Elementary abelian 2-subgroups of compact Lie groups

Abstract

We classify elementary abelian 2-subgroups of compact simple Lie groups of adjoint type. This finishes the classification of elementary abelian $p$-subgroups of compact simple Lie groups (equivalently, complex linear algebraic simple groups) of adjoint type started in Griess (Geom Dedicata 39(3):253–305, 1991).

1 Introduction

For a positive integer $m$, let $C_{m}=\mathbb{Z }/m\mathbb{Z }$ be the cyclic group of order $m$. For a prime $p$ and a positive integer $n$, an elementary abelian $p$-group of rank $n$ is a finite group isomorphic to

$$\begin{aligned} (C_{p})^{n}=\bigoplus _{1}^{n}C_{p}. \end{aligned}$$

The goal of this paper is to study elementary abelian $p$-subgroups of compact simple Lie groups of adjoint type. Precisely, we focus on the case of $p=2$. Here, we say a compact Lie group $G$ is simple if its Lie algebra $\mathfrak{g }_0=\mathrm{Lie}G$ is simple; and say it is of adjoint type if the adjoint homomorphism $\pi : G\longrightarrow \mathrm{Aut}(\mathfrak{g }_0)$ is an injective map. For a compact simple Lie algebra ${\mathfrak{u }}_0$ and any compact simple Lie group of adjoint type $G$ with Lie algebra $\mathrm{Lie}G\cong {\mathfrak{u }}_0$, the adjoint homomorphism

$$\begin{aligned} \pi : G\longrightarrow \mathrm{Aut}({\mathfrak{u }}_0) \end{aligned}$$

is injective, so it suffices to study elementary abelian $2$-subgroups of the compact Lie group $G=\mathrm{Aut}({\mathfrak{u }}_0)$.

The structure of elementary abelian $p$-subgroups of a compact group $G$ is related to the topology of $G$ and its classifying space (cf. [2, 4, 10]). In the 1950s, Borel made an observation that, for a compact connected Lie group $G$, the cohomology ring $H^{*}(G,\mathbb{Z })$ has non-trivial $p$-torsion if and only if $G$ has a non-toral elementary abelian $p$-subgroup (cf. [4, 10]). Recall that, a subgroup of a compact Lie group $G$ is called toral if it is contained in a maximal torus of $G$, otherwise it is called non-toral (cf. [10]). We call a prime $p$ a torsion prime of a compact (not necessary connected) Lie group $G$ if $G$ has a non-toral elementary abelian $p$-subgroup. This definition is a bit different with that in [10]. For $G=\mathrm{Aut}({\mathfrak{u }}_0)$ (the automorphism group of ${\mathfrak{u }}_0$) with ${\mathfrak{u }}_0$ a compact simple Lie algebra, the torsion primes are as in Table 1. From Table 1 we see that: the prime $2$ is a torsion prime of $\mathrm{Aut}({\mathfrak{u }}_0)$ for any compact simple Lie algebra ${\mathfrak{u }}_0$; when ${\mathfrak{u }}_0$ is a compact simple exceptional Lie algebra, any prime $p>5$ is not a torsion prime and $5$ is a torsion prime only when ${\mathfrak{u }}_0$ is of type $\mathbf E_8$.

Table 1 Torsion primes

The study of elementary abelian $p$-subgroups began at 1950s (or even earlier) by the famous mathematicians Borel, Serre, et al. In the 1990s, Griess [5] got a classification of maximal elementary abelian $p$-subgroups of linear algebraic simple groups (of adjoint type) defined over an algebraic closed field of characteristic 0. Since there exists a one-one correspondence between conjugacy classes of compact subgroups of a complex semisimple Lie algebraic group and such subgroups of (any of) its maximal compact subgroup (cf. Appendix of [1]), so we also have a classification of maximal elementary abelian $p$-subgroups of compact simple Lie groups of adjoint type. For odd primes $p$, non-toral elementary abelian $p$-subgroups are more or less well understood from [5] and [2]. Precisely, when ${\mathfrak{u }}_0$ is a compact exceptional simple Lie algebra, non-toral elementary abelian $p$-subgroups of $\mathrm{Aut}({\mathfrak{u }}_0)$ are classified up to conjugacy. When $p>5$, such subgroups don’t exist; when $p=5$, there is a unique conjugacy class in $\mathrm{Aut}({\mathfrak{e }}_8)$ (cf. [5]); when $p=3$, there are some conjugacy classes when ${\mathfrak{u }}_0$ is of type $\mathrm{E}_6,\mathrm{E}_7,\mathrm{E}_8$ or $\mathrm{F}_4$ (cf. [2, 5]). But a complete classification is impossible when ${\mathfrak{u }}_0$ is a classical simple Lie algebra, since some complicated combinatorial problem will arise.

In this paper, we will first study elementary abelian $2$-subgroups of $\mathrm{Aut}({\mathfrak{u }}_0)$ for compact classical simple Lie algebras ${\mathfrak{u }}_0$ systematically. The method is to define and use linear algebraic structures on them (a bilinear form $m$ or a bilinear form $m$ together with a function $\mu$, all with values in $\mathbb F _2=\mathbb{Z }/2\mathbb{Z }$). For any compact exceptional simple Lie algebra ${\mathfrak{u }}_0$, we classify elementary abelian $2$-subgroups of $\mathrm{Aut}({\mathfrak{u }}_0)$ up to conjugation and calculate their automizer groups (cf. Definition 3.4). A simple account of this classification is as follows. Theorem 1.1 follows by combining Corollaries 4.2, 5.3, 6.8, 7.24 and 8.13.

Theorem 1.1

For ${\mathfrak{u }}_0=\mathfrak{e }_6,\mathfrak{e }_7,\mathfrak{e }_8,\mathfrak{f }_4,\mathfrak{g }_2$, there are $51,78,66,12,4$ conjugacy classes of elementary abelian $2$-subgroups in $\mathrm{Aut}({\mathfrak{u }}_0)$ respectively.

This paper is organized as follows. In Sect. 2, we do the classification for classical simple Lie algebras, which amounts to classify elementary abelian 2-subgroups of the groups $\mathrm{PU}(n)\rtimes \langle \tau _{0} \rangle , \, \mathrm{O}(n)/\langle -I \rangle , \, \mathrm{Sp}(n)/\langle -I\rangle$. Here, $\mathrm{PU}(n)=\mathrm{U}(n)/Z_{n}$ ($Z_{n}=\{\lambda I_{n}:|\lambda |=1\}$) is the projective unitary group and $\tau _0=\mathrm complex conjugation$. We have $\tau _0^{2}=1$ and

$$\begin{aligned} \tau _0[A]\tau _0^{-1}=[\overline{A}],\quad \forall A\in \mathrm{U}(n). \end{aligned}$$

In the first case, we will separate the discussion of subgroups contained in $\mathrm{PU}(n)$ and those not contained in it.

For an elementary abelian 2-subgroup $F$ of $\mathrm{PU}(n)$, define a map

$$\begin{aligned} m: F\times F\longrightarrow \{\pm {1}\} \end{aligned}$$

by $m(x,y)=\lambda$ if $x=[A], \, y=[B]$ and $ABA^{-1}B^{-1}=\lambda I$. We show that $m$ is a bilinear form when $F$ is viewed as a vector space over $\mathbb F _2=\mathbb{Z }/2\mathbb{Z }$ and $\{\pm {1}\}$ is identified with $\mathbb F _2$. We also prove that $\ker m$ is diagonalizable and the conjugacy class of $F$ is determined by the conjugacy class of $\ker m$ and the number $\mathrm{rank}(F/\ker m)$. This gives $F$ a structure we called symplectic vector space. For an elementary abelian 2-subgroup $F$ of $\mathrm{O}(n)/\langle -I\rangle$ or $\mathrm{Sp}(n)/\langle -I\rangle$, we define a bilinear map $m: F\times F\longrightarrow \{\pm {1}\}$ and a function

$$\begin{aligned} \mu : F\longrightarrow \{\pm {1}\}. \end{aligned}$$

The definition of $m$ is similar as in the $\mathrm{PU}(n)$ case; $\mu (x)=\lambda$ if $x=[A]$ and $A^2=\lambda I$. The bilinear map $m$ and the function $\mu$ satisfy a compatibility relation ($m(x,y)=\mu (x)\mu (y)\mu (xy)$). The compatible pair $(m,\mu )$ gives $F$ a structure we called symplectic metric space and we get invariants $r,s,\epsilon ,\delta$ from the structure of a symplectic metric space. We show that the conjugacy class of $F$ is determined by the conjugacy class of the subgroup $A_{F}=\ker (\mu |_{\ker m})$ and the numbers $s,\epsilon ,\delta$. The consideration of elementary abelian 2-subgroups of the group $\mathrm{PU}(n)\rtimes \langle \tau _0\rangle$ is reduced to consideration of elementary abelian 2-subgroups of the above three groups.

In Sect. 2.4, we discuss a class of elementary abelian 2-subgroups of the groups $\mathrm{O}(n)/\langle -I\rangle$ and $\mathrm{Sp}(n)/\langle -I\rangle$ and introduce the notions of symplectic vector space and symplectic metric space and study their automorphism groups (Definition 3.4). They will play an important role in later sections.

In Sects. 4–8, we classify elementary abelian 2-subgroups of the automorphism group of any compact exceptional simple Lie algebra. A detailed account of the method is presented in Sect. 3. The study of some of these elementary abelian 2-subgroups is reduced to consideration of the class of subgroups of $\mathrm{Sp}(n)/\langle -I\rangle$ discussed in Sect. 2.4. Moreover, their automizer groups are described in terms of the automorphism groups of symplectic vector spaces or symplectic metric spaces.

Notation and conventions. Let $Z(G)$ ($\mathfrak z (\mathfrak g )$) denote the center of a Lie group $G$ (Lie algebra $\mathfrak{g }$) and $G_0$ denote the connected component of $G$ containing identity element. For Lie groups $H\subset G$ (or Lie algebras $\mathfrak{h }\subset \mathfrak{g }$), let $C_{G}(H)$ ($C_{\mathfrak{g }}(\mathfrak{h })$) denote the centralizer of $H$ in $G$ ($\mathfrak{h }$ in $\mathfrak{g }$) and let $N_{G}(H)$ ($C_{\mathfrak{g }}(\mathfrak{h })$) denote the normalizer of $H$ in $G$ ($\mathfrak{h }$ in $\mathfrak{g }$). For an element $x$ in $G$ (or an automorphism of $G$), we also write $G^{x}$ for the centralizer of $x$ in $G$, so $G^{x}=C_{G}(x)$ when $x$ is an element of $G$.

For any two elements $x,y\in G$, the notation $x\sim y$ means $x,y$ are conjugate in $G$, i.e., $y=gxg^{-1}$ for some $g\in G$; and for a subgroup $H\subset G$, the notation $x\sim _{H} y$ means $y=gxg^{-1}$ for some $g\in H$. For two subsets $X_1,X_2\subset G$, the notation $X_1\sim X_2$ means $X_2=gX_1 g^{-1}$ for some $g\in G$; and for a subgroup $H\subset G$, the notation $X_1 \sim _{H} X_2$ means $X_2=gX_1 g^{-1}$ for some $g\in H$.

For a quotient group $G=H/N$, let $[x]=xN$ ($x\in H$) denote a coset.

All adjoint homomorphisms in this paper are denoted as $\pi$. This causes no ambiguity, as the reader can understand it is the adjoint homomorphism for which group everywhere $\pi$ appears in this paper.

For a compact semisimple Lie algebra ${\mathfrak{u }}_0$, let $\mathrm{Aut}({\mathfrak{u }}_0)$ be the group of automorphisms of ${\mathfrak{u }}_0$ and let $\mathrm{Int}({\mathfrak{u }}_0)=\mathrm{Aut}({\mathfrak{u }}_0)_0$. The elements in $\mathrm{Int}({\mathfrak{u }}_0)$ are called inner automorphisms of ${\mathfrak{u }}_0$ and the elements in $\mathrm{Aut}({\mathfrak{u }}_0)-\mathrm{Int}({\mathfrak{u }}_0)$ are called outer automorphisms of ${\mathfrak{u }}_0$.

We denote by $\mathfrak{e }_6$ the compact simple Lie algebra of type $\mathbf E_6$. Let $\mathrm{E}_6$ be the connected and simply connected Lie group with Lie algebra $\mathfrak{e }_6$. Let $\mathfrak{e }_6(\mathbb{C })$ and $\mathrm{E}_6(\mathbb{C })$ denote their complexifications. Similar notations will be used for other types. In the case of $G=\mathrm{E}_6$ or $\mathrm{E}_7$, let $c$ denote a non-trivial element in $Z(G)$. In the case of ${\mathfrak{u }}_{0}={\mathfrak{e }}_7$, let

$$\begin{aligned} H^{\prime }_0=\frac{H^{\prime }_2+H^{\prime }_5+H^{\prime }_{7}}{2}\in i{\mathfrak{e }}_7\subset {\mathfrak{e }}_7(\mathbb{C }) \end{aligned}$$

(cf. Sect. 3.1).

Let $V=\mathbb{R }^{n}$ be an Euclidean linear space of dimension $n$ with an orthogonal basis $\{e_1,e_2,\ldots ,e_{n}\}$ and $\mathrm{Pin}(n)$ ($\mathrm{Spin}(n)$) be the Pin (Spin) group of degree $n$ associated to $V$. Write

$$\begin{aligned} c=e_1e_2\ldots e_{n}\in \mathrm{Pin}(n). \end{aligned}$$

Then $c$ is in $\mathrm{Spin}(n)$ if and only if $n$ is even, in this case $c\in Z(\mathrm{Spin}(n))$. If $n$ is odd, then $\mathrm{Spin}(n)$ has a Spinor module $M$ of dimension $2^{\frac{n-1}{2}}$. If $n$ is even, then $\mathrm{Spin}(n)$ has two Spinor modules $M_{+}, \, M_{-}$ of dimension $2^{\frac{n-2}{2}}$. We distinguish $M_{+}$ and $M_{-}$ by requiring that $c$ acts on $M_{+}$ and $M_{-}$ by scalar 1 and $-1$ respectively when $4|n$; and by $-i$ and $i$ respectively when $4|n-2$.

For a prime $p$, let $\mathbb F _{p}=\mathbb{Z }/p\mathbb{Z }$ be the finite field with $p$ elements. In particualr, for $p=2, \, \mathbb F _{2}=\mathbb{Z }/p\mathbb{Z }$ is a field with $2$ elements. We have an isomorphism $\mathbb F _2\cong \{\pm {1}\}$ between the additive group $\mathbb F _2$ and the multiplicative group $\{\pm {1}\}$.

Let $I_{n}$ be the $n\times n$ identity matrix. We define the following matrices,

$$\begin{aligned} I_{p,q}=\left(\begin{array}{cc} -I_{p}&0\\ 0&I_{q}\\ \end{array}\right), J_{n}&= \left(\begin{array}{cc} 0&I_{n}\\ -I_{n}&0\\ \end{array}\right),\ \mathrm{J}^{\prime }_{n}=\left(\begin{array}{cc} 0&\mathrm{I}_{n}\\ \mathrm{I}_{n}&0\\ \end{array}\right),\\ I^{\prime }_{p,q}&= \left(\begin{array}{cccc} -I_{p}&0&0&0 \\ 0&I_{q}&0&0 \\ 0&0&-I_{p}&0\\ 0&0&0&I_{q} \\ \end{array}\right), \\ J_{p,q}&= \left( \begin{array}{cccc} 0&I_{p}&0&0 \\ -I_{p}&0&0&0\\ 0&0&0&I_{q}\\ 0&0&-I_{q}&0 \\ \end{array} \right), \\ K_{p}&= \left(\begin{array}{cccc} 0&0&0&I_{p}\\ 0&0&-I_{p}&0\\ 0&I_{p}&0&0 \\ -I_{p}&0&0&0 \\ \end{array}\right). \end{aligned}$$

And we define the following groups,

$$\begin{aligned} \mathrm{Z}_{m}&= \{\lambda I_{m}|\lambda ^{m}=1\},\\ \mathrm{Z}^{\prime }&= \{(\epsilon _1,\epsilon _2,\epsilon _3,\epsilon _4)| \epsilon _{i}=\pm {1}, \epsilon _1\epsilon _2\epsilon _3\epsilon _4=1\}, \\ \Gamma _{p,q,r,s}&= \left\langle \left( \begin{array} {cccc} -I_{p}&0&0&0 \\ 0&-I_{q}&0&0 \\ 0&0&I_{r}&0\\ 0&0&0&I_{s}\\ \end{array} \right), \left( \begin{array}{cccc} -I_{p}&0&0&0 \\ 0&I_{q}&0&0 \\ 0&0&-I_{r}&0 \\ 0&0&0&I_{s} \\ \end{array} \right) \right\rangle . \end{aligned}$$

 

2 Matrix groups

Let $\mathrm{M}_{n}(\mathbb{R }), \, \mathrm{M}_{n}(\mathbb{C }), \, \mathrm{M}_{n}(\mathbb{H })$ be the set of $n\times n$ matrices with entries in the field $\mathbb{R }, \, \mathbb{C }, \, \mathbb{H }$ respectively. Let

$$\begin{aligned}&\mathrm{O}(n)=\{X\in \mathrm{M}_{n}(\mathbb{R })|XX^{t}=\mathrm{I}\},\quad \mathrm{SO}(n)=\{X\in \mathrm{O}(n)|\det X=1\},\\&\mathrm{U}(n)=\{X\in \mathrm{M}_{n}(\mathbb{C })|XX^{*}=\mathrm{I}\},\quad \mathrm{SU}(n)=\{X\in \mathrm{U}(n)|\det X=1\}, \\&\mathrm{Sp}(n)=\{X\in \mathrm{M}_{n}(\mathbb{H })|XX^{*}=\mathrm{I}\}. \end{aligned}$$

Defined as sets in this way, $\mathrm{O}(n), \,\mathrm{SO}(n),\,\mathrm{U}(n),\,\mathrm{SU}(n), \, \mathrm{Sp}(n)$ are actually Lie groups, i.e., groups with a smooth manifold structure. Moreover, they are compact Lie groups, i.e., the underlying manifolds are compact. Also let

$$\begin{aligned} \mathrm{PO}(n),\ \mathrm{PSO}(n),\ \mathrm{PU}(n),\ \mathrm{PSU}(n) \end{aligned}$$

be the quotients of the groups $\mathrm{O}(n),\,\mathrm{SO}(n),\,\mathrm{U}(n),\, \mathrm{SU}(n)$ modulo their centers (so $\mathrm{PU}(n)\cong \mathrm{PSU}(n)$, which is the projective unitary group). Let

$$\begin{aligned}&\mathfrak so (n)=\{X\in M_{n}(\mathbb{R })|X+X^{t}=0\},\\&\mathfrak su (n)=\{X\in M_{n}(\mathbb{C })|X+X^{*}=0, \mathrm{tr}X=0\},\\&\mathfrak sp (n)=\{X\in M_{n}(\mathbb{H })|X+X^{*}=0\}, \end{aligned}$$

where $X^{t}$ denotes the transposition of a matrix $X$ and $X^{*}$ denotes the conjugate transposition of $X$. Then $\mathfrak so (n),\,\mathfrak su (n),\,\mathfrak sp (n)$ are Lie algebras of $\mathrm{SO}(n),\,\mathrm{SU}(n),\,\mathrm{Sp}(n)$ respectively. They represent all isomorphism classes of compact classical simple Lie algebras.

2.1 Projective unitary groups

Let $G=\mathrm{PU}(n)=\mathrm{U}(n)/\mathrm{Z}_{n}$. Then

$$\begin{aligned} G\cong \mathrm{Int}(\mathfrak su (n)). \end{aligned}$$

Any involution $x\in G$ is of the form $x=[A], \, A\in \mathrm{U}(n)$ with $A^{2}=I$. Then

$$\begin{aligned} A\sim I_{p,n-p}= \left(\begin{array}{cc} -I_{p}&0\\ 0&I_{n-p}\\ \end{array} \right) \end{aligned}$$

for some $p, \, 1\le p \le n-1$. One has

$$\begin{aligned} (\mathrm{U}(n)/\mathrm{Z}_{n})^{[I_{p,n-p}]}=(\mathrm{U}(p)\times \mathrm{U}(n-p))/\mathrm{Z}_{n}\mathrm if p\ne \frac{n}{2} \end{aligned}$$

and

$$\begin{aligned} (\mathrm{U}(n)/\mathrm{Z}_{n})^{[I_{\frac{n}{2},\frac{n}{2}}]}= ((\mathrm{U}(n/2)\times \mathrm{U}(n/2))/\mathrm{Z}_{n})\rtimes \langle [J^{\prime }_{n}]\rangle . \end{aligned}$$

Let $F\subset G$ be an elementary abelian 2-subgroup. For any $x,y\in F$, choose $A,B\in \mathrm{U}(n)$ with $A^{2}=B^{2}=I$ representing $x,y$ (that is, $x=[A]$ and $y=[B]$). Then

$$\begin{aligned} 1=xyx^{-1}y^{-1}=(ABA^{-1}B^{-1})Z_{n}/Z_{n}\Longrightarrow [A,B]=\lambda _{A,B} I \end{aligned}$$

for some $\lambda _{A,B}\in \mathbb{C }$. It is clear that $\lambda _{A,B}\in \mathbb{C }$ doesn’t depend on the choice of $A$ and $B$. Moreover, since $x^2=y^{2}=1$, we have $\lambda _{A,B}=\pm {1}$.

For any $x,y\in F$, define

$$\begin{aligned} m(x,y)=m_{F}(x,y)=\lambda _{A,B}. \end{aligned}$$

Lemma 2.1

For any $x, y, z \in F, \, m(x,x)=m(x,y) m(y,x)=1$ and $m(xy,z)=m(x,z)m(y,z)$.

Proof

$m(x,x)=m(x,y) m(y,x)=1$ is clear. Choose $A,B,C\in \mathrm{U}(n)$ with $A^{2}=B^{2}=C^{2}=I$ representing $x,y,z$. Let $[A,C]=\lambda _1 I, \, [B, C]=\lambda _2 I$ for some numbers $\lambda _1,\lambda _2=\pm {1}$. We have

$$\begin{aligned}{}[AB,C]=A[B,C]A^{-1}[A,C]=A(\lambda _2 I)A^{-1}(\lambda _1 I)=(\lambda _1 \lambda _2)I. \end{aligned}$$

So $m(xy,z)=m(x,z)m(y,z)$. $\square$

If we regard $F$ as a vector space on $\mathbb F _2=\mathbb{Z }/2\mathbb{Z }$ and identify $\{\pm {1}\}$ with $\mathbb F _2=\mathbb{Z }/2\mathbb{Z }$, Lemma 2.1 just said $m$ is an anti-symmetric bilinear $2$-form on $F$. Let

$$\begin{aligned} \ker m=\{x\in F| m(x,y)=1,\forall y \in F\}. \end{aligned}$$

Then it is a subgroup of $F$.

For an even $n$, let $\Gamma _0=\langle [I_{\frac{n}{2},\frac{n}{2}}],[J^{\prime }_{\frac{n}{2}}]\rangle$, then any non-identity element of $\Gamma _0$ is conjugate to $I_{\frac{n}{2},\frac{n}{2}}$ and

$$\begin{aligned} (\mathrm{U}(n)/\mathrm{Z}_{n})^{\Gamma _0}\cong (\mathrm{U}(n/2)/\mathrm{Z}_{\frac{n}{2}})\times \Gamma _0. \end{aligned}$$

Lemma 2.2

For a Klein four subgroup $F\subset G$, if $m_{F}$ is non-trivial, then $F$ is conjugate to $\Gamma _0$.

