# A construction of the Shephard–Todd group G32 through the Weyl group of type E6

Cédric Bonnafé

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Cédric Bonnafé. A construction of the Shephard–Todd group G32 through the Weyl group of type E6. Portugaliae Mathematica, 2024, 82 (1), pp.63-70. 10.4171/PM/2119. hal-05067063

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# A construction of the Shephard–Todd group $G_{32}$ through the Weyl group of type $E_6$

Cédric Bonnafé

**Abstract.** It is well known that the quotient of the derived subgroup of the Shephard–Todd complex reflection group $G_{32}$ (which has rank 4) by its center is isomorphic to the derived subgroup of the Weyl group of type $E_6$. We show that this isomorphism can be realized through the second exterior power, and take the opportunity to propose an alternative construction of the group $G_{32}$.

Let $G_{32}$ denote the complex reflection group constructed by Shephard–Todd [5] and let $E_6$ be a Weyl group of type $E_6$ (which is denoted by $G_{35}$ in the Shephard–Todd classification). Let $Sp_4(\mathbb{F}_3)$ (resp. $SO_5(\mathbb{F}_3)$) denote the symplectic (resp. orthogonal) group of dimension 4 (resp. 5) over the finite field with three éléments $\mathbb{F}_3$. Let $\Omega_5(\mathbb{F}_3)$ be the image of $Sp_4(\mathbb{F}_3)$ in $SO_5(\mathbb{F}_3)$ through the natural morphism $Sp_4(\mathbb{F}_3) \to SO_5(\mathbb{F}_3)$ induced by the second exterior power: this is the normal subgroup of index 2 of $SO_5(\mathbb{F}_3)$. Finally, if $G$ is a group, let $D(G)$ and $Z(G)$ denote respectively its derived subgroup and its center and, if $d$ is a non-zero natural number, let $\mu_d$ be the group of $d$-th roots of unity in $\mathbb{C}$.

It is shown in [4, Thms. 8.43 and 8.54] that $G_{32} \simeq \mu_3 \times Sp_4(\mathbb{F}_3)$ and that $E_6 \simeq SO_5(\mathbb{F}_3)$. In particular,

$$D(G_{32})/\mu_2 \simeq \Omega_5(\mathbb{F}_3) \simeq D(E_6). \quad (*)$$

The purpose of this note is to present a direct elementary explanation of the isomorphism $D(G_{32})/\mu_2 \simeq D(E_6)$, which in fact allows us to construct the complex reflection group $G_{32}$ from the rational reflection group $E_6$. This construction uses the classical morphism $SL_4(\mathbb{C}) \to SO_6(\mathbb{C})$, and follows the same lines as our previous paper [1] (in which we constructed the complex reflection group $G_{31}$ from the Weyl group of type $B_6$).

<footnote>Mathematics Subject Classification 2020: 20F55.</footnote>

<footnote>Keywords: reflection groups, primitive groups.</footnote>
This note does not pretend to prove a deep result: it is just a nice example of the application of the classical theory of reflection groups (invariants, Springer theory, ...).

**Remark.** It is shown in [4, Thm. 8.53] that $G_{33} \simeq \mu_2 \times \Omega_5(\mathbb{F}_3)$. In particular, this gives an indirect isomorphism $D(G_{33}) \simeq D(E_6)$. However, we do not know of any construction of this isomorphism which would be in the same spirit as above.

# 1. The morphism $SL_4(\mathbb{C}) \to SO_6(\mathbb{C})$

We recall here the construction of this morphism, using some notation from [1]. Let us fix a complex vector space $V$ of dimension 4 and let

$$ \Lambda : \mathrm{GL}(V) \to \mathrm{GL}(\Lambda^2 V), \\ g \mapsto \Lambda^2 g $$

be the natural morphism of algebraic groups. Note that $\Lambda^2 V$ has dimension 6 and that

$$ \mathrm{Ker} \Lambda \simeq \mu_2 = \{\pm \mathrm{Id}_V\}. $$

The lists of eigenvalues of elements of $\mathrm{GL}(V)$ or $\mathrm{GL}(\Lambda^2 V)$ will always be given with multiplicities. If $g \in \mathrm{GL}(V)$ admits $a, b, c$ and $d$ as eigenvalues, then

$$ \Lambda(g) \text{ admits } ab, ac, ad, bc, bd \text{ and } cd \text{ as eigenvalues.} \qquad (1.1) $$

