Surjectivity of certain word maps on ${{\mathrm{PSL}}}(2,\mathbb {C})$ and ${{\mathrm{SL}}}(2,\mathbb {C})$

Abstract

Let $n \geqslant 2$ be an integer and $F_n$ the free group on n generators, its first and second derived subgroups. Let K be an algebraically closed field of characteristic zero. We show that if , then the corresponding word map is surjective. We also describe certain word maps that are surjective on ${{\mathrm{SL}}}(2,\mathbb {C})$.

1 Introduction

The surjectivity of word maps on groups became recently a vivid topic: the review on the latest activities may be found in [3, 17, 19, 22].

Let $w\in F_n$ be an element of the free group $F_n$ on $n>1$ generators $g_1,\dots ,g_n$:

$$\begin{aligned} w =\prod _{i=1}^{k}g_{n_i}^{m_i}, \qquad 1\leqslant {n_i}\leqslant n. \end{aligned}$$

For a group G by the same letter w we shall denote the corresponding word map defined as a non-commutative product by the formula

$$\begin{aligned} w(x_1,\dots , x_n)=\prod _{i=1}^{k}x_{n_i}^{m_i}. \end{aligned}$$

(1)

We call $w (x_1,\dots , x_n)$ both a word in n letters if considered as an element of a free group and a word map in n letters if considered as the corresponding map . We assume that it is reduced, i.e. ${n_i}\ne {n_{i+1}}$ for every $1\leqslant i\leqslant k-1$ and $m_i\ne 0$ for $1\leqslant i\leqslant k$.

Let K be a field and H a connected semisimple linear algebraic group that is defined over K. If w is not the identity then, by the Borel theorem [6], the regular map of (affine) K-algebraic varieties

is dominant, i.e., its image is a Zariski dense subset of H. Let us consider the group $G=H(K)$ and the image

We say that a word (word map) w is surjective on G if $w_G=G$.

In [18, Problem 7], [19, Question 2.1 (i)], the following question is formulated: Assume that w is not a power of another reduced word and $G=H(K)$. Is w surjective when $K = \mathbb {C}$ is a field of complex numbers and H is of adjoint type?

According to [19], Question 2.1 (i) is still open, even in the simplest case $G ={{\mathrm{PSL}}}(2,\mathbb {C})$, even for words in two letters.

We consider word maps on groups $G={{\mathrm{SL}}}(2,K)$ and $\widetilde{G}={{\mathrm{PSL}}}(2,K)$. Put

As usual, ${\mathbb Z},{\mathbb Q},{\mathbb R},{\mathbb C}$ stand for the ring of integers and fields of rational, real and complex numbers respectively. $\mathbb {A}(K)^m_{x_1,\dots ,x_m}$ or, simply, $\mathbb {A}^m$, stands for the m-dimensional affine space over a field K with coordinates $x_1,\dots ,x_m$. If $K=\mathbb {C}$, we use the notation $\mathbb {C}^m_{x_1,\dots ,x_m}$.

Let $w\in F$. For the corresponding word map on $G={{\mathrm{SL}}}(2,K)$ we check the following properties of the image $w_G$.

Properties 1.1

                                                                       

• (a) $w_G$ contains all semisimple elements x with ${{\mathrm{tr}}}(x)\ne 2$;

• (b) $w_G$ contains all unipotent elements x with ${{\mathrm{tr}}}(x)= 2$;

• (c) $w_G$ contains all minus unipotent elements x with and $x\ne -\mathrm{id}$;

• (d) $w_G$ contains $-\mathrm{id}$.

The word map w is surjective on $G={{\mathrm{SL}}}(2,K)$ if all Properties 1.1 are satisfied. For the surjectivity on ${\widetilde{G}}={{\mathrm{PSL}}}(2,K)$ it is sufficient that only Properties 1.1 (a), (b) are valid.

Definition 1.2

(cf. [2]) We say that the word map w is almost surjective on $G={{\mathrm{SL}}}(2,K)$ if it has Properties 1.1 (a)–(c), i.e.

The goal of the paper is to describe certain words $w\in F$ such that the corresponding word maps are surjective or almost surjective on G and/or ${\widetilde{G}}$. Assume that the field K is algebraically closed. If $w(x_1,\dots ,x_d)=x_i^n$ then w is surjective on G if and only if n is odd (see [10, 11]). Indeed, the element

is not a square in ${{\mathrm{SL}}}(2,K)$. Since only the elements with may be outside $w_G$ [10, 11], the induced by w word map ${\widetilde{w}}$ is surjective on ${\widetilde{G}}$.

Consider a word map (1). For an index $j\leqslant n$ let $S_j=\sum _{n_i=j}m_i$. If, say, $S_1\ne 0,$ then $w(x_1,\mathrm{id},\dots ,\mathrm{id})=x_1^{\scriptscriptstyle S_1}$, hence the word w is surjective on ${{\mathrm{PSL}}}(2,K)$. If $S_j=0$ for all $1\leqslant j\leqslant n$, then . In Sect. 5 we prove (see Corollary 5.4) the following:

The word map defined by a word is surjective on ${{\mathrm{PSL}}}(2,K)$ if K is an algebraically closed field with .

The proof makes use of a variation on the Magnus Embedding Theorem, which is stated in Sect. 3 and proven in Sect. 4.

In Sects. 6–8, we consider words in two variables, i.e. $n=2$. In this case we give explicit formulas for w(x, y), where $x,y\in {{\mathrm{SL}}}(2,\mathbb {C})$ are upper triangular matrices. Using explicit formulas, in Sects. 7–8 we provide criteria for surjectivity and almost surjectivity of a word map on $G={{\mathrm{SL}}}(2,\mathbb {C})$. In Sect. 7, these criteria are formulated in terms of exponents ${a_i}, {b_i}$, $i=1\dots ,k$, of the word

where $a_i\ne 0$ and $b_i\ne 0$ for all $i=1,\dots ,k$. A sample of such criteria is (Corollary 7.4)

If all $b_i$ are positive, then the word map w is either surjective or the square of another word $v\ne \mathrm{id}$.

In Sect. 8, we connect the almost surjectivity of a word map with a property of the corresponding trace map. The last sections contain explicit examples.

2 Semisimple elements

Let K be an algebraically closed field with ${{\mathrm{char}}}(K)=0$, and $G={{\mathrm{SL}}}(2,K)$. Consider a word map defined by (1). We consider G as an affine set

The following lemma is, may be, known, but the authors do not have a proper reference.

