4
$\begingroup$

When $q$ is a power of some odd prime, is $1\neq a\in Z(2.E_7(q))\cong Z_2$ a square element in $2.E_7(q)$?

A Lie algebra is a vector space $L$ over a field $K$ on which a product operation $[xy]$ is defined satisfying the following axioms:

(i) $[xy]$ is bilinear for $x, y\in L$.

(ii) $[xx]=0$ for $x\in L$.

(iii) $[[xy]z]+[[yz]x]+[[zx]y]=0$ for $x, y, z\in L$.

for each element $x$ of a Lie algebra $L$ we define a map ${\rm ad}~x$ of $L$ into itself by $${\rm ad}~x.y=[xy],~~~y\in L.$$

For each $x,y\in L$ we define the scalar product $(x,y)=tr(ad~x.ad~y)$, which is called the Killing form.

The dimension of the Cartan subalgebras $H$ of $L$ is called the rank of $L$, and will usually be denoted by $l$.

Although the roots are defined as elements of the dual space of $H$ they can, by considering the Killing form, be regarded as elements of $H$ itself.

Each element of the dual space of $H$ is expressible in the form $h\rightarrow (x, h)$ for a unique element $x\in H$. The element $x$ is associated with the map $h\rightarrow r(h)$ may be identified with the root $r$. Thus $r$ can be regarded either as an element of $H$ or an element of its dual space; the relation between these two being given by $$r(h)=(r, h),~~~h\in H.$$

We now define the Dynkin diagram of the Lie algebra $L$. This is a graph with $l$ nodes, one associated with each fundamental root $p_i$, such that the $i$th node is joined to the $j$th node by a bond of strength $n_{ij}$.

The Dynkin diagram of simple Lie algebra $E_7$ is as follows:

Let $L$ be a Lie algebra over a field of characteristic $0$ and $\delta$ be a derivation of $L$ which is nilpotent, i.e. satisfies $\delta^n=0$ for some $n$. Then $${\rm exp}~\delta=1+\delta+\frac{\delta^2}{2!}+...+\frac{\delta^{n-1}}{(n-1)!}$$ is an sutomorphism of $L$.

We write $x_r(\zeta)={\rm exp}(\zeta ad~e_r)$ for $\zeta\in \mathbb{C}$.

We shall write $h_r$ for $\bar{h}_r$, $e_r$ for $\bar{e}_r$, $x_r(t)$ for $\bar{x}_r(t)$, and $A_r(t)$ for $\bar{A}_r(t)$. This omission of the bars will not lead to confusion or inconsistency since the objects originaly called $h_r$, $e_r$, $x_r(t)$, $A_r(t)$ are special cases of $\bar{h}_r$, $\bar{e}_r$, $\bar{x}_r(t)$, $\bar{A}_r(t)$ when $K=\mathbb{C}$.

The Chevalley group of type $L$ over the field $K$, denoted by $L(K)$, is defined to be the group of automorphisms of the Lie algebra $L_K$ generated by the $x_r(t)$ for all $r\in \Phi$, $t\in K$.

We now consider the special case in which the base field $K$ is the finite field $GF(q)$ with $q$ elements, where $q$ is an arbitrary prime power. $G$ is then a group of non-singular linear transformations of a space over a finite field, so is a finite group. The Chevalley group of type $L$ over $GF(q)$ will be denoted by $L(q)$.

$|E_7(q)|=q^{63}(q^18-1)(q^{14}-1)(q^{12}-1)(q^{10}-1)(q^8-1)(q^6-1)(q^2-1)/{(2,q-1)}.$

The points over a finite field with $q$ elements of the (split) algebraic group $E_7$, whether of the adjoint (centerless) or simply connected form (its algebraic universal cover), give a fnite Chevalley group. This closely connected to the group written $E_7(q)$, however there is ambiguity in this notation, which can stand for several thing:

  1. The finite group consisting of the points over $F_q$ of the simply connected form of $E_7$ (for clarity, this can be written $E_{7, sc}(q)$ and is known as the "universal" Chevalley group of type $E_7$ over $F_q$)

  2. (rarely) the finite group consisting of the points over $F_q$ of the adjoint form of $E_7$ (for clarity, this can be written $E_{7, ad}(q)$, and is known as the "adjoint" Chevalley group of type $E_7$ over $F_q$), or

  3. the finite group which is the image of the natural map from the former to the latter: this is what will be denoted by $E_7(q)$ in the following, as is most commmon in texts dealing with finite groups.

