in{\rm M}_{2}(F)^{n}$ denote by ${\rm d}(U)={\rm det}(u_{1})+\dots+{\rm det}(u_{n})$. From direct calculations follows that

[TABLE]

Let $X\in{\rm M}_{2n}(F)$ be a matrix from the formulation of the theorem. Express $X$ as a matrix over ${\rm M}_{2}(F)$ ($X\in{\rm M}_{2n}(F)={\rm M}_{n}({\rm M}_{2}(F))$) and let $X_{1},\dots,X_{n}$ be the rows of the matrix $X$ (the entries of $X_{1},\dots,X_{n}$ are $2\times 2$ matrices). Denote by $Y_{1},\dots,Y_{n}$ the following elements from ${\rm M}_{2}(F)^{n}$.

[TABLE]

The element $Y_{1}$ is uniquely defined in (8). In order the element $Y_{2}$ be well defined we need to check that ${\rm d}(Y_{1})\neq 0$. If ${\rm d}(Y_{1})\neq 0$, then in order $Y_{3}$ be well defined we need to check that ${\rm d}(Y_{2})\neq 0$. So, in order formulas (8) be well defined, we need subsequently check that ${\rm d}(Y_{k})\neq 0$ in $F$ for all $k=1,\dots,n-1$.

Since the entries of $X$ are algebraically independent over $\mathbb{Q}$, we can think about these entries as about variables, and we can think about ${\rm d}(Y_{1})$ as about rational function from $\mathbb{Q}^{2n\times 2n}$ to $\mathbb{Q}$. Denote this function by $f_{1}(X)={\rm d}(Y_{1})$. Since

[TABLE]

the function $f_{1}(X)$ is not constantly equal to zero, therefore ${\rm d}(Y_{1})\neq 0$ as an element from $F$, and $Y_{2}$ is well defined. In a similar way, we can think about ${\rm d}(Y_{2})$ as about rational function from $\mathbb{Q}^{2n\times 2n}$ to $\mathbb{Q}$. Denote this function by $f_{2}(X)={\rm d}(Y_{2})$. Since

[TABLE]

the function $f_{2}(X)$ is not constantly equal to zero, therefore ${\rm d}(Y_{2})\neq 0$ as an element from $F$, and $Y_{3}$ is well defined. In the same way we prove that ${\rm d}(Y_{k})\neq 0$ for all $k=1,\dots,n-1$ and therefore formulas (1) uniquely define $Y_{1},\dots,Y_{n}$ from $X_{1},\dots,X_{n}$.

From equalities (6), (7) follows that

[TABLE]

Assuming that $\langle Y_{i},Y_{j}\rangle=0$ for all $1\leq j<i$, using direct calculations similar to (9) we can show that $\langle Y_{i},Y_{j}\rangle=0$ for all $1\leq j<i+1$. It means, that for all $1\leq j\neq i\leq n$ we have $\langle Y_{i},Y_{j}\rangle=0$.

Denote by $Y$ the matrix from ${\rm M}_{n}({\rm M}_{2}(F))$ with the first line $Y_{1}$, the second line $Y_{2}$, $\dots$, the $n$-th line $Y_{n}$. From equalities (8) follows that $Y=TX$, where $T$ is a lower block triangular with $2\times 2$ blocks matrix. For $k=1,\dots,n$ denote by $Q_{k}=Y_{k}\sqrt{{\rm d}(Y_{k})^{-1}}$ (we can take square roots since $F$ is algebraically closed). Since $\langle Y_{i},Y_{j}\rangle=0$ for all $1\leq j<i\leq n$, we have $\langle Q_{i},Q_{j}\rangle=0$ for all $1\leq j<i\leq n$. Moreover, from equalities (6), (7) follows that $\langle Q_{i},Q_{i}\rangle=J$ for all $i=1,\dots,n$. Therefore the matrix $Q$ with the first line $Q_{1}$, the second line $Q_{2}$, $\dots$, the $n$-th line $Q_{n}$, belongs to ${\rm Sp}_{2n}(F)$. Since $Q_{k}=Y_{k}\sqrt{{\rm d}(Y_{k})^{-1}}$ for $k=1,\dots,n$, we have $Q=DY=DTX$ for a diagonal $2n\times 2n$ matrix $D$. Denoting by $L=(DT)^{-1}$, we conclude the proof of the lemma. $\square$

Lemma** 4****.**

Let $S$ be a countable set of variables, $F=\overline{\mathbb{Q}(S)}$, $n\geq 2$ be an even integer, and $\varphi$ be an automorphism of $F$ introduced in Theorem 1. Let $D_{1},\dots,D_{n/2}$ be matrices from ${\rm SL}_{2}(F)$ and $D=D_{1}\oplus\dots\oplus D_{n/2}$. Then $D$ is $\varphi_{n}$-conjugated in ${\rm Sp}_{n}(F)$ to ${\rm I}_{n}$.