Proof

Choose $A,B\in \mathrm{U}(n)$ with $A^{2}=B^{2}=I$ and $F=\langle [A],[B]\rangle$. Since $m_{F}$ is non-trivial, we have $[A,B]=-I$. Since $A^{2}=I$, we may assume that $A=I_{p,n-p}$ for some $1\le p\le \frac{n}{2}$. From $[A,B]=-I$, we get $ABA^{-1}=-B$. Then $B$ is of the form

$$\begin{aligned} B=\left(\begin{array}{cc}&B_1\\ B_2^{t}&\end{array}\right) \end{aligned}$$

for some $B_1,B_2\in M_{p,n-p}$. Since $B$ is invertible, we get $p=\frac{n}{2}$. Since $B^{2}=I$, we have $B_1B_2^{t}=I$. Let $S=\mathrm{diag}\{I_{n/2},B_1\}$. Then

$$\begin{aligned} (SAS^{-1},SBS^{-1})=\left(\left( \begin{array}{cc} -I_{\frac{n}{2}}&0 \\ 0&I_{\frac{n}{2}}\\ \end{array}\right), \left( \begin{array}{cc} 0&I_{\frac{n}{2}}\\ I_{\frac{n}{2}}&0\\ \end{array} \right)\right). \end{aligned}$$

$\square$

For any $m,k\ge 1$ and $A\in \mathrm{U}(m)$, let

$$\begin{aligned} \mathrm{D}(A)=\mathrm{diag}\{A,A,\ldots ,A\}. \end{aligned}$$

Then $D:\mathrm{U}(m)\longrightarrow \mathrm{U}(km)$ is the diagonal homomorphism.

Lemma 2.3

For any two closed subgroups $S_1,S_2\subset \mathrm{U}(m)$,

$$\begin{aligned} \mathrm{D}(S_1)\sim \mathrm{D}(S_2)\Leftrightarrow S_1\sim S_2. \end{aligned}$$

Proof

Since $S_1,S_2$ are closed subgroups of $\mathrm{U}(m)$, so they are compact groups. Then by character theory of representations of compact groups, both conditions in the lemma are equivalent to the existence of an isomorphism $\phi : S_1\rightarrow S_2$ such that $\mathrm{tr}(\phi (x))=\mathrm{tr}(x), \, \forall x\in S_1$. Thus these two conditions are equivalent. $\square$

Proposition 2.4

Let $F$ be an elementary abelian 2-subgroup of $G$,  

• (1) when $\ker m=1$, the conjugacy class of $F$ is determined by $\mathrm{rank}F$;

• (2) in general, $\ker m$ is diagonalizable and the conjugacy class of $F$ is determined by the conjugacy class of $\ker m$ and the number $\mathrm{rank}F$.

Proof

For $(1)$, we prove by induction on $n$. Since $\ker m=1$, so $\mathrm{rank}F$ is even. When $\mathrm{rank}F\ge 2$, choose any $x_1,x_2\in F$ with $m(x_1,x_2)=-1$. By Lemma 2.2, we have

$$\begin{aligned} \langle x_1,x_2\rangle \sim \Gamma _0. \end{aligned}$$

We may and do assume that $\langle x_1,x_2\rangle =\Gamma _0$, then

$$\begin{aligned} F\subset (\mathrm{U}(n)/\mathrm{Z}_{n})^{\Gamma _0}= \Delta (\mathrm{U}(n/2)/\mathrm{Z}_{\frac{n}{2}})\times \Gamma _0. \end{aligned}$$

And so $F=\Delta (F^{\prime })\times \Gamma _0$ for some $F^{\prime }\subset \mathrm{U}(n/2)/\mathrm{Z}_{\frac{n}{2}}$. We also have $\ker m_{F^{\prime }}=1$. By induction, the conjugacy class of $F^{\prime }$ is determined by $\mathrm{rank}F^{\prime }$, so the conjugcay class of $F$ is determined by $\mathrm{rank}F$.

For $(2)$, we have that $\pi ^{-1}(\ker m)$ is abelian by the definition of $m$ and $\ker m$, where $\pi$ is the natural projection from $\mathrm{U}(n)$ to $\mathrm{U}(n)/\mathrm{Z}_{n}$. So $\pi ^{-1}(\ker m)$ is diagonalizable. Then $\ker m$ is diagonalizable. We may write $F$ as $F=\ker m \times F^{\prime }$ with $m(\ker m,F^{\prime })=1$ and $m|_{F^{\prime }}$ non-degenerate. By $(1)$, the conjugcay class of $F^{\prime }$ is determined by

$$\begin{aligned} \mathrm{rank}F^{\prime }=\mathrm{rank}F-\mathrm{rank}(\ker m). \end{aligned}$$

Moreover, it is clear that

$$\begin{aligned} (\mathrm{U}(n)/\mathrm{Z}_{n})^{F^{\prime }}=\Delta (\mathrm{U}(n^{\prime })/\mathrm{Z}_{n^{\prime }})\times F^{\prime }, \end{aligned}$$

where $n^{\prime }=n/2^{\frac{\mathrm{rank}F^{\prime }}{2}}$. So $\ker m=\Delta (F^{\prime \prime })$ for some $F^{\prime \prime }\subset \mathrm{U}(n^{\prime })/\mathrm{Z}_{n^{\prime }}$. Fix $F^{\prime }$, by Lemma 2.3, the conjugacy class of $\ker m$ in $\mathrm{U}(n)/\mathrm{Z}_{n}$ and the conjugacy class of $F^{\prime \prime }$ in $\mathrm{U}(n^{\prime })/\mathrm{Z}_{n^{\prime }}$ determine each other. Since the conjugacy of $F$ is determined by $F^{\prime }$ and the class of $\ker m$ in $(\mathrm{U}(n)/\mathrm{Z}_{n})^{F^{\prime }}$, we get the last statement of $(2)$. $\square$

2.2 Projective orthogonal and projective symplectic groups

Let $G=\mathrm{PO}(n)=\mathrm{O}(n)/\langle -I\rangle , \, n\ge 2$. Let $F$ be an elementary abelian 2-subgroup of $G$. For any $x\in F$, choose $A\in \mathrm{O}(n)$ representing $x$, then $A^{2}=\lambda _{A} I$ for some $\lambda _{A}=\pm {1}$. For any $x,y\in F$, choose $A,B\in \mathrm{O}(n)$ representing $x,y$, then $[A,B]=\lambda _{A,B} I$ for some $\lambda _{A,B}=\pm {1}$. The values of $\lambda _{A},\lambda _{A,B}$ don’t depend on the choice of $A$ and $B$. For any $x,y\in F$, define

$$\begin{aligned} \mu (x)=\mu _{F}(x)=\lambda _{A} \end{aligned}$$

and

$$\begin{aligned} m(x,y)=m_{F}(x,y)=\lambda _{A,B}. \end{aligned}$$

Lemma 2.5

For any $x,y,z \in F, \, m(x,x)=1, \, m(xy,z)=m(x,z)m(y,z), \, \mu (1)=1$ and $m(x,y)=\mu (x)\mu (y)\mu (xy)$.

Proof

The equalities $m(x,x)=1$ and $\mu (1)=1$ are clear.

The proof for $m(xy,z)=m(x,z)m(y,z)$ is similar as that for 2.1.

Choose $A,B\in \mathrm{O}(n)$ representing $x,y$. Then

$$\begin{aligned}{}[A,B]&= ABA^{-1}B^{-1}\\&= (AB)^2(B^{2})^{-1}B(A^{2})^{-1}B^{-1}\\&= (\mu (xy)I)(\mu (y)I)^{-1}B(\mu (x)I)^{-1}B^{-2}\\&= \mu (x)\mu (y)\mu (xy I). \end{aligned}$$

So $m(x,y)=\mu (x)\mu (y)\mu (xy)$. $\square$

Lemma 2.6

For an even $n, \, \mathfrak su (n)$ has two conjugacy classes of outer involutive automorphisms with representatives $\tau _0=\mathrm complex conjugation$ and $\tau ^{\prime }_0=\tau _0\mathrm{Ad}(J_{n/2})$.

For an odd $n, \, \mathfrak su (n)$ has a unique conjugacy class of outer involutive automorphisms with representative $\tau _0=\mathrm complex conjugation$.

Proof

This follows from Cartan ’s classification of compact Riemannian symmetric pairs (cf. [7, Pages 451–455]). $\square$

Lemma 2.7

Let $F$ be an elementary abelian 2-subgroup of $G$.

For $x\in F, \, \mu (x)=-1$ if and only if $x\sim [J_{\frac{n}{2}}]$.

For $x,y\in F$ with $m(x,y)=-1$,  

• (1) when $\mu (x)=\mu (y)=-1$, we have $(x,y)\sim ([J_{\frac{n}{2}}],[K_{\frac{n}{4}}])$;

• (2) when $\mu (x)=\mu (y)=1$, we have $(x,y)\sim ([I_{\frac{n}{2},\frac{n}{2}}], [J^{\prime }_{\frac{n}{2}}])$.

Proof

If $\mu (x)=-1$, then $x=[A]$ for some $A\in \mathrm{O}(n)$ with $A^{2}=-I$. Then $A\sim J_{\frac{n}{2}}$, so $x\sim [J_{\frac{n}{2}}]$.

The proof of $(2)$ is the same as that for Lemma 2.2.

For $(1)$, first we may and do assume that $x=[J_{\frac{n}{2}}]$ by the first statement proved above. Then $\mathfrak so (n)^{x}=\mathfrak u (n/2)$. By Lemma 2.6, after replace $y$ by some $gyg^{-1}$ with $g\in G^{x}$, we may assume that

$$\begin{aligned} (\mathfrak u (n/2))^{y}=\mathfrak so (n/2)\mathrm or \mathfrak sp (n/4). \end{aligned}$$

Then a little more argument shows that $y=[K_{\frac{n}{4}}]$. $\square$

Definition 2.8

For an elementary abelian 2-groupsub $F\subset G$, define

$$\begin{aligned} A_{F}=\ker (\mu |_{\ker m}) \end{aligned}$$

and

$$\begin{aligned} \mathrm{defe}F=|\{x\in F: \mu (x)=1\}|-|\{x\in F:\mu (x)=-1\}|. \end{aligned}$$

We call $\mathrm{defe}F$ the defect index of $F$.

Define $(\epsilon _{F},\delta _{F})$ as follows,

• when $\mu |_{\ker m}\ne 1$, define $(\epsilon _{F},\delta _{F})=(1,0)$;

• when $\mu |_{\ker m}=1$ and $\mathrm{defe}F<0$, define $(\epsilon _{F},\delta _{F})=(0,1)$;

• when $\mu |_{\ker m}=1$ and $\mathrm{defe}F>0$, define $(\epsilon _{F},\delta _{F})=(0,0)$.

Define $r_{F}=\mathrm{rank}A_{F}$ and $s_{F}=\frac{1}{2}\mathrm{rank}(F/\ker m)-\delta _{F}$.

We will see in the proof of Proposition 2.12 that $\mathrm{defe}F=0$ if and only if $\mu |_{\ker m}\ne 1$. It is clear that $\epsilon _{F},\delta _{F},r_{F},s_{F}$ and the conjugcay class of $A_{F}$ are determined by the conjugacy class of $F$.

Let $\Gamma _1=\langle [I_{\frac{n}{2},\frac{n}{2}}],[J^{\prime }_{\frac{n}{2}}]\rangle$ and $\Gamma _2=\langle [J_{\frac{n}{2}}],[K_{\frac{n}{4}}\rangle ]$. Then $\mathrm{defe}\Gamma _1=2, \, \mathrm{defe}\Gamma _2=-2$,

$$\begin{aligned} (\mathrm{O}(n)/\langle -I\rangle )^{\Gamma _1}= \Delta (\mathrm{O}(\frac{n}{2})/\langle -I\rangle )\times \Gamma _1 \end{aligned}$$

and

$$\begin{aligned} (\mathrm{O}(n)/\langle -I\rangle )^{\Gamma _2}=\Delta (\mathrm{Sp}(\frac{n}{4}) /\langle -I\rangle )\times \Gamma _2. \end{aligned}$$

Lemma 2.9

Let $F$ be a non-trivial elementary abelian 2-subgroup of $\mathrm{O}(n)/\langle -I\rangle$, if $\mathrm{rank}(F/\ker m)>2$, then there exists a Klein four subgroup $F^{\prime }\subset F$ such that $F^{\prime }\sim \Gamma _1$.

Proof

Choose a subgroup $F^{\prime \prime }\subset F$ such that $F=\ker m \times F^{\prime \prime }$, then $\ker (m_{F^{\prime \prime }})=1$ and $\mathrm{rank}F^{\prime \prime }>2$. Replace $F$ by $F^{\prime \prime }$, we may assume that $\ker m=1$ and $\mathrm{rank}F>2$.

We first show that, there exists $1\ne x\in F$ with $\mu (x)=1$. From $\mathrm{rank}F>2$, we get $\mathrm{rank}F\ge 4$ since it is even ($m_{F}$ is non-degenerate). Suppose that any $1\ne x\in F$ has $\mu (x)=-1$. Then for any distinct non-trivial elements $x,y\in F$, we have

$$\begin{aligned} m(x,y)=\mu (x)\mu (y)\mu (xy)=-1 \end{aligned}$$

by Lemma 2.5. This contradicts that $m$ is bilinear on $F$.

Upon we get $1\ne x\in F$ with $\mu (x)=1$, choose any $z\in F$ with $m(x,z)=-1$. Then

$$\begin{aligned} \mu (xz)\mu (z)=m(x,z)\mu (x)=-1. \end{aligned}$$

So exactly one of $\mu (z),\mu (xz)$ is equal to $-1$. By Lemma 2.7, we have $\langle x,z\rangle \sim \Gamma _1$. $\square$

Lemma 2.10

Let $F$ be a non-trivial elementary abelian 2-subgroup of $\mathrm{O}(n)/\langle -I\rangle$. If $\mathrm{rank}(\ker m/A_{F})=1$ and $\mathrm{rank}(F/A_{F})>1$, then there exists a Klein four subgroup $F^{\prime }\subset F$ with $F^{\prime }\sim \Gamma _1$.

Proof

Choose a subgroup $F^{\prime \prime }\subset F$ such that $F=A_{F} \times F^{\prime \prime }$, then $\mathrm{rank}(\ker (m_{F^{\prime \prime }}))=1, \, A_{F^{\prime \prime }}=1$ and $\mathrm{rank}F^{\prime \prime }>1$. Replace $F$ by $F^{\prime \prime }$, we may assume that $A_{F}=1, \, \mathrm{rank}(\ker m)=1$ and $\mathrm{rank}F>1$.

The subgroup $F$ is of the form $F=\ker m\times F^{\prime \prime }$ with $m(\ker m, F^{\prime \prime })=1, \, \mathrm{rank}F^{\prime \prime }\ge 2$, and $m_{F^{\prime \prime }}$ non-degenerate. When $\mathrm{rank}F^{\prime \prime }>2$ or $F^{\prime \prime }\sim \Gamma _1$, there exists $F^{\prime }\subset F^{\prime }$ with $F^{\prime }\sim \Gamma _1$ by Lemma 2.9. Otherwise $F^{\prime \prime }\sim \Gamma _2$. Choose $x,y\in F^{\prime \prime }$ generating $F^{\prime \prime }$ and $1\ne z\in \ker m$, then

$$\begin{aligned} F^{\prime }=\langle xz,yz\rangle \sim \Gamma _1 \end{aligned}$$

since $(\mu (xz),\mu (yz),\mu (xy))=(1,1,-1)$. $\square$

For any $n\ge 1$, let $T:\mathrm{O}(n)\hookrightarrow \mathrm{U}(n), \, T^{\prime }:\mathrm{Sp}(n/2)\hookrightarrow \mathrm{U}(n)$) be the natural inclusions.

Lemma 2.11

For any two closed subgroups $S_1,S_2\subset \mathrm{O}(n)$ or $S^{\prime }_1,S^{\prime }_2\subset \mathrm{Sp}(n/2)$

$$\begin{aligned} T(S_1)\sim T(S_2)\Leftrightarrow S_1\sim S_2 \end{aligned}$$

and

$$\begin{aligned} T^{\prime }(S^{\prime }_1)\sim T^{\prime }(S^{\prime }_2)\Leftrightarrow S^{\prime }_1\sim S^{\prime }_2. \end{aligned}$$

Proof

These follow from [6] Theorem 2.3 and [1] Theorem 8.1. $\square$

Proposition 2.12

Let $F$ be an elementary abelian 2-subgroup of $\mathrm{O}(n)/\langle -I\rangle$,  

• (1) when $\ker m=1$, the conjugacy class of $F$ is determined by $\delta _{F}$ and $s_{F}$;

• (2) in general, $\ker m$ is diagonalizable and the conjugacy class of $F$ is determined by the conjugacy class of $A_{F}$ and the invariants $(\epsilon _{F},\delta _{F},s_{F})$.

• (3) we have $\mathrm{defe}(F)=(1-\epsilon _{F})(-1)^{\delta _{F}}2^{r_{F}+s_{F}+\delta _{F}}$.

Proof

For $(1)$, since $\ker m=1$, so $\mathrm{rank}F$ is even. When $\mathrm{rank}F=2, \, F\sim \Gamma _1\mathrm or \Gamma _2$ by Lemma 2.2. When $\mathrm{rank}F\ge 2$, there exists a Klein four subgroup $F^{\prime }\subset F$ with $F^{\prime }\sim \Gamma _1$ by Lemma 2.9. We may and do assume that $\Gamma _1\subset F$. Then

$$\begin{aligned} F\subset \left(\mathrm{O}(n)/\langle -I\rangle \right)^{\Gamma _1}= \Delta \left(\mathrm{O}(\frac{n}{2})/\langle -I\rangle \right)\times \Gamma _1. \end{aligned}$$

So $F=\Delta (F^{\prime })\times \Gamma _1$ for some $F^{\prime }\subset \mathrm{O}(\frac{n}{2})/\langle -I\rangle$. By induction, we can show $\mathrm{defe}F\ne 0$ and the conjugacy class of $F$ is determined by $\delta _{F}$ and $s_{F}$.

For $(2), \, \ker m$ is diagonalizable since $\pi ^{-1}(\ker m)$ is abelian by the definition of $m$, where $\pi$ is the natural projection

$$\begin{aligned} \pi :\mathrm{O}(n)\longrightarrow \mathrm{O}(n)/\langle -I\rangle . \end{aligned}$$

We break the proof into two parts according to the value of $\epsilon _{F}$. When $\epsilon _{F}=1$, by Lemma 2.10, $F$ is of the form $F=\ker m\times F^{\prime }$ with $m_{F^{\prime }}$ non-degenerate and $\mathrm{defe}F^{\prime }>0$. By $(1)$, the conjugacy class of $F^{\prime }$ is determined by $s_{F}=\frac{\mathrm{rank}F^{\prime }}{2}$. We have

$$\begin{aligned} (\mathrm{O}(n)/\langle -I\rangle )^{F^{\prime }}=\Delta (\mathrm{O}(n^{\prime })/\langle -I\rangle ) \times F^{\prime }, \end{aligned}$$

where $n^{\prime }=\frac{n}{2^{\frac{\mathrm{rank}F^{\prime }}{2}}}$. Fixing $F^{\prime }$, by Lemmas 2.3 and 2.11, the conjugacy class of $\ker m$ in $\mathrm{O}(n)/\langle -I\rangle$ determines the conjugacy class of it in $(\mathrm{O}(n)/\langle -I\rangle )^{F^{\prime }}$. Moreover, as $\epsilon _{F}=1$ is given, the conjugacy class of $\ker m$ is determined by the conjugacy class of $A_{F}=\ker \mu |_{\ker m}$. So the conjugacy class of $F$ is determined by that of $A_{F}$ and the invariants $(\delta _{F},s_{F})$. When $\epsilon _{F}=0$, it is similar as the above proof for $\epsilon _{F}=1$ case to show that the conjugacy class of $F$ is determined by the conjugacy class of $A_{F}$ and the invariants $(\delta _{F},s_{F})$.

(3) follows from Lemma 2.7 and (2). $\square$

The classification of elementary abelian 2-subgroup of $\mathrm{Sp}(n)/\langle -I\rangle$ is similar as that of $\mathrm{O}(n)/\langle -I\rangle$. We give the definitions and results below but omit the proofs.

Let $F$ be an elementary abelian 2 subgroup of $\mathrm{Sp}(n)/\langle -I\rangle ,n\ge 2$. For any $x\in F$, choose $A\in \mathrm{Sp}(n)$ representing $x$, then $A^{2}=\lambda _{A} I$ for some $\lambda _{A}=\pm {1}$. For any $x,y\in F$, choose $A,B\in \mathrm{Sp}(n)$ representing $x,y$, then $[A,B]=\lambda _{A,B} I$ for some $\lambda _{A,B}=\pm {1}$. The values of $\lambda _{A},\lambda _{A,B}$ don’t depend on the choice of $A,B$. For any $x,y\in F$, define

$$\begin{aligned} \mu (x)=\mu _{F}(x)=\lambda _{A} \end{aligned}$$

and

$$\begin{aligned} m(x,y)=m_{F}(x,y)=\lambda _{A,B}. \end{aligned}$$

Lemma 2.13

Let $F$ be an elementary abelian 2-subgroup of $\mathrm{Sp}(n)/\langle -I\rangle$. For any $x,y,z\in F, \, m(x,x)=1, \, m(xy,z)=m(x,z)m(y,z), \, \mu (1)=1$ and

$$\begin{aligned} m(x,y)=\mu (x)\mu (y)\mu (xy). \end{aligned}$$

Lemma 2.14

Let $F$ be an elementary abelian 2-subgroup of $\mathrm{Sp}(n)/\langle -I\rangle$.

For $x\in F, \, \mu (x)=-1$ if and only if $x\sim [J_{\frac{n}{2}}]$.

For $x,y\in F$ with $m(x,y)=-1$,  

• (1) when $\mu (x)=\mu (y)=-1$, we have $(x,y)\sim ([{{\varvec{i}}}I],[{{\varvec{j}}}I])$;

• (2) when $\mu (x)=\mu (y)=1$, we have $(x,y)\sim ([I_{\frac{n}{2},\frac{n}{2}}], [J^{\prime }_{\frac{n}{2}}])$.

Definition 2.15

For an elementary abelian 2-subgroup $F\subset \mathrm{Sp}(n)/\langle -I\rangle$, define

$$\begin{aligned} A_{F}=\ker (\mu |_{\ker m}) \end{aligned}$$

and the defect index

$$\begin{aligned} \mathrm{defe}F=|\{x\in F: \mu (x)=1\}|-|\{x\in F:\mu (x)=-1\}|. \end{aligned}$$

Define $(\epsilon _{F},\delta _{F})$ as follows,

• when $\mu |_{\ker m}\ne 1$, define $(\epsilon _{F},\delta _{F})=(1,0)$;

• when $\mu |_{\ker m}=1$ and $\mathrm{defe}F<0$, define $(\epsilon _{F},\delta _{F})=(0,1)$;

• when $\mu |_{\ker m}=1$ and $\mathrm{defe}F>0$, define $(\epsilon _{F},\delta _{F})=(0,0)$.