Lemma 2.1

A regular non-constant function on $G^n$ omits no values in K.

Proof

Since all sets are affine, a function f regular on $G^k$ is a restriction of a polynomial $P_f$ onto $G^k$. We use induction on k.

Step 1. $k=1$. G is an irreducible quadric. Assume that $f\in K[G]$ omits a value. Let $p:G\rightarrow \mathbb {A}^1_{a}$ be a projection defined by $p(a,b,c,d)=a$. If $a\ne 0$ then the fiber $F_a=p^{-1}(a)\cong \mathbb {A}^2_{b,c}$ is an affine space with coordinates b, c because $d=(1+bc)/a$. Since f omits a value, the restriction is constant for every $a\ne 0$. Therefore it is constant on every fiber (note that the fiber $a=0$ is connected). On the other hand, f has to be constant along the curve

Since the curve $C\subset G$ intersects every fiber $F_a$ of projection p, the function f is constant on G.

Step 2. Assume that the statement of the lemma is valid for all $k\leqslant n$. Let $f\in K[G^n]$ omit a value. We have , where $M=G^{n-1}$ and $N=G$. Let be a natural projection. Then, by induction assumption, f is constant along every fiber of this projection. Take $x\in M$ and consider the set . Then and C intersects every fiber of p. Hence, f is constant. $\square$

Proposition 2.2

For every word $w(x_1,\dots , x_k)\ne \mathrm{id}$ the image $w_G$ contains every element $z\in G$ with .

Proof

We consider as the product of

$1\leqslant i\leqslant n$. The function $f(a_1,b_1,c_1,d_1,\dots , a_n,b_n,c_n,d_n)={{\mathrm{tr}}}(w(x_1,\dots ,x_n))$ is a polynomial in 4n variables with integer coefficients, i.e., $f\in K[G^n]$. According to Lemma 2.1, it takes on all values in K. Thus for every value $A\in K$ there is an element $u= w(y_1,\dots ,y_n)\in w_G$ such that ${{\mathrm{tr}}}(u)=A$. Let now $z\in G$, . Since ${{\mathrm{tr}}}(z)={{\mathrm{tr}}}(u)$, z is conjugate to u, i.e., there is $v\in G$ such that $vuv^{-1}=z$. Hence .$\square$

It follows that in order to check whether the word map w is surjective on G (or on ${\widetilde{G}}$) it is sufficient to check whether the elements z with (or the elements z with ${{\mathrm{tr}}}(z)= 2$, respectively) are in the image. For that we need a version of the Embedding Magnus Theorem.

3 Variation on the Magnus Embedding Theorem: statements

Let $n \geqslant 2$ be an integer and be the ring of Laurent polynomials in n independent variables $t_1, \dots , t_n$ over $\mathbb {Z}$. Let $F=F_n$ be a free group of rank n with generators . Recall: we write $F^{(1)}$ for the derived subgroup of F and $F^{(2)}$ for the derived subgroup of $F^{(1)}$. We have

both $F^{(1)}$ and $F^{(2)}$ are normal subgroups in F. The quotient is a free abelian group of rank n with (standard) generators where each $e_i$ is the image of $g_i$, $1 \leqslant i \leqslant n$. The group ring of A is canonically isomorphic to ${\Lambda }_n$: under this isomorphism each goes to

We write $R_n$ for the ring of polynomials

in n independent variables $s_1, \dots , s_n$ over ${\Lambda }_n$. If R is a commutative ring with 1 then we write for the group of invertible matrices of the form

$$\begin{aligned} \begin{bmatrix} a&0\\ b&1 \end{bmatrix} \end{aligned}$$

with $a \in R^{*}$, $b \in R$ and for the group of unimodular matrices of the form

$$\begin{aligned} \begin{bmatrix} a&0\\ b&a^{-1} \end{bmatrix} \end{aligned}$$

with $a \in R^{*}$, $b \in R$. We have

Every homomorphism $R \rightarrow R'$ of commutative rings (with 1) induces the natural group homomorphisms

which are injective if $R \rightarrow R'$ is injective.

The following assertion (that is based on the properties of the famous Magnus embedding [20]) was proven in [26, Lemma 2].

Theorem 3.1

The assignment

$$\begin{aligned} g_i \mapsto \begin{bmatrix} t_i&0\\ s_i&t_i^{-1} \end{bmatrix}, \qquad 1 \leqslant i \leqslant n, \end{aligned}$$

extends to a group homomorphism with kernel $F^{(2)}$ and therefore defines an embedding

It follows from Theorem 3.1 that if K is a field of characteristic zero, whose transcendence degree over $\mathbb {Q}$ is, at least, 2n then there is an embedding

(In particular, it works for $K=\mathbb {R}, \mathbb {C}$ or the field $\mathbb {Q}_p$ of p-adic numbers [26].) The aim of the following considerations is to replace in this statement the lower bound 2n by n.

Theorem 3.2

The assignment

$$\begin{aligned} g_i \mapsto \begin{bmatrix} t_i&0\\ 1&t_i^{-1} \end{bmatrix},\qquad 1 \leqslant i \leqslant n, \end{aligned}$$

extends to a group homomorphism with kernel $F^{(2)}$ and therefore defines an embedding

Remark 3.3

Let

be the ${\Lambda }_n$-algebra homomorphism that sends all $s_i$ to 1 and let

be the group homomorphism induced by $\mathrm {ev}_1$. Then $\mu _1$ coincides with the composition

Corollary 3.4

Let K be a field of characteristic zero. Suppose that the transcendence degree of K over $\mathbb {Q}$ is, at least, n. Then there is a group embedding

The proof of Theorem 3.2 is based on the following observation.

Lemma 3.5

Let K be a field of characteristic zero. Suppose that the transcendence degree of K over $\mathbb {Q}$ is, at least, n and let be an n-tuple of algebraically independent elements (over $\mathbb {Q})$. Let $\mathbb {Q}(u_1, \dots , u_n)$ be the subfield of K generated by and let us consider K as the $\mathbb {Q}(u_1, \dots , u_n)$-vector space. Let be an n-tuple that is linearly independent over $\mathbb {Q}(u_1, \dots , u_n)$. Let R be the subring of K generated by 3n elements .