Proof. For $k=1,\dots,n/2$ denote by $A_{k}=D_{k}\oplus{\rm I}_{n-2}$. By Theorem 1 there exists a matrix $X_{k}\in{\rm GL}_{n}(F)$ such that the entries of $X_{k}$ are algebraically independent over $\mathbb{Q}$ and $\varphi_{n}(X_{k})=X_{k}A_{k}$. Let

[TABLE]

where $P_{k}$ is a $2\times 2$ matrix, $Q_{k}$ is a $2\times(n-2)$ matrix, $R_{k}$ is a $(n-2)\times 2$ matrix, and $S_{k}$ is $(n-2)\times(n-2)$ matrix. From the equality $\varphi_{n}(X_{k})=X_{k}A_{k}$ and the fact that $A_{k}=D_{k}\oplus{\rm I}_{n-2}$ follows, that $\varphi_{2}(P_{k})=P_{k}D_{k}$. Since the entries of $X_{k}$ are algebraically independent over $\mathbb{Q}$, and ${\rm det}(P_{k})$ is a polynomial over the entries of $X$ with integer coefficients, ${\rm det}(P_{k})\neq 0$ and $P_{k}$ is invertible. Therefore we have the equality

[TABLE]

for $k=1,\dots,n/2$. From this equality follows that

[TABLE]

Denote by $Y_{k}=\sqrt{{\rm det}(P_{k})^{-1}}P_{k}$ (we can take square roots since $F$ is algebraically closed). Then ${\rm det}(Y_{k})=1$ and

[TABLE]

for $k=1,\dots,n/2$. Denoting by $Y=Y_{1}\oplus\dots\oplus Y_{n/2}$ we have $D=Y^{-1}\varphi_{n}(Y)$, what concludes the proof. $\square$

Theorem** 3****.**

Let $S$ be a countable set of variables, $F=\overline{\mathbb{Q}(S)}$, $n\geq 2$ be an even integer, and $\varphi$ be an automorphism of $F$ introduced in Theorem 1. Then $\varphi$ induces an automorphism $\varphi_{n}$ of ${\rm Sp}_{n}(F)$ with $R(\varphi_{n})=1$.

Proof. Let $A$ be a matrix from ${\rm Sp}_{n}(F)$. By Theorem 1 there exists a matrix $X\in{\rm GL}_{n}(F)$ such that the entries of $X$ are algebraically independent over $\mathbb{Q}$ and $A=X^{-1}\varphi_{n}(X)$. From Lemma 3 follows that $X$ can be presented as $X=LQ$, where $L$ is a lower block triangular with $2\times 2$ blocks matrix, and $Q\in{\rm Sp}_{n}(F)$. Since $A\in{\rm Sp}_{n}(F)$, we have

[TABLE]

Since $Q\in{\rm Sp}_{n}(F)$, from equality (10) follows that $\Omega_{n/2}=L^{-1}\varphi_{n}(L)\Omega_{n/2}\varphi_{n}(L)^{T}(L^{-1})^{T}$ or

[TABLE]

Since $L$ is a lower block triangular with $2\times 2$ blocks matrix, and $\Omega_{n/2}$ is a block diagonal with $2\times 2$ blocks matrix, the matrix on the left in equality (11) is a lower block triangular with $2\times 2$ blocks, and the matrix on the right in equality (11) is an upper block triangular with $2\times 2$ blocks. It is possible only in the case when $L^{-1}\varphi_{n}(L)=D$ is block diagonal with $2\times 2$ blocks. Therefore we have the following equality.