In particular,

$$ \det \Lambda(g) = (\det g)^3. $$

Let us fix now a generator $\varepsilon$ of the one-dimensional vector space $\Lambda^4 V$. The choice of this generator allows to identify $\mathbb{C}$ and $\Lambda^4 V$ and to define a bilinear form

$$ \beta_{\Lambda} : \Lambda^2 V \times \Lambda^2 V \to \mathbb{C}, \\ (x, y) \mapsto x \wedge y. $$

This bilinear form is symmetric and non-degenerate. By definition of the determinant,

$$ \beta_{\Lambda}(\Lambda(g)(x), \Lambda(g)(y)) = (\det g) \beta_{\Lambda}(x, y) $$

for all $g \in \mathrm{GL}(V)$ and $x, y \in \Lambda^2 V$. For dimension and connectedness reasons, the image of $\mathrm{GL}(V)$ through $\Lambda$ is the neutral component $\mathrm{CO}(\Lambda^2 V)^{\circ}$ of the conformal orthogonal group $\mathrm{CO}(\Lambda^2 V) = \mathrm{CO}(\Lambda^2 V, \beta_{\Lambda})$ and $\Lambda$ induces an isomorphism of algebraic groups

$$ \mathrm{SL}(V)/\mu_2 \simeq \mathrm{SO}(\Lambda^2 V). $$
The next elementary lemma will be useful (in this paper, we denote by $j$ a primitive third root of unity).

**Lemma 1.6.** Let $g \in \mathbf{SO}(\wedge^2 V)$ having $j, j, j, j^2, j^2$ and $j^2$ as eigenvalues. Then there exists a unique element $\tilde{g} \in \mathbf{SL}(V)$, having 1, $j, j$ and $j$ as eigenvalues, and such that

$$ \Lambda^{-1}(\{g, g^{-1}\}) = \{\pm \tilde{g}, \pm \tilde{g}^{-1}\}. $$

*Proof.* Let $h \in \mathbf{SL}(V)$ be such that $\Lambda(h) = g$. Then $\Lambda^{-1}(\{g, g^{-1}\}) = \{\pm h, \pm h^{-1}\}$. Let $a, b, c$ and $d$ be the eigenvalues of $h$. By (1.1), by reordering if necessary the eigenvalues of $h$, we may assume that $ab = ac = j$. In particular, $b = c$. So $bd = cd$, which implies that $bd = cd = j^2$ (for otherwise $j$ would be an eigenvalue of $g$ with multiplicity $\ge 4$). Still by (1.1), $(ad, bc) = (j, j^2)$ or $(ad, bc) = (j^2, j)$.

If $(ad, bc) = (j, j^2)$, then $b = c = d$ and $b^2 = j^2$, so $b = c = d = \eta j$ and $a = \eta$, for some $\eta \in \{\pm 1\}$. By replacing $h$ by $-h$ if necessary, the eigenvalues of $h$ are then 1, $j, j, j$ and so $h$ is indeed the unique element of $\Lambda^{-1}(\{g, g^{-1}\})$ admitting this list of eigenvalues.

If $(ad, bc) = (j^2, j)$, then $d = jb =jc$ and $b^2 = j$, which implies that $a = b = c = \eta j^2$ and $d = \eta$, for some $\eta \in \{\pm 1\}$. By replacing $h$ by $-h$ if necessary, the eigenvalues of $h$ are then $j^2, j^2, j^2$ and so $h^{-1}$ is indeed the unique element of $\Lambda^{-1}(\{g, g^{-1}\})$ admitting 1, $j, j, j$ as list of eigenvalues. ■

**Corollary 1.7.** Let $g \in \mathbf{SO}(\wedge^2 V)$ having $j, j, j, j^2, j^2$ and $j^2$ as eigenvalues. Then $g$ and $g^{-1}$ are not conjugate in $\mathbf{SO}(\wedge^2 V)$.