Define $r_{F}=\mathrm{rank}A_{F}$ and $s_{F}=\frac{1}{2}\mathrm{rank}(F/\ker m)-\delta _{F}$.

Proposition 2.16

Let $F$ be an elementary abelian 2-subgroup of $\mathrm{Sp}(n)/\langle -I\rangle$,  

• (1) when $\ker m=1$, the conjugacy class of $F$ is determined by $\delta _{F}$ and $s_{F}$;

• (2) in general, $\ker m$ is diagonalizable and the conjugacy class of $F$ is determined by the conjugacy class of $A_{F}$ and the invariants $(\epsilon _{F},\delta _{F},s_{F})$.

• (3) we have $\mathrm{defe}(F)=(1-\epsilon _{F})(-1)^{\delta _{F}}2^{r_{F}+s_{F}+\delta _{F}}$.

2.3 Twisted projective unitary groups

For $n\ge 3$, let $G=\mathrm{Aut}(\mathfrak su (n))$, which has two connected components and $G_0= \mathrm{Int}(\mathfrak su (n))=\mathrm{PU}(n)=\mathrm{U}(n)/\mathrm{Z}_{n}$. When $n$ is even, $G$ has two conjugacy classes of outer involutions with representatives $\tau _0=complex\ conjugation$ and $\tau _0\mathrm{Ad}(J_{n/2})$; when $n$ is odd, $G$ has a unique conjugacy class of outer involutions with representative $\tau _0$. We have (cf. [8, Table 2])

$$\begin{aligned} \mathrm{Int}(\mathfrak su (n))^{\tau _0}=\mathrm{O}(n)/\langle -I\rangle \end{aligned}$$

and

$$\begin{aligned} \mathrm{Int}(\mathfrak su (n))^{\tau _0\mathrm{Ad}(J_{n/2})}=\mathrm{Sp}(n/2)/\langle -I\rangle . \end{aligned}$$

Let $F$ be an elementary abelian 2-subgroup of $G$. For the subgroup $F\cap \mathrm{Int}(\mathfrak su (n))$ of $\mathrm{Int}(\mathfrak su (n))=\mathrm{PU}(n)$, we have a bilinear form

$$\begin{aligned} m: F\cap \mathrm{Int}(\mathfrak su (n))\times F\cap \mathrm{Int}(\mathfrak su (n))\longrightarrow \{\pm {1}\}. \end{aligned}$$

Moreover, we define a function

$$\begin{aligned} \mu : F-F\cap \mathrm{Int}(\mathfrak su (n))\longrightarrow \{\pm {1}\} \end{aligned}$$

by $\mu (z)=1$ if $z\sim \tau _0$, and $\mu (z)=-1$ if $z\sim \tau _0\mathrm{Ad}(J_{n/2})$. On the other hand, for any $z\in F-\mathrm{Int}(\mathfrak su )(n)$, define $\mu _{z}: F\cap \mathrm{Int}(\mathfrak su (n))\longrightarrow \{\pm {1}\}$ and

$$\begin{aligned} m_{z}:(F\cap \mathrm{Int}(\mathfrak su (n)))\times (F\cap \mathrm{Int}(\mathfrak su (n))) \longrightarrow \{\pm {1}\} \end{aligned}$$

from the inclusion

$$\begin{aligned} F\cap \mathrm{Int}(\mathfrak su (n))\subset \mathrm{Int}(\mathfrak su (n))^{z}\cong \mathrm{O}(n)/\langle -I\rangle \mathrm or \mathrm{Sp}(n/2)/\langle -I\rangle . \end{aligned}$$

Definition 2.17

For an elementary abelian 2-subgroup $F\subset \mathrm{Aut}(\mathfrak su (n))$, define

$$\begin{aligned} A_{F}=\{x\in F\cap \mathrm{Int}(\mathfrak su (n))|z\sim zx, \forall z\in F- F\cap \mathrm{Int}(\mathfrak su (n))\} \end{aligned}$$

and the defect index

$$\begin{aligned} \mathrm{defe}F=|\{x\in F: x\sim \tau _0\}|-|\{x\in F: x\sim \tau _0\mathrm{Ad}(J_{n/2})\}|. \end{aligned}$$

Define $(\epsilon _{F},\delta _{F})$ as follows,

• when $\mathrm{defe}F=0$, define $(\epsilon _{F},\delta _{F})=(1,0)$;

• when $\mathrm{defe}F>0$, define $(\epsilon _{F},\delta _{F})=(0,0)$;

• when $\mathrm{defe}F<0$, define $(\epsilon _{F},\delta _{F})=(0,1)$.

Define $r_{F}=\mathrm{rank}A_{F}$ and $s_{F}=\frac{1}{2}\mathrm{rank}(F/\ker m)-\delta _{F}$.

Lemma 2.18

Let $F$ be an elementary abelian 2-subgroup of $\mathrm{Aut}(\mathfrak su (n))$. Then for any $z\in F-F\cap \mathrm{Int}(\mathfrak su (n))$, we have $m_{z}=m$ on $F\cap \mathrm{Int}(\mathfrak su (n))$.

Proof

For $z\in F-\mathrm{Int}(\mathfrak su (n))$ and $x,y\in F\cap \mathrm{Int}(\mathfrak su (n))$, by Lemma 2.2, 2.7 and 2.14, we have

$$\begin{aligned} m(x,y)=-1\Leftrightarrow \langle x,y\rangle \sim \langle [I_{\frac{n}{2},\frac{n}{2}}],[J^{\prime }_{\frac{n}{2}}]\rangle \Leftrightarrow m_{z}(x,y)=-1. \end{aligned}$$

So $m_{z}(x,y)=m(x,y)$. $\square$

Lemma 2.19

For any $z\in F-F\cap \mathrm{Int}(\mathfrak su (n))$ and $x\in F\cap \mathrm{Int}(\mathfrak su (n)), \, \mu _{z}(x)=\mu (z)\mu (zx)$.

Proof

We may and do assume that $z=\tau _0$ or $\tau _0\mathrm{Ad}(J_{n/2})$.

In the case of $z=\tau _0$ and $\mu _{z}(x)=1$, we may assume that $x=[I_{p,n-p}]\in \mathrm{O}(n)/\langle -I\rangle =\mathrm{Int}(\mathfrak su (n))^{z}$ for some $0\le p\le n$. Let $u=[\mathrm{diag}\{iI_{p},I_{n-p}\}]$, then

$$\begin{aligned}&uzu^{-1}=z(z^{-1}uz)u^{-1}=z(\overline{u})u^{-1}\\&= z[\mathrm{diag}\{-iI_{p},I_{n-p}\}] [\mathrm{diag}\{-iI_{p},I_{n-p}\}]\\&= z[I_{p,n-p}]=zx. \end{aligned}$$

So $zx\sim z$. And so $1=\mu _{z}(x)=\mu (z)\mu (zx)$.

In the case of $z=\tau _0$ and $\mu _{z}(x)=-1$, we may assume that $x=[J_{n/2}]\in \mathrm{O}(n)/\langle -I\rangle =\mathrm{Int}(\mathfrak su (n))^{z}$. Then $zx=\tau _0\mathrm{Ad}(J_{n/2})=\tau ^{\prime }_0$. So $-1=\mu _{z}(x)=\mu (z)\mu (zx)$.

The proof in the case of $z=\tau ^{\prime }_0=\tau _0\mathrm{Ad}(J_{n/2})$ is similar. $\square$

Lemma 2.20

For any elementary abelian 2-subgroup $F\subset \mathrm{Aut}(\mathfrak su (n))$, we have $A_{F}\subset \ker m$ and $A_{F}=\ker (\mu _{z}|_{\ker m})$ for any $z\in F-F\cap \mathrm{Int}(\mathfrak su (n))$.

Proof

Choose any $x\in A_{F}$ and choose an element $z\in F-F\cap \mathrm{Int}(\mathfrak su (n))$. By the definition of $A_{F}$, for any $y\in F\cap \mathrm{Int}(\mathfrak su (n))$, we have $\mu (zy)=\mu (zyx)$. In particular for $y=1$, we have $\mu (z)=\mu (zx)$. Then

$$\begin{aligned} m(x,y)=m_{z}(x,y)=\mu _{z}(x)\mu _{z}(y)\mu _{z}(xy)=\mu (z) \mu (zx)\mu (zy)\mu (zxy)=1. \end{aligned}$$

So $A_{F}\subset \ker m$.

On the other hand, for any $x\in \ker m=\ker m_{z}, \, x\in A_{F}$ if and only if $\forall y\in F\cap \mathrm{Int}(\mathfrak su (n)), \mu (zy)=\mu (zyx)$. Since

$$\begin{aligned} \mu (zy)\mu (zyx)=\mu _{z}(y)\mu _{z}(xy)=m_{z}(x,y)\mu _{z}(x)=\mu _{z}(x), \end{aligned}$$

we get that $A_{F}=\ker (\mu _{z}|_{\ker m})$. $\square$

Proposition 2.21

For an elementary abelian 2-subgroup $F$ of $\mathrm{Aut}(\mathfrak su (n))$ which is not contained in $\mathrm{Int}(\mathfrak su (n)), \, \ker m$ is diagonalizable and the conjugacy class of $F$ is determined by the conjugacy class of $A_{F}$ and the invariants $(\epsilon _{F},\delta _{F},\mathrm{rank}F)$.

Proof

We break the proof into two cases.

When there exists $z\in F$ with $z\sim \tau _0$, we may and do assumen that $z=\tau _0\in F$. Then

$$\begin{aligned} F\subset \mathrm{Aut}(\mathfrak su (n))^{z}=(\mathrm{O}(n)/\langle -I\rangle )\times \langle z\rangle . \end{aligned}$$

By Lemma 2.18, we get that $m_{z}=m$. Then $(\epsilon _{F},\delta _{F})$ coincides with $(\epsilon _{F^{\prime }},\epsilon _{F^{\prime }})$ when $F^{\prime }=F\cap \mathrm{Int}(\mathfrak su (n))$ is considered as a subgroup of $\mathrm{O}(n)/\langle -I\rangle$. Then the conclusion follows from Proposition 2.12 and Lemma 2.11.

Otherwise, for any $z\in F-F\cap \mathrm{Int}(\mathfrak su (n))$, we have that $z\sim \tau ^{\prime }_0= \tau _0\mathrm{Ad}(J_{n/2})$. We may and do assumen that $z=\tau ^{\prime }_0\in F$. Then

$$\begin{aligned} F\subset \mathrm{Aut}(\mathfrak su (n))^{z}=(\mathrm{Sp}(n/2)/\langle -I\rangle )\times \langle z\rangle . \end{aligned}$$

And we have $\mu _{z}\equiv 1$ since all elements in $F-F\cap \mathrm{Int}(\mathfrak su (n))$ are conjugate to $\tau ^{\prime }_0$. Then the conjugacy class of $F$ is determined by $\mathrm{rank}F$ by Proposition 2.16. Moreover, in this case, we have $(\epsilon _{F},\delta _{F})=(0,1)$ and $\mathrm{rank}A_{F}=\mathrm{rank}F-1$. Then the tuple of invariants $(\epsilon _{F},\delta _{F},\mathrm{rank}F, \mathrm{rank}A_{F})$ is different from that for any subgroup considered in the first case. The reason is: if a subgroup $F$ in the first case satisfies $\mathrm{rank}A_{F}=\mathrm{rank}F-1$, then its elements in $F-F\cap \mathrm{Int}(\mathfrak su (n))$ are all conjugate to $\tau _0$, by which we have $(\epsilon _{F},\delta _{F})=(0,0)$. $\square$

2.4 A class of elementary abelian 2-subgroups and symplectic metric spaces

The elementary abelian 2-subgroups $F$ of $\mathrm{O}(n)/\langle -I\rangle$ (or $\mathrm{Sp}(n)/\langle -I\rangle$) with non-identity elements all conjugate to $[I_{\frac{n}{2},\frac{n}{2}}], [J_{\frac{n}{2}}]$ (or $[I_{\frac{n}{2},\frac{n}{2}}],[\mathbf{i}I]$) have a particular nice shape.

Proposition 2.22

For an elementary abelian 2-subgroup $F$ of $\mathrm{O}(n)/\langle -I\rangle$ (or $\mathrm{Sp}(n)/\langle -I\rangle$), any non-identity element of $F$ is conjugate to $[I_{\frac{n}{2},\frac{n}{2}}], [J_{\frac{n}{2}}]$ (or $[I_{\frac{n}{2},\frac{n}{2}}],[{{\varvec{i}}}I]$) if and only if any non-identity element of $A_{F}$ is conjugate to $[I_{\frac{n}{2},\frac{n}{2}}]$.

Proof

Since elements in $F-A_{F}$ are all conjugate to $[I_{\frac{n}{2},\frac{n}{2}}], [J_{\frac{n}{2}}]$ (or $[I_{\frac{n}{2},\frac{n}{2}}],[\mathbf{i}I]$) and any element of $A_{F}$ is not conjugate to $[J_{\frac{n}{2}}]$ (or $[\mathbf{i}I]$), the conclusion follows. $\square$

Regard $A_{F}$ as a subgroup of $G^{\prime }=\mathrm{O}(n^{\prime })/\langle -I\rangle , \, \mathrm{U}(n^{\prime })/\langle -I\rangle$ or $\mathrm{Sp}(n^{\prime })/\langle -I\rangle$, where $n^{\prime }=\frac{n}{2^{s+k}}$ ($k=2,1,0$). Then the condition of any non-identity element of $A_{F}$ is conjugate to $[I_{\frac{n}{2},\frac{n}{2}}]$ in $G$ is equivalent to any non-identity element of $A_{F}$ is conjugate to $[I_{\frac{n^{\prime }}{2},\frac{n^{\prime }}{2}}]$ in $G^{\prime }$.

Let $F^{*}=\mathrm{Hom}(F,\mathbb F _2)$ be the dual group of an elementary abelian 2-group.

For $n=2^{m}s$ with $s$ odd, let

$$\begin{aligned} K=\{\pm {1}\}^{n}/\langle \underbrace{(-1,\ldots ,-1)} \rangle . \end{aligned}$$

This is an elementary abelian 2-group of rank $n-1$. We want to characterize subgroups $F$ of $K$ such that any non-identity element $x\in F$ is of the form $x=[(x_1,x_2,\ldots ,x_{n})]$ with $x_{i}=-1$ for $\frac{n}{2}$ indices $i$ and $x_{i}=1$ for the other $\frac{n}{2}$ indices $i$.

Lemma 2.23

For a subgroup $F$ of $K$ as above, let $r$ be the rank of $F$ as an elementary abelian 2-group. Then we can divide $J=\{1,2,\ldots ,n\}$ into a disjoint union of $2^{r}$ subsets

$$\begin{aligned} \{J_{\alpha }: \alpha \in F^{*}\} \end{aligned}$$

with each $J_{\alpha }$ of cardinality $\frac{n}{2^{r}}=2^{m-r}s$ such that any element $x\in F$ is of the form

$$\begin{aligned} x=\underbrace{[(t_1,t_2,\ldots ,t_{n})]},\ t_{i}=\alpha (x),\quad \forall i\in J_{\alpha }. \end{aligned}$$

Proof

Choose a subgroup $F^{\prime }$ of $\{\pm {1}\}^{n}$ such that its projection to $K$ has image equal to $F$ and the projection map onto $F$ is an isomorphism. Then any non-identity element $x\in F^{\prime }$ is of the form $x=(x_1,x_2,\dots ,x_{n})$ with $x_{i}=-1$ for $\frac{n}{2}$ indices $i$ and $x_{i}=1$ for the other $\frac{n}{2}$ indices $i$. For any $x\in F^{\prime }$, let $x=(t_{x,1},t_{x,2},\ldots ,t_{x,n}), \, t_{x,i}=\pm {1}$. For an index $i$, the map $x\mapsto t_{x,i}$ is an homomorphism from $F^{\prime }$ to $\pm {1}$, so there exists $\alpha _{i}\in F^{\prime }{^{*}}$ such that

$$\begin{aligned} t_{x,i}=\alpha _{i}(x),\forall x\in F^{\prime }. \end{aligned}$$

For any $\alpha \in F^{*}$, define

$$\begin{aligned} J_{\alpha }=\{1\le i\le n|\alpha _{i}=\alpha \}. \end{aligned}$$

Then $J=\{1,2,\ldots ,n\}$ is a disjoint union of $2^{r}$ subsets $\{J_{\alpha }: \alpha \in F^{*}\}$ and any element $x\in F^{\prime }$ is of the form $x=(t_1,t_2,\ldots ,t_{n}), \, t_{i}=\alpha (x), \, \forall i\in J_{\alpha }$. We show that the cardinarity of each $J_{\alpha }$ is $\frac{n}{2^{r}}=2^{m-r}s$.

Let $\alpha _0=0\in F^{\prime }{^{*}}$ be the zero element. For any $\alpha \ne \alpha _0$, count the number of pairs $(x,i)$ with $\alpha (x)=-1$ and $t_{x,i}=-1$. For a fixed $x\in F$, when $x\not \in \ker \alpha$, there are $\frac{n}{2}$ such $(x,i)$; when $x\in \ker \alpha$, there are no such $(x,i)$. For a fixed $i, 1\le i\le n$, when $x\not \in J_{\alpha _0}\cup J_{\alpha }$, i.e., $\alpha _{i}\ne \alpha _0$ and $\alpha _{i}\ne \alpha$, there are $2^{r-2}$ such $(x,i)$; when $i\in J_{\alpha }$, i.e., $\alpha _{i}=\alpha$, there are $2^{r-1}$ such $(x,i)$; when $i\in J_{\alpha _0}$, i.e., $\alpha _{i}=\alpha _0$, there are no such $(x,i)$. Count the number of pairs $(x,i)$ with $\alpha (x)=-1$ and $t_{x,i}=-1$ in two ways, we get an equality

$$\begin{aligned} 2^{r-1}\frac{n}{2}=(n-|J_{\alpha }|-|J_{\alpha _0}|)2^{r-2} +|J_{\alpha }|2^{r-1}. \end{aligned}$$

This implies that $|J_{\alpha }|=|J_{\alpha _0}|$. Then the cardinarity of each $J_{\alpha }$ is $\frac{n}{2^{r}}=2^{m-r}s$.

Since the projection map from $F^{\prime }$ to $F$ is an isomorphism, we can identify $F^{\prime }{^{*}}$ and $F^{*}$. Then we get the conclusion of the lemma. $\square$

Proposition 2.24

For an elementary abelian 2-subgroup $F$ of $\mathrm{O}(n)/\langle -I\rangle$ (or $F\subset \mathrm{Sp}(n)/\langle -I\rangle$) with non-identity elements all conjugate to $[I_{\frac{n}{2},\frac{n}{2}}], \, [J_{\frac{n}{2}}]$ (or $[I_{\frac{n}{2},\frac{n}{2}}], \, [{{\varvec{i}}}I]$), the conjugacy class of $F$ is determined by the tuple $(\epsilon _{F},\delta _{F},r_{F},s_{F})$.

Proof

This follows from Propositions 2.12, 2.16, 2.22 and Lemma 2.23. $\square$

Let $F_{r,s,\epsilon ,\delta }$ be an elementary abelian 2-subgroup of $\mathrm{O}(n)/\langle -I\rangle$ (or $\mathrm{Sp}(n)/\langle -I\rangle$) satisfying the properties in Proposition 2.24 and with invariants $(\epsilon ,\delta ,r,s)$, which is unique up to conjugation.

Definition 2.25

A finite-dimensional vector space $V$ over the field $\mathbb F _2=\mathbb Z /2\mathbb Z$ is called a symplectic vector space if it is associated with a map $m: V\times V\longrightarrow \mathbb F _2$ such that $m(x,x)=0, \, m(x,y)=m(y,x)$ and $m(x+y,z)=m(x,z)m(y,z)$ for any $x,y,x\in F$.

Moreover, it is called a symplectic metric space if there is another map $\mu : V\longrightarrow \mathbb F _2$ such that $\mu (0)=0$ and $m(x,y)=\mu (x)+\mu (y)+\mu (x+y)$ for any $x,y\in V$.

Two symplectic vector spaces $(V,m)$ and $(V^{\prime },m^{\prime })$ are called isomorphic if there exists a linear space isomorphism $f: V\longrightarrow V^{\prime }$ transferring $m$ to $m^{\prime }$.

Two symplectic metric spaces $(V,m,\mu )$ and $(V^{\prime },m^{\prime },\mu ^{\prime })$ are called isomorphic if there exists a linear space isomorphism $f: V\longrightarrow V^{\prime }$ transferring $(m,\mu )$ to $(m^{\prime },\mu ^{\prime })$.

The following proposition is clear.

Proposition 2.26

The isomorphism class of a symplectic vector space $(V,m)$ is determined by the dimensions $(\dim _\mathbb{F _2} V,\dim _\mathbb{F _2}\ker m)$.

Definition 2.27

For a symplectic metric space $V$, define $A_{V}=\ker \mu |_{\ker m}$ and the defect index $\mathrm{defe}V=|\{x\in V: \mu (x)=1\}|-|\{x\in V:\mu (x)=-1\}|$.

Define $(\epsilon _{V},\delta _{V})$ as follows,

• When $\mu |_{\ker m}\ne 1$, define $(\epsilon _{V},\delta _{V})=(1,0)$;

• when $\mu |_{\ker m}=1$ and $\mathrm{defe}V<0$, define $(\epsilon _{V},\delta _{V})=(0,1)$;

• when $\mu |_{\ker m}=1$ and $\mathrm{defe}V>0$, define $(\epsilon _{V},\delta _{V})=(0,0)$.

Define $r_{V}=\dim _\mathbb{F _2} A_{V}, \, s_{V}=\frac{1}{2}\dim _\mathbb{F _2}(V/\ker m)- \delta _{V}$.

Remark 2.28

When $m$ is non-degenerate, $\mu$ is a non-degenerate quadratic form, in this case $\delta _{V}$ is the Arf invariant of $\mu$.

The following proposition is an analogue of Proposition 2.24. And it also can be proved by the same method.

Proposition 2.29

The isomorphism class of a symplectic metric space is determined by the invariants $(r_{V},s_{V},\epsilon _{V},\delta _{V})$.

We have $\mathrm{defe}V=(1-\epsilon )(-1)^{\delta }2^{r+s+\delta }$.

Proposition 2.30

For a vector space $V$ over $\mathbb{F }_2$ of rank 3 with a map $\mu : V\longrightarrow \mathbb{F }_2$ satisfying $\mu (0)=0$, let $m(x,y)=\mu (x)+\mu (y)+\mu (x+y)$. Then $(V,m, \mu )$ is a symplectic metric space if and only if $m$ is bilinear, if and only if there are even number of elements in $V$ with non-trivial values of the function $\mu$.

Proof

With the definiton of $m$ and the property $\mu (0)=0$, we get the compatability relation and the property $m(x,x)=0$, then $(V,m,\mu )$ is a symplectic metric if and only of $m$ is a bilinear form. This is the first statement.

