A nonstandard rational permutation representation $PE_3(\mathbb Z)\to\operatorname{GL}_7(\mathbb Q)$

Problem

Describe, at least in particular cases, homomorphisms from projective elementary groups that fail the standard Weyl-image condition. Can such a homomorphism be given explicitly over the integers?

$$\rho_0:PE_3(\mathbb Z)\longrightarrow\operatorname{GL}_{\mathbb Q}(W).$$

The obstruction below compares nonidentity involutory Weyl images with signed Weyl matrices of order four. It is an explicit example, not a classification of all homomorphisms.

The representation and its obstruction

For a commutative ring $R$, let $E_n(R)\leq\operatorname{GL}_n(R)$ be generated by the transvections $t_{ij}(r)$, and put

$$t_{ij}(r)=1+r e_{ij},\qquad a_{ij}=t_{ij}(1)t_{ji}(-1)t_{ij}(1) \quad(i\ne j,\ r\in R).$$

We write $PE_n(R)$ for the image of $E_n(R)$ in $\operatorname{GL}_n(R)/Z(\operatorname{GL}_n(R))$, and $\bar x$ for the projective image of $x\in E_n(R)$.

Problem 10.43 of the Kourovka Notebook (1) asks, at least in particular cases, for homomorphisms outside a Weyl-image condition. That condition identifies all images of $\bar a_{ij}$ with signed Weyl matrices over an endomorphism ring, through an injective group homomorphism and a module isomorphism. The standard-description theory was developed by Petechuk (2); trivial Weyl-square images also occur in the analysis of Petechuk and Petechuk (3). The example below fails the printed condition for every permitted embedding.

Set $\Omega=\mathbb F_2^3\setminus\{0\}$ and $W=\mathbb Q^{\Omega}$, with coordinate basis $(b_v)_{v\in\Omega}$. For $x\in\operatorname{GL}_3(\mathbb Z)$, write $\widetilde x$ for its reduction modulo two and define

$$\tag{2} \rho(x)b_v=b_{\widetilde x v}\qquad(v\in\Omega).$$

For a rational vector space $P$, put $D_P=\operatorname{End}_{\mathbb Q}(P)$, with $1_{D_P}$ the identity of $P$, and define

$$\tag{3} a_P^*=\begin{pmatrix}0&1&0\\-1&0&0\\0&0&1\end{pmatrix} \in\operatorname{GL}_3(D_P).$$

Theorem 1. The representation (2) induces a homomorphism

$$\rho_0:PE_3(\mathbb Z)\longrightarrow\operatorname{GL}_{\mathbb Q}(W), \qquad \rho_0(\bar x)=\rho(x),$$

where $|\Omega|=7$. For every $i\ne j$,

$$\tag{5} \rho_0(\bar a_{ij})^2=1,\qquad \rho_0(\bar a_{ij})\ne1.$$

For any rational vector spaces $P,T$, any linear isomorphism $g:W\to P^3\oplus T$, and any injective homomorphism

$$\tau:\operatorname{GL}_3(D_P)\longrightarrow \operatorname{GL}_{\mathbb Q}(P^3\oplus T),$$

one has

$$\tag{7} g\rho_0(\bar a_{12})g^{-1}\ne\tau(a_P^*).$$

Consequently $\rho_0$ fails the condition in Problem 10.43 and is not induced by a standard homomorphism.

No finiteness, freeness, or nonzero hypothesis is imposed on $P$ or $T$. The theorem concerns this particular representation.

Descent and Weyl images

Reduction modulo two sends invertible matrices to invertible matrices and preserves products. Thus each $\widetilde x$ permutes $\Omega$. The convention in (2) gives

$$\rho(x)\rho(y)b_v=b_{\widetilde x\widetilde y v} =\rho(xy)b_v,$$

so $\rho$ is a homomorphism. There are $2^3-1=7$ basis vectors.

If $z=(z_{kl})$ belongs to the centre of $\operatorname{GL}_3(\mathbb Z)$, then

$$z(1+e_{ij})=(1+e_{ij})z, \qquad z_{ki}\delta_{jl}=\delta_{ki}z_{jl}\quad(i\ne j).$$

Taking $l=j$ and $k\ne i$ gives $z_{ki}=0$; taking $k=i$ and $l=j$ gives $z_{ii}=z_{jj}$. Thus $z=cI_3$ for some $c\in\mathbb Z$. Its reduction $\widetilde cI_3$ is invertible, so $\widetilde c\ne0$ in $\mathbb F_2$ and hence $\widetilde c=1$. Therefore $\rho(z)=1$, and the restriction of $\rho$ to $E_3(\mathbb Z)$ descends to its image in the quotient by the ambient centre, giving $\rho_0$.

For $i\ne j$, direct multiplication of the three defining transvections gives

$$\widetilde a_{ij}^{\,2}=I_3,\qquad \widetilde a_{ij}e_i=e_j.$$

The first identity implies $\rho(a_{ij})^2=1$; the second gives $\rho(a_{ij})b_{e_i}=b_{e_j}\ne b_{e_i}$. This proves (5) for all six ordered pairs.

Failure of the standard formula

Lemma 2. Let $P$ be a rational vector space, $H$ a group, and $\tau:\operatorname{GL}_3(D_P)\to H$ an injective homomorphism. If $y\in H$ satisfies $y^2=1$ and $y\ne1$, then $\tau(a_P^*)\ne y$.

Apply Lemma 2 with $H=\operatorname{GL}_{\mathbb Q}(P^3\oplus T)$ and $y=g\rho_0(\bar a_{12})g^{-1}$. Conjugation preserves the two properties in (5), and (7) follows.

To exclude the standard formula, factor its unital coefficient homomorphism through its image as $\mathbb Z\xrightarrow{\delta}B\xrightarrow{\iota}D_P$, with $\delta$ surjective and $\iota$ injective. Let $f\in B$ be the central idempotent and $\nu:B\to B^{\mathrm{op}}$ the antiisomorphism. Writing $\widehat\nu(b)$ for the underlying element of $B$ corresponding to $\nu(b)$, the coefficient matrix has entries

$$\tag{11} M(x)_{ij}=\iota\!\left( \delta(x_{ij})f+ \widehat\nu\bigl(\delta((x^{-1})_{ji})\bigr)(1-f) \right).$$

All the coefficient maps preserve $0,1,-1$. For $x=a_{12}$, one has

$$x=\begin{pmatrix}0&1&0\\-1&0&0\\0&0&1\end{pmatrix}, \qquad x^{-1}=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix}.$$

Evaluating (11) gives $M(a_{12})=a_P^*$: on each nonzero entry the two branches recombine using $f+(1-f)=1$. Standardness would imply $g\rho_0(\bar a_{12})g^{-1}=\tau(a_P^*)$, contrary to (7). This completes the proof of Theorem 1.

References

Preprint · Lean (GitHub)

1. E. I. Khukhro and V. D. Mazurov (eds.), The Kourovka Notebook: Unsolved problems in group theory, 21st ed., September 2026 update, Problem 10.43, arXiv:1401.0300v46.

2. V. M. Petechuk, Homomorphisms of linear groups over rings, Mathematical Notes 45 (1989), 144–151, doi:10.1007/BF01158061.

3. V. M. Petechuk and Yu. V. Petechuk, Images of formal matrices of elements of matrix groups over associative rings, Scientific Bulletin of Uzhhorod University. Series of Mathematics and Informatics 36(1) (2020), 16–29, doi:10.24144/2616-7700.2020.1(36).16-29.