Then the assignment

extends to a group homomorphism with kernel $F^{(2)}$ and therefore defines an embedding

Example 3.6

Let K be the field $\mathbb {Q}(t_1, \dots , t_n)$ of rational functions in n independent variables $t_1, \dots , t_n$ over $\mathbb {Q}$. One may view ${\Lambda }_n$ as the subring of K generated by 2n elements . By definition, the n-tuple is algebraically independent (over $\mathbb {Q}$). Clearly, the n-tuple

is also algebraically independent. Then the n elements $y_1=t_1, \dots , y_i=t_i, \dots , y_n=t_n$ are linearly independent over the (sub)field $\mathbb {Q}(t_1^2, \dots , t_n^2)=\mathbb {Q}(u_1, \dots , u_n)$. Indeed, if a rational function

where all $f_i \in \mathbb {Q}(t_1^2, \dots , t_n^2)$ then

This proves that if $f=0$ then all $f_i$ are also zero, i.e., the set is linearly independent over $\mathbb {Q}(t_1^2, \dots t_n^2)$.

By definition, R coincides with the subring of K generated by 3n elements $\dots , t_n^2, t_n^{-2}; t_1, \dots , t_n$. This implies easily that $R={\Lambda }_n$. Applying Lemma 3.5, we conclude the example by the following statement.

The assignment

extends to a group homomorphism with kernel $F^{(2)}$ and therefore defines an embedding

We prove Lemma 3.5, Theorem 3.2 and Corollary 3.4 in Sect. 4.

4 Variation on the Magnus Embedding Theorem: proofs

Proof

of Lemma 3.5 Let ${\Lambda } \subset \mathbb {Q}(u_1, \dots , u_n)\subset K$ be the subring generated by 2n elements . Since $u_i$ are algebraically independent over $\mathbb {Q}$, the assignment

extends to a ring isomorphism ${\Lambda }_n \cong {\Lambda }$. The linear independence of $y_i$ over $\mathbb {Q}(u_1, \dots ,$ $u_n)$ implies that is a free ${\Lambda }$-module of rank n. On the other hand, let

be the multiplicative (sub)group generated by all $u_i$. The assignment $g_i\mapsto u_i$ extends to the surjective group homomorphism $\delta :F \twoheadrightarrow U$ with kernel $F^{(1)}$ and gives rise to the group isomorphism $A \cong U$, which sends $e_i$ to $u_i$ and allows us to identify the group ring of U with . Notice that M carries the natural structure of free -module of rank n defined by

We have

It follows from [27, Lemma 1 (c), p. 175] that coincides with the derived subgroup of . Since , we conclude that and we are done.$\square$

Proof

of Theorem 3.2 Let us return to the situation of Example 3.6. In particular, the group homomorphism (we know its kernel, thanks to already proven Lemma 3.5) is defined by

for all $g_i$. Let us consider the group homomorphism

$$\begin{aligned} \kappa :F \rightarrow {\Lambda }_n^{*}, \qquad g_i \mapsto t_i. \end{aligned}$$

Since $t_i$ are algebraically independent, they are multiplicatively independent and . We claim that coincides with the group homomorphism

Indeed, we have for all $g_i$

which proves our claim. Recall that we need to check that . In order to do that, first notice that $\mu _1(g)$ is of the form

$$\begin{aligned} \begin{bmatrix} \kappa (g)&0\\ *&\kappa (g)^{-1} \end{bmatrix} \end{aligned}$$

for all $g \in F$ just because it is true for all $g=g_i$. This implies that . But $\mu =\mu _1$ on $F^{(1)}$. This implies that . However, as we have seen in Example 3.6, . This implies that

and we are done. $\square$

Proof

of Corollary 3.4 There exists an n-tuple that is algebraically independent over $\mathbb {Q}$. The assignment

extends to an injective ring homomorphism

This implies that is isomorphic to a subgroup of . Thanks to Theorem 3.2, $F/F^{(2)}$ is isomorphic to a subgroup of . This implies that $F/F^{(2)}$ is isomorphic to a subgroup of . $\square$

Similar arguments prove the following generalization of Theorem 3.2.

Theorem 4.1

Let be an n-tuple of non-zero integers. Then the assignment

$$\begin{aligned} g_i \mapsto \begin{bmatrix} t_i&0\\ b_i&t_i^{-1} \end{bmatrix}, \qquad 1 \leqslant i \leqslant n, \end{aligned}$$

extends to a group homomorphism with kernel $F^{(2)}$.

5 Word maps and unipotent elements

Lemma 5.1

Let w be an element of $F^{(1)}$ that does not belong to $F^{(2)}$. Then there exists a non-zero Laurent polynomial

such that

$$\begin{aligned} \mu _1(w)= \begin{bmatrix} 1&0\\ \mathscr {L}_w&1 \end{bmatrix}. \end{aligned}$$

Proof

We have seen in the course of the proof of Theorem 3.2 that for all $g \in F$

This means that there exists a Laurent polynomial $\mathscr {L}_g \in {\Lambda }_n$ such that

$$\begin{aligned} \mu _1(g)= \begin{bmatrix} \kappa (g)&0\\ \mathscr {L}_g&\kappa (g)^{-1} \end{bmatrix}. \end{aligned}$$

We have also seen that if $g\in F^{(1)}$ then $\kappa (g)=1$. Since $w\in F^{(1)}$,

$$\begin{aligned} \mu _1(w)= \begin{bmatrix} 1&0\\ \mathscr {L}_w&1 \end{bmatrix} \end{aligned}$$

with $\mathscr {L}_w \in {\Lambda }_n$. On the other hand, by Theorem 3.2, . Since $w\not \in F^{(2)}$, $\mathscr {L}_w \ne 0$ in ${\Lambda }_n$. $\square$

Corollary 5.2

Let w be an element of $F^{(1)}$ that does not belong to $F^{(2)}$. Suppose that is an n-tuple of non-zero rational numbers such that $c=\mathscr {L}_w(a_1, \dots , a_n) \ne 0$. (Since $\mathscr {L}_w \ne 0$, such an n-tuple always exists.) Let us consider the group homomorphism

Then

$$\begin{aligned} \gamma _{\mathbf a}(w)= \begin{bmatrix} 1&0\\ c&1 \end{bmatrix}=w(Z_1,\dots ,Z_n). \end{aligned}$$

is a unipotent matrix that is not the identity matrix.

Proof

One has only to notice that $\gamma _{\mathbf a}$ is the composition of $\mu _1$ and the group homomorphism induced by the ring homomorphism .$\square$

Corollary 5.3

Let w be an element of $F^{(1)}$ that does not belong to $F^{(2)}$. Let K be a field of characteristic zero. Then for every unipotent matrix $X \in {{\mathrm{SL}}}(2,K)$ there exists a group homomorphism $\psi _{w,X}:F \rightarrow {{\mathrm{SL}}}(2,K)$ such that $\psi _{w,X}(w)=X$. In other words, there exist $Z_1,\dots ,Z_n\in {{\mathrm{SL}}}(2,K)$ such that $w(Z_1,\dots ,Z_n)=X$.