For any $x,y,z\in V$, when $x,y,z$ are linearly dependent, the equality $m(x+y,z)=m(x,z)+m(y,z)$ follows from the definiton of $m$ and the property $\mu (0)=0$. When $x,y,z$ are linearly independent, they consist in a basis of $V$. By the definition of $m$ and the property $\mu (0)=0$, we have that the equality $m(x+y,z)=m(x,z)+m(y,z)$ holds if and only if the sum of the values of $\mu$ over all elements of $V$ is 0. That is also equivalent to there are even elements in $V$ with $\mu$-value $1$. So the second statement follows. $\square$

Let $V_{r,s;\epsilon ,\delta }$ be a symplectic metric space with the prescribed invariants, which is unique up to isomorphism. Let $\mathrm{Sp}(r,s;\epsilon ,\delta )$ be the group of automorphisms of $V_{r,s,\epsilon ,\delta }$ preserving $m$ and $\mu$. Let $V_{s;\epsilon ,\delta }=V_{0,s;\epsilon ,\delta }$ and $\mathrm{Sp}(s;\epsilon ,\delta )=Sp(0,s;\epsilon ,\delta )$. It is clear that

$$\begin{aligned} \mathrm{Sp}(r,s;\epsilon ,\delta )=\mathrm{Hom}(V_{s;\epsilon ,\delta },\mathbb F _2^{r})\rtimes (\mathrm{Sp}(s;\epsilon ,\delta )\times \mathrm{GL}(\mathbb F _2^{r})). \end{aligned}$$

Let $\mathrm{Sp}(s)=\mathrm{Sp}(s,\mathbb F _2)$ be the degree-$s$ symplectic group over the field $\mathbb F _2$.

Proposition 2.31

We have the following formulas for the orders of $\mathrm{Sp}(s;\epsilon ,\delta )$,

$$\begin{aligned}&|\mathrm{Sp}(s;0,0)|=\left(\prod _{1\le i\le s-1}(2^{i+1}-1)(2^{i}+1)\right)\cdot 2^{s^{2}-s+1},\\&|\mathrm{Sp}(s-1;0,1)|=3\cdot \left(\prod _{1\le i\le s-1}(2^{i}-1)(2^{i+1}+1)\right)\cdot 2^{s^{2}-s+1},\\&|\mathrm{Sp}(s;1,0)|=|\mathrm{Sp}(s)|=\left(\prod _{1\le i\le s}(2^{i}-1)(2^{i}+1)\right)2^{s^{2}}.\\ \end{aligned}$$

Proof

When $s=1$ or 0, these are clear. So we just need to calculate

$$\begin{aligned} |\mathrm{Sp}(s;\epsilon ,\delta )|/ |\mathrm{Sp}(s-1;\epsilon ,\delta )|. \end{aligned}$$

We calculate it for the case $\epsilon =\delta =0$, the other cases are similar.

$\mathrm{Sp}(s;0,0)$ permutes the non-identity elements $x\in V_{s;0,0}$ with $\mu (x)=0$, there are $\frac{2^{2s}+2^{s}}{2}-1=(2^{s}-1)(2^{s-1}+1)$ such elements. Fix two distinct non-identity elements $x_1,x_2\in V_{s,0,0}$ with $\mu (x_1)=\mu (x_2)=0$ and $m(x_1,x_2)=1$. For any other $x$ with $\mu (x)=0$ and $(x_1,x)=1, \, (x_1,x)$ is transformed to $(x_1,x_2)$ under some transformation in $\mathrm{Sp}(s;0,0)$. Fixing $x_1$, there are $2^{2s-2}$ such elements $x$. Moreover, the subgroup of $\mathrm{Sp}(s;0,0)$ consisting of elements fixing $x_1$ and $x_2$ is isomorphic to $\mathrm{Sp}(s-1;0,0)$. So So we have $|\mathrm{Sp}(s;\epsilon ,\delta )|/|\mathrm{Sp}(s-1;\epsilon ,\delta )|= (2^{s}-1)(2^{s-1}+1)2^{2s-2}$. $\square$

Since we have

$$\begin{aligned} V_{s;0,0}\oplus V_{0;1,0}\cong V_{s-1;0,1}\oplus V_{0;1,0}\cong V_{s;1,0}, \end{aligned}$$

so we can regard $\mathrm{Sp}(s;0,0)$ and $\mathrm{Sp}(s-1;0,1)$ as subgroups of $\mathrm{Sp}(s;1,0)$.

Proposition 2.32

$\mathrm{Sp}(s;1,0)\cong \mathrm{Sp}(s)$.

Proof

Since $V_{s;1,0}/\ker m=\mathbb F _2^{2s}$ is a symplectic vector space of dimension $2s$, by restriction we get a natural homomorphism $p:\mathrm{Sp}(s;1,0)\longrightarrow \mathrm{Sp}(s)$.

Let $z$ be the unique non-identity element in $\ker m$. Suppose that $p(f)=1$ for some $f\in \mathrm{Sp}(s;1,0)$, then for any $x\in V_{s;1,0}, \, f(x)=x$ or $f(x)=xz$. Since $\mu (xz)= \mu (x)+\mu (z)+m(x,z)=\mu (x)+1$, so $f(x)\ne xz$. Then $f(x)=x$ for any $x\in V_{s;1,0}$. Thus $p$ is injective.

Moreover, by Proposition 2.31 we have $|\mathrm{Sp}(s;1,0)|=|\mathrm{Sp}(s)|$. So $p$ is an isomorphism. $\square$

Since an element in $\mathrm{Sp}(s;0,0)$ or $\mathrm{Sp}(s-1;0,1)$ preserves the symplectic form $m$ on $V=\mathbb{F }_2^{2s}$, so we have inclusions $\mathrm{Sp}(s;0,0)\subset \mathrm{Sp}(s)$ and $\mathrm{Sp}(s-1;0,1)\subset \mathrm{Sp}(s)$.

Proposition 2.33

We have

$$\begin{aligned}{}[\mathrm{Sp}(s):\mathrm{Sp}(s;0,0)]=2^{s-1}(2^{s}+1) \end{aligned}$$

and

$$\begin{aligned}{}[\mathrm{Sp}(s):\mathrm{Sp}(s-1;0,1)]=2^{s-1}(2^{s}-1). \end{aligned}$$

Proof

This follows from Proposition 2.31 directly. $\square$

Define the groups $\mathrm{Sp}(s;t)$ ($s,t\ge 0$) as the automorphism group a symplecic vector space $(V,m)$ over $\mathbb F _2$ with $\mathrm{rank}V=2s+t$ and $\mathrm{rank}\ker m=t$. It is clear that $\mathrm{Sp}(s;0)=\mathrm{Sp}(s)$ and

$$\begin{aligned} \mathrm{Sp}(s;t)=\mathrm{Hom}(\mathbb F _2^{2s},\mathbb F _2^{t})\rtimes (\mathrm{GL}(t,\mathbb F _2)\times \mathrm{Sp}(s)). \end{aligned}$$

 

3 Exceptional compact simple Lie groups (algebras)

3.1 Complex semi-simple Lie algebra and a specific compact real form

Let $\mathfrak{g }$ be a complex semisimple Lie algebra and $\mathfrak{h }$ be a Cartan subalgebra of $\mathfrak{g }$. Then $\mathfrak{g }$ has a root-space decomposition

$$\begin{aligned} \mathfrak{g }=\mathfrak{h }\oplus \left(\bigoplus _{\alpha \in \Delta }\mathfrak{g }_{\alpha }\right), \end{aligned}$$

where $\Delta =\Delta (\mathfrak{g },\mathfrak{h })$ is the root system of $\mathfrak{g }$ and $\mathfrak{g }_{\alpha }$ is the root space of a root $\alpha \in \Delta$. Let $B$ be the Killing form on $\mathfrak{g }$. It is a non-degenerate symmetric form. The restriction of $B$ to $\mathfrak{h }$ is also non-degenerate. Let $\mathfrak{h }^{*}$ be the dual complex vector space of $\mathfrak{h }$. For any $\lambda \in \mathfrak{h }^{*}$, let $H_{\lambda }\in \mathfrak{h }$ be the element in $\mathfrak{h }$ determined uniquely by

$$\begin{aligned} B(H_{\lambda }, H)=\lambda (H),\quad \forall H\in \mathfrak{h }. \end{aligned}$$

For any $\lambda ,\mu \in \mathfrak{h }^{*}$, let $\langle \lambda ,\mu \rangle :=B(H_{\lambda },H_{\mu })$. Then $\langle \cdot ,\cdot \rangle$ is an inner product on $\mathfrak{h }^{*}$.

For any root $\alpha$, we have

$$\begin{aligned} H_{\alpha }\in \mathfrak{h }. \end{aligned}$$

(1)

Define

$$\begin{aligned} H^{\prime }_{\alpha }=\frac{2}{\alpha (H_{\alpha })}H_{\alpha }, \end{aligned}$$

(2)

which is called a co-root; let

$$\begin{aligned} 0\ne X_{\alpha }\in \mathfrak{g }_{\alpha } \end{aligned}$$

(3)

be any non-zero vector (recall that $\dim \mathfrak{g }_{\alpha }=1$), which is called a root vector of the root $\alpha$. The notations $H_{\alpha }, \, H^{\prime }_{\alpha }, \, X_{\alpha }$ will be used frequently in this paper.

Note that, for any $\alpha ,\beta \in \Delta$,

$$\begin{aligned} \langle \alpha ,\beta \rangle&= B(H_{\alpha },H_{\beta }) =\beta (H_{\alpha })=\alpha (H_{\beta })\in \mathbb{R }, \\ \langle \alpha ,\alpha \rangle&= B( H_{\alpha },H_{\alpha })=\alpha (H_{\alpha })\not = 0, \end{aligned}$$

and $2\langle \alpha ,\beta \rangle /\langle \beta , \beta \rangle \in \mathbb{Z }$. We also note that

$$\begin{aligned} \mathrm{span}_{\mathbb{R }}\{\alpha |\alpha \in \Delta \}\subset \mathfrak{h }^{*} \end{aligned}$$

is a real vector space of dimension equal to $r=\mathrm{rank}\mathfrak{g }=\dim _{\mathbb{C }}\mathfrak{h }$ (cf. [9, Pages 140–162]).

We set

$$\begin{aligned} A_{\alpha ,\beta }=2\langle \alpha , \beta \rangle /\langle \beta ,\beta \rangle = \alpha (H^{\prime }_{\beta }). \end{aligned}$$

Then

$$\begin{aligned}{}[H^{\prime }_{\alpha },X_{\beta }]=\beta (H^{\prime }_{\alpha })X_{\beta }= \frac{2\langle \alpha ,\beta \rangle }{\langle \alpha , \alpha \rangle }X_{\beta }= A_{\beta ,\alpha }X_{\beta }. \end{aligned}$$

Choose a lexicography order of $\mathrm{span}_{\mathbb{R }}\{\alpha |\alpha \in \Delta \}$ to get a positive system $\Delta ^{+}$ and a simple system $\Pi$. Let

$$\begin{aligned} \Pi =\{\alpha _1,\alpha _2, \ldots ,\alpha _{r}\}. \end{aligned}$$

(4)

For brevity, we write

$$\begin{aligned} H_{i},\ H^{\prime }_{i} \end{aligned}$$

(5)

instead of $H_{\alpha _{i}}, \, H^{\prime }_{\alpha _{i}}$ for a simple root $\alpha _{i}$.

Draw $A_{\alpha ,\beta } A_{\beta ,\alpha }$ edges to connect any two distinct simple roots $\alpha$ and $\beta$, and draw an arrow from $\alpha$ to $\beta$ if $\langle \alpha ,\alpha \rangle >\langle \beta , \beta \rangle$, we get a graph. This graph is connected if and only if $\mathfrak{g }$ is a simple Lie algebra, in this case it is called the Dynkin diagram of $\mathfrak{g }$. We always follow the Bourbaki numbering to order the simple roots (cf. [8, Page 3]).

Let $\mathrm{Aut}(\mathfrak{g })$ be the group of all complex linear automorphisms of $\mathfrak{g }$ and $\mathrm{Int}(\mathfrak{g })$ be the subgroup of inner automorphisms. We define

$$\begin{aligned} \mathrm{Out}(\mathfrak{g }):=\mathrm{Aut}(\mathfrak{g })/\mathrm{Int}(\mathfrak{g }). \end{aligned}$$

The exponential map $\exp : \mathfrak{g }\longrightarrow \mathrm{Aut}(\mathfrak{g })$ is given by

$$\begin{aligned} \exp (X)=\exp (\mathrm{ad}(X)),\quad \forall X\in \mathfrak{g }=\mathrm{Lie}(\mathrm{Aut}(\mathfrak{g })), \end{aligned}$$

where $\mathrm{ad}(X)\in \mathfrak gl (\mathfrak{g })$ is defined by $\mathrm{ad}(X)(Y)=[X,Y], \, \forall Y\in \mathfrak{g }$.

One can normalize the root vectors $\{X_{\alpha }, X_{-\alpha }\}$ so that $B(X_{\alpha },X_{-\alpha })=2/\alpha (H_{\alpha })$. Then $[X_{\alpha }, X_{-\alpha }]=H^{\prime }_\alpha$. Moreover, one can normalize $\{X_{\alpha }\}$ appropriately, such that

$$\begin{aligned} {\mathfrak{u }}_0=\mathrm{span}_{\mathbb{R }}\{X_{\alpha }-X_{-\alpha }, i(X_{\alpha }+X_{-\alpha }), i H_{\alpha }: \alpha \in \Delta ^{+}\} \end{aligned}$$

(6)

is a compact real form of $\mathfrak{g }$ ([9, Pages 348–354]). Define

$$\begin{aligned} \theta (X+iY):=X-iY,\quad \forall X,Y\in {\mathfrak{u }}_0. \end{aligned}$$

Then $\theta$ is a Cartan involution of $\mathfrak{g }$ (as a real semisimple Lie algebra) and ${\mathfrak{u }}_0=\mathfrak{g }^{\theta }$ is a maximal compact subalgebra of $\mathfrak{g }$. Any other compact real form of $\mathfrak{g }$ is conjugate to ${\mathfrak{u }}_0$. In the below, whenever we discuss a compact real form of $\mathfrak{g }$, we always use this compact real form ${\mathfrak{u }}_0$ in  (6).

Let $\mathrm{Aut}({\mathfrak{u }}_0)$ be the group of automorphisms of ${\mathfrak{u }}_0$ and $\mathrm{Int}({\mathfrak{u }}_0)=\mathrm{Aut}({\mathfrak{u }}_0)_0$ be the subgroup of inner automorphisms. Any automorphism of ${\mathfrak{u }}_0$ extends uniquely to a holomorphic automorphism of $\mathfrak{g }$, so $\mathrm{Aut}({\mathfrak{u }}_0)\subset \mathrm{Aut}(\mathfrak{g })$. Similarly we have $\mathrm{Int}({\mathfrak{u }}_0)\subset \mathrm{Int}(\mathfrak{g })$. Define

$$\begin{aligned} \Theta (f):=\theta f \theta ^{-1},\quad \forall f\in \mathrm{Aut}(\mathfrak{g }). \end{aligned}$$

Then it is a Cartan involution of $\mathrm{Aut}(\mathfrak{g })$ with differential $\theta$. It follows that $\mathrm{Aut}({\mathfrak{u }}_0)=\mathrm{Aut}(\mathfrak{g })^{\Theta }$ and $\mathrm{Int}({\mathfrak{u }}_0)=\mathrm{Int}(\mathfrak{g })^{\Theta }$ are maximal compact subgroups of $\mathrm{Aut}(\mathfrak{g })$ and $\mathrm{Int}(\mathfrak{g })$ respectively. We also have

$$\begin{aligned} \mathrm{Out}({\mathfrak{u }}_0):=\mathrm{Aut}({\mathfrak{u }}_0)/\mathrm{Int}({\mathfrak{u }}_0)\cong \mathrm{Out}(\mathfrak{g })\cong \mathrm{Aut}(\Pi ), \end{aligned}$$

where $\mathrm{Aut}(\Pi )$ is the symmetry group of the graph $\Pi$ consisting of permutations of vertices preserving the multiples of edges and directions of arrows.

3.2 Involutions

Let ${\mathfrak{u }}_0$ be a compact simple Lie algebra and $G=\mathrm{Aut}({\mathfrak{u }}_0)$ be its automorphism group. The conjugacy classes of involutions in $G$ are in one-one correspondence with the isomorphism classes of real forms of the complexified Lie algebra $\mathfrak{g }={\mathfrak{u }}_0\otimes _{\mathbb{R }}\mathbb{C }$, and also in one-one correspondence with compact irreducible Riemannian symmetric pairs $({\mathfrak{u }}_0,\mathfrak{h }_0)$. These objects were classified by Élie Cartan in 1920s. We give representatives of conjugacy classes of involutions in the automorphism group $G=\mathrm{Aut}({\mathfrak{u }}_0)$ for each compact simple exceptional Lie algebra ${\mathfrak{u }}_0$. The following are from [8, Pages 5–6]. In particular, as explained in [8], the notation ${\mathfrak{e }}_{6,-2}$ denotes a real simple Lie algebra with a Cartan decomposition ${\mathfrak{u }}_0=\mathfrak{k }_0+\mathfrak{p }_0$ such that $\mathfrak{g }={\mathfrak{u }}_0\otimes _{\mathbb{R }}\mathbb{C }$ is a complex simple Lie algebra of type $\mathbf E_6$ and $\dim \mathfrak{k }_0-\dim \mathfrak{p }_0=-2$, and similarly for the notations of other real simple Lie algebras.  

• (i) Type $\mathbf E_6$. For ${\mathfrak{u }}_0={\mathfrak{e }}_6$, let $\tau$ be a specific diagram involution defined by

  $$\begin{aligned}&\tau (H_{\alpha _1})=H_{\alpha _6}, \quad \tau (H_{\alpha _6})=H_{\alpha _1}, \quad \tau (H_{\alpha _3})=H_{\alpha _5}, \quad \tau (H_{\alpha _5})=H_{\alpha _3},\\&\tau (H_{\alpha _2})=H_{\alpha _2}, \quad \tau (H_{\alpha _4})=H_{\alpha _4}, \quad \tau (X_{\pm {\alpha _1}})=X_{\pm {\alpha _6}}, \quad \tau (X_{\pm {\alpha _6}})=X_{\pm {\alpha _1}},\\&\tau (X_{\pm {\alpha _3}})=X_{\pm {\alpha _5}}, \quad \tau (X_{\pm {\alpha _5}})=X_{\pm {\alpha _3}}, \quad \tau (X_{\pm {\alpha _2}})=X_{\pm {\alpha _2}}, \quad \tau (X_{\pm {\alpha _4}})=X_{\pm {\alpha _4}}. \end{aligned}$$

  Let

  $$\begin{aligned} \sigma _1=\exp (\pi i H^{\prime }_2), \sigma _2=\exp (\pi i (H^{\prime }_1+H^{\prime }_6)), \sigma _3=\tau , \sigma _4=\tau \exp (\pi i H^{\prime }_2). \end{aligned}$$

  Then $\sigma _1,\sigma _2,\sigma _3,\sigma _4$ represent all conjugacy classes of involutions in $\mathrm{Aut}({\mathfrak{u }}_0)$, which correspond to Riemannian symmetric spaces of type EII, EIII, EIV, EI and the corresponding real forms are ${\mathfrak{e }}_{6,-2},{\mathfrak{e }}_{6,14},{\mathfrak{e }}_{6,26},{\mathfrak{e }}_{6,-6}$. $\sigma _1,\sigma _2$ are inner automorphisms, $\sigma _3,\sigma _4$ are outer automorphisms.

• (ii) Type $\mathbf E_7$. For ${\mathfrak{u }}_0={\mathfrak{e }}_7$, let

  $$\begin{aligned}&\sigma _1=\exp \left(\pi i H^{\prime }_2\right), \quad \sigma _2=\exp \left(\pi i \frac{H^{\prime }_2+H^{\prime }_5+H^{\prime }_7}{2}\right),\\&\quad \sigma _3=\exp \left(\pi i \frac{H^{\prime }_2+H^{\prime }_5+H^{\prime }_7+2H^{\prime }_1}{2}\right). \end{aligned}$$

  Then $\sigma _1,\sigma _2,\sigma _3$ represent all conjugacy classes of involutions in $\mathrm{Aut}({\mathfrak{u }}_0)$, which correspond to Riemannian symmetric spaces of type EVI, EVII, EV and the corresponding real forms are ${\mathfrak{e }}_{7,3},{\mathfrak{e }}_{7,25},{\mathfrak{e }}_{7,-7}$.

• (iii) Type $\mathbf E_8$. For ${\mathfrak{u }}_0={\mathfrak{e }}_8$, let

  $$\begin{aligned} \sigma _1=\exp (\pi i H^{\prime }_2), \sigma _2=\exp (\pi i (H^{\prime }_2+H^{\prime }_1)). \end{aligned}$$

  Then $\sigma _1,\sigma _2$ represent all conjugacy classes of involutions in $\mathrm{Aut}({\mathfrak{u }}_0)$, which correspond to Riemannian symmetric spaces of type EIX, EVIII and the corresponding real forms are ${\mathfrak{e }}_{8,24},{\mathfrak{e }}_{8,-8}$.

• (iv) Type $\mathbf F_4$. For ${\mathfrak{u }}_0=\mathfrak{f }_4$, let

  $$\begin{aligned} \sigma _1=\exp (\pi i H^{\prime }_1), \sigma _2=\exp (\pi i H^{\prime }_4). \end{aligned}$$

  Then $\sigma _1,\sigma _2$ represent all conjugacy classes of involutions in $\mathrm{Aut}({\mathfrak{u }}_0)$, which correspond to Riemannian symmetric spaces of type FI, FII and the corresponding real forms are $\mathfrak{f }_{4,-4},\mathfrak{f }_{4,20}$.

• (v) Type $\mathbf G_2$. For ${\mathfrak{u }}_0=\mathfrak{g }_2$, let $\sigma =\exp (\pi H^{\prime }_1)$, which represents the unique conjugacy class of involutions in $\mathrm{Aut}({\mathfrak{u }}_0)$, corresponds to Riemannian symmetric space of type G and the corresponding real form is $\mathfrak{g }_{2,-2}$.

We remark that, in types $\mathbf E_8,\mathbf F_4,\mathbf G_2$, the automorphism groups of the simple Lie algebras, $\mathrm{Aut}({\mathfrak{e }}_8),\mathrm{Aut}(\mathfrak{f }_4),\mathrm{Aut}(\mathfrak{g }_2)$, are connected and simply connected. In type $\mathbf{{E_6}},\, \mathrm{Aut}({\mathfrak{e }}_6)$ is not connected and $\mathrm{Int}({\mathfrak{e }}_6)$ is not simply connected, the image of the adjoint homomorphism $\pi : \mathrm{E}_6\longrightarrow \mathrm{Aut}({\mathfrak{e }}_6)$ is $\mathrm{Int}({\mathfrak{e }}_6)$ and the kernel of $\pi$ (i.e., $Z(\mathrm{E}_6)$) is of order 3. Since $\mathrm{Int}({\mathfrak{e }}_6)$ has two conjugacy classes of involutions, so $\mathrm{E}_6$ has two conjugacy classes of involutions. Their representatives $\sigma ^{\prime }_1=\exp (\pi i H^{\prime }_2), \, \sigma ^{\prime }_2=\exp (\pi i(H^{\prime }_1+H^{\prime }_6))$. Here $\exp : {\mathfrak{e }}_6\longrightarrow \mathrm{E}_6$ is the exponential map for the Lie group $\mathrm{E}_6$. In type $\mathbf{{E_7}}, \, \mathrm{Aut}({\mathfrak{e }}_7)$ is connected but not simply connected, the adjoint homomorphism $\pi : \mathrm{E}_7\longrightarrow \mathrm{Aut}({\mathfrak{e }}_7)$ is surjective and the kernel of $\pi$ (i.e., $Z(\mathrm{E}_7)$) is of order 2. The preimages of $\sigma _2,\sigma _3\in \mathrm{Aut}({\mathfrak{e }}_7)$ in $\mathrm{E}_7$ are elements of order 4; and the preimages of $\sigma _1$ are two non-conjugate involutions. So $\mathrm{E}_7$ has two conjugacy classes of involutions. Their representatives are $\sigma ^{\prime }_1=\exp (\pi i H^{\prime }_2)$ and $\sigma ^{\prime }_2=\exp (\pi i(H^{\prime }_1+H^{\prime }_6))$. Here $\exp : {\mathfrak{e }}_7\longrightarrow \mathrm{E}_7$ is the exponential map for the Lie group $\mathrm{E}_7$.