[TABLE]

Since $D$ is block diagonal with $2\times 2$ blocks, $D=D_{1}\oplus\dots\oplus D_{n/2}$ for $D_{1},\dots,D_{n/2}\in{\rm M}_{2}(F)$. Since $A,Q\in{\rm Sp}_{n}(F)$, from equality (12) follows that $D\in{\rm Sp}_{n}(F)$, therefore $D_{k}\in{\rm SL}_{2}(F)$ for all $k=1,\dots,n/2$. From Lemma 4 follows that $D$ is $\varphi_{n}$-cojugated in ${\rm Sp}_{n}(F)$ to $I_{n}$, therefore $A$ is $\varphi_{n}$-conjugated in ${\rm Sp}_{n}(F)$ to ${\rm I}_{n}$. $\square$

5 Orthogonal and symplectic groups in general case

Let us recall some necessary facts from model theory. The signature $\Sigma$ is a triple $(\mathcal{F},\mathcal{P},\rho)$, where $\mathcal{F}$ and $\mathcal{P}$ are disjoint sets not containing basic logical symbols, called, respectively, the set of function symbols and the set of predicate symbols, and $\rho:\mathcal{F}\cup\mathcal{P}\to\{0\}\cup\mathbb{N}$ is a function, called arity, which assigns a non-negative integer to every function or predicate symbol. A function symbol $f$ is called $n$-ary if $\rho(f)=n$. A [math]-ary function symbol is called a constant symbol.

The alphabet of a first-order logic of a signature $\Sigma=(\mathcal{F},\mathcal{P},\rho)$ consists of symbols of variables (usually $x$, $y$, $z$, $\dots$), logical operations (negation $\neg$, conjunction $\wedge$, disjunction $\vee$, implication $\to$), quantifiers (existential $\exists$, universal $\forall$), functional symbols from $\mathcal{F}$ (usually $f$, $g$, $\dots$ for symbols with positive arity, and $a$, $b$, $\dots$ for [math]-ary symbols), predicate symbols from $\mathcal{P}$ (usually $p$, $q$, $\dots$), parentheses, brackets and other punctuation symbols.

The set of terms of a first-order logic of a signature $\Sigma$ is inductively defined by the following rules: any variable is a term; for a functional symbol $f\in\mathcal{F}$ with $\rho(f)=n$ and terms $t_{1}$, $\dots$, $t_{n}$ the expression $f(t_{1},\dots,t_{n})$ is a term.

The set of formulas of a first-order logic of a signature $\Sigma$ is inductively defined by the following rules: for a predicate symbol $p\in\mathcal{P}$ with $\rho(p)=n$ and for terms $t_{1}$, $\dots$, $t_{n}$ the expression $p(t_{1},\dots,t_{n})$ is a formula; for terms $t_{1}$, $t_{2}$ the expression $t_{1}=t_{2}$ is a formula; if $x$ is a variable and $\varphi$, $\psi$ are formulas, then $\varphi\wedge\psi$, $\varphi\vee\psi$, $\varphi\to\psi$, $\neg\varphi$, $\exists x\leavevmode\nobreak\ \varphi$, $\forall x\leavevmode\nobreak\ \varphi$ are formulas. A set $\mathcal{A}$ of some formulas of a first-order logic of a signature $\Sigma$ is called a theory of a signature $\Sigma$.

Let $\Sigma=(\mathcal{F},\mathcal{P},\rho)$ be a signature, $M$ be a non-empty set and $\sigma$ be a function which maps every functional symbol $f$ from $\mathcal{F}$ with $\rho(f)=n$ to $n$-ary function $\sigma(f):M^{n}\to M$, and maps every predicate symbol $p$ from $\mathcal{P}$ to $n$-ary relation $\sigma(p)\subseteq M^{n}$. Denote by $\mathcal{M}$ the pair $(M,\sigma)$. Let $s$ be a function which maps any variable to some element from $M$. The interpretation $[[t]]_{s}$ of a term $t$ in $M$ with respect to $s$ is inductively defined by the following rules: $[[x]]_{s}=s(x)$ if $x$ is a variable and $[[f(t_{1},\dots,t_{n})]]_{s}=\sigma(f)([[t_{1}]]_{s},\dots,[[t_{n}]]_{s})$ for a functional symbol $f\in\mathcal{F}$ with $\rho(f)=n$ and terms $t_{1}$, $\dots$, $t_{n}$. The truth of a formula $\varphi$ in $\mathcal{M}$ with respect to $s$ (we write $\mathcal{M}\models_{s}\varphi$ if $\varphi$ is true in $\mathcal{M}$ with respect to $s$) is inductively defined by the following rules:

• •

$\mathcal{M}\models_{s}p(t_{1},\dots,t_{n})$ if an only if $\left([[t_{1}]]_{s},\dots,[[t_{n}]]_{s}\right)\in\sigma(p)$,

• •

$\mathcal{M}\models_{s}t_{1}=t_{2}$ if and only if $[[t_{1}]]_{s}=[[t_{2}]]_{s}$,

• •

$\mathcal{M}\models_{s}\varphi\wedge\psi$ if and only if $\mathcal{M}\models_{s}\varphi$ and $\mathcal{M}\models_{s}\psi$,

• •

$\mathcal{M}\models_{s}\varphi\vee\psi$ if and only if $\mathcal{M}\models_{s}\varphi$ or $\mathcal{M}\models_{s}\psi$,

• •

$\mathcal{M}\models_{s}\varphi\to\psi$ if and only if $\mathcal{M}\models_{s}\varphi$ implies $\mathcal{M}\models_{s}\psi$,

• •

$\mathcal{M}\models_{s}\neg\varphi$ if and only if $\mathcal{M}\models_{s}\varphi$ is not true,

• •

$\mathcal{M}\models_{s}\exists x\leavevmode\nobreak\ \varphi$ if and only if $\mathcal{M}\models_{s^{\prime}}\varphi$ for some $s^{\prime}$ with $s^{\prime}(y)=s(y)$ for all $y\neq x$,

• •

$\mathcal{M}\models_{s}\forall x\leavevmode\nobreak\ \varphi$ if and only if $\mathcal{M}\models_{s^{\prime}}\varphi$ for all $s^{\prime}$ with $s^{\prime}(y)=s(y)$ for all $y\neq x$.

We say that $\mathcal{M}=(M,\sigma)$ is a model for a theory $\mathcal{A}$ if $\mathcal{M}\models_{s}\varphi$ for all formulas $\varphi\in\mathcal{A}$ and all functions $s$. A theory can have no models, one model or several models (even an infinite number).

Theorem** 4**** (Löwenheim-Skolem Theorem).**

If a countable theory $\mathcal{A}$ of a signature $\Sigma$ has an infinite model, then for every infinite cardinal number $\kappa$ it has a model $\mathcal{M}=(M,\sigma)$ with $|M|=\kappa$.

Theorem** 5****.**

Let $F$ be an algebraically closed field of zero characteristic with infinite transcendence degree over $\mathbb{Q}$. Then there exists an automorphism $\varphi$ of $F$ which induces an automorphism $\varphi_{n}$ of ${\rm O}_{n}(F)$ with $R(\varphi_{n})=2$.

Proof. If the transcendence degree of $F$ over $\mathbb{Q}$ is countable, then $F=\overline{\mathbb{Q}(S)}$, where $S$ is a countable set of variables, and the result follows from Theorem 2. So let the transcendence degree of $F$ over $\mathbb{Q}$ be uncountable.

Let $\Sigma=(\mathcal{F},\mathcal{P},\rho)$ be a signature with $\mathcal{\mathcal{F}}=\{+,\cdot,-,^{-1},f,0,1\}$, $\mathcal{P}=\varnothing$ and $\rho(+)=\rho(\cdot)=2$, $\rho(-)=\rho(^{-1})=\rho(f)=1$, $\rho(0)=\rho(1)=0$. For terms $t_{1}$, $t_{2}$ of the signature $\Sigma$ we will write $t_{1}+t_{2}$, $t_{1}\cdot t_{2}$, $-t_{1}$, $t_{1}^{-1}$ instead of $+(t_{1},t_{2})$, $\cdot(t_{1},t_{2})$, $-(t_{1})$, ${}^{-1}(t_{1})$, respectively, and we will write $t_{1}\neq t_{2}$ instead of $\neg(t_{1}=t_{2})$. Let $\mathcal{A}$ be the theory consisting of the following formulas.

$\forall x\leavevmode\nobreak\ \forall y\leavevmode\nobreak\ \forall z\leavevmode\nobreak\ [x+y=y+x]\wedge[x+(y+z)=(x+y)+z]$, 2. 2.