Proof

We have

$$\begin{aligned} \mathbb {Q}\subset K, \qquad {{\mathrm{SL}}}(2,\mathbb {Q})\subset {{\mathrm{SL}}}(2,K)\vartriangleleft {{\mathrm{GL}}}(2,K). \end{aligned}$$

We may assume that X is not the identity matrix. Let and $\gamma _{\mathbf a}$ be as in Corollary 5.2. Recall that $c=\mathscr {L}_w(a_1, \dots , a_n) \ne 0$. Then there exists a matrix $S \in {{\mathrm{GL}}}(2,K)$ such that

Let us consider the group homomorphism . Then $\psi _{w,X}$ sends w to

Corollary 5.4

Let w be an element of $F^{(1)}$ that does not belong to $F^{(2)}$. Let K be an algebraically closed field of characteristic zero. Then the word map w is surjective on ${{\mathrm{PSL}}}(2,K)$.

Proof

Consider w as a word map on $G={{\mathrm{SL}}}(2,K)$. Due to Corollary 5.3, the image $w_G$ contains all unipotents. According to Proposition 2.2, the image contains all semisimple elements as well. Thus, the word map w has Properties 1.1 (a) and (b). It follows that it is surjective on ${{\mathrm{PSL}}}(2,K)$. $\square$

Remark 5.5

In [12], the words from are proved to be surjective on for an infinite set of integers n.

Theorem 5.6

Let w be an element of $F^{(1)}$ that does not belong to $F^{(2)}$. Let G be a connected semisimple linear algebraic group of positive dimension over a field K of characteristic zero. If $u \in G(K)$ is a unipotent element then there exists a group homomorphism $F \rightarrow G(K)$ such that the image of w coincides with u. In other words, there exist $Z_1,\dots ,Z_n\in G(K)$ such that $w(Z_1,\dots ,Z_n)=u$.

Proof

Let , $\gamma _{\mathbf a}$ and $c=\mathscr {L}_w(a_1, \dots , a_n) \ne 0$ be as in Corollary 5.2. By Lemma 5.7 below, there exists a group homomorphism such that $u=\phi (\mathbf {u}_1)$ for

Now the result follows from Corollary 5.2: the desired homomorphism is the composition .$\square$

Lemma 5.7

Let K be a field of characteristic zero, G a connected semisimple linear algebraic K-group of positive dimension, and u a unipotent element of G(K). Then for every non-zero $c\in K$ there is a group homomorphism such that u is the image of

Proof

Let us identify the additive algebraic K-group ${\mathbb G}_\mathrm{a}$ with the closed subgroup H of all matrices of the form

$$\begin{aligned} v(t)=\begin{bmatrix} 1&0\\ t&1 \end{bmatrix} \end{aligned}$$

in ${{\mathrm{SL}}}(2)$. Its Lie subalgebra $\mathrm {Lie}(H)$ is the one-dimensional K-vector subspace of $\mathfrak {sl}_2(K)$ generated by the matrix

$$\begin{aligned} \mathbf {x}_0= \begin{bmatrix} 0&0\\ 1&0 \end{bmatrix} \subset \mathfrak {sl}_2(K). \end{aligned}$$

Here we view the K-Lie algebra $\mathfrak {sl}_2(K)$ of traceless matrices as the Lie algebra of the algebraic K-group ${{\mathrm{SL}}}(2)$. Moreover, for all $\lambda \in K$.

We may view G as a closed algebraic K-subgroup of the linear algebraic group ${{\mathrm{GL}}}(N)={{\mathrm{GL}}}(V)$, where V is an N-dimensional K-vector space for a suitable positive integer N. Then

$$\begin{aligned} u \in G(K)\subset \mathrm {Aut}_{K}(V)={{\mathrm{GL}}}(N,K). \end{aligned}$$

Thus the K-Lie algebra $\mathrm {Lie}(G)$ becomes a certain semisimple Lie subalgebra of $\mathrm {End}_{K}(V)$. Here we view $\mathrm {End}_{K}(V)$ as the Lie algebra $\mathrm {Lie}({{\mathrm{GL}}}(V))$ of the K-algebraic group ${{\mathrm{GL}}}(V)$. As usual, we write

$$\begin{aligned} \mathrm {Ad}:G(K) \rightarrow \mathrm {Aut}_K(\mathrm {Lie}(G)) \end{aligned}$$

for the adjoint action of G. We have

$$\begin{aligned} \mathrm {Ad}(g)(u)=g u g^{-1} \end{aligned}$$

for all $g \in G(K)\subset \mathrm {Aut}_K(V)$ and $u \in \mathrm {Lie}(G)\subset \mathrm {End}_K(V)$. Since u is a unipotent element, the linear operator $u-1:V \rightarrow V$ is nilpotent. Let us consider the nilpotent linear operator

([7, Section 7, p. 106], [24, Section 23, p. 336]) and the corresponding homomorphism of algebraic K-groups

In particular, since $\mathbf {u}_1=v(1)$, $\varphi _u(\mathbf {u}_1)=u$. Clearly, the differential of $\varphi _u$

$$\begin{aligned} d\varphi _u:\mathrm {Lie}(H)\rightarrow \mathrm {Lie}({{\mathrm{GL}}}(V))=\mathrm {End}_K(V) \end{aligned}$$

is defined as

$$\begin{aligned} d\varphi _u (\lambda \mathbf {x}_0)=\lambda x\qquad \text {for all}\quad \lambda \in K, \end{aligned}$$

and sends $\mathbf {x}_0$ to $x\in \mathrm {Lie}({{\mathrm{GL}}}(V))$. Since for all integers m and G is closed in ${{\mathrm{GL}}}(V)$ in Zariski topology, the image $\varphi _u(H)$ of H lies in G and therefore one may view $\varphi _u$ as a homomorphism of algebraic K-groups $\varphi _u:H \rightarrow G$. This implies

$$\begin{aligned} d\varphi _u(\mathrm {Lie}(H)) \subset \mathrm {Lie}(G); \end{aligned}$$

in particular, $x \in \mathrm {Lie}(G)$.