*Proof.* Assume that we have found $w \in \mathbf{SO}(\wedge^2 V)$ such that $wgw^{-1} = g^{-1}$. Let $\omega$ and $\gamma$ be respective preimages of $w$ and $g$ in $\mathbf{SL}(V)$. Then $\Lambda(\omega\gamma\omega^{-1}) = g^{-1} = \Lambda(\gamma^{-1})$. This shows that $\gamma^{-1}$ is conjugate to $\gamma$ or to $-\gamma$, which is impossible by examining the possible lists of eigenvalues of $\gamma$ obtained in the proof of Lemma 1.6. ■

## 2. Construction of $G_{32}$

Let us see $E_6$ as a finite subgroup of $\mathbf{O}(\wedge^2 V)$ generated by reflections. Set

$$ \mathbf{jSO}(\wedge^2 V) = \langle \mathbf{SO}(\wedge^2 V), j \operatorname{Id}_{\wedge^2 V} \rangle $$

and

$$ \sqrt[3]{\mathbf{SL}(V)} = \{g \in \mathbf{GL}(V) \mid \det(g)^3 = 1\} = \langle \mathbf{SL}(V), j \operatorname{Id}_V \rangle. $$

Then

$$ \Lambda^{-1}(\mathbf{jSO}(\wedge^2 V)) = \sqrt[3]{\mathbf{SL}(V)}. $$
We then define

$$W = \Lambda^{-1}(\langle D(E_6), j \operatorname{Id}_{\Lambda^2 V} \rangle).$$

It is a subgroup of $\sqrt[3]{SL}(V)$. The aim of this note is to show that $W$ is isomorphic to the complex reflection group $G_{32}$ of Shephard–Todd. Note first that

$$\mu_6 \subset W.$$

## 2.A. Reflections in W

The list of degrees of $E_6$ is 2, 5, 6, 8, 9, 12 while its list of codegrees is 0, 3, 4, 6, 7, 10 (see [2, Table A.3]). In particular, exactly 3 of the degrees are divisible by 3, which shows [6, Thm. 3.4] that $E_6$ contains an element $w_3$ admitting the eigenvalue $j$ with multiplicity 3. We will denote by $C_3$ the conjugacy class of $w_3$ in $W$. Since also exactly 3 of the codegrees are divisible by 3, this implies, for instance by [3, Thm. 1.2], that $w_3$ is regular in the sense of Springer [6, §4] (that is, admits an eigenvector for the eigenvalue $j$ whose stabilizer in $E_6$ is trivial). Since $E_6$ is a rational group, $w_3$ also admits $j^2$ as an eigenvalue with multiplicity 3. Hence, the eigenvalues of $w_3$ are $j, j, j, j^2, j^2$ and $j^2$. In particular, $\det w_3 = 1$ and so

$$w_3 \in D(E_6) = E_6 \cap SO(\Lambda^2 V).$$

By [6, Thm. 4.2 (iii)], the centralizer of $w_3$ in $E_6$ has order $6 \cdot 9 \cdot 12$, which shows that

$$|C_3| = 2 \cdot 5 \cdot 8 = 80. \qquad (2.1)$$

Moreover [6, Thm. 4.2 (iv)], if $w \in E_6$, then

$$w \in C_3 \text{ if and only if } \dim \ker(w - j \operatorname{Id}_V) = 3. \qquad (2.2)$$

Hence, if $w \in C_3$, then $w^{-1} \in C_3$ but $w^{-1}$ is not conjugate to $w$ in $D(E_6)$ (by Corollary 1.7).

Now, let $\mathrm{Ref}(W)$ denote the set of reflections of $W$. If $s \in \mathrm{Ref}(W)$, the eigenvalues $\Lambda(s)$ are 1, 1, 1, $\det(s)$, $\det(s)$, $\det(s)$. Since $\det(s) \in \{j, j^2\}$, the eigenvalues of $\det(s)\Lambda(s)$ are then $j, j, j, j^2, j^2, j^2$. Since the characteristic polynomials of elements of $D(E_6)$ have coefficients in $\mathbb{Q}$, this implies that $\Lambda(s)$ and $\det(s)^2\Lambda(s)$ do not belong to $D(E_6)$. So $\det(s)\Lambda(s) \in D(E_6)$ and it follows from (2.2) that $\det(s)\Lambda(s) \in C_3$. This defines a map

$$\lambda : \mathrm{Ref}(W) \to C_3, \\ s \mapsto \det(s)\Lambda(s).$$

The next result will be very useful.
**Lemma 2.5.** The map $\lambda$ is bijective.

