A theorem of Tits type for automorphism groups of projective varieties in arbitrary characteristic

Abstract

We prove a theorem of Tits type for automorphism groups of projective varieties over an algebraically closed field of arbitrary characteristic, which was first conjectured by Keum, Oguiso and Zhang for complex projective varieties.

Appendix A: An example of a solvable subgroup of an automorphism group

1.1 Tomohide Terasoma⁴

This is an example of a finitely generated solvable subgroup of the automorphism group of a projective variety whose homological trivial part is infinitely generated.

1.2 A.1. Free groups generated by two elements

Let $F_2=\langle a, b\rangle$ be a free group generated by two elements a, b. Let

$$\begin{aligned} F'_2 :=[F_2,F_2] \quad \text {and}\quad F''_2 :=[F'_2,F'_2] \end{aligned}$$

be the first and second derived subgroup of $F_2$, respectively. Then the maximal abelian quotient group $F_2^{{\text {ab}}} :=F_2/F'_2$ is isomorphic to the free abelian group of rank 2 generated by the image $\bar{a}, \bar{b}$ of a, b. The group $F_2$ acts on the abelian group $(F'_2)^{{\text {ab}}} :=F_2'/F_2''$ by the adjoint action ${\text {ad}}(g) \in {\text {Aut}}((F'_2)^{{\text {ab}}})$ for $g\in F_2$ defined by

$$\begin{aligned} {\text {ad}}(g)(s)=gsg^{-1}. \end{aligned}$$

Since

$$\begin{aligned}{}[{\text {ad}}(g),{\text {ad}}(h)]={\text {ad}}(g){\text {ad}}(h) {\text {ad}}(g)^{-1}{\text {ad}}(h)^{-1}={\text {ad}}([g,h]) \end{aligned}$$

and

$$\begin{aligned}{}[g,h]w[g,h]^{-1}=w \ \mathrm{mod }\ F_2'' \end{aligned}$$

for any $w\in F_2'$, we have

$$\begin{aligned} {\text {ad}}(g){\text {ad}}(h)={\text {ad}}(h){\text {ad}}(g)\in {\text {Aut}}((F'_2)^{{\text {ab}}}). \end{aligned}$$

Thus the above adjoint action of $F_2$ on $(F'_2)^{{\text {ab}}}$ induces an adjoint action of $F_2^{{\text {ab}}}$ on $(F'_2)^{{\text {ab}}}$. Therefore, $(F'_2)^{{\text {ab}}}$ becomes a module over a (commutative) Laurent polynomial ring $\mathbb {Z}[{\text {ad}}(\bar{a})^{\pm 1},{\text {ad}}(\bar{b})^{\pm 1}]$ generated by two elements ${\text {ad}}(\bar{a}),{\text {ad}}(\bar{b})$.

Lemma A.1

The group $(F'_2)^{{\text {ab}}}=F_2'/F_2''$ is generated by [a, b] over $\mathbb {Z}[{\text {ad}}(\bar{a})^{\pm 1},{\text {ad}}(\bar{b})^{\pm 1}]$.

Proof

We have

$$\begin{aligned}{}[gh,x]= ghxh^{-1}g^{-1}x^{-1}= ghxh^{-1}x^{-1}g^{-1}gxg^{-1}x^{-1}={\text {ad}}(g)([h,x])\cdot [g,x] \end{aligned}$$

and $[g,h]=[h,g]^{-1}$. By writing the product of $(F'_2)^{{\text {ab}}}$ additively, we have

$$\begin{aligned}{}[gh,x]={\text {ad}}(g)([h,x])+[g,x] \text { and } [g,h]=-[h,g]. \end{aligned}$$