There exists a cocharacter ${\Phi }:{\mathbb G}_m \rightarrow G\subset {{\mathrm{GL}}}(V)$ of K-algebraic group G such that for each $\beta \in K^{*}={\mathbb G}_m(K)$

(see [21, Section 6, pp. 402–403], here ${\mathbb G}_m$ is the multiplicative algebraic K-group). This means that for all $\lambda \in K$

which implies that

It follows that

Recall that is a semi-direct product of its normal subgroup H(K) and the torus

In addition,

It follows from [8, Chapter III, Proposition 27, p. 240] that there is a group homomorphism that sends each $\bigl ({\begin{matrix} 1&{}0\\ \lambda &{} 1 \end{matrix}}\bigr )$ to and each $\bigl ({\begin{matrix} \beta ^{-1}&{}0\\ 0&{} \beta \end{matrix}}\bigr )$ to ${\Phi }(\beta )$. Clearly, $\phi$ sends $\mathbf {u}_1=\bigl ({\begin{matrix} 1&{}0\\ c&{} 1 \end{matrix}}\bigr )$ to . $\square$

6 Words in two letters on ${{\mathrm{SL}}}(2,\mathbb {C})$

In this section we consider words in two letters. We provide the explicit formulas for w(x, y), where x, y are upper triangular matrices. This enables us to extract some additional information on the image of words in two letters.

Consider a word map , where $a_i\ne 0$ and $b_i\ne 0$ for all $i=1,\dots ,k$. Let $A(w)=\sum _{i=1}^k a_i$ and $B(w)=\sum _{i=1}^k b_i$.

If $A(w)=B(w)=0$, then . Since $F^{(1)}$ is a free group generated by elements , $n\ne 0$, $m\ne 0$ [23, Chapter 1, Section 1.3], the word w with $A(w)=B(w)=0$ may be written as a (non-commutative) product (with $s_i\ne 0$)

(2)

Moreover, the shortest (reduced) representation of this kind is unique. We denote by $S_w(n,m)$ the number of appearances of $w_{n,m}$ in representation (2) of w and by $R_w(n,m)$ the sum of exponents at all appearances. We denote by ${{\mathrm{Supp}}}(w)$ the set of all pairs (n, m) such that $w_{n,m}$ appears in the product. For example, if , then

The subgroup

Example 6.1

The Engel word

belongs to (see also [12]). Indeed, the direct computation shows that

It follows that

(3)

Let us prove by induction that $|R_{e_n}(1,n)|=1$, $S_{e_n}(1,n)=1$ and $S_{e_n}(r,m)=0$ if $r\ne 1$ or $m>n$, i.e.

(4)

for some integers $t\geqslant 0$, $s\geqslant 0$, $m_i<n$, , and where .

Indeed $e_1=w_{1,1}$. Assume that the claim is valid for all $k\leqslant n$. We have $e_{n+1}=e_nye_n^{-1}y^{-1}$. Using (4), we get

Applying (3) to every factor of this product, we obtain that $e_{n+1}$ has the needed form.

Thus the claim will remain valid for $n+1$. Since $|R_{e_n}(1,n)|=1$, $e_n\not \in F^{(2)}$. The surjectivity of the Engel words on simple algebraic groups was studied in [2, 12, 16]. There is a beautiful proof of surjectivity of $e_n$ on ${{\mathrm{PSL}}}(2,\mathbb {C})$ in [18, Corollary 4].

Let us take

$$\begin{aligned} x&=\begin{pmatrix} \lambda &{} c\\ 0 &{} 1/{\lambda }\end{pmatrix}, \end{aligned}$$

(5)

$$\begin{aligned} y&=\begin{pmatrix} \mu &{} d\\ 0 &{} 1/\mu \end{pmatrix}. \end{aligned}$$

(6)

Then

Here ${{\mathrm{sgn}}}$ is the signum function, and (see [1, Lemma 5.2]) for $n\geqslant 1$

Note that $h_n( 1)=n$. Direct computations show that

(7)

where

Hence,

where

$$\begin{aligned} F_w(c,d,\lambda ,\mu )=\sum _{i=1}^r s_i f(c,d,n_i,m_i)=c{\Phi }_w(\lambda ,\mu )+d{\Psi }_w(\lambda ,\mu ) \end{aligned}$$

and

$$\begin{aligned} \mathrm{\Phi }_w(\lambda ,\mu )= & {} \sum _{(\alpha ,\beta )\in {{\mathrm{Supp}}}(w)} \!\!\!\!\! R_w(\alpha ,\beta ){{\mathrm{sgn}}}(\alpha )(1-\mu ^{2\beta })\, \frac{(\lambda ^{2|\alpha |}-1)\lambda ^{\alpha }}{\lambda ^{|\alpha |-1}\,(\lambda ^{2}-1)}, \end{aligned}$$

(8)

$$\begin{aligned} \mathrm{\Psi }_w(\lambda ,\mu )= & {} \sum _{(\alpha ,\beta )\in {{\mathrm{Supp}}}(w)} \!\!\!\!\!R_w(\alpha ,\beta ){{\mathrm{sgn}}}(\beta )(\lambda ^{2\alpha }-1)\, \frac{(\mu ^{2|\beta |}-1)\mu ^{\beta }}{\mu ^{|\beta |-1}\,(\mu ^{2}-1)}. \end{aligned}$$

(9)

(Since the order of factors in w is not relevant for (8) and (9), we use here $\alpha ,\beta$ instead of $n_i, m_i$ to simplify the formulas).

Proposition 6.2

Rational functions ${\Phi }(\lambda ,\mu )$ and ${\Psi }(\lambda ,\mu )$ are non-zero linearly independent rational functions.

Remark 6.3

It is evident from the Magnus Embedding Theorem that at least one of functions ${\Phi }(\lambda ,\mu )$ and ${\Psi }(\lambda ,\mu )$ is not identical zero. It follows from Proposition 6.2 that the same is valid for both of them.

Proof

The proof is based on the following

Claim 6.4

If ${\Phi }_w(\lambda ,\mu )\equiv 0$ then $R_w(\alpha ,\beta )=0$ for all $(\alpha ,\beta )\in {{\mathrm{Supp}}}(w)$.

Proof

We use induction by the number of elements in ${{\mathrm{Supp}}}(w)$ for the word w. If ${{\mathrm{Supp}}}(w)$ contains only one pair $(\alpha ,\beta )$, then there is nothing to prove, because

Assume that for words v with it is proved. Let w be such a word that . Let .

Case 1. $n>0$. We have

It means that the coefficient of $\lambda ^{2|n|-1}$ in the rational function ${\Phi }_w(\lambda ,\mu )$ is

Hence, if ${\Phi }_w(\lambda ,\mu )\equiv 0$, then $p(\mu ) \equiv 0$, and all $R_w(n,\beta )=0$ for all $\beta$.