There is an ascending sequence

$$\begin{aligned} \mathrm{F}_4\subset \mathrm{E}_6\subset \mathrm{E}_7\subset \mathrm{E}_8, \end{aligned}$$

we observe that under these inclusions, the involutions $\sigma _2$ in $\mathrm{F}_4$ ($\sigma _2^{\prime }$ in $\mathrm{E}_6$, or $\sigma ^{\prime }_2$ in $\mathrm{E}_7$) is mapped to conjugate element of the involution $\sigma ^{\prime }_2$ in $\mathrm{E}_6$ ($\sigma _2^{\prime }$ in $\mathrm{E}_7$, or $\sigma _2$ in $\mathrm{E}_8$). The conjugacy class containing $\sigma _2$ (or $\sigma ^{\prime }_2$) in each type is particularly important to us as we will use them to define the translation subgroup $A_{F}$ for an elementary abelian 2-subgroup $F$.

The following Table 2 is from Tables 1 and 2 in [8], which describes the isomorphism type of the symmetric subgroup $\mathrm{Aut}({\mathfrak{u }}_0)^{\theta }$ and the isotropic module $\mathfrak{p }=\mathfrak{g }^{-\theta }$ for each pair $({\mathfrak{u }}_0,\theta )$ with ${\mathfrak{u }}_0$ a compact exceptional simple Lie algebra and $\theta$ an involtution in $\mathrm{Aut}({\mathfrak{u }}_0)$.

Table 2 Symmetric subgroups and isotropic modules

Remark 3.1

We apologize that we use $\sigma _{i}$ to represent the conjugacy classes of involutions in the automorphism groups $\mathrm{Aut}({\mathfrak{u }}_0)$ in all types (as well as use $\sigma ^{\prime }_{i}$ to represent the conjugacy classes of involutions in the connected and simply connected compact Lie groups $\mathrm{E}_6$ and $\mathrm{E}_7$). But this causes no ambiguity as we always specify in which group we are talking about conjugacy classes.

3.3 Klein four subgroups

In [8, Section 4], we constructed some Klein four subgroups of $\mathrm{Aut}({\mathfrak{u }}_0)$ and described the conjugacy classes of involutions in them, it is showed that they represent all conjugacy classes of Klein four subgroups. These Klein fours subgroups, as well as their fixed point subalgebras and their involutions types (cf. Definition 3.3) are listed in Table 3.

Table 3 Klein four subgroups in $\mathrm{Aut}({\mathfrak{u }}_0)$ for exceptional case

From Table 3, we see that, the groups $\mathrm{Aut}({\mathfrak{e }}_6), \, \mathrm{Aut}({\mathfrak{e }}_7), \, \mathrm{Aut}({\mathfrak{e }}_8), \, \mathrm{Aut}(\mathfrak{f }_4), \, \mathrm{Aut}(\mathfrak{g }_2)$ have 8, 8, 4, 3, 1 conjugacy classes of Klein four subgroups in them respectively. Most of these conjugacy classes are distinguished by their involution types (Definition 3.3) except that the Klein four subgroups $\Gamma _1,\Gamma _2$ of $\mathrm{Aut}({\mathfrak{e }}_7)$ have the same involution type [both are $(\sigma _1,\sigma _1,\sigma _1)$]. The Klein four subgroups $\Gamma _1,\Gamma _2\subset \mathrm{Aut}({\mathfrak{e }}_7)$ can be characterized in this way: a Klein four subgroup $F\subset \mathrm{E}_7$ with $\pi (F)=\Gamma _1$ [or $\pi (F)=\Gamma _2$] have an odd number of elements (or an even number of elements) conjugate to $\sigma ^{\prime }_2$, where $\pi : \mathrm{E}_7\longrightarrow \mathrm{Aut}({\mathfrak{e }}_7)$ is the adjoint homomorphism, which is a double covering. That is equivalent to say, we can choose a Klein four subgroup $F\subset \mathrm{E}_7$ with $\pi (F)=\Gamma _1$ (or $\pi (F)=\Gamma _2$) such that all of its involutions are conjugate to $\sigma ^{\prime }_1$ (or $\sigma ^{\prime }_2$).

Given a Klein four subgroup $F\subset G$, we have six different pairs $(\theta ,\sigma )$ generating $F$, but some of them may be conjugate.

Theorem 3.2

[8, Theorem 5.2] Let $(\theta ,\sigma ), \, (\theta ^{\prime },\sigma ^{\prime })$ be two pairs of commuting involutions in $\mathrm{Aut}({\mathfrak{u }}_0)$ for ${\mathfrak{u }}_0$ a compact exceptional simple Lie algebra, then they are conjugate if and only if

$$\begin{aligned} \theta \sim \theta ^{\prime }, \sigma \sim \sigma ^{\prime }, \theta \sigma \sim \theta ^{\prime }\sigma ^{\prime } \end{aligned}$$

and the Klein four subgroups $\langle \theta ,\sigma \rangle , \langle \theta ^{\prime },\sigma ^{\prime }\rangle$ are conjugate.

We remark that, $\mathrm{Aut}({\mathfrak{e }}_7)$ has two non-conjugate Klein four subgroups with involutions all conjugate to $\sigma _1$, so the condition of “the Klein four subgroups $\langle \theta ,\sigma \rangle , \, \langle \theta ^{\prime },\sigma ^{\prime }\rangle$ are conjugate” is necessary. By Theorem 3.2, Table 3 also classifies conjugacy classes of ordered pairs of commuting involutions in $\mathrm{Aut}({\mathfrak{u }}_0)$. Which is another approach to Berger ’s classification of semisimple symmetric pairs (cf. [3]).

3.4 An outline of the method of the classification

In type $\mathbf G_2$, it turns out the conjugacy class of an elementary abelian 2-subgroup of $\mathrm{Aut}(\mathfrak{g }_2)$ is determined by its rank and the rank is at most 3. So we have four conjugacy classes of elementary abelian 2-subgroups in total.

In type $\mathbf F_4$, by Table 3, we see that $\mathrm{Aut}(\mathfrak{f }_4)$ does not possess any Klein four subgroup with involution type $(\sigma _1,\sigma _2,\sigma _2)$. That implies, the subset consisting of the identity element and all elements conjugate to $\sigma _2$in an elementary abelian 2-subgroup $F$ of $\mathrm{Aut}(\mathfrak{f }_4)$ is a subgroup of $F$. Let $A_{F}$ be this subgroup. Then $r=\mathrm{rank}A_{F}$ and $s=\mathrm{rank}F/A_{F}$ are conjugate invariant. We show that the conjugacy class of $F$ is determined by the pair $(r,s)$ and the range of the pairs is $\{(r,s): r\le 2,s\le 3\}$. So we have twelve conjugacy classes of elementary abelian 2-subgroups in total.

In type $\mathbf E_6$, we divide the elementary abelian 2-subgroups $F$ of $\mathrm{Aut}({\mathfrak{e }}_6)$ into four disjoint and exhausting classes:  

• Class 1, $F$ contains an involution conjugate to $\sigma _3$;

• Class 2, $F$ doesn’t contain any element conjugate to $\sigma _3$, but contains one conjugate to $\sigma _4$;

• Class 3, $F\subset \mathrm{Int}({\mathfrak{e }}_6)$ and it contains no Klein four subgroups conjugate to $\Gamma _3$;

• Class 4, $F\subset \mathrm{Int}({\mathfrak{e }}_6)$ and it contains a Klein four subgroup conjugate to $\Gamma _3$.

As $\mathrm{Int}({\mathfrak{e }}_6)^{\sigma _3}\cong \mathrm{F}_4$ and $\mathrm{Int}({\mathfrak{e }}_6)^{\sigma _4}\cong \mathrm{Sp}(4)/\langle (-I,-1)\rangle$, the classification for subgroups in Class 1 reduces to the classification in $\mathbf F_4$ case; the classification for subgroups in Class 2 reduces to the classification of subgroups of $\mathrm{Sp}(4)/\langle (-I,-1)\rangle$, but only those subgroups with any involution conjugate to $\mathbf{i}I$ or $\mathrm{diag}\{-I_2,I_2\}$ are concerned (cf. Sect. 6 for the reason). Our representatives of conjugacy classes in Class 1 are denoted as $\{F_{r,s}: r\le 2,s\le 3\}$ and representatives of conjugacy classes in Class 1 are denoted as $\{F_{\epsilon ,\delta ;r,s}:\epsilon +\delta \le 1, r+s\le 2\}$. Two important observations are: any subgroup in Class 3 is of the form $F\cap \mathrm{Int}({\mathfrak{e }}_6)$ for a subgroup $F$ in Class 1; and any subgroup in Class 4 is of the form $F\cap \mathrm{Int}({\mathfrak{e }}_6)$ for a subgroup $F$ in Class 2 satisfying some additional condition. Our representatives of conjugacy classes in Class 3 are denoted as $\{F^{\prime }_{r,s}: r\le 2,s\le 3\}$ and representatives of conjugacy classes in Class 1 are denoted as $\{F^{\prime }_{\epsilon ,\delta ;r,s}:\epsilon +\delta \le 1, r+s\le 2, s\ge 1\}$. In total, we have $3\times 4+3\times 6+3\times 4+3\times 3=51$ conjugacy classes of elementary abelian 2-subgroups.

In type $\mathbf E_7$, we divide the elementary abelian 2-subgroups $F$ of $\mathrm{Aut}({\mathfrak{e }}_7)$ into three disjoint and exhausting classes:  

• Class 1, $F$ contains an involution conjugate to $\sigma _2$;

• Class 2, $F$ doesn’t contain any element conjugate to $\sigma _2$, but contains one conjugate to $\sigma _3$;

• Class 3, any involution in $F$ is conjugate to $\sigma _1$.

From Table 2, we have that

$$\begin{aligned} \mathrm{Aut}({\mathfrak{e }}_7)^{\sigma _2}\cong ((\mathrm{E}_6\times \mathrm{U}(1))/\langle (c,e^{\frac{2\pi i}{3}})\rangle \rtimes \langle \omega \rangle , \end{aligned}$$

where $1\ne c\in Z_{\mathrm{E}_6}, \, \omega ^{2}=1, \, ({\mathfrak{e }}_6\oplus i\mathbb{R })^{\omega }=\mathfrak{f }_4\oplus 0, \, \sigma _2=(1,-1)$. Modulo $\mathrm{U}(1)$, we have a homomorphism $\pi : \mathrm{Aut}({\mathfrak{e }}_7)^{\sigma _2}\longrightarrow \mathrm{Aut}({\mathfrak{e }}_6)$. It turns out there is a bijection between conjugacy classes of elementary abelian 2-subgroups of $\mathrm{Aut}({\mathfrak{e }}_7)$ in Class 1 and elementary abelian 2-subgroups of $\mathrm{Aut}({\mathfrak{e }}_6)$. So we have fifty-one conjugacy classes in Class 1.

From Table 2, we have that

$$\begin{aligned} \mathrm{Aut}({\mathfrak{e }}_7)^{\sigma _3}\cong (\mathrm{SU}(8)/\langle iI\rangle )\rtimes \langle \omega _0\rangle , \end{aligned}$$

where $\omega _0^{2}=1, \, \omega _0 X\omega _0^{-1}=\overline{X}$ for any $X\in \mathrm{SU}(8), \, \sigma _3=\frac{1+i}{\sqrt{2}}I$. So we have a homomorphism

$$\begin{aligned} \pi : \mathrm{Aut}({\mathfrak{e }}_7)^{\sigma _3}\longrightarrow \mathrm{Aut}(\mathfrak su (8)) =(\mathrm{U}(8)/\mathrm{Z}_{8})\rtimes \langle \omega _0\rangle . \end{aligned}$$

There is a bijection between conjugacy classes of elementary abelian 2-subgroups of $\mathrm{Aut}({\mathfrak{e }}_7)$ in Class 2 and elementary abelian 2 subgroups of $\mathrm{Aut}(\mathfrak su (8))$ whose inner involutions are all conjugate to $I_{4,4}=\mathrm{diag}\{I_4,-I_4\}$ and outer involutions all conjugate to $\omega _0$. These subgroups are classified by Propositions 2.4, 2.12 and Lemma 2.23. We get fourteen conjugacy classes in Class 2.

For an elementary abelian 2-subgroup $F$ of $\mathrm{Aut}({\mathfrak{e }}_7)$ in Class 3, we show either $F$ is toral or it contains a rank 3 subgroup whose Klein four subgroups are all conjugate to $\Gamma _1$. In the first case, we can find an involution $\theta \in \mathrm{Aut}({\mathfrak{e }}_7)^{F}$ such that elements in $\theta F$ are all conjugate to $\sigma _3$. In the second case, we can find a Klein four subgroup $F^{\prime }\subset \mathrm{Aut}({\mathfrak{e }}_7)^{F}$ conjugate to $\Gamma _6$. Then $F$ is a canonical subgroup of some well-chosen subgroup in Class 2 or Class 1. We get thirteen conjugacy classes in Class 3. In total, we have $51+14+13=78$ conjugacy classes of elementary abelian 2-subgroups.

In type $\mathbf{{E_8}}, \, \mathrm{Aut}({\mathfrak{e }}_8)=\mathrm{E}_8$ has two conjugacy classes of involutions with representatives $\sigma _1,\sigma _2$. A nice observation is: for an elementary abelian 2-subgroup $F$ of $\mathrm{Aut}({\mathfrak{e }}_8)$ and any element $x$ of $F$ conjugate to $\sigma _1$, the subset

$$\begin{aligned} H_{x}=\{y\in F|xy\not \sim y\} \end{aligned}$$

is a subgroup. We define $H_{F}$ as the subgroup generated by elements of $F$ conjugate to $\sigma _1$ and define

$$\begin{aligned} A_{F}=\{1\}\cup \{x\in F|x\sim \sigma _2, \mathrm and \forall y\in F-\{1,x\}, xy\sim y\}. \end{aligned}$$

Then $A_{F}\subset H_{F}$ if $H_{F}\ne 1$.

By [8, Table 6], we have that

$$\begin{aligned} \mathrm{Aut}({\mathfrak{e }}_8)^{\Gamma _1}\cong ((\mathrm{E}_6\times \mathrm{U}(1)\times \mathrm{U}(1))/ \langle (c,e^{\frac{2\pi i}{3}},1)\rangle \rtimes \langle \omega \rangle , \end{aligned}$$

where $1\ne c\in Z_{\mathrm{E}_6}, \, \omega ^{2}=1, \, ({\mathfrak{e }}_6\oplus i\mathbb{R }\oplus i\mathbb{R })^{\omega }=\mathfrak{f }_4\oplus 0\oplus 0, \, \Gamma _1=\langle (1,1,-1),(1,-1,1)\rangle$. Modulo $\mathrm{U}(1)\times \mathrm{U}(1)$, we have a homomorphism $\pi : \mathrm{Aut}({\mathfrak{e }}_8)^{\sigma _2}\longrightarrow \mathrm{Aut}({\mathfrak{e }}_6)$. It turns out, $\pi$ does not give a bijection between conjugacy classes of elementary abelian 2-subgroups of $\mathrm{Aut}({\mathfrak{e }}_8)$ containing a Klein four subgroup conjugate to $\Gamma _1$ and elementary abelian 2-subgroups of $\mathrm{Aut}({\mathfrak{e }}_6)$ and we find an explicit relation between these two kinds of conjugacy class and so get a classification of elementary abelian 2-subgroups of $\mathrm{Aut}({\mathfrak{e }}_8)$ containing a Klein four subgroup conjugate to $\Gamma _1$. We have 48 conjugacy classes of elementary abelian 2-subgroups of $\mathrm{Aut}({\mathfrak{e }}_8)$ containing a Klein four subgroup conjugate to $\Gamma _1$

When $F$ doesn’t contain any Klein four subgroup conjugate to $\Gamma _1$ and $H_{F}\ne 1$, we show that $\mathrm{rank}(H_{F}/A_{F})=1, \, \mathrm{rank}A_{F}\le 3$ and $\mathrm{rank}(F/H_{F})\le 2$. Moreover, the conjugacy class of $F$ is determined by the numbers $\mathrm{rank}A_{F}$ and $\mathrm{rank}(F/H_{F})$. We have 12 conjugacy classes of elementary abelian 2-subgroups of $\mathrm{Aut}({\mathfrak{e }}_8)$ of this type.

When $H_{F}=1$, we have $\mathrm{rank}F\le 5$ and the conjugacy class of $F$ is determined by $\mathrm{rank}F$. So we have 6 conjugacy classes of elementary abelian 2-subgroups of $\mathrm{Aut}({\mathfrak{e }}_8)$ of this type. In total, we have $48+12+6=66$ conjugacy classes of elementary abelian 2-subgroups.

3.5 Some notions

Definition 3.3

(Involution type) For an elementary abelian 2 subgroup $F$ of a compact Lie group $G$, we call the distribution of conjugacy classes of involutions in $F$ the involution type of $F$.

Definition 3.4

(Automizer group) For an elementary abelian 2 subgroup $F$ of a compact Lie group $G$, we call $W(F)=N_{G}(F)/C_{G}(F)$ the automizer group of $F$.

$W(F)$ is also called Weyl group in Literature, e.g, [2]. The name of automizer is suggested by Professor R. Griess. We determine the automizer group $W(F)$ for each elementary abelian 2-subgroup $F$ of $\mathrm{Aut}({\mathfrak{u }}_0)$ with ${\mathfrak{u }}_0$ a compact exceptional simple Lie algebra. Conjugation action gives us an inclusion

$$\begin{aligned} W(F)\subset \mathrm{Aut}(F)=\mathrm{GL}(\mathrm{rank}F,\mathbb F _2). \end{aligned}$$

Then we need to determine which automorphisms of $F$ can be realized as $\mathrm{Ad}(g)$ for some $g\in G$.

We also introduce other notions like translation subgroup, defect index, residual rank in the following sections. As the definitions of these notions depend on the types of the Lie algebras (or Lie groups), we give the precise definitions in each section below. These notions help us to show the subgroups we constructed in different classes or in the same class but with different parameters are non-conjugate to each other.

4 $\mathrm{G}_2$

For $G=\mathrm{Aut}(\mathfrak{g }_2)$, by Table 2 we know $G$ has a unique conjugacy class of involution and we have $G^{\sigma }\cong \mathrm{Sp}(1)\times \mathrm{Sp}(1)/\langle (-1,-1)\rangle$ for any involution $\sigma \in G$.

Proposition 4.1

The conjugacy class of an elementary abelian 2-subgroup $F$ of $G$ is determined by $\mathrm{rank}F$ and the possible values of $\mathrm{rank}F$ are $\{0,1,2,3\}$.

Proof

We first prove that, for any $r\le 3$, there exists a unique conjugacy class of ordered tuples $\{x_1,\ldots ,x_{r}\}$ such that they generate an elementary abelian 2-subgroup of $G$ with rank $r$. When $r=1$, this follows from the classification of involutions in $G$, moreover we have

$$\begin{aligned} G^{x_1}\cong \mathrm{Sp}(1)\times \mathrm{Sp}(1)/\langle (-1,-1)\rangle \end{aligned}$$

for any involution $x_1\in G$. Let $x_2\in G^{x_1}$ be an involution different from $x_1$. Then $x_2\sim _{G^{x_1}} [(\mathbf {i},\mathbf {i})]$. This proves the statement when $r=2$. Moreover we have (when $x_2=[(\mathbf {i},\mathbf {i})]$, we can take $t=[(\mathbf {j},\mathbf {j})]$ below)

$$\begin{aligned} G^{x_1,x_2}\cong ((\mathrm{U}(1)\times \mathrm{U}(1))/\langle (-1,-1)\rangle )\rtimes \langle t \rangle , \end{aligned}$$

where $t^{2}=1$ and $t(z_1,z_2)t^{-1}=(z_1^{-1},z_2^{-1}), \, \forall z_1,z_2 \in \mathrm{U}(1)$. Let $x_3\in G^{x_1,x_2}$ be an involution not in $\langle x_1,x_2\rangle$. Then $x_3\sim _{G^{x_1,x_2}} t$. This proves the statement when $r=3$.

Moreover, we have $G^{x_1,x_2,x_3}=\langle x_1,x_2,x_3\rangle$, so $\langle x_1,x_2,x_3\rangle$ is not properly contained in any abelian subgroup of $G$. So an elementary abelian 2-subgroup $F$ of $G$ has rank at most 3. Then the proposition is proved.$\square$

Corollary 4.2

$G$ has 4 conjugacy classes of an elementary abelian 2-subgroup.

Proposition 4.3

For $0\le r\le 3$, for any elementary abelian 2-subgroup $F_r$ of $G=\mathrm{G}_2$ with $\mathrm{rank}F_{r}=r$, we have $W(F_{r})\cong \mathrm{GL}(r,\mathbb F _2)$.

Proof

This follows from the following statement: for any $r\le 3$, there exists a unique conjugacy class of ordered tuples $\{x_1,\ldots ,x_{r}\}$ such that they generate an elementary abelian 2-subgroup of $G$ with rank $r$. This is proved during the above proof for Proposition 4.1.$\square$

5 $\mathrm{F}_4$

Let $G=\mathrm{Aut}(\mathfrak{f }_4)$. From Table 2, we see that $G$ has two conjugacy classes of involutions with representatives $\sigma _1,\sigma _2$ such that

$$\begin{aligned} G^{\sigma _1}\cong \mathrm{Sp}(3)\times \mathrm{Sp}(1)/\langle (-I,-1)\rangle \end{aligned}$$

and

$$\begin{aligned} G^{\sigma _2} \cong \mathrm{Spin}(9). \end{aligned}$$

From [8, Page 18], we see that $G^{\sigma _1}$ has three conjugacy classes of involutions except $\sigma _1=(-I,1)=(I,-1)$ with representatives $(\mathbf{i} I,\mathbf{i}), \, \left(\left(\begin{array}{ccc}-1&\,&\\&1&\\&\,&1 \\ \end{array}\right),1 \right), \, \left(\left(\begin{array}{ccc}-1&\,&\\&-1&\\&\,&1 \\ \end{array}\right), 1\right)$. Moreover in $G$, we have the conjugacy relations

$$\begin{aligned}&(\mathbf{i}I,\mathbf{i}\big )\sim \sigma _1, \\&\left(\left(\begin{array}{ccc}-1&\,&\\&1&\\&\,&1 \\ \end{array}\right),1 \right)\sim \sigma _1, \\&\left(\left(\begin{array}{ccc}-1&\,&\\&-1&\\&\,&1\\ \end{array}\right), 1\right)\sim \sigma _2. \end{aligned}$$

And $G^{\sigma _2}$ has two conjugacy classes of involutions except $\sigma _2=-1$ with representatives $e_1e_2e_3e_4, \, e_1e_2e_3e_4e_5e_6e_7e_8$. And in $G$, we have the conjugacy relations

$$\begin{aligned} e_1e_2e_3e_4\sim \sigma _1 \end{aligned}$$

and

$$\begin{aligned} e_1e_2 \ldots e_8\sim \sigma _2. \end{aligned}$$

In $G^{\sigma _1}=\mathrm{Sp}(3)\times \mathrm{Sp}(1)/\langle (-I,-1)\rangle$, let $x_1=(I,-1), \, x_2= (\mathbf{i}I,\mathbf{i}), \, x_3=(\mathbf{j}I,\mathbf{j})$,

$$\begin{aligned} x_4=\left(\begin{array}{ccc} -1&\,&\\&-1&\\&\,&1 \\ \end{array}\right),\quad x_5=\left(\begin{array}{ccc} -1&\,&\\&1&\\&\,&-1\\ \end{array} \right). \end{aligned}$$

For $0\le r\le 2$ and $0\le s\le 3$, define

$$\begin{aligned} F_{r,s}=\langle x_{1},\ldots ,x_{s},x_{4}, \ldots ,x_{3+r}\rangle \end{aligned}$$

and $A_{r}=\langle x_{4},\ldots ,x_{3+r}\rangle$.