*Proof.* First, if $\lambda(s) = \lambda(s')$, then there exists $\xi \in \mathbb{C}^\times$ such that $s' = \xi s$. Since $s$ and $s'$ are reflections, this is possible only if $\xi = 1$, and so $s = s'$. This shows that $\lambda$ is injective.

Let us now show the surjectivity. Let $w \in C_3$. By Corollary 1.7, there exists a unique $\tilde{w} \in \Lambda^{-1}(\{w, w^{-1}\})$ admitting 1, j, j, j as eigenvalues. Then $j^2\tilde{w}$ and $j\tilde{w}^{-1}$ are reflections satisfying

$$ \lambda(j^2\tilde{w}) = \det(j^2\tilde{w})\Lambda(j^2\tilde{w}) = j^8 \cdot j^4\Lambda(\tilde{w}) = \Lambda(\tilde{w}) $$

and

$$ \lambda(j\tilde{w}^{-1}) = \det(j\tilde{w}^{-1})\Lambda(j\tilde{w}^{-1}) = j^4 \cdot j^2\Lambda(\tilde{w}^{-1}) = \Lambda(\tilde{w})^{-1}. $$

So $w \in \{\lambda(j^2\tilde{w}), \lambda(j\tilde{w}^{-1})\}$, which shows that $\lambda$ is surjective. ■

We then deduce from (2.1) and Lemma 2.5 that

$$ |\mathrm{Ref}(W)| = 80. \qquad (2.6) $$

## 2.B. Structure of W

Our main result is Theorem 2.7 below: the proof we propose here uses neither known properties of the group $G_{32}$ nor the classification of complex reflection groups and so might be viewed as an alternative construction of $G_{32}$ starting from $E_6$ (however, note that we use properties of $E_6$).

**Theorem 2.7.** The group W

(a) has order 155 520;

(b) is generated by reflections of order 3;

(c) is irreducible and primitive;

(d) admits 12, 18, 24, 30 as list of degrees.

*Proof.* Set $E_6^\# = \langle D(E_6), j \mathrm{Id}_{\Lambda^2 V} \rangle$ and $W^+ = W \cap \mathrm{SL}(V)$. Recall that $|E_6| = 51 840$. So

$$ |D(E_6)| = 25 920, \quad |E_6^\#| = 77 760, \quad |W| = 155 520 \quad \text{and} \quad |W^+| = 51 840. $$

This shows (a). Moreover,

$$ Z(W) = \mu_6 \quad \text{and} \quad W/\mu_6 \simeq D(E_6). $$

Let $\mathcal{R} = \{\det(s)^{-1}s \mid s \in \mathrm{Ref}(W)\}$. Set

$$ G = \langle \mathrm{Ref}(W) \rangle \quad \text{and} \quad H = \langle \mathcal{R} \rangle. $$
Statement (b) is equivalent to the following one:

$$W = G. \qquad (\#)$$

First, $\Lambda(\mathcal{R}) = C_3$ and so $\Lambda(H) = D(E_6)$ (as this last group is simple and $C_3$ is a conjugacy class). In particular, $W = H \cdot \mu_6$ and so, since $H \subset G \cdot \mu_3$, we obtain

$$W = G \cdot \mu_6,$$

which is almost the expected result (#). Before showing (#), note that, since $\Lambda(H) = D(E_6)$, we have that $H$ (and so $G$) acts irreducibly on $V$ (indeed, if $H$ did not act irreducibly on $V$, then the representation of $D(E_6)$ on $\Lambda^2 V$ would split as a direct sum of four submodules, which is impossible since $D(E_6)$ has index 2 in $E_6$). Moreover, if $G$ is not primitive, then $G$ (and so $H$) would be monomial [4, Lem. 2.12], which would imply that $\Lambda(H) = D(E_6)$ is monomial, which is false. So

G is irreducible and primitive. (♣)

Statement (c) follows.