That means that ${\Phi }_w(\lambda ,\mu )={\Phi }_v(\lambda ,\mu )$, where v is such a word that may be obtained from $w(x,y)=\prod _{1}^r w_{n_i,m_i}^{s_i}(x,y)$ by taking away every appearance of $w_{n,\beta }$:

But and by the induction assumption $R_v(\alpha ,\beta )=0$ for all $(\alpha ,\beta )\in {{\mathrm{Supp}}}(v)$. Thus claim is valid for w in this case.

Case 2. $n<0$. Let . We proceed as in Case 1 with instead of n: the coefficient of $\lambda ^{-2n'+1}$ is

If ${\Phi }_w(\lambda ,\mu )\equiv 0$, then $q(\mu ) \equiv 0$, and all for all $\beta$. Once more, we may replace w by a word v with . $\blacksquare$

Clearly, the similar statement is valid for ${\Psi }_w(\lambda ,\mu )$. The functions ${\Phi }$ and ${\Psi }$ are linearly independent, because ${\Phi }$ is odd with respect to $\lambda$ and even with respect to $\mu$, while ${\Psi }$ has opposite properties. $\square$

Proposition 6.5

Assume that the word and that ${\Phi }_w(1,i)\ne 0$, where . Then $-\mathrm{id}\in w_G,$ where $G={{\mathrm{SL}}}(2,\mathbb {C})$.

Proof

Assume that ${\Phi }(1,i)\ne 0$. From (8) we get

Take

Then

Thus, if , then

where . Choose a such that . Then $w(x,y)=-\mathrm{id}$. $\square$

Remark 6.6

The case ${\Psi }(i,1)\ne 0$ may be treated in a similar way, one should only exchange roles of x and y.

Remark 6.7

Let , let , s odd. Put $\mu _j=m_j/k$ and . Note that some of $\mu _j$ are odd. Let $z\in {{\mathrm{SL}}}(2,\mathbb {C})$ be such that

Then $w(x,z)=u(x,y)$, hence, if ${\Phi }_u(1,i)\ne 0$, then $-\mathrm{id}\in w_G$.

7 Surjectivity on ${{\mathrm{SL}}}(2,\mathbb {C})$

We keep the notation of Sect. 6.

Lemma 7.1

Assume that , $a_i\ne 0$, $b_i\ne 0$, $i=1,\dots ,k$, $A=\sum a_i\ne 0$ or $B=\sum b_i\ne 0$ and x, y are defined by (5), (6) respectively. Then

where and

$$\begin{aligned} \widetilde{\mathrm{\Phi }}_w(\lambda ,\mu )= & {} \sum _{i=1}^k {{\mathrm{sgn}}}(a_i)h_{|a_i|}(\lambda )\,\frac{\lambda ^{\scriptscriptstyle \sum _{j<i}a_j} \mu ^{\scriptscriptstyle \sum _{j<i}b_j}}{\lambda ^{\scriptscriptstyle \sum _{j>i}a_j} \mu ^{\scriptscriptstyle \sum _{j\geqslant i}b_j}}, \end{aligned}$$

(10)

$$\begin{aligned} \widetilde{\mathrm{\Psi }}_w(\lambda ,\mu )= & {} \sum _{i=1}^k {{\mathrm{sgn}}}(b_i)h_{|b_i|}(\mu )\,\frac{\lambda ^{\scriptscriptstyle \sum _{j\leqslant i}a_j} \mu ^{\scriptscriptstyle \sum _{j<i}b_j}}{\lambda ^{\scriptscriptstyle \sum _{j>i}a_j} \mu ^{\scriptscriptstyle \sum _{j> i}b_j}}. \end{aligned}$$

(11)

Proof

We use induction on the complexity k of the word w. Using (7), we get

Thus for $k=1$ the lemma is valid. Assume that it is valid for . Let and . By the induction assumption,

From (7) we get

Multiplying matrices u and , we get

$$\begin{aligned} \widetilde{F}_{w}(c,d,\lambda ,\mu )= \lambda ^{A-a_k}\mu ^{B-b_k} \bigl (d{\cdot }\lambda ^{a_k}\!{{\mathrm{sgn}}}(b_k)h_{|b_k|}(\mu )+c{\cdot }{{\mathrm{sgn}}}(a_k)h_{|a_k|}(\lambda )\mu ^{-b_k} \bigr )\\+ \widetilde{F}_u(c,d,\lambda ,\mu )\lambda ^{-a_k}\mu ^{-b_k}. \end{aligned}$$

Thus, the induction assumption implies that

Denote

$$\begin{aligned} A_i=\sum _{j\leqslant i}a_i, \qquad B_i=\sum _{j<i}b_i, \end{aligned}$$

and let C be a curve .

Multiplying (10) and (11) by $\lambda ^{A}\mu ^B$, we see that on C the following relations are valid:

In particular, on C

(12)

(13)

Lemma 7.2

Assume that $A\ne 0$ and the word map w is not surjective. Then

for every root $\gamma$ of equation .

If $B\ne 0$ and the word map w is not surjective, then

for every root $\delta$ of equation .

Proof

The matrices z with ${{\mathrm{tr}}}(z)=2$ are in the image, because $w(x,\mathrm{id})=x^A$, $w(\mathrm{id},y)=y^B$. It is evident that $-\mathrm{id}$ is in the image: one may take $c=d=0$. Assume now that for complex numbers $\varkappa \ne 0$ the matrices

are not in the image. This implies that $\widetilde{\Phi }_w(\lambda ,\mu )\equiv 0$ and $\widetilde{\Psi }_w(\lambda ,\mu )\equiv 0$ on the defined above curve .

If $A\ne 0$ or $B\ne 0$, then the pair $(\gamma ,1)$ or $(1,\delta )$ respectively belongs to the curve C. We have to use only (12), (13), respectively. $\square$

Corollary 7.3

Let $2B_i=k_iB+T_i$, where $k_i$ are integers and $0\leqslant T_i<B\ne 0$. If w is not surjective, then for every $0\leqslant T<B$

Proof

Indeed in this case

for any root $\delta$ of equation . Since p(z) has no multiple roots, it implies that p(z) divides the polynomial

But since degree of polynomial p(z) is bigger than degree of $p_1(z)$ that can be only if $p_1(z)\equiv 0$.