$\forall x\leavevmode\nobreak\ \forall y\leavevmode\nobreak\ \forall z\leavevmode\nobreak\ [x\cdot y=y\cdot x]\wedge[x\cdot(y\cdot z)=(x\cdot y)\cdot z]$, 3. 3.

$\forall x\leavevmode\nobreak\ \forall y\leavevmode\nobreak\ \forall z\leavevmode\nobreak\ x\cdot(y+z)=x\cdot y+x\cdot z$, 4. 4.

$\forall x\leavevmode\nobreak\ [x+0=x]\wedge[x\cdot 1=x]$, 5. 5.

$\forall x\leavevmode\nobreak\ [x+(-x)=0]\wedge[(x\neq 0)\to(x\cdot x^{-1}=1)]$, 6. 6.

$1\neq 0$, $1+1\neq 0$, $1+1+1\neq 0$, $\dots$ 7. 7.

$\forall y_{1}\leavevmode\nobreak\ \forall y_{0}\leavevmode\nobreak\ [(y_{1}\neq 0)\to(\exists x\leavevmode\nobreak\ y_{1}\cdot x+y_{0}=0)]$,

$\forall y_{2}\leavevmode\nobreak\ \forall y_{1}\leavevmode\nobreak\ \forall y_{0}\leavevmode\nobreak\ [(y_{2}\neq 0)\to(\exists x\leavevmode\nobreak\ y_{2}\cdot x\cdot x+y_{1}\cdot x+y_{0}=0)]$,

$\dots$ 8. 8.

$\forall x\leavevmode\nobreak\ \forall y\leavevmode\nobreak\ \exists z\leavevmode\nobreak\ [(f(x)=f(y))\to(x=y)]\wedge[f(z)=x]$, 9. 9.

$\forall x\leavevmode\nobreak\ \forall\leavevmode\nobreak\ y\leavevmode\nobreak\ [f(x+y)=f(x)+f(y)]\wedge[f(x\cdot y)=f(x)\cdot f(y)]$, 10. 10.

$\forall x_{1,1}\leavevmode\nobreak\ \forall x_{1,2}\dots\forall x_{1,n}\leavevmode\nobreak\ \forall x_{2,1}\dots\forall x_{2,n}\dots\forall x_{n,n}\\ \exists y_{1,1}\leavevmode\nobreak\ \exists y_{1,2}\dots\exists y_{1,n}\leavevmode\nobreak\ \exists y_{2,1}\dots\exists y_{2,n}\dots\exists y_{n,n}$

$[XX^{T}=I_{n}]\to[(YY^{T}=I_{n})\wedge((f(Y)=YX)\vee(((-1)\oplus{\rm I_{n-1}}))f(Y)=YX))]$.

where $X=(x_{i,j})$, $Y=(y_{i,j})$ and $f(Y)=(f(y_{i,j}))$.

Note that in formula 10 the expression $XX^{T}={\rm I}_{n}$ (multiplication of matrices) meas the conjunction of $n^{2}$ formulas (in terms of entries of matrices), and each of these $n^{2}$ formulas can be uniquely written in terms of $\cdot$, $+$ and $-$ (we write $XX^{T}={\rm I}_{n}$ for the simplicity of denotation). Similarly, formulas $YY^{T}=I_{n}$, $f(Y)=YX$, $((-1)\oplus{\rm I}_{n-1})f(Y)=YX$ can be written as conjunctions of $n^{2}$ formulas. So, what is written in 10 is a formula.