The lemma follows from these relations. $\square$

1.3 A.2. A group homomorphism from $F_2$ to $\mathrm {PGL}(2,\mathbb {K})$

Let $\mathbb {K}$ be an algebraically closed field whose transcendental degree over the prime field $\mathbb {k}$ is greater than or equal to 2. Let $\alpha , \beta$ be elements algebraically independent over $\mathbb {k}$. We set

$$\begin{aligned} \widetilde{A}= \left( \begin{matrix} \alpha &{}\quad 1 \\ 0 &{}\quad 1 \end{matrix}\right) , \quad \widetilde{B}= \left( \begin{matrix} 1 &{}\quad 0 \\ 0 &{}\quad \beta \end{matrix} \right) . \end{aligned}$$

The class of $\widetilde{A}, \widetilde{B}$ in $\mathrm {PGL}(2,\mathbb {K})={\text {Aut}}(\mathbf {P}_\mathbb {K}^1)$ is denoted by A, B. We define a group homomorphism

$$\begin{aligned} \varphi :F_2 \longrightarrow \mathrm {PGL}(2,\mathbb {K})\quad \text {via } \varphi (a)=A, \varphi (b)=B. \end{aligned}$$

One can easily check the following lemma.

Lemma A.2

The image of $F_2'$ under $\varphi$ is contained in

$$\begin{aligned} N=\left\{ \left( \begin{matrix} 1 &{}\quad n \\ 0 &{}\quad 1 \end{matrix}\right) : n\in \mathbb {K}\right\} . \end{aligned}$$

Since N is an abelian group, the image of $F_2''$ under $\varphi$ is the identity group. Thus we have a homomorphism between abelian groups:

$$\begin{aligned} \varphi ' :(F_2')^{{\text {ab}}} = F_2'/F_2'' \longrightarrow N. \end{aligned}$$

(A.1)

We introduce an $R=\mathbb {Z}[\xi ^{\pm 1},\eta ^{\pm 1}]$-module structure on the abelian group N by setting

$$\begin{aligned} \xi ^i\eta ^j \left( \begin{matrix} 1 &{}\quad n \\ 0 &{}\quad 1 \end{matrix}\right) = \left( \begin{matrix} 1 &{}\quad \alpha ^i\beta ^{-j}n \\ 0 &{}\quad 1 \end{matrix}\right) . \end{aligned}$$

By a direct calculation, we have the following lemma.

Lemma A.3

Via the identification between $\mathbb {Z}[{\text {ad}}(\bar{a})^{\pm 1},{\text {ad}}(\bar{b})^{\pm 1}]$ and $R=\mathbb {Z}[\xi ^{\pm 1},\eta ^{\pm 1}]$ given by

$$\begin{aligned} \xi ={\text {ad}}(\bar{a}),\quad \eta ={\text {ad}}(\bar{b}), \end{aligned}$$

the homomorphism $\varphi '$ becomes an R-homomorphism.

Then the image of the commutator $[a,b] \in F_2'$ under the homomorphism $\varphi$ is equal to

$$\begin{aligned}{}[A,B]=\left( \begin{matrix} 1 &{}\quad 1-\beta ^{-1} \\ 0 &{}\quad 1 \end{matrix}\right) . \end{aligned}$$

By Lemma A.1, $(F_2')^{{\text {ab}}}$ is generated by [a, b] as a $\mathbb {Z}[{\text {ad}}(\bar{a})^{\pm 1},{\text {ad}}(\bar{b})^{\pm 1}]$-module. Therefore, the image of $\varphi '$ in the abelian group N is equal to

$$\begin{aligned}&\mathbb {Z}[\xi ^{\pm 1},\eta ^{\pm 1}]\left( \begin{matrix} 1 &{}\quad 1-\beta ^{-1} \\ 0 &{}\quad 1 \end{matrix}\right) \nonumber \\&\quad = \left\{ \left( \begin{matrix} 1 &{}\quad n \\ 0 &{}\quad 1 \end{matrix}\right) : n\in \mathbb {Z}[\alpha ^{\pm 1},\beta ^{\pm 1}](1-\beta ^{-1}) \right\} , \end{aligned}$$

(A.2)

which is not finitely generated since our $\alpha$ and $\beta$ are assumed to be algebraically independent over $\mathbb {k}$.