$\square$

Corollary 7.4

If all $b_i$ are positive, then the word map w is either surjective or the square of another word $v\ne \mathrm{id}$.

Proof

In this case $0\leqslant 2B_i<2B$ and the sequence $B_i$ is increasing. If w is not surjective, $p_1(z)\equiv 0$, by Corollary 7.3. Thus for every $B_i$ there is $B_j$ such that $2B_i=2B_j+B$ and $a_i-a_j=0$.

Thus, the sequence of $2B_i$ looks like

It follows that $k=2s$ and

$$\begin{aligned} b_{s+1}&=B_{s+2}-B_{s+1}=B_{2}-B_{1}=b_1,\\ b_{s+2}&=B_{s+3}-B_{s+2}=B_{3}-B_{2}=b_2,\\ b_{2s-1}&=B_{2s}-B_{2s-1}=B_{s}-B_{s-1}=b_{s-1},\\ b_k&=b_{2s}=B_{2s+1}-B_{2s}=B_{s+1}-B_s=b_s. \end{aligned}$$

Thus,

$$\begin{aligned} b_i=b_{i+s}, \quad 2B_i=2B_{i+s}+B,\quad a_i=a_{i+s}, \qquad i=1,\dots , s. \end{aligned}$$

Therefore the word is the square of .$\square$

Corollary 7.5

If all $b_i$ are negative, then the word map of the word w is either surjective or the square of another word $v\ne \mathrm{id}$.

Proof

We may change y to $z=y^{-1}$ and apply Corollary 7.4 to the word w(x, z). $\square$

Corollary 7.6

If all $a_i$ are positive, then the word map of the word w is either surjective or the square of another word $v\ne \mathrm{id}$.

Proof

Consider $v=x^{-1}$, $z=y^{-1}$, a word

and apply Corollary 7.4 to the word $w'(z,v)$. $\square$

8 Trace criteria of almost surjectivity

For every word map there are defined the trace polynomials $P_w(s,t,u)={{\mathrm{tr}}}(w(x,y))$ and $Q_w={{\mathrm{tr}}}(w(x,y)y)$ in three variables $s={{\mathrm{tr}}}(x)$, $t={{\mathrm{tr}}}(y)$, and $u={{\mathrm{tr}}}(xy)$ [13–15, 25].

In other words, the maps

may be included into the following commutative diagram:

Moreover, $\pi$ is a surjective map [15]. For details, one can be referred to [3, 5].

Since the coordinate t is invariant under $\psi _w$, for every fixed value $t=a\in \mathbb {C}$ we may consider the restriction $\psi _a(s,u)=(P_w(s,a,u),Q_w(s,a,u))$ of morphism $\psi _w$ to the plane .

Definition 8.1

We say that $\psi _a(s,u)$ is Big if the image , where $T_a$ is a finite set. We say that the trace map $\psi _w$ of a word $w\in F$ is Big if there is a value a such that $\psi _a(s,u)$ is Big.

Proposition 8.2

If the trace map $\psi _w$ of a word $w\in F$ is Big then the word map is almost surjective.

Proof

Let a be such a value of t that the map $\psi _a$ is Big. Let . Consider lines and in $\mathbb {C}^2_{s,u}$. Let and . Since $S_a$ is finite, $B_+\ne \varnothing$, $B_-\ne \varnothing$. Moreover, since these curves are outside $S_a,$ we have: $D_+=\psi ^{-1}(B_+)\ne \varnothing$, $D_-=\psi ^{-1}( B_-)\ne \varnothing$. Take $(s_0,u_0)\in D_+$ and $(s_1,u_1)\in D_-$. Then $\psi _w(s_0,a,u_0)=(2,a,b)$ with $a\ne b$ and with . The projection is surjective, thus there is a pair $(x_0,y_0)\in G^2$ such that ${{\mathrm{tr}}}(x_0)=s_0$, ${{\mathrm{tr}}}(y_0)=a$, ${{\mathrm{tr}}}(x_0y_0)=u_0.$ Then $\pi (w(x_0,y_0))=\psi _w(s_0,a,u_0)=(2,a,b)$. Hence, ${{\mathrm{tr}}}(w(x_0,y_0))=2$, but $w(x_0,y_0)\ne \mathrm{id}$, since ${{\mathrm{tr}}}(w(x_0,y_0)y_0)=b\ne a={{\mathrm{tr}}}(y_0)$. Similarly, there is a pair $(x_1,y_1)\in G^2$ such that ${{\mathrm{tr}}}(x_1)=s_1$, ${{\mathrm{tr}}}(y_1)=a$, ${{\mathrm{tr}}}(x_1y_1)=u_1$. Then . Hence, , but $w(x_1,y_1)\ne -\mathrm{id}$, since . It follows that all the elements $z\ne -\mathrm{id}$ with trace 2 and are in the image of the word map w. $\square$

Corollary 8.3

Assume that the trace map $\psi _w$ of a word w is Big. Consider a sequence of words defined recurrently in the following way:

$$\begin{aligned} v_1(x,y)=w(x,y), \qquad v_{n+1}(x,y)=w(v_n(x,y),y). \end{aligned}$$

Then the word map is almost surjective for all $n\geqslant 1$.

Proof

The trace map $\psi _n=\psi _{v_n}$ of the word map $v_n$ is the $n^\mathrm{th}$ iteration $\psi ^{(n)}_1$ of the trace map $\psi _1=\psi _w$ (see [3, 5]). Let us show by induction, that all maps $\psi _{n}$ are Big. Indeed $\psi _1$ is Big by assumption, hence ${(\psi _1)}_a (\mathbb {C}^2_{s,u})= \mathbb {C}^2_{s,u}-T_a$ for some value a and some finite set $T_a$. Assume now that $\psi _{n-1}$ is Big. Let for a value a of t the image for some finite set N. Hence

Thus ${(\psi _n)}_a$ is Big as well for the same value a.

According to Proposition 8.2, the word map $v_n$ is almost surjective. $\square$

Example 8.4

Consider the word and the corresponding sequence

This is one of the sequences that were used for characterization of finite solvable groups (see [3, 5, 9]).

We have [5, Section 5.1]

We want to show that for a general value $t=a$ the system of equations

$$\begin{aligned} f_1(s,a,u)=A,\qquad f_2(s,a,u)=B \end{aligned}$$

(14)

has solutions for all pairs , where $T_a$ is a finite set.