If there exists a model $\mathcal{M}=(M,\sigma)$ for $\mathcal{A}$, then formulas 1 and 2 say that addition and multiplication are commutative and associative, formula 3 describes distributive law between addition and multiplication, formula 4 says that [math] and $1$ are neutral elements with respect to addition and multiplication respectively, formula 5 says that $-x$ is an opposite to $x$ element and that $x^{-1}$ is an inverse to non-zero element $x$ element. So, all together formulas 1-5 say that $(M,\sigma(+),\sigma(\cdot))$ is a field with the zero element $\sigma(0)$ and the unit element $\sigma(1)$. The countable set of formulas 6 says that the field $M$ has zero characteristic. The countable set of formulas 7 says that $M$ is an algebraically closed field. Formula 8 says that $\sigma(f)$ is a bijection on $M$ and formula 9 says that this bijection respects addition and multiplication. So, formulas 8 and 9 together say that $\sigma(f)$ is an automorphism of the field $M$. Finally, formula 10 says that for every orthogonal matrix $X=(x_{i,j})$ with coefficients from $M$ there exists an orthogonal matrix $Y=(y_{i,j})$ with coefficients from $M$ such that either $(\sigma(f)(y_{i,j}))=YX$ or $((-1)\oplus{\rm I}_{n-1})(\sigma(f)(y_{i,j}))=YX$. In other words all together formulas 1-10 say that $M$ is an algebraically closed field of zero characteristic and $\sigma(f)$ is an automorphism of $M$ which induced an automorphism $\varphi$ of ${\rm O}_{n}(M)$ with $R(\varphi)=2$ (every orthogonal matrix is $\varphi$-conjugated either to ${\rm I}_{n}$ or to $(-1)\oplus{\rm I}_{n-1}={\rm diag}(-1,1,\dots,1)$).

The previous paragraph is written in assumption that the theory $\mathcal{A}$ has a model, but by Theorem 2 it has an infinite model. Since $\mathcal{A}$ is a countable theory, by Löwenheim-Skolem theorem it has a model $\mathcal{M}=(M,\sigma)$ with $|M|=\kappa=|F|$, where $F$ is a field from the formulation of the theorem. Since $|M|=\kappa$ is uncountable, the transcendence degree of $M$ over $\mathbb{Q}$ is infinite and is equal to $\kappa$. Since every algebraically closed field is completely determined by its characteristic and transcendence degree over prime subfield, we have $M\cong F$. Therefore there exists an automorphism $\varphi$ of the field $F$ which induces an automorphism of ${\rm O}_{n}(F)$ with Reidemeister number equals to $2$. $\square$

Corollary** 1****.**

Let $F$ be an algebraically closed field of zero characteristic with infinite transcendence degree over $\mathbb{Q}$. Then there exists an automorphism $\varphi$ of $F$ which induces an automorphism $\varphi_{n}$ of ${\rm SO}_{n}(F)$ with $R(\varphi_{n})=1$.

Proof. Let $\varphi$ be an automorphism from Theorem 5, and $A$ be an arbitrary matrix from ${\rm SO}_{n}(F)$. From the proof of Theorem 1 follows that $A$ is $\varphi_{n}$ conjugated in ${\rm O}_{n}(F)$ either to ${\rm I}_{n}$ or to ${\rm diag}(-1,1,\dots,1)$. Since ${\rm det}(A)=1$, the matrix $A$ cannot be $\varphi_{n}$-conjugated in ${\rm O}_{n}(F)$ to ${\rm diag}(-1,1,\dots,1)$, therefore $A$ is $\varphi_{n}$-conjugated in ${\rm O}_{n}(F)$ to ${\rm I}_{n}$, i. e. there exists a matrix $X\in{\rm O}_{n}(F)$ such that $A=X^{-1}\varphi_{n}(X)$. Denote by $Y=(({\rm det}(X))\oplus{\rm I}_{n-1})X$. Then $Y\in{\rm SO}_{n}(F)$ and $Y^{-1}\varphi_{n}(Y)=X^{-1}\varphi_{n}(X)=A$, i. e. an arbitrary matrix $A\in{\rm SO}_{n}(F)$ is $\varphi_{n}$-conjugated in ${\rm SO}_{n}(F)$ to ${\rm I}_{n}$. $\square$

Slightly modifying the set of formulas in the proof of Theorem 5 we get the following result.

Theorem** 6****.**

Let $F$ be an algebraically closed field of zero characteristic with infinite transcendence degree over $\mathbb{Q}$, and $n$ be an even integer. Then there exists an automorphism $\varphi$ of $F$ which induces an automorphism $\varphi_{n}$ of ${\rm Sp}_{n}(F)$ with $R(\varphi_{n})=1$.

Proof. The proof is the same as the proof of Theorem 5 changing formula 10 from the proof of Theorem 5 to the formula

[TABLE]

where $X=(x_{i,j})$, $Y=(y_{i,j})$ and $f(Y)=(f(y_{i,j}))$ and using Theorem 3 instead of Theorem 2 to provide an infinite model which we need in order to use Löwenheim-Skolem theorem. $\square$

6 Chevalley groups

The following statement is almost obvious, it can be found it [10, Lemmas 2.1].