$E_7(q)$ is simple for any $q$ and $E_{7,sc}$ is its Schur cover, and we often write $E_{7,sc}(q)$ as $2.E_7(q)$ when $q$ is odd.

$\bar{K}$ denotes a semisimple algebraic group, with maximal torus $\bar{T}$ and root system $\sum$.

If $\sum=E_7$, then genertors of $Z(\bar{K})$ are $h=h_{\alpha_4}(-1)h_{\alpha_5}(-1)h_{\alpha_7}(-1)$.

if $\sum=D_{2m}$, then the generators of $Z(\bar{K})$ are $h_1=h_{\alpha_1}(-1)h_{\alpha_3}(-1)...h_{\alpha_{2m-1}}(-1)$ and $h_2=h_{\alpha_{2m-1}}(-1)h_{\alpha_{2m}}(-1)$

For a $\mathbb{C}G$-module $V$ with irreducible character $\chi$ we have the Frobenius-Schur indicator $$\nu(\chi) = \frac{1}{|G|} \sum_{g \in G} \chi(g^2),$$

and $\nu(\chi)$ takes one of the values {+1, -1, 0}, as $\chi$ is afforded by a real representation or is real-valued but not afforded by a real representation or is not real-valued, respectively.

Theorem 12.1.1 Let $L$ be a simple Lie algebra with $L\neq A_1$ and let $K$ be a field. For each root $r$ of $L$ and each element $t$ of $K$ introduce a symbol $\bar{x}_r(t)$. Let $\bar{G}$ be the abstract group generated by the elements $\bar{x}_r(t)$ subject to relations $$\bar{x}_r(t_1)\bar{x}_r(t_1)=\bar{x}_r(t_1+t_2),$$ $$[\bar{x}_s(u),\bar{x}_r(t)]=\prod_{i,j>0}\bar{x}_{ir+js}(C_{ijrs}(-t)^iu^j),$$ $$\bar{h}_r(t_1)\bar{h}_r(t_2)=\bar{h}_r(t_1t_2),~~~t_1t_2\neq 0,$$ and $$\bar{n}_r(t)=\bar{x}_r(t)\bar{x}_{-r}(-t^{-1})\bar{x}_{r}{(t)}.$$ Let $\bar{Z}$ be the centre of $\bar{G}$. Then $\bar{G}/{\bar{Z}}$ is isomorphic to the Chevalley group $G=L(K)$.

Let $S$ be a Sylow 2-subgroup of $E_7^u(q)$ (universal Chevalley group). Then $$Z(S)=\langle h_e(-1), h_{s_3}(-1)h_{s_5}(-1)h_{s_7}(-1), h_{s_2}(-1)h_{s_3}(-1)\rangle.$$ Since $$h_e(-1)=h_{s_2}(-1)h_{s_5}(-1)h_{s_7}(-1),$$ we conclude that $$Z(S)=\langle h_{s_3}(-1)h_{s_5}(-1)h_{s_7}(-1), h_{s_2}(-1)h_{s_3}(-1)\rangle\cong C_2\times C_2.$$ Recall that the center of $Z(E_7^u(q))$ is $Z_0=Z(E_7^u(q))=\langle h_{s_3}(-1)h_{s_5}(-1)h_{s_7}(-1)\rangle$. It follows that $\bar{S}=S/{Z_0}$ is a Sylow $2$-subgroup of $E_7(q)$.

The center of $D_6^u(q)$ is $\langle h_{s_3}(-1)h_{s_5}(-1)h_{s_7}(-1), h_{s_2}(-1)h_{s_3}(-1)\rangle$.

$D_m(q)\cong P\Omega_{2m}^+(q)$ for $m\geq 3$.

If $n=2m$ and $q^m\equiv -\epsilon~{\rm mod}~4$, then $\Omega_n^\epsilon(q)$ is already simple, and the spin group has the structure $2\cdot \Omega_n^\epsilon (q)$. If $n=2m$ and $q^m\equiv \epsilon~{\rm mod}~4$, then $\Omega_n^\epsilon(q)$ has a centre of order $2$, and the spin group has the structure $4.P\Omega_n^\epsilon(q)$ if $m$ is odd, and the structure $2^2.P\Omega_n^{\epsilon}(q)~$(necessarily with $\epsilon +$) if $m$ is even.