Definition 5.1

For an elementary abelian 2-subgroup $F\subset G$, define

$$\begin{aligned} A_{F}=\{x\in F: x\sim \sigma _2\}\cup \{1\}. \end{aligned}$$

Proposition 5.2

For an elementary abelian 2-subgroup $F$ of $G, \, A_{F}$ is a subgroup of $F$ and we have $\mathrm{rank}{A_{F}}\le 2$ and $\mathrm{rank}(F/A_{F})\le 3$.

For each $(r,s)$ with $0\le r\le 2$ and $0\le s\le 3$, there exists a unique conjugacy class of elementary abelian 2-subgroups $F$ of $G$ such that $\mathrm{rank}{A_{F}}=r$ and $\mathrm{rank}(F/A_{F})=s$.

Proof

Let $F\subset G$ be an elementary abelian $2$-subgroup. By Table Table 3, we see that there are no Klein four subgroups of $G$ with involutions type $(\sigma _1,\sigma _2,\sigma _2)$. Then for any distinct non-identity elements $x,y\in F$ with $x\sim y\sim \sigma _2$, we have $xy\sim \sigma _2$. So $A_{F}$ is a subgroup.

In $G^{\sigma _2}=\mathrm{Spin}(9)$, besides $\sigma _2=-1$, the elements conjugate to $\sigma _2$ in $G$ are all conjugate to $e_1e_2\ldots e_8$ in $G^{\sigma _2}$. There does not exist $x,y\in \mathrm{Spin}(9)$ with $x,y,xy$ all conjugate to $e_1e_2\ldots e_8$, so $\mathrm{rank}A_{F}\le 2$.

In $G^{\sigma _1}=\mathrm{Sp}(3)\times \mathrm{Sp}(1)/\langle (-I,-1)\rangle$, the elements $x$ with $x,\sigma _1 x=(-I,1)x$ both conjugate to $\sigma _1$ in $G$ are all conjugate to $[(\mathbf{i} I,\mathbf{i})]$ in $G^{\sigma _1}$. By this, it is clear that any elementary abelian 2-subgroup of $G$ whose non-identity elements all conjugate to $\sigma _1$ has rank at most 3 ($\langle \sigma _1,[(\mathbf{i} I,\mathbf{i})],[(\mathbf{j} I,\mathbf{j})]\rangle$ is an example of rank 3). Since non-identity elements of a complement of $A_{F}$ in $F$ are all conjugate to $\sigma _1$, so $\mathrm{rank}F/A_{F}\le 3$.

In $F=F_{r,s}$, we have $A_{F}=A_{r}$ is of rank $r$ and $F/A_{F}=F_{r,s}/A_{r}$ is of rank $s$, so $F=F_{r,s}$ satisfies $\mathrm{rank}A_{F}=r$ and $\mathrm{rank}F/A_{F}=s$.

When $s=0$, the uniqueness of the conjugacy class is showed in the proof for $r\le 2$ above. When $s=1$, we may and do assume that $\sigma _1\in F$, then

$$\begin{aligned} F\subset G^{\sigma _1}=\mathrm{Sp}(3)\times \mathrm{Sp}(1)/\langle (-I,-1)\rangle . \end{aligned}$$

The elements in $G^{\sigma _1}$ which are conjugate to $\sigma _2$ in $G$ are conjugate to $[(I_{2,1},1)]$ in $G^{\sigma _1}$. Moreover, any pair $(x_1,x_2)$ with $x_1,x_2\in G^{\sigma _1}$ are distinct and both conjugate to $\sigma _2$ in $G$ is conjugate to $([(I_{2,1},1)],[(I_{1,2},1)])$ in $G^{\sigma _1}$. This proves the uniqueness of the conjugacy classes when $s=1$. When $s=2$, we may and do assume that $\sigma _1\in F$ and $[(\mathbf{i} I,\mathbf{i})]\in F\cap G^{\sigma _1}$. Then

$$\begin{aligned} F\subset (G^{\sigma _1})^{[(\mathbf{i} I,\mathbf{i})]}=(\mathrm{U}(3)\times \mathrm{U}(1)/\langle (-I,-1)\rangle ) \rtimes \langle t\rangle , \end{aligned}$$

where $t=(\mathbf{j} I,\mathbf{j})$. Similarly as $s=1$ case, we get the uniqueness of the conjugacy classes when $s=2$. When $s=3$, we may and do assume that $\sigma _1\in F$ and $[(\mathbf{i} I,\mathbf{i})],[(\mathbf{j} I,\mathbf{j})]\in F\cap G^{\sigma _1}$. Then

$$\begin{aligned} F\subset (G^{\sigma _1})^{[(\mathbf{i} I,\mathbf{i})],[(\mathbf{j} I,\mathbf{j})]}= (\mathrm{SO}(3)\times \mathrm{SO}(1)\times \langle \sigma _1,[(\mathbf{i} I,\mathbf{i})], [(\mathbf{j} I,\mathbf{j})]\rangle . \end{aligned}$$

We have $\mathrm{SO}(1)=1$ and the elements in $\mathrm{SO}(3)$ which are conjugate to $\sigma _2$ in $G$ are conjugate to $I_{2,1}$ in $\mathrm{SO}(3)$. Moreover any pair $(x_1,x_2)$ with $x_1,x_2\in \mathrm{SO}(3)$ are distinct and both conjugate to $I_{2,1}$ in $\mathrm{SO}(3)$ is conjugate to $(I_{2,1},I_{1,2})$ in $\mathrm{SO}(3)$, so we get the uniqueness of the conjugacy classes when $s=3$. $\square$

Corollary 5.3

We have 12 conjugacy classes of elementary abelian 2-subgroups in $G$.

Proof

Since $3\times 4=12$, by Proposition 5.2, we get that there are 12 conjugacy classes of elementary abelian 2-subgroups in $G$. $\square$

Proposition 5.4

For two elementary abelian 2-subgroups $F,F^{\prime }\subset G$, if $f: F\longrightarrow F^{\prime }$ is an isomorphism such that $f(x)\sim x, \, \forall x\in F$, then there exists $g\in G$ such that $f=\mathrm{Ad}(g)$.

Proof

This is proved in the proof of Proposition 5.2 $\square$

Proposition 5.5

For any $r\le 2, \, s\le 3, \, W(F_{r,s})\cong P(r,s, \mathbb F _2)$, where $P(r,s,\mathbb F _2)$ is the group of $(r,s)$ block wise upper triangular matrices in $GL(r+s,\mathbb F _2)$.

Proof

For $F=F_{r,s}$, we have $A_{F}=A_{r}$ and any $g\in N_{G}(F)$ satisfies $g A_{r} g^{-1}=A_{r}$. By Proposition 5.4, we get $W(F)=N_{G}(F)/C_{G}(F)\cong P(r,s,\mathbb F _2)$. $\square$

6 $\mathrm{E}_6$

Let $G=\mathrm{Aut}({\mathfrak{e }}_6)$. By Table 2, $G$ has four conjugacy classes of involutions, two of them consist of inner automorphisms with representatives $\sigma _1,\sigma _2$ and the other two consist of outer automorphisms with representatives $\sigma _3,\sigma _4$. We have

$$\begin{aligned}&(G_0)^{\sigma _1}\cong \mathrm{SU}(6)\times \mathrm{Sp}(1)/\langle (e^{\frac{2\pi i}{3}}I,1),(-I,-1)\rangle , \\&(G_0)^{\sigma _2} \cong \mathrm{Spin}(10)\times \mathrm{U}(1)/\langle (c,i)\rangle , c=e_1e_2\ldots e_{10}, \\&(G_0)^{\sigma _3}\cong \mathrm{F}_4 \end{aligned}$$

and

$$\begin{aligned} (G_0)^{\sigma _4}\cong \mathrm{Sp}(4)/\langle -I\rangle . \end{aligned}$$

From [8, Page 15], we see that $(G_0)^{\sigma _1}$ has four conjugacy classes of involutions except $\sigma _1$. Their representatives and their conjugacy classes in $G$ are as follows,

$$\begin{aligned}&\left(\left(\begin{array}{cc} -I_{4}&\\&I_{2}\\ \end{array}\right),1\right) \sim \sigma _2,\quad \left(\left(\begin{array}{cc} -I_{2}&\\&I_{4}\\ \end{array}\right),1\right)\sim \sigma _1, \\&\left(\left(\begin{array}{cc} iI_{5}&\\&-i\\ \end{array}\right),i\right)\sim \sigma _2,\quad \left(\left(\begin{array}{cc} iI_{3}&\\&-iI_{3}\\ \end{array}\right), i\right)\sim \sigma _1. \end{aligned}$$

And $(G_0)^{\sigma _2}$ has four conjugacy classes of involutions except $\sigma _2$. Their representatives and their conjugacy classes in $G$ are as follows,

$$\begin{aligned}&\big (e_1e_2e_3e_4,1\big )\sim \sigma _1,\quad \big (e_1e_2\ldots e_8,1\big )\sim \sigma _2, \\&\big (\Pi ,\frac{1+i}{\sqrt{2}}\big )\sim \sigma _2,\quad \big (-\Pi ,\frac{1+i}{\sqrt{2}}\big )\sim \sigma _1, \end{aligned}$$

where

$$\begin{aligned} \Pi =\frac{1+e_1e_2}{\sqrt{2}}\frac{1+e_3e_4}{\sqrt{2}} \ldots \frac{1+e_9e_{10}}{\sqrt{2}}. \end{aligned}$$

Definition 6.1

For an elementary abelian $2$-subgroup $F\subset G$, define

$$\begin{aligned} \mu : F\cap G_0\longrightarrow \{\pm {1}\} \end{aligned}$$

by $\mu (y)=-1$ if $y\sim \sigma _1$; and $\mu (y)=1$ if $y\sim \sigma _2$.

And define

$$\begin{aligned} m: (F\cap G_0)\times (F\cap G_0)\longrightarrow \{\pm {1}\} \end{aligned}$$

by $m(y_1,y_2)=\mu (y_1y_2)\mu (y_1)\mu (y_2)$.

Here $m$ is not always a bilinear form.

Definition 6.2

For an elementary abelian $2$-subgroup $F\subset G$, define the translation subgroup

$$\begin{aligned} A_{F}= \{x\in H\cap G_0: \mu (x)=1\mathrm and m(x,y)=1,\forall y\in F\cap G_0\} \end{aligned}$$

and define the defect index

$$\begin{aligned} \mathrm{defe}(F)=|\{y\in F\cap G_0: \mu (y)=1\}|-|\{y\in F\cap G_0: \mu (y)=-1\}|. \end{aligned}$$

The subgroup $A_{F}$ the has an equivalent definition as

$$\begin{aligned} A_{F}=\{1\}\cup \{x\in F|x\sim \sigma _2, \mathrm and y\sim xy \mathrm for any y\in F-\langle x\rangle , \}, \end{aligned}$$

this is why the name of “translation subgroup” arises.

6.1 Subgroups from $\mathrm{F}_4$

In $(G_0)^{\sigma _3}\cong \mathrm{F}_4$, let $\tau _1,\tau _2$ be involutions such that

$$\begin{aligned} \mathfrak{f }_4^{\tau _1} \cong \mathfrak sp (3)\oplus \mathfrak sp (1),\quad \mathfrak{f }_4^{\tau _2}\cong \mathfrak so (9). \end{aligned}$$

From [8, Page 15], we see that $\tau _1,\tau _2,\sigma _3\tau _1,\sigma _3\tau _2$ represent all conjugacy classes of involutions in $G^{\sigma _3}$ except $\sigma _3$ and in we have the conjugacy relations in $G$,

$$\begin{aligned}&\tau _1\sim \sigma _1,\quad \tau _2\sim \sigma _2, \\&\sigma _3\tau _1\sim \sigma _4,\quad \sigma _3\tau _2\sim \sigma _3. \end{aligned}$$

We have $((G_0)^{\sigma _3})^{\tau _1}\cong \mathrm{Sp}(3)\times \mathrm{Sp}(1)/\langle (-I,-1)\rangle$. Let

$$\begin{aligned}&x_0=\sigma _3,\quad x_1=\tau _1=[(I,-1)], \\&x_2=[(\mathbf{i}I,\mathbf{i})],\quad x_3=[(\mathbf{j}I,\mathbf{j})], \\&x_4=\left[\left(\left(\begin{array}{ccc}-1&\,&\\&-1&\\&\,&1\\ \end{array}\right),1\right)\right],\quad x_5=\left[\left(\left(\begin{array}{ccc} -1&\,&\\&1&\\&\,&-1\\ \end{array}\right),1\right)\right]. \end{aligned}$$

For a pair $(r,s)$ with $r\le 2$ and $s\le 3$, define

$$\begin{aligned} F_{r,s}=\langle x_0,x_1,\ldots , x_{s},x_{4},\ldots ,x_{3+r}\rangle \end{aligned}$$

and

$$\begin{aligned} F^{\prime }_{r,s}=\langle x_1,\ldots , x_{s},x_{4}, \ldots ,x_{3+r} \rangle . \end{aligned}$$

Proposition 6.3

For an elementary abelian $2$-subgroup $F\subset G$, if $F$ contains an element conjugate to $\sigma _3$, then $F\sim F_{r,s}$ for some $(r,s)$ with $r\le 2, s\le 3$; if $F\subset G_0$ and it contains no Klein four subgroups conjugate to $\Gamma _3$, then $F\sim F^{\prime }_{r,s}$ for some pair $(r,s)$ with $r\le 2$ and $s\le 3$.

Proof

For the first statement, we may and do assume that $\sigma _3\in F$, then $F\subset G^{\sigma _3}=F_4\times \langle \sigma _3\rangle$. Then $F\sim F_{r,s}$ ($r\le 2, s\le 3$) by Proposition 5.2.

For the latter statement, since we assume that $F$ does not contain any Klein four subgroup conjugate to $\Gamma _3$, so $F$ does not contain any Klein four subgroup of involutions type $(\sigma _1,\sigma _2,\sigma _2)$. Then we have $A_{F}=\{1\}\cup \{x\in F|x\sim \sigma _2\}$. Prove in the same line as the proof for Proposition 5.2, we can show that $\mathrm{rank}A_{F}\le 2, \, \mathrm{rank}(F/A_{F})\le 3$ and the conjugacy class of $F$ is uniquely determined by $\mathrm{rank}A_{F}$ and $\mathrm{rank}(F/A_{F})$. Then we have $F\sim F^{\prime }_{r,s}$ ($r\le 2, s\le 3$) since $\mathrm{rank}A_{F^{\prime }_{r,s}}=r$ and $\mathrm{rank}(F/A_{F^{\prime }_{r,s}})=s$. $\square$

Lemma 6.4

For an elementary abelian 2-subgroup $F$ in Proposition 6.3,

$$\begin{aligned} m(x,y)=-1\Leftrightarrow x,y\in (F\cap G_0)-A_{F},\quad \forall x,y\in F\cap G_0. \end{aligned}$$

Proof

This follows from the equality $A_{F}=\{1\}\cup \{x\in F|x\sim \sigma _2\}$. $\square$

6.2 Subgroups from $\mathrm{Sp}(4)/\langle -I\rangle$

In $(G_0)^{\sigma _4}\cong \mathrm{Sp}(4)/\langle -I\rangle$, let $\tau _1=\mathbf{i}I, \, \tau _2=\left(\begin{array}{cc} -I_{2}&\\&I_{2}\\ \end{array}\right),\, \tau _3=\left(\begin{array}{cc} -1&\\&I_{3}\\ \end{array}\right)$. From [8, Pages 15–16], we see that $\tau _1,\tau _2,\tau _3,\sigma _4\tau _1,\sigma _4\tau _2,\sigma _4\tau _3$ represent all conjugacy classes of involutions in $G^{\sigma _4}$ except $\sigma _4$ and we have the following conjugacy relations in $G$,

$$\begin{aligned} \tau _1\sim \sigma _1,\ \tau _2\sim \sigma _2,\ \tau _3\sim \sigma _1 \end{aligned}$$

and

$$\begin{aligned} \sigma _4\tau _1\sim \sigma _4,\ \sigma _4\tau _2\sim \sigma _4,\ \sigma _4\tau _3\sim \sigma _3. \end{aligned}$$

Let $x_0=\sigma _4, \, x_1=\mathbf{i}I, \, x_2=\mathbf{j}I$,

$$\begin{aligned} x_3=\left(\begin{array}{cc}-I_{2}&\\&I_{2}\\ \end{array}\right), x_4=\left(\begin{array}{cc} 0&I_{2}\\ I_{2}&0\\ \end{array}\right), \end{aligned}$$

$$\begin{aligned} x_5=\left(\begin{array}{cccc} 1&0&\,&\\ 0&-1&\,&\\&\,&1&0\\&\,&0&-1\\ \end{array}\right), x_6=\left(\begin{array}{cccc} 0&1&\,&\\ 1&0&\,&\\&\,&0&1\\&\,&1&0\\ \end{array}\right). \end{aligned}$$

For any $(\epsilon ,\delta ,r,s)$ with $\epsilon +\delta \le 1, r+s\le 2$, define

$$\begin{aligned} F_{\epsilon ,\delta ,r,s}= \langle x_0,x_1,\ldots ,x_{\epsilon +2\delta },x_3,\ldots ,x_{r+2s}\rangle \end{aligned}$$

and

$$\begin{aligned} F^{\prime }_{\epsilon ,\delta ,r,s}=\langle x_1,\ldots ,x_{\epsilon +2\delta }, x_3,\ldots ,x_{r+2s}\rangle . \end{aligned}$$

Proposition 6.5

For an elementary abelian $2$-subgroup $F\subset G$, if $F\not \subset G_0$ and it contains no elements conjugate to $\sigma _3$, then $F\sim F_{\epsilon ,\delta ,r,s}$ for some $(\epsilon ,\delta ,r,s)$ with $\epsilon +\delta \le 1$ and $r+s\le 2$; if $F\subset G_0$ and it contains a Klein four subgroup conjugate to $\Gamma _3$, then $F\sim F^{\prime }_{\epsilon ,\delta ,r,s}$ for some $(\epsilon ,\delta ,r,s)$ with $\epsilon +\delta \le 1, \, r+s\le 2$ and $s\ge 1$.

Proof

For the first statement, we may assume that $\sigma _4\in F$, then

$$\begin{aligned} F\cap G_0\subset G_0^{\sigma _4}\cong \mathrm{Sp}(4)/\langle -I\rangle . \end{aligned}$$

Any involution in $\mathrm{Sp}(4)/\langle -I\rangle$ is conjugate to one of

$$\begin{aligned} \tau _1=[\mathbf{i}I], \tau _2=[\mathrm{diag}\{I_2,-I_2\}], \tau _3=[\mathrm{diag}\{1,-I_{3}\}]. \end{aligned}$$

Since $\sigma _4\tau _3\sim \sigma _3$ in $G$ and we assume that $F$ contains no elements conjugate to $\sigma _3$, so any non-identity element of $F\cap G_0$ is conjugate to $\tau _1$ or $\tau _2$ in $\mathrm{Sp}(4)/\langle -I\rangle$. Then $F\cap G_0\subset \mathrm{Sp}(4)/\langle -I\rangle$ is in the subclass discussed in Sect. 2.4. Then $F\sim F_{\epsilon ,\delta ,r,s}$ for some $(\epsilon ,\delta ,r,s)$ with $\epsilon +\delta \le 1$ and $r+s\le 2$ by Proposition 2.24.

For the second statement, we may and do assume that $\Gamma _3\subset F$, then

$$\begin{aligned} F\subset (G_0)^{\Gamma _3}\cong (\mathrm{U}(5)\times \mathrm{U}(1))/\langle (-I,-1),(e^{\frac{2\pi i}{3}},1)\rangle \cong (\mathrm{U}(5)/\langle e^{\frac{2\pi i}{3}}\rangle )\times \mathrm{U}(1). \end{aligned}$$

Here we use that the map $(A,\lambda )\longmapsto (\lambda A,\lambda ^{2})$ gives an isomorphism

$$\begin{aligned} (\mathrm{U}(5)\times \mathrm{U}(1))/ \langle (-I,-1) \rangle \cong \mathrm{U}(5)\times \mathrm{U}(1). \end{aligned}$$

Since any abelian subgroup of $\mathrm{U}(5)\times \mathrm{U}(1)$ is total, so $F\subset G_0$ is total. We may and do assume that $F\subset \exp (\mathfrak{h }_0)$ for a Cartan subalgebra $\mathfrak{h }_0$ of ${\mathfrak{u }}_0={\mathfrak{e }}_6$. Choose a Chevelley involution $\theta$ of ${\mathfrak{u }}_0$ with respect to $\mathfrak{h }_0$. Then $\theta$ commutes with all elements $x\in \exp (\mathfrak{h }_0)$ satisfying $x^2=1$. Moreover, we have $\theta \sim \sigma _4$ (since $\dim {\mathfrak{u }}_0^{\theta }=63$) and $\theta x\sim \theta$ for any $x\in \exp (\mathfrak{h }_0)$. Then $\langle F,\theta \rangle$ is an elementary abelian $2$-subgroup without elements conjugate to $\sigma _3$. By the first statement, we get that $\langle F,\theta \rangle \sim F_{\epsilon ,\delta ,r,s}$ for some $(\epsilon ,\delta ,r,s)$ with $\epsilon +\delta \le 1$ and $r+s\le 2$. Then $F\sim F^{\prime }_{\epsilon ,\delta ,r,s}$. Since we assume that $F$ contains a Klein four subgroup conjugate to $\Gamma _3$, so we have $s\ge 1$. $\square$

Let $F$ be an elementary abelian $2$-subgroup of $G$ without elements conjugate to $\sigma _3$ and containing an element conjugate to $\sigma _4$. For any $x\in F$ with $x\sim \sigma _4$, we have $F\cap G_0\subset (G_0)^{\sigma _4}\cong \mathrm{Sp}(4)/\langle -I\rangle$. With this inclusion we have a function $\mu _{x}:F\cap G_0\longrightarrow \{\pm {1}\}$ and a map $m_{x}:(F\cap G_0)\times (F\cap G_0)\longrightarrow \{\pm {1}\}$ (cf. Sect. 2.2).