Consider the system

(15)

Note that with respect to u the leading coefficients of $h_1$ and $h_2$ are 1 and respectfully. The MAGMA computations show that the resultant (elimination of u) of $h_1-C$ and $h_2-D$ is of the form

It has a non-zero root $s\ne 0$ at any point (a, C, D), where at least two of three polynomials $p_1,p_2,p_3$ do not vanish. The MAGMA computations show that the ideals , generated, respectively, by $p_1(a,C,D)$ and $p_2(a,C,D),$ by $p_1(a,C,D)$ and $p_3(a,C,D),$ by $p_2(a,C,D)$ and $p_3(a,C,D)$, are one-dimensional. It follows that for a general value of a the set

is a finite subset $N_a\subset \mathbb {C}_{C,D}$. On the other hand, at any point (C, D) outside $N_a$ the polynomial $R_a(s)=R(s,a,C,D)$ has a non-zero root, and, therefore system (15) has a solution. Thus, outside the finite set of points , system (14) has a solution as well. Thus, $\psi _w=(f_1,t,f_2)$ is Big and all word maps $v_n$ are almost surjective on G. Let us cite the MAGMA computations for $t=a=1,$ where $p=h_1-C$, $q=h_2-D$ and R is the resultant of p, q with respect to u.

Clearly every pair among polynomials $p_1,p_2,p_3$ has only finite number of common zeros. For example, $p_1=p_3=0$ implies $D^2-5D+1=0$ or . Computations show also that the word w(x, y) takes on value $-\mathrm{id}$. For example, one make take

$$\begin{aligned} x=\begin{pmatrix}- 1&{}1\\ -2 &{}1\end{pmatrix}, \qquad y=\begin{pmatrix} 1&{}t\\ 0&{}1\end{pmatrix}, \end{aligned}$$

where . Therefore, the word $v_1$ is surjective. Here are computations:

Therefore, $t^2=-1/2$ implies that , $q_{12}=q_{21}=0$.

9 The word

In this section we provide an example of a word v that is surjective though it belongs to $F^{(2)}$. The interesting feature of this word is the following: if we consider it as a polynomial in the Lie algebra $\mathfrak {sl}_2$ ( being the Lie bracket) then it is not surjective [4, Example 4.9].

Theorem 9.1

The word is surjective on ${{\mathrm{SL}}}(2,\mathbb {C})$ (and, consequently, on ${{\mathrm{PSL}}}(2,\mathbb {C}))$.

Proof

As it was shown in Proposition 2.2, for every $z\in {{\mathrm{SL}}}(2,\mathbb {C})$ with there are $x,y\in {{\mathrm{SL}}}(2,\mathbb {C})^2$ such that $v(x,y)=z$.

Assume now that . We have to show that $-\mathrm{id}$ is in the image and that there are matrices x, y in ${{\mathrm{SL}}}(2,\mathbb {C})$ such that

$$\begin{aligned} v(x,y)=\begin{pmatrix} q_{11} &{}q_{12}\\ q_{21} &{} q_{22}\end{pmatrix} \end{aligned}$$

has the following properties:

• ,

• $q_{12}\ne 0$.

We may look for these pairs among the matrices $x=\bigl ({\begin{matrix} 0 &{}b\\ c &{} d\end{matrix}}\bigr )$ and $y=\bigl ({\begin{matrix} 1 &{}t \\ 0 &{} 1\end{matrix}}\bigr )$.

In the following MAGMA calculations , , , . The ideal I in the polynomial ring is defined by conditions , ${{\mathrm{tr}}}(A)=2$. The ideal J in the polynomial ring is defined by conditions , . Let $T_+\subset {{\mathrm{SL}}}(2)^2$ and $T_-\subset {{\mathrm{SL}}}(2)^2$ be, respectively, the corresponding affine subsets in the affine variety ${{\mathrm{SL}}}(2)^2$. The computations show that $q_{12}(b,c,d,t)$ does not vanish identically on $T_+$ or $T_-$.

It follows that the function $q_{12}(b,c,d,t)$ does not vanish identically on the sets $T_+$ and $T_-$, hence, there are pairs with ${{\mathrm{tr}}}(v(x,y))=2$, $v(x,y)\ne \mathrm{id}$, and , $v(x,y)\ne - \mathrm{id}$.

In order to produce the explicit solutions for $v(x,y)=-\mathrm{id}$ and $v(x,y) =z$, $z\ne -\mathrm{id}$, , consider the following matrices depending on one parameter d:

Since the images of the commutator word on ${{\mathrm{GL}}}(2,\mathbb {C})$ and ${{\mathrm{SL}}}(2,\mathbb {C})$ are the same, we do not require that or . We only assume that and . Let

$$\begin{aligned} A=v(x,y)=\begin{pmatrix} q_{11}(d) &{}\ q_{12}(d)\\ q_{21}(d) &{}\ q_{22}(d)\end{pmatrix} \end{aligned}$$

and $TA={{\mathrm{tr}}}(A)$. The MAGMA computations show that

$$\begin{aligned} q_{11}(d)+1&=N_{11}\biggl (d^2 - d + \frac{1}{3}\biggr )H_{11}(d),\\\ q_{22}(d)+1&=N_{22}\biggl (d^2 - d + \frac{1}{3}\biggr )H_{22}(d),\\ q_{21}(d)&=N_{21}\biggl (d-\frac{2}{3}\biggr )^2\biggl (d - \frac{1}{2}\biggr )^{\!3}\biggl (d - \frac{1}{3}\biggr )^{\!2}\\ {}&\cdot \biggl (d^2 - d - \frac{2}{3}\biggr )\biggl (d^2 - d + \frac{1}{3}\biggr )H_{21}(d),\\ q_{12}(d)&=N_{21}\biggl (d-\frac{2}{3}\biggr )^{\!2}\biggl (d - \frac{1}{2}\biggr )^{\!3}\biggl (d - \frac{1}{3}\biggr )^{\!2}\\ {}&\cdot \biggl (d^2 - d - \frac{2}{3}\biggr )\biggl (d^2 - d + \frac{1}{3}\biggr )H_{12}(d),\\ TA+2&=N\biggl (d^2 - d + \frac{1}{3}\biggr )H(d), \end{aligned}$$

where $N_{ij}$ and N are non-zero rational numbers; $H_{ij}$ and H are polynomials with rational coefficients that are irreducible over $\mathbb {Q}$. Moreover $\deg H_{21}=\deg H_{12}=25$ and $\deg H=38$. It follows that if $d^2 - d + 1/3=0$ then $A=-\mathrm{id}$. If d is a root of H that is not a root of $H_{21}$, then A is a minus unipotent (which is not $-\mathrm{id}$). $\square$