When $m$ is an even integer and $q$ a power of a odd prime, then $q^m\equiv 1~{\rm mod~4}$.

Analysis:

$$\pi: 2.E_7(q)\rightarrow E_7(q)$$

The following websites may be useful to my quesion:

https://math.stackexchange.com/questions/785603/what-do-sylow-2-subgroups-of-finite-simple-groups-look-like

Kernel of a double cover of group as stem extension

Square roots of elements in a finite group and representation theory

$\endgroup$
3
  • 1
    $\begingroup$ The expression you have for $z=h_e(-1)$ is valid in the "diagram" $D_6^u(q)$ subgroup of $E_7^u(q)$. Hence, if you can show that $z$ is a square iin $D_6^u(q)$ it will a fortiori be a square in $E_7^u(q)$. This should be possible, perhaps by using the fact that $\Omega_{12}^+(q)$ contains the direct product of three copies of $\Omega_4^+(q)$. $\endgroup$ May 3, 2020 at 17:28
  • $\begingroup$ @Richard Lyons, Thank you very much! I think if we should check $h_{s_3}(-1)h_{s_5}(-1)h_{s_7}(-1)$? $\endgroup$
    – Yi Wang
    May 4, 2020 at 9:57
  • $\begingroup$ 27th version of this question! $\endgroup$ May 6, 2020 at 12:32

2 Answers 2

3
$\begingroup$

The answer is always, yes. Note that there are three classes of involutions in the simply connected version of the algebraic group $E_7$: the central involution $a$, an involution $t$ with centralizer of type $A_1D_6$, and the product $at$. If $a$ were not a square, then in the simple group $E_7(q)$, we would only see involutions with centralizer type $A_1D_6$. However, in the adjoint group we find centralizers of type $E_6T_1$ and $A_7$. This website: http://www.math.rwth-aachen.de/~Frank.Luebeck/chev/23elts.html lists the classes of involutions in the simply connected and adjoint groups, or consult the 3rd volume of Gorenstein-Lyons-Solomon.

Notice that this yields explicit elements of the simple group that power to the centre in the central extension.

$\endgroup$
1
  • $\begingroup$ Thank you very much! $\endgroup$
    – Yi Wang
    May 11, 2020 at 1:46
3
$\begingroup$

Here is a general remark about whether a central involution $z$ in a finite group $G$ is a square : It is well known, and easy to derive from the orthogonality relations for group characters and properties of the Frobenius-Schur indicator $\nu$ that $z$ is a square in $G$ if and only if $\sum_{ \chi \in {\rm Irr}(G)} \nu(\chi) \chi(z) > 0.$ Since $\nu$ vanishes on irreducible characters which are not real-valued, the sum may be restricted to the real-valued complex irreducible characters of $G$. Note that the set $S$ of real-valued irreducible characters which make a positive contribution to the sum contains those $\chi$ which have $z$ in their kernel and $\nu(\chi) = 1,$ (contribution $\chi(1)$) and those $\chi$ which do not contain $z$ in their kernel and $\nu(\chi) = -1$ (contribution also $\chi(1)$). Any real-valued irreducible character $\chi$ of $G$ which lies outside $S$ makes a contribution $- \chi(1)$ to the sum. Hence $z$ is a square in $G$ if and only if $\sum_{ \chi \in S} \chi(1) > \sum_{ \chi \in {\rm Irr}_{\mathbb{R}}(G) \backslash S } \chi(1)$, where $Irr_{\mathbb{R}}(G)$ denoted the set of real-valued complex irreducible characters of $G$, and $S$ denotes the set of real-valued irreducible characters $\chi$ of $G$ with $\nu(\chi) \chi(z) = \chi(1)$.

However, I am not sure whether enough information about the character table and Frobenius-Schur indicators is available for the groups you are considering.

$\endgroup$
1
  • $\begingroup$ Thank you very much! $\endgroup$
    – Yi Wang
    Apr 8, 2020 at 12:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.