Lemma 6.6

We have $\mu _{x}=\mu$ and $m_{x}=m$.

Proof

We may assume that $x=\sigma _4$, then $F\cap G_0\subset (G_0)^{\sigma _4}\cong \mathrm{Sp}(4)/\langle -I\rangle$. Since $F$ does not have any element conjugate to $\sigma _3$, from the proof for Proposition 6.5 we see that any element of $F\cap G_0$ is conjugate to $\tau _1=\mathbf{i}I$ or $\tau _2=\left(\begin{array}{cc} -I_{2}&\\&I_{2}\\ \end{array}\right)$ in $(G_0)^{\sigma _4}\cong \mathrm{Sp}(4)/\langle -I\rangle$. Since $\tau _1\sim _{G}\sigma _1$ and $\tau _2\sim _{G}\sigma _2$, so we have $\mu _{x}=\mu$. Then we have $m_{x}=m$ as well. $\square$

6.3 Automizer groups

Proposition 6.7

We have the following formulas for $\mathrm{rank}A_{F}$ and $\mathrm{defe}F$,  

• (1) for $F=F_{r,s}, \, r\le 2, \, s\le 3, \, \mathrm{rank}A_{F}=r, \, \mathrm{defe}F=2^{r}(2-2^{s})$;

• (2) for $F=F^{\prime }_{r,s}, \, r\le 2, \, s\le 3, \, \mathrm{rank}A_{F}=r, \, \mathrm{defe}F=2^{r}(2-2^{s})$;

• (3) for $F=F_{\epsilon ,\delta ,r,s}, \, \epsilon +\delta \le 1, \, r+s\le 2, \, \mathrm{rank}A_{F}=r, \, \mathrm{defe}F=(1-\epsilon )(-1)^{\delta }2^{r+s+\delta }$;

• (4) for $F=F^{\prime }_{\epsilon ,\delta ,r,s}, \, \epsilon +\delta \le 1, \, r+s\le 2, \, s\ge 1, \, \mathrm{rank}A_{F}=r, \, \mathrm{defe}F= (1-\epsilon )(-1)^{\delta } 2^{r+s+\delta }$.

Proof

They follow from Lemmas 6.4 and 6.6. $\square$

Corollary 6.8

We have 51 conjugacy classes of elementary abelian 2-subgroups of $\mathrm{Aut}({\mathfrak{e }}_6)$.

Proof

By the formulas of $\mathrm{rank}A_{F}$ and $\mathrm{defe}F$ in Proposition 6.7, we see that the subgroups in each family with different parameters are non-conjugate. And the subgroups in different families are clearly non-conjugate, so these subgroups are non-conjugate to each other. In total, we have $3\times 4+3\times 4+3\times 6+3\times 3=51$ conjugacy classes. $\square$

Proposition 6.9

For two elementary abelian 2-subgroups $F,F^{\prime }\subset G$, if an isomorphism $f: F\longrightarrow F^{\prime }$ has the property that $f(x)\sim x$ for any $x\in F$, then $f=\mathrm{Ad}(g)$ for some $g\in G$.

Proof

We may and do assume that $F=F^{\prime }$ and they are equal to one of

$$\begin{aligned} F_{r,s},F^{\prime }_{r,s}, F_{\epsilon ,\delta ,r,s},F^{\prime }_{\epsilon ,\delta ,r,s}. \end{aligned}$$

When $F=F^{\prime }=F_{r,s}$, we may and do assume that $f(\sigma _3)=\sigma _3$, then $F\cap G_0=F^{\prime }\cap G_0 \subset (G_0)^{\sigma _3}=F_4$. By the proof of Proposition 5.2, we get some $g\in (G_0)^{\sigma _3}$ such that $f=\mathrm{Ad}(g)$.

When $F=F^{\prime }=F_{r,s}$, similar as the proof for Proposition 5.2, we find some $g\in G_0$ such that $f=\mathrm{Ad}(g)$.

When $F=F^{\prime }=F_{\epsilon ,\delta ,r,s}$, we may and do assume that $f(\sigma _4)=\sigma _4$, then $F\cap G_0=F^{\prime }\cap G_0\subset (G_0)^{\sigma _4}=\mathrm{Sp}(4)/\langle -I\rangle$ and non-identity elements of $F\cap G_0=F^{\prime }\cap G_0$ are all conjugate to $\mathbf{i}I$ or $[I_{2,2}]$ in $\mathrm{Sp}(4)/\langle -I\rangle$. Then $f=\mathrm{Ad}(g)$ for some $g\in G_0^{\sigma _4}$ by Proposition 2.24.

When $F=F^{\prime }=F^{\prime }_{\epsilon ,\delta ,r,s}$, since $F^{\prime }_{\epsilon ,\delta ,r,s}\subset (G_0)^{\sigma _4}= \mathrm{Sp}(4)/\langle -I\rangle$ and non-identity elements of $F=F^{\prime }$ are all conjugate to $\mathbf{i}I$ or $[I_{2,2}]$ in $\mathrm{Sp}(4)/\langle -I\rangle$. Then $f=\mathrm{Ad}(g)$ for some $g\in G_0^{\sigma _4}$ by Proposition 2.24. $\square$

Proposition 6.10

We have the following description for the automizer groups,  

• (1) $r\le 2, \, s\le 3, \, W(F_{r,s})\cong (\mathbb{F }_2)^{r}\rtimes P(r,s,\mathbb{F }_2)$;

• (2) $r\le 2, \, s\le 3, \, W(F^{\prime }_{r,s})\cong P(r,s,\mathbb{F }_2)$;

• (3) $\epsilon +\delta \le 1, \, r+s\le 2$,

  $$\begin{aligned} W(\mathbb{F }_{\epsilon ,\delta ,r,s})\cong \mathbb F _2^{r+2s+\epsilon +2\delta }\rtimes \big (\mathrm{Hom}(\mathbb{F }_2^{\epsilon +2\delta +2s},\mathbb{F }_2^{r}) \rtimes (\mathrm{GL}(r,\mathbb{F }_2)\times \mathrm{Sp}(s;\epsilon ,\delta ))\big ); \end{aligned}$$

• (4) $\epsilon +\delta \le 1, \, r+s\le 2, \, s\ge 1$,

  $$\begin{aligned} W(F^{\prime }_{\epsilon ,\delta ,r,s}) \cong \mathrm{Hom}(\mathbb{F }_2^{\epsilon +2\delta +2s},\mathbb{F }_2^{r}) \rtimes (\mathrm{GL}(r,\mathbb{F }_2)\times \mathrm{Sp}(s;\epsilon ,\delta )). \end{aligned}$$

Proof

The action of any $w\in W(F)$ preserves $\mu$ and $m$ on $F\cap G_0$ and the conjugacy classes of elements in $F-(F\cap G_0)$. By Proposition 6.9, an automorphism of $F$ preserves these data is actually the action of some $w\in W(F)$ on $F$. Then by Lemmas 6.4 and 6.6, we get these automizer groups. $\square$

7 $\mathrm{E}_7$

Let $G=\mathrm{Aut}({\mathfrak{e }}_7)$. By Table 2 we see that there are three conjugacy classes of involutions in $G$ with representatives $\sigma _1,\sigma _2,\sigma _3$ and we have

$$\begin{aligned}&G^{\sigma _1}\cong (\mathrm{Spin}(12)\times \mathrm{Sp}(1))/\langle (c,1),(-c,-1)\rangle ,\\&G^{\sigma _2}\cong ((\mathrm{E}_{6}\times \mathrm{U}(1))/\langle (c^{\prime },e^{\frac{2\pi i}{3}})) \rtimes \langle \omega \rangle ,\\&G^{\sigma _3}\cong (\mathrm{SU}(8)/\langle iI\rangle ) \rtimes \langle \omega \rangle , \end{aligned}$$

where $c=e_1e_2\ldots e_{12}, \, 1\ne c^{\prime }\in Z_{E_6}, \, \omega ^{2}=1$, and

$$\begin{aligned} ({\mathfrak{e }}_6\oplus i\mathbb{R })^{\omega }=\mathfrak{f }_4\oplus 0, \mathfrak su (8)^{\omega }\cong \mathfrak sp (4). \end{aligned}$$

Definition 7.1

For an elementary abelian 2-subgroup $F$ of $G$, define

$$\begin{aligned} H_{F}=\{1\}\cup \{x\in F|x\sim \sigma _1\}; \end{aligned}$$

define

$$\begin{aligned} m: H_{F}\times H_{F}\longrightarrow \{\pm {1}\} \end{aligned}$$

by $m(x,y)=-1$ if $\langle x,y\rangle \sim \Gamma _1$, and $m(x,y)=1$ otherwise.

Definition 7.2

Define the translation subgroup

$$\begin{aligned} A_{F}:=\{x\in H_{F}: \forall y\in F-H_{F},y\sim xy;\mathrm and \forall y\in H_{F}, m(x,y)=1\} \end{aligned}$$

and the defect index

$$\begin{aligned} \mathrm{defe}(F)=|\{x\in F: x\sim \sigma _2\}|-|\{x\in F: x\sim \sigma _3\}|. \end{aligned}$$

For any $x\in F$ with $x\sim \sigma _2$, let

$$\begin{aligned} H_{x}=\{y\in H_{F}|xy\sim \sigma _2\}, \end{aligned}$$

which is not always a subgroup.

Lemma 7.3

$H_{F}$ is a subgroup of $F$ and we have $\mathrm{rank}(F/H_{F})\le 2$.

Proof

Since the product of any two distinct elements in $F$ conjugate to $\sigma _1$ is also conjugate to $\sigma _1$, so $H_{F}$ is a subgroup.

Suppose that $\mathrm{rank}(F/H_{F})\ge 3$, then there exists a rank 3 subgroup $F^{\prime }\subset F$ with $H_{F^{\prime }}=1$. For any $1\ne x\in F^{\prime }, \, G^{x}\sim G^{\sigma _2}$ or $G^{\sigma _3}$ has only two connected components, so $\mathrm{rank}(F^{\prime }\cap (G^{x})_0)\ge 2$. Choose $y\in F^{\prime }\cap (G^{x})_0-\langle x\rangle$, then $\langle x,y\rangle$ is a toral Klein four subgroup of $G$. By Table 3, at least one of $x,y,xy$ is conjugate to $\sigma _1$, which contradicts that $H_{F^{\prime }}=1$. $\square$

7.1 Subgroups from $\mathrm{E}_6$

By Table 2, we have that

$$\begin{aligned} G^{\sigma _2}\cong \big ((\mathrm{E}_{6}\times \mathrm{U}(1))/\langle (c,e^{\frac{2\pi i}{3}})\big )\rtimes \langle \omega \rangle , \end{aligned}$$

where $\omega ^2=1$ and $({\mathfrak{e }}_6\oplus i\mathbb{R })^{\omega }=\mathfrak{f }_4\oplus 0$. Let $\tau _1,\tau _2\in \mathrm{E}_6$ be involutions with

$$\begin{aligned} {\mathfrak{e }}_6^{\tau _1}\cong \mathfrak su (6)\oplus \mathfrak sp (1),\ {\mathfrak{e }}_6^{\tau _2}\cong \mathfrak so (10)\oplus i\mathbb{R }. \end{aligned}$$

Let $\eta _1, \eta _2\in \mathrm{E}_6^{\omega }\cong \mathrm{F}_4$ be involutions with

$$\begin{aligned} \mathfrak{f }_4^{\eta _1}\cong \mathfrak sp (3)\oplus \mathfrak sp (1), \mathfrak{f }_4^{\eta _2}\cong \mathfrak so (9). \end{aligned}$$

Let $\tau _3=\omega , \, \tau _4=\eta _1\omega$. From [8, Page 16], we see that $\tau _1, \, \tau _2, \, \sigma _2\tau _1, \, \sigma _2\tau _2, \, \tau _3, \, \tau _4$ represent all conjugacy classes of involutions in $G^{\sigma _2}$ except $\sigma _2$ and we have the following conjugacy relations in $G$,

$$\begin{aligned}&\tau _1 \sim \tau _2\sim \sigma _1, \\&\sigma _2\tau _1\sim \sigma _3, \sigma _2\tau _2\sim \sigma _2, \\&\tau _3\sim \sigma _2\tau _3\sim \sigma _2, \\&\tau _4\sim \sigma _2\tau _4\sim \sigma _3. \end{aligned}$$

Lemma 7.4

In $G^{\sigma _2}$, we have the conjugacy relations

$$\begin{aligned} \eta _1\sim _{\mathrm{E}_6}\tau _1,\ \eta _2\sim _{\mathrm{E}_6}\tau _2,\ \eta _2\omega \sim _{\mathrm{E}_6}\omega . \end{aligned}$$

Proof

This follows from [8, Page 15] (for $\mathrm{Aut}({\mathfrak{e }}_6)^{\sigma _3}$), the elements $\tau _1, \, \tau _2, \, \omega , \, \eta _1\omega , \, \eta _1, \, \eta _2$ correspond to the elements $\sigma _1, \, \sigma _2, \, \sigma _3, \, \sigma _4, \, \tau _1, \, \tau _2$ there. $\square$

Let $L_1,L_2,L_3,L_4$ be Klein four subgroups of $\mathrm{E}_6$ of involution types $(\tau _1,\tau _1,\tau _1), (\tau _1,\tau _1,\tau _2), (\tau _1,\tau _2,\tau _2), \, (\tau _2,\tau _2,\tau _2)$ respectively. By Table 3, we have that

$$\begin{aligned}&({\mathfrak{e }}_7^{\sigma _2})^{L_1}\cong \mathfrak su (3)^{2}\oplus (i\mathbb{R })^{3}, \\&({\mathfrak{e }}_7^{\sigma _2})^{L_2}\cong \mathfrak su (4)\oplus \mathfrak su (2)^{2}\oplus (i\mathbb{R })^{2},\\&({\mathfrak{e }}_7^{\sigma _2})^{L_3}\cong \mathfrak su (5)\oplus (i\mathbb{R })^{3}, \\&({\mathfrak{e }}_7^{\sigma _2})^{L_4}\cong \mathfrak so (8)\oplus (i\mathbb{R })^{3}. \end{aligned}$$

Lemma 7.5

In $G$, we have $L_1\sim L_3\sim \Gamma _1$ and $L_2\sim L_4\sim \Gamma _2$.

Proof

First since $\tau _1\sim \tau _2\sim \sigma _1$ in $G$, so each of $L_1,L_2,L_3,L_4$ is conjugate to $\Gamma _1$ or $\Gamma _2$. Since $\mathfrak su (3)^{2}\oplus (i\mathbb{R })^{3}, \mathfrak su (5)\oplus (i\mathbb{R })^{3}$ are not symmetric subalgebras of ${\mathfrak{e }}_7^{\Gamma _2}\cong \mathfrak so (8)\oplus \mathfrak su (2)^{3}$ and $\mathfrak su (4)\oplus \mathfrak su (2)^{2}\oplus (i\mathbb{R })^{2},\mathfrak so (8)\oplus (i\mathbb{R })^{3}$ are not symmetric subalgebras of ${\mathfrak{e }}_7^{\Gamma _1}\cong \mathfrak su (6)\oplus (i\mathbb{R })^{2}$, so $L_1,L_3$ are conjugate to $\Gamma _1$ and $L_2,L_4$ are conjugate to $\Gamma _2$. $\square$

Let $F\subset G$ be an elementary abelian 2-subgroup containing an element conjugate to $\sigma _2$, we may an do assume that $\sigma _2\in F$, then

$$\begin{aligned} F\subset G^{\sigma _2}\cong ((\mathrm{E}_{6}\times \mathrm{U}(1))/\langle (c,e^{\frac{2\pi i}{3}})) \rtimes \langle \omega \rangle , \end{aligned}$$

where $c$ is a non-trivial central element of $\mathrm{E}_6, \, c^{3}=1, \, \omega ^{2}=1$ and $({\mathfrak{e }}_6\oplus i\mathbb{R })^{\omega }=\mathfrak{f }_4\oplus 0$. Let $G_{\sigma _2}=(\mathrm{E}_6\times 1)\ltimes \langle \omega \rangle$ be the subgroup generated by $\mathrm{E}_6$ ($=\mathrm{E}_6\times 1$) and $\omega$. This definition of $G_{\sigma _2}$ is not quite canonical, another choice is to define it as $(\mathrm{E}_6\times 1)\ltimes \langle \sigma _2\omega \rangle$, but these are conjugate since

$$\begin{aligned} (1,i)\omega (1,i)^{-1}&= \omega (\omega ^{-1}(1,i)\omega )(1,i)^{-1}\\&= \omega (1,-i)(1,-i)=\omega (1,-1)MYAMP]=&\omega \sigma _2. \end{aligned}$$

And so they are equivalent,

Lemma 7.6

For an elementary abelian 2-subgroup $F\subset G$ containing $\sigma _2$, in the inclusion $F\subset G^{\sigma _2}\cong ((\mathrm{E}_{6}\times \mathrm{U}(1)) /\langle (c,e^{\frac{2\pi i}{3}}))\rtimes \langle \omega \rangle$, we have $H_{F}=F\cap \mathrm{E}_6$.

Moreover, the map $m: H_{F}\times H_{F}\longrightarrow \{\pm {1}\}$ is equal to the the similar map when $H_{F}$ is viewed as a subgroup of $\mathrm{E}_6$ (or $\mathrm{E}_6/\langle c\rangle =\mathrm{Int}({\mathfrak{e }}_6)$).

Proof

$H_{F}=F\cap \mathrm{E}_6$ follows from the comparison of conjugacy classes of involutions in $G^{\sigma _2}$ and in $G$. The two maps $m$ are equal follows from Lemma 7.5. $\square$

Let $\pi : G_{\sigma _2}\longrightarrow \mathrm{Aut}({\mathfrak{e }}_6)$ be the adjoint homomorphism and $p: G_{\sigma _2}\longrightarrow G^{\sigma _2}$ be the inclusion map. For any elementary abelian 2-subgroup $K$ of $\mathrm{Aut}({\mathfrak{e }}_6), \, p(\pi ^{-1}K)\times \langle \sigma _2\rangle$ is the direct product of its unique Sylow 2-subgroup $F$ and $\langle (c,1)\rangle$. Let

$$\begin{aligned}&\{F_{r,s}:r\le 2,s\le 3\},\\&\{F^{\prime }_{r,s}:r\le 2,s\le 3\},\\&\{F_{\epsilon ,\delta ,r,s}:\epsilon +\delta \le 1, r+s\le 2\},\\&\{F^{\prime }_{\epsilon ,\delta ,r,s}:\epsilon +\delta \le 1, r+s\le 2, s\ge 1\} \end{aligned}$$

be elementary abelian 2-subgroups of $G^{\sigma _2}\subset G$ obtained from elementary abelian 2-subgroups of $\mathrm{Aut}(\mathfrak e _6)$ with the corresponding notation in this way.

Proposition 7.7

Any elementary abelian 2-subgroup of $G$ with an element conjugate to $\sigma _2$ is conjugate to one of $F_{r,s}, \, F^{\prime }_{r,s}, \, F_{\epsilon ,\delta ,r,s}, \, F^{\prime }_{\epsilon ,\delta ,r,s}$.

Proof

We may and do assume that $\sigma _2\in F$, then

$$\begin{aligned} F\subset G^{\sigma _2}\cong ((\mathrm{E}_{6}\times U(1))/ \langle (c,e^{\frac{2\pi i}{3}}))\rtimes \langle \omega \rangle . \end{aligned}$$

By Lemma 7.3, we have $\mathrm{rank}(F/H_{F})\le 2$. When $\mathrm{rank}(F/H_{F})=1$, we have $F\subset \mathrm{E}_6\times \langle \sigma _2\rangle$. When $\mathrm{rank}(F/H_{F})=2$, we may an do assume that $\omega \in F$ or $\tau _4=\eta _1\omega \in F$, then $F\subset (\mathrm{E}_6\rtimes \langle \omega \rangle )\times \langle \sigma _2\rangle$. Then the conclusion follows from Propositions 6.3 and 6.5. $\square$

Proposition 7.8

The four families have the following characterization, so subgroups in different families are not conjugate to each other.  

• (1) $F$ is conjugate to some $F_{r,s}$ if and only if $F$ contains a subgroup conjugate to $\Gamma _6$;

• (2) $F$ is conjugate to some $F^{\prime }_{r,s}$ if and only if $\mathrm{rank}(F/H_{F})=1, \, F$ contains an element $x$ conjugate to $\sigma _2$ and $H_{x}$ is a subgroup;

• (3) $F$ is conjugate to some $F_{\epsilon ,\delta ,r,s}$ if and only if $\mathrm{rank}(F/H_{F})=2, \, F$ contains an element conjugate to $\sigma _2$ but contains no subgroups conjugate to $\Gamma _6$;

• (4) $F$ is conjugate to some $F^{\prime }_{\epsilon ,\delta ,r,s}$ if and only if $\mathrm{rank}(F/H_{F})=1, \, F$ contains an element conjugate to $\sigma _2$ and $H_{x}$ is not a subgroup.

Proof

(1) and (3) are clear. (2) and (4) follow from the comparison of conjugacy classes of involutions in $G^{\sigma _2}$ and in $G$ and the classification of elementary abelian 2-subgroups of $\mathrm{Int}({\mathfrak{e }}_6)$ (by Propositions 6.3 and 6.5). $\square$

We make a remark that, for a subgroup $F$ in case (2) or (4), if $H_{x}$ for some $x\in F$ with $x\sim \sigma _2$ is an subgroup, then the $H_{x^{\prime }}$ for any other $x^{\prime }\in F$ with $x^{\prime }\sim \sigma _2$ is a subgroup; conversely, if $H_{x}$ for $x\in F$ with $x\sim \sigma _2$ is not a subgroup, then the $H_{x^{\prime }}$ for any other $x^{\prime }\in F$ with $x^{\prime }\sim \sigma _2$ is not a subgroup.

Proposition 7.9

We have the following formulas for $\mathrm{rank}A_{F}$ and $\mathrm{defe}F$.  

• (1) For $F=F_{r,s}, \, r\le 2,\,s\le 3,\,\mathrm{rank}A_{F}=r,\,\mathrm{defe}F=3\cdot 2^{r}(2-2^{s})$;

• (2) For $F=F^{\prime }_{r,s},\,r\le 2,\,s\le 3,\,\mathrm{rank}A_{F}=r,\,\mathrm{defe}F=2^{r}(2-2^{s})$;

• (3) For $F=F_{\epsilon ,\delta ,r,s},\,\epsilon +\delta \le 1,\,r+s\le 2,\,\mathrm{rank}A_{F}=r,\,\mathrm{defe}F=(1-\epsilon )(-1)^{\delta }2^{r+s+\delta } -2^{1+r+\epsilon +2s+2\delta }$;

• (4) For $F=F^{\prime }_{\epsilon ,\delta ,r,s}, \, \epsilon +\delta \le 1,\,r+s\le 2,\,s\ge 1, \, \mathrm{rank}A_{F}=r, \, \mathrm{defe}F=(1-\epsilon )(-1)^{\delta } 2^{r+s+\delta }$.

Proof

These follows from Lemma 7.6 and Proposition 6.7. $\square$

Proposition 7.10

Any two of the subgroups $\{F_{r,s}\},\,\{F^{\prime }_{r,s}\},\, \{F_{\epsilon ,\delta ,r,s}\},\, \{F^{\prime }_{\epsilon ,\delta ,r,s}\}$ are non-conjugate.