Let us conclude this proof by showing simultaneously (b) and (d). Let $d_1, d_2, d_3, d_4$ be the degrees of $G$. It follows from (2.6) and for instance from [2, Theo. 4.1] that

$$\begin{cases} d_1 d_2 d_3 d_4 = |G|, \\ d_1 + d_2 + d_3 + d_4 = 84. \end{cases} \qquad (\diamond)$$

The morphism $\det : G \to \mu_3$ is surjective (since $\det(s) \in \{j, j^2\}$ for any $s \in \text{Ref}(W)$), and this implies that $\mu_3 \subset G$ (because $G/(G \cap \mu_6) \simeq D(E_6)$ is simple). It remains to show that $|G| \neq |W|/2 = 77760$.

So assume that $|G| = 77760$. Since $\mu_3 \subset G$, all the $d_i$'s are divisible by 3 and, since $\mu_6 \not\subset G$, at least one of them (say $d_4$) is not divisible by 6. Write $e_i = d_i/3$. Then

$$\begin{cases} e_1 e_2 e_3 e_4 = 960 = 2^6 \cdot 3 \cdot 5, \\ e_1 + e_2 + e_3 + e_4 = 28, \\ e_4 \text{ is odd.} \end{cases}$$

From the second equality, we deduce that at least one more of the $e_i$'s (say $e_3$) is odd. From the first equality, we deduce that at least one of the $e_i$'s (say $e_2$) is even and so $e_1$ is also even. The first equality shows that $e_1$ or $e_2$ (say $e_2$) is divisible by 8. So $e_2 \in \{8, 16, 24\}$. A quick inspection of the possibilities shows that $e_1 = 4$, $e_2 = 16$ and $\{e_3, e_4\} = \{3, 5\}$. The degrees of $G$ are then 9, 12, 15 and 48. Since 16 divides one of the degrees, it then follows from [6, Thm. 3.4 (i)] that $G = H \cdot \mu_3$ contains an element of order 16 and so $H$ contains an element of order 16. Therefore,
$D(E_6) = \Lambda(H)$ contains an element of order 8, which is impossible (see Remark 2.11 below for a proof of this fact, based only on Springer theory). This contradicts the fact that $|G| = 77760$. So we have shown (#), that is,

$$W = \langle \mathrm{Ref}(W) \rangle = G.$$

This is statement (b). In particular, $\mu_6 \subset W$ and so all the $d_i$'s are divisible by 6. Set $a_i = d_i/6$. Then (♦) implies that

$$\begin{cases} a_1 a_2 a_3 a_4 = 120 = 2^3 \cdot 3 \cdot 5, \\ a_1 + a_2 + a_3 + a_4 = 14. \end{cases}$$

By the same argument as before, since $\mu_{12} \not\subset W$, we may assume that $a_3$ and $a_4$ are odd and that $a_1$ and $a_2$ are even. A quick inspection of the possibilities shows that $\{a_1, a_2\} = \{2, 4\}$ and $\{e_3, e_4\} = \{3, 5\}$. This concludes the proof of (d). $\blacksquare$

**Corollary 2.10.** The group $W$ is isomorphic to the reflection group $G_{32}$ of Shephard–Todd.

*Proof.* This follows from Theorem 2.7 and from the classification of complex reflection groups [5]. $\blacksquare$

Hence, Corollary 2.10 gives an explanation for the fact, mentioned in the introduction, that $D(G_{32})/\mu_2 \simeq D(E_6)$: the isomorphism is realized by $\Lambda$.

**Remark 2.11.** In the proof of Theorem 2.7, we have used the fact that $D(E_6)$ does not contain any element of order 8. This fact can be easily obtained by a computer calculation for instance, but we propose here a proof using only Springer theory. Let $w \in E_6$ be an element of order 8. Then $w$ necessarily admits an eigenvalue which is a primitive 8-th root of unity $\zeta$. As only one of the degrees of $E_6$ and only one of the codegrees of $E_6$ is divisible by 8, this implies that $w$ is a regular element in the sense of Springer [3, Thm. 1.2]. Then, by [6, Thm. 4.2 (v)] the list of eigenvalues of $w$ is $\zeta^{-1}, \zeta^{-4}, \zeta^{-5}, \zeta^{-7}, \zeta^{-8}, \zeta^{-11}$, and so $\det(w) = \zeta^{-36} = -1$. So $w \notin D(E_6)$.

## References

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Received 13 November 2023; revised 28 December 2023.

Cédric Bonnafé

IMAG, Université de Montpellier, Montpellier, France; cedric.bonnafe@umontpellier.fr