1.4 A.3. Homological trivial part of a finitely generated solvable subgroup of automorphisms

Let $\mathbb {K}$ be the field as in Sect. A.2. Let Y be a smooth projective K3 surface over $\mathbb {K}$ such that the automorphism group ${\text {Aut}}(Y)$ of Y contains a subgroup M isomorphic to $\mathbb {Z}^2$. We choose a generator $\{p,q\}$ of M. We set $X = \mathbf {P}_\mathbb {K}^1\times Y$. We define the homologically trivial part ${\text {Aut}}^0(X)$ of ${\text {Aut}}(X)$ by

$$\begin{aligned} {\text {Ker}}(\rho :{\text {Aut}}(X) \longrightarrow \mathrm {GL}(H^*(X,\mathbb {Q}))). \end{aligned}$$

Then ${\text {Aut}}^0(X)$ is isomorphic to ${\text {Aut}}(\mathbf {P}_\mathbb {K}^1)=\mathrm {PGL}(2,\mathbb {K})$. Let G be the subgroup of ${\text {Aut}}(X)$ generated by

$$\begin{aligned} s=(A, p),\quad t=(B, q) \in {\text {Aut}}(\mathbf {P}_\mathbb {K}^1)\times {\text {Aut}}(Y), \end{aligned}$$

where A, B are elements in $\mathrm {PGL}(2,\mathbb {K})$ defined in Sect. A.2. Then G becomes a finitely generated solvable subgroup of ${\text {Aut}}(X)$. In particular, the solvable length of G is 2.

Proposition A.4

The subgroup

$$\begin{aligned} G^0 :={\text {Aut}}^0(X)\cap G = {\text {Ker}}(G \longrightarrow \mathrm {GL}(H^*(X,\mathbb {Q}))) \end{aligned}$$

of G is not finitely generated.

Proof

We define homomorphisms $\psi :F_2 \rightarrow G$, $\pi :F_2 \rightarrow M$ and $j :M \longrightarrow \mathrm {GL}(H^*(X))$ by

$$\begin{aligned} \psi (a)=s,\quad \psi (b)=t, \quad \pi (a)=p, \quad \pi (b)=q, \end{aligned}$$

and

$$\begin{aligned} j(m)=\left( {\text {id}}_{\mathbf {P}_\mathbb {K}^1},m\right) ^*. \end{aligned}$$

Then the morphism $\psi$ is surjective and j is injective. We consider the diagram

Since the actions of $A, B \in {\text {Aut}}(\mathbf {P}_\mathbb {K}^1)$ on $H^*(\mathbf {P}_\mathbb {K}^1)$ are trivial, we have $\rho \circ \psi = j \circ \pi$. The morphism $\pi$ will be identified with the abelianization of $F_2$.

We show that the group $G^0$ is equal to the image $\psi (F_2')$ of $F_2'$ under $\psi$. Let k be an element of $G^0$. Using the surjectivity of $\psi$, we choose an element $\widetilde{k}$ of $F_2$ such that $\psi (\widetilde{k})=k$. Since $(j\circ \pi )(\widetilde{k})=(\rho \circ \psi )(\widetilde{k})=\rho (k)={\text {id}}$, and j is injective, we have

$$\begin{aligned} \widetilde{k} \in {\text {Ker}}(F_2 \xrightarrow {\ \pi \ } M)=F_2'. \end{aligned}$$

Thus we have $k\in \psi (F_2')$.

We set $G'=[G,G]$. Since $\psi$ is surjective, the group $\psi (F_2')=G^0$ is identified with $G'$. Thus we have $G'=G^0\subset {\text {Aut}}(\mathbf {P}_\mathbb {K}^1)$. Also, the image of $F''_2$ in $G'$ under the morphism $\psi$ is trivial. The composite homomorphism

is nothing but the homomorphism $\varphi '$ defined in (A.1). Therefore, $G^0 = {\text {Im}}(\varphi ')$ is equal to the group (A.2) by Lemma A.1, which is not finitely generated. $\square$