Proof

This follows from Propositions 7.7 and 7.9. $\square$

7.2 Subgroups from $\mathrm{SU}(8)$ or $\mathrm{SO}(8)$

By Table 2, we have that

$$\begin{aligned} G^{\sigma _3}\cong (\mathrm{SU}(8)/\langle iI\rangle )\rtimes \langle \omega \rangle , \end{aligned}$$

where $\omega ^{2}=1, \, ({\mathfrak{u }}_0^{\sigma _3})^{\omega }=\mathfrak sp (4)$ and $\mathfrak{p }\cong \wedge ^{4}(\mathbb{C }^{8})$. Let $\tau _1=[I_{2,6}]$ and $\tau _2=[I_{4,4}]$.

Let $\omega _0=\omega \left(\begin{array}{cc} 0&I_{4}\\ -I_{4}&0 \\ \end{array}\right)$. Then $\omega _0^{2}=1$ and $(\mathrm{SU}(8)/\langle iI\rangle )^{\omega _0} =(\mathrm{SO}(8)/\langle -I\rangle )\times \langle \sigma _3\rangle$. In $((\mathrm{SU}(8)/\langle iI\rangle )^{\omega _0})_0=\mathrm{SO}(8)/\langle -I\rangle$, let

$$\begin{aligned}&\eta _1=\left(\begin{array}{cc} 0&I_{4}\\ -I_{4}&0\\ \end{array}\right), \quad \eta _2=\left(\begin{array}{cc} -I_{4}&\\&I_{4}\\ \end{array}\right), \\&\eta _3=\left(\begin{array}{cc} -I_{2}&\\&I_{6}\\ \end{array}\right), \quad \eta _4=\left(\begin{array}{cc} 0&I_{1,3}\\ -I_{1,3}&0\\ \end{array}\right), \end{aligned}$$

where $I_{1,3}=\mathrm{diag}\{-1,1,1,1\}$. Let $\tau _3=\omega _0, \, \tau _4=\eta _1\omega _0$. From [8, Pages 16–17], we see that $\tau _1, \, \tau _2, \, \sigma _3\tau _1, \, \sigma _3\tau _2, \, \tau _3, \, \tau _4, \, \sigma _3\tau _4$ represent all conjugacy classes of involutions in $G^{\sigma _3}$ except $\sigma _3=\frac{1+i}{\sqrt{2}}I$ and we have the following conjugacy relations in $G$,

$$\begin{aligned}&\tau _1\sim \tau _2\sim \sigma _1, \\&\sigma _3\tau _1\sim \sigma _2,\ \sigma _3\tau _2\sim \sigma _3, \\&\tau _3\sim \sigma _3,\ \tau _4\sim \sigma _2, \\&\tau _4\sigma _3\sim \sigma _3. \end{aligned}$$

Lemma 7.11

In $G^{\sigma _3}$, we have the conjugacy classes

$$\begin{aligned}&\eta _3\sim _{G^{\sigma _3}}\tau _1, \\&\eta _1\sim _{G^{\sigma _3}}\eta _2\sim _{G^{\sigma _3}}\eta _4\sim _{G^{\sigma _3}}\tau _2, \\&\eta _2\omega _0\sim _{G^{\sigma _3}}\eta _3\omega _0\sim _{G^{\sigma _3}}\omega _0=\tau _3, \\&\eta _4\omega _0\sim _{G^{\sigma _3}}\eta _1\omega _0\sigma _3=\tau _4\sigma _2. \end{aligned}$$

Proof

These conjugacy relations can be prove by matrix calculation in the group $(\mathrm{SU}(8)/\langle iI\rangle )\rtimes \langle \omega _0\rangle$. We show the relation $\eta _4\omega _0\sim _{G^{\sigma _3}}\eta _1\omega _0\sigma _3$ here, which is the most complicated one among them.

Let $y=e^{\frac{\pi i}{8}}\mathrm{diag}\{I_7,-1\}\in \mathrm{SU}(8)/\langle iI\rangle$, then

$$\begin{aligned} y(\eta _4\omega _0)y^{-1}&= (y\eta _4y^{-1})\omega _0 (\omega _0^{-1}y\omega _0)y^{-1} \\&= \eta _1\omega _0y^{-1}y^{-1}=\eta _1\omega _0e^{\frac{-\pi i}{4}} \\&= \eta _1\omega _0\sigma _2, \end{aligned}$$

in the last equality we use $e^{\frac{-\pi i}{4}}I=(e^{\frac{\pi i}{4}}I)(iI)^{-1}=e^{\frac{\pi i}{4}}I=\sigma _3$ in $\mathrm{SU}(8)/\langle iI\rangle$. $\square$

Let

$$\begin{aligned}&M_1=\left\langle \left(\begin{array}{cc} -I_{4}&\\&I_{4}\\ \end{array}\right), \left(\begin{array}{cc} 0_{4}&I_{4}\\ I_{4}&0_{4}\\ \end{array}\right)\right\rangle ,\\&M_2=\langle \mathrm{diag}\{-I_{4},I_{4}\},\mathrm{diag}\{-I_{2},I_{2},-I_{2},I_{2}\} \rangle , \end{aligned}$$

then $({\mathfrak{e }}_7^{\sigma _3})^{M_1}\cong \mathfrak su (4)$ and $({\mathfrak{e }}_7^{\sigma _3})^{M_2}\cong (\mathfrak sp (1))^{4}\oplus (i\mathbb{R })^{3}$.

Lemma 7.12

In $G$, we have $M_1\sim \Gamma _1$ and $M_2\sim \Gamma _2$.

Proof

First since $M_1,M_2$ are pure $\sigma _1$ subgroups, so each of them is conjugate to $\Gamma _1$ or $\Gamma _2$. Since $\mathfrak su (4)$ is not a symmetric subalgebra of ${\mathfrak{e }}_7^{\Gamma _2}\cong \mathfrak so (8)\oplus (\mathfrak sp (1))^{3}$ and $(\mathfrak sp (1))^{4}\oplus (i\mathbb{R })^{3}$ is not a symmetric subalgebra of ${\mathfrak{e }}_7^{\Gamma _1}\cong \mathfrak su (6)\oplus (i\mathbb{R })^{2}$, so we have $M_1\sim \Gamma _1$ and $M_2\sim \Gamma _2$.$\square$

Let $F$ be an elementary abelian 2-subgroup of $G$ with $\sigma _3\in F$. Then

$$\begin{aligned} F\subset G^{\sigma _3}\cong (\mathrm{SU}(8)/\langle iI\rangle )\rtimes \langle \omega _0\rangle , \end{aligned}$$

where $\omega _0^{2}=1$ and $\mathfrak su (8)^{\omega _0}=\mathfrak so (8)$. If $F$ has no elements conjugate to $\sigma _2$, from the description of conjugacy classes of involutions in $G^{\sigma _3}$ as above, we get that $x\sim \tau _2=\mathrm{diag}\{-I_{4},I_{4}\}$ for any $1\ne x\in H_{F}$.

Lemma 7.13

For an elementary abelian 2-subgroup $F$ of $G$ containing $\sigma _3$ and without elements conjugate to $\sigma _2$, in the inclusion $F\subset G^{\sigma _3}\cong (\mathrm{SU}(8)/\langle iI\rangle ) \rtimes \langle \omega _0\rangle$, we have $H_{F}\subset \mathrm{SU}(8)/\langle iI\rangle$ and the homomorphism

$$\begin{aligned} H_{F}\longrightarrow \mathrm{SU}(8)/\left\langle iI,\sigma _3\right\rangle = \mathrm{SU}(8)\left.\left\langle \frac{1+i}{\sqrt{2}}I\right\rangle \right.=\mathrm{PSU}(8) \end{aligned}$$

is injective. Moreover, the map

$$\begin{aligned} m: H_{F}\times H_{F}\longrightarrow \{\pm {1}\} \end{aligned}$$

is equal to the similar map when $H_{F}$ is regarded as a subgroup of $\mathrm{PSU}(8)$.

Proof

We have $H_{F}\subset \mathrm{SU}(8)/\langle iI\rangle$ since any involution in $\omega _0\mathrm{SU}(8)/\langle iI\rangle$ is conjugate to $\sigma _2$ or $\sigma _3$. The map $H_{F}\longrightarrow \mathrm{PSU}(8)$ is injective since $\sigma _3\not \in H_{F}$. The two maps $m$ are equal follows from Lemma 7.12. $\square$

In $(G^{\sigma _3})_0\cong \mathrm{SU}(8)/\langle iI\rangle$, let $y_{1}=\mathrm{diag}\{-I_{4},I_{4}\}$,

$$\begin{aligned} y_{2}&= \mathrm{diag}\{-I_{2},I_{2},-I_{2},I_{2}\}, \\ y_{3}&= \mathrm{diag}\{-1,1,-1,1,-1,1,-1,1\}, \\ y_{4}&= \left(\begin{array}{cc} 0_{4}&I_{4}\\ I_{4}&0_{4}\\ \end{array}\right), \\ y_{5}&= \left(\begin{array}{cccc} 0_{2}&I_{2}&\,&\\ I_{2}&0_{2}&\,&\\&\,&0_{2}&I_{2}\\&\,&I_{2}&0_{2}\\ \end{array}\right), \\ y_{6}&= \left(\begin{array}{cccccccc} 0&1&\,&\,&\,&\\ 1&0&\,&\,&\,&\\&\,&0&1&\,&\,&\\&\,&1&0&\,&\,&\\&\,&\,&0&1&\,&\\&\,&\,&1&0&\,&\\&\,&\,&\,&0&1 \\&\,&\,&\,&1&0 \\ \end{array}\right). \end{aligned}$$

For each $(r,s)$ with $r+s\le 3$, let $F^{\prime \prime }_{r,s}=\langle \sigma _3,y_1,y_2,\ldots ,y_{r+s}, y_{4},\ldots ,y_{3+s}\rangle$.

In $(G^{\sigma _3})^{\omega _0}=(\mathrm{SO}(8)/\langle -I\rangle )\times \langle \sigma _3,\omega _0\rangle$, let $x_{1}=\mathrm{diag}\{-I_{4},I_{4}\}$,

$$\begin{aligned} x_{2}&= \mathrm{diag}\{-I_{2},I_{2},-I_{2},I_{2}\}, \\ x_{3}&= \mathrm{diag}\{-1,1,-1,1,-1,1,-1,1\}. \end{aligned}$$

For each $r\le 3$, let $F^{\prime }_{r}=\langle \sigma _2,\omega _0,x_1,\ldots ,x_{r}\rangle$.

Proposition 7.14

For an elementary abelian 2-group $F\subset G$, if $F$ contains an element conjugate to $\sigma _3$ but contains no elements conjugate to $\sigma _2$, then $F$ is conjugate to one of $\{F^{\prime \prime }_{r,s}: r+s\le 3\}, \, \{F^{\prime }_{r}: r\le 3\}$.

Proof

We may and do assume that $\sigma _3\in F$, then

$$\begin{aligned} F\subset G^{\sigma _3}\cong (\mathrm{SU}(8)/\langle iI\rangle )\rtimes \langle \omega _0\rangle . \end{aligned}$$

By Lemma 7.3, we have $\mathrm{rank}(F/H_{F})\le 2$.

When $\mathrm{rank}(F/H_{F})=1, \, F\subset (G^{\sigma _3})_0\cong \mathrm{SU}(8)/\langle iI\rangle$. As $F$ has no elements conjugate to $\sigma _2$, so any element of $F$ is conjugate to $\tau _2$ or $\sigma _3\tau _2$ in $\mathrm{SU}(8)/\langle iI\rangle$, where $\tau _2= \left(\begin{array}{cc}-I_{4}&\\&I_{4}\\ \end{array}\right)$. Then $F\sim F^{\prime \prime }_{r,s}$ for some $r,s\ge 0$ with $r+s\le 3$ by Proposition 2.24.

When $\mathrm{rank}(F/H_{F})=2$, we may and do assume that $\omega _0\in F$ as well, then

$$\begin{aligned} F\subset (G^{\sigma _3})^{\omega _0}=(\mathrm{SO}(8)/\langle -I\rangle )\times \langle \sigma _3,\omega _0\rangle . \end{aligned}$$

We have $H_{F}=F\cap \mathrm{SO}(8)/\langle -I\rangle$. Since $F$ contains no elements conjugate to $\sigma _2$, so any involution in $H_{F}$ is conjugate to $\eta _2=\mathrm{diag}\{-I_4,I_4\}$ in $\mathrm{SO}(8)/\langle -I\rangle$. Then $F\sim F^{\prime }_{r}$ for some $r\le 3$ by Proposition 2.24. $\square$

Proposition 7.15

We have $\mathrm{rank}A_{F^{\prime \prime }_{r,s}}=\mathrm{rank}A_{F^{\prime }_{r}}=r$ and any two groups in $\{F^{\prime \prime }_{r,s}:r+s\le 3\},\{F^{\prime }_{r}:r\le 3\}$ are non-conjugate.

Proof

By Lemma 7.13, we get $\mathrm{rank}A_{F^{\prime \prime }_{r,s}}=\mathrm{rank}A_{F^{\prime }_{r}}=r$. Then the conjugacy class of any group $F$ in $\{F^{\prime \prime }_{r,s}\},\{F^{\prime }_{r}\}$ is determined by the numbers

$$\begin{aligned} (\mathrm{rank}(F/H_{F}),\mathrm{rank}A_{F},\mathrm{rank}F). \end{aligned}$$

$\square$

7.3 Pure $\sigma _1$ subgroups

A subgroup $F$ of $G$ is called a pure $\sigma _1$ subgroup if any of its non-identity element is conjugate to $\sigma _1$.

By Table 2, we have $G^{\sigma _1}\cong (\mathrm{Spin}(12)\times \mathrm{Sp}(1))/ \langle (c,1),(-c,-1)\rangle$, where $c=e_1e_2\ldots e_{12}$. From [8, Page 16], we see that $(e_1e_2e_3e_4,1), \, (e_1e_2,\mathbf{i}), \, (e_1e_2e_3e_4e_5e_6,\mathbf{i}), (\Pi ,1), \, (\Pi ,-1)$ represent the conjugacy classes of involutions in $G^{\sigma _1}$ except $\sigma _1=(1,-1)$ and we have the following conjugacy classes in $G$,

$$\begin{aligned}&(e_1e_2e_3e_4,1)\sim \sigma _1, \\&(e_1\Pi e_1,i)\sim \sigma _1,\ (e_1e_2,\mathbf{i})\sim \sigma _2, \\&(e_1e_2e_3e_4e_5e_6,\mathbf{i}) \sim \sigma _3, \\&(\Pi ,1)\sim \sigma _2,\ (\Pi ,-1)\sim \sigma _3. \end{aligned}$$

Here

$$\begin{aligned} \Pi = \frac{1+e_1e_2}{\sqrt{2}} \frac{1+e_3e_4}{\sqrt{2}} \frac{1+e_5e_6}{\sqrt{2}}\frac{1+e_7e_8}{\sqrt{2}} \frac{1+e_9e_{10}}{\sqrt{2}}\frac{1+e_{11}e_{12}}{\sqrt{2}}\in \mathrm{Spin}(12). \end{aligned}$$

Let

$$\begin{aligned} K_1&= \langle (e_1\Pi e_1^{-1},\mathbf{i}),(e_1\Pi ^{\prime }e_1^{-1},\mathbf{j})\rangle , \\ K_2&= \langle (e_1e_2e_3e_4,1),(e_5e_6e_7e_8,1)\rangle , \\ K_3&= \langle (e_1\Pi e_1^{-1},\mathbf{i}),(-e_1e_2e_3e_4,1)\rangle ,\\ K_4&= \langle (e_1\Pi e_1^{-1}, \mathbf{i}), (e_1e_2e_3e_4,1)\rangle ,\\ K_5&= \langle (e_1e_2e_3e_4,1),(e_1e_2e_5e_6,1)\rangle , \end{aligned}$$

where

$$\begin{aligned} \Pi ^{\prime }=\frac{1+e_1e_3}{\sqrt{2}} \frac{1+e_4e_2}{\sqrt{2}} \frac{1+e_5e_7}{\sqrt{2}} \frac{1+e_8e_6}{\sqrt{2}}\frac{1+e_9e_{11}}{\sqrt{2}} \frac{1+e_{12}e_{10}}{\sqrt{2}}. \end{aligned}$$

Lemma 7.16

We have $\Pi ^{2}=\Pi ^{\prime }{^{2}}=[\Pi ,\Pi ^{\prime }]=c$.

Proof

$\Pi ^{2}=\Pi ^{\prime }{^{2}}=c$ is clear. Calculation shows that

$$\begin{aligned} \Pi \Pi ^{\prime }=\frac{1+e_1e_4}{\sqrt{2}} \frac{1+e_2e_3}{\sqrt{2}}\frac{1+e_5e_8}{\sqrt{2}} \frac{1+e_6e_7}{\sqrt{2}} \frac{1+e_9e_{12}}{\sqrt{2}}\frac{1+e_{10}e_{11}}{\sqrt{2}}, \end{aligned}$$

so $(\Pi \Pi ^{\prime })^{2}=c$. Then

$$\begin{aligned}{}[\Pi ,\Pi ^{\prime }]=\Pi \Pi ^{\prime }\Pi ^{-1}\Pi ^{\prime }{^{-1}}=\Pi \Pi ^{\prime }(c\Pi )(c\Pi ^{\prime })= (\Pi \Pi ^{\prime })^{2}=c. \end{aligned}$$

$\square$

Lemma 7.17

In $G$, we have $K_1\sim K_3\sim K_5\sim \Gamma _1$ and $K_2\sim K_4\sim \Gamma _2$.

Proof

Since $({\mathfrak{u }}_0^{\sigma _1})^{K_1}\cong \mathfrak sp (3)$ is not a symmetric subgroup of ${\mathfrak{u }}_0^{\Gamma _2}\cong \mathfrak so (8)\oplus (\mathfrak sp (1))^{2}$ and $({\mathfrak{e }}_7^{\sigma _1})^{K_2}\cong (\mathfrak sp (1))^{7}$ is not a symmetric subgroup of ${\mathfrak{u }}_0^{\Gamma _1}\cong \mathfrak su (6)\oplus (i\mathbb{R })^{2}$, we get that $K_1\sim \Gamma _1$ and $K_2\sim \Gamma _2$.

Choose a Cartan subalgebra $\mathfrak{h }_0$ of ${\mathfrak{e }}_7$, we may assume that $\sigma _1= \exp (\pi i H^{\prime }_2)$. Then $\mathfrak{g }^{\sigma _1}$ has a simple root system

$$\begin{aligned} \{\alpha _2,\alpha _4,\alpha _5,\alpha _6,\beta ,\alpha _7\}(\mathrm Type \mathrm{D}_6) \bigsqcup \{\alpha _1\}, \end{aligned}$$

where $\beta =\alpha _1+2\alpha _3+2\alpha _4+\alpha _2+\alpha _5$. By identifying conjugacy (classes of) elements in $\exp (\mathfrak{h }_0)$ and in $\mathrm{Spin}(12)\times \mathrm{Sp}(1)/\langle (c,1),(-c,-1)\rangle$, we get the conjugacy relations

$$\begin{aligned} (\exp (\pi i H^{\prime }_1),\exp (\pi i H^{\prime }_3),\exp (\pi i H^{\prime }_2))\sim \left(\sigma _1, (e_1 \Pi e_1^{-1},\mathbf{i}), e_1e_2e_3e_4\right) \end{aligned}$$

and

$$\begin{aligned} (\exp (\pi i H^{\prime }_1), \exp (\pi i H^{\prime }_2),\exp (\pi i H^{\prime }_4))\sim (\sigma _1,e_1e_2e_3e_4,e_1e_2e_5e_6). \end{aligned}$$

Then we have $K_3\sim K_5\sim \Gamma _1$ and $K_4\sim \Gamma _2$. $\square$

Let $\pi :\mathrm{Spin}(12)\longrightarrow \mathrm{SO}(12)$ be the natural projection.

Lemma 7.18

In $\mathrm{Spin}(12)$, we have $\Pi \sim \Pi ^{-1}, \, \Pi \not \sim -\Pi$ and $\Pi \not \sim \pm {}e_1\Pi e_1^{-1}$.

Proof

We have $(e_1e_3e_5e_7e_9e_{11})\Pi (e_1e_3e_5e_7e_9e_{11})^{-1}=\Pi ^{-1}$, so $\Pi \sim \Pi ^{-1}$. Since

$$\begin{aligned} \mathrm{SO}(12)^{\pi (\Pi )}=\{g\in \mathrm{Spin}(12)|g\Pi g^{-1}=\pm {\Pi }\}/\langle -1\rangle , \end{aligned}$$

$-1\in \{g\in \mathrm{Spin}(12)|g\Pi g^{-1}=\Pi \}$ and $\mathrm{SO}(12)^{\pi (\Pi )}=\mathrm{U}(6)$ is connected, so we must have $\Pi \not \sim -\Pi$. We have $\pi (\Pi )=J_{6}\in \mathrm{SO}(12)$ and $\pi (\pm {}e_1\Pi e_1^{-1})= I_{1,11}J_{6}I_{1,11}^{-1}$. Since $J_{6}\not \sim _{\mathrm{SO}(12)}I_{1,11}J_{6}I_{1,11}^{-1}$, so $\Pi \not \sim _{\mathrm{Spin}(12)}\pm {}e_1\Pi e_1^{-1}$. $\square$

Lemma 7.19

We have $\mathrm{Aut}({\mathfrak{e }}_7)^{\Gamma _1}=(\mathrm{Aut}({\mathfrak{e }}_7)^{\Gamma _1})_0\rtimes \langle (e_1\Pi ^{\prime }e_1,\mathbf{j}) \rangle$ and

$$\begin{aligned} (\mathrm{Aut}({\mathfrak{e }}_7)^{\Gamma _1})_0\cong (\mathrm{SU}(6)\times \mathrm{U}(1)\times \mathrm{U}(1))/\langle (\omega I,\omega ^{-1},1),(-I,1,1)\rangle . \end{aligned}$$

Proof

First we calculate $\mathrm{Spin}(12)^{\Pi }$. We have

$$\begin{aligned} \mathrm{SO}(12)^{\pi (\Pi )}\cong \mathrm{U}(6)= (\mathrm{SU}(6)\times \mathrm{U}(1))/\langle \eta I,\eta ^{-1}\rangle , \end{aligned}$$

where $\eta =e^{\frac{2\pi i}{6}}$. Then $\mathrm{Spin}(12)^{\Pi }=(\mathrm{SU}(6)\times A)/Z$, where

$$\begin{aligned} A=\{\prod _{1\le jleq 6}(\cos \theta +\sin \theta e_{2j-1}e_{2j}):\theta \in \mathbb{R }\}\cong \mathrm{U}(1) \end{aligned}$$

and $Z\subset Z(\mathrm{SU}(6))\times A$. The isomorphism $\mathrm{U}(1)\cong A$ maps $-1\in \mathrm{U}(1)$ to $c\in A$, and $\pi (c)=-I\in \mathrm{SO}(12)$, so

$$\begin{aligned} \pi : \mathrm{Spin}(12)^{\Pi }\longrightarrow \mathrm{SO}(12)^{\pi (\Pi )} \end{aligned}$$

is an isomorphism when it is restricted to $\mathrm{SU}(12)$ or $A$.

We show that $-c\in \mathrm{SU}(