Lemma** 5****.**

Let $G$ be a group, $\varphi$ be an automorphism of $G$ and $N$ be a $\varphi$ admissible normal subgroup of $G$. Denote by $\overline{\varphi}$ an automorphism of $G/N$ induced by $\varphi$. Then $R(\overline{\varphi})\leq R(\varphi)$.

Theorem** 7****.**

Let $F$ be an algebraically closed field of zero characteristic which has infinite transcendence degree over $\mathbb{Q}$. Then Chevalley groups of types $B_{n},C_{n},D_{n}$ over $F$ do not possess the $R_{\infty}$-property.

Proof. We separately consider the root systems $B_{n},D_{n}$ and the root systems $C_{n}$.

Case 1: The root system has the type $B_{n}$ or $D_{n}$. Let $G$ be a Chevalley group of type $B_{n}$ or $D_{n}$ over $F$, and let $Z(G)$ be the center of $G$ (it is known to be finite). The quotient group $G/Z(G)$ is isomorphic to ${\rm P}\Omega_{k}(F)={\rm PSO}_{k}(F)$, where $k=2n+1$ if the root system has the type $B_{n}$, and $k=2n$ if the root system has the type $D_{n}$ (see [2, §11.3, §12.1]). Denote by $\varphi$ the automorphism of $F$ from Corollary 1 (changing $n$ by $k$ in Corollary 1). Denote by $\psi$, $\overline{\psi}$, $\theta$, $\overline{\theta}$, respectively, the automorphisms of $G$, $G/Z(G)$, ${\rm SO}_{k}(F)$, ${\rm PSO}_{k}(F)$ induced by $\varphi$. Since $G/Z(G)={\rm PSO}_{k}(F)$, we have $R(\overline{\theta})=R(\overline{\psi})$. From Corollary 1 we have $R(\theta)=1$. From Lemma 5 follows that $R(\overline{\theta})=1$, therefore $R(\overline{\psi})=1$, i. e. every element from $G/Z(G)$ is $\overline{\psi}$-conjugated to the unit element in $G/Z(G)$. Hence every element from $G$ is $\psi$-conjugated to some element from $Z(G)$. Since $Z(G)$ is finite, the number of $\psi$-conjugacy classes in $G$ is finite, i. e. $G$ does not possess the $R_{\infty}$-property.

Case 2: The root system has the type $C_{n}$. Let $G$ be a Chevalley group of type $C_{n}$ over $F$, and let $Z(G)$ be the center of $G$ (it is known to be finite). The quotient group $G/Z(G)$ is isomorphic to ${\rm PSp}_{2n}(F)$ (see [2, §11.3, §12.1]). Denote by $\varphi$ the automorphism of $F$ from Theorem 6 (changing $n$ by $2n$ in Theorem 6). Denote by $\psi$, $\overline{\psi}$, $\theta$, $\overline{\theta}$, respectively, the automorphisms of $G$, $G/Z(G)$, ${\rm Sp}_{2n}(F)$, ${\rm PSp}_{2n}(F)$ induced by $\varphi$. Since $G/Z(G)={\rm PSp}_{2n}(F)$, we have $R(\overline{\theta})=R(\overline{\psi})$. From Theorem 6 we have $R(\theta)=1$. From Lemma 5 follows that $R(\overline{\theta})=1$, therefore $R(\overline{\psi})=1$, i. e. every element from $G/Z(G)$ is $\overline{\psi}$-conjugated to the unit element in $G/Z(G)$. Hence every element from $G$ is $\psi$-conjugated to some element from $Z(G)$. Since $Z(G)$ is finite, the number of $\psi$-conjugacy classes in $G$ is finite, i. e. $G$ does not possess the $R_{\infty}$-property. $\square$

Theorem** 8****.**

Let $F$ be an algebraically closed field of zero characteristic. Then Chevalley groups of types $A_{n},B_{n},C_{n},D_{n}$ over $F$ possess the $R_{\infty}$-property if and only if the transcendence degree of $F$ over $\mathbb{Q}$ is finite.

Proof. The necessity condition follows from Theorem 7 and [14, Theorem 7]. The sufficiency condition follows from [5, Theorem 3.2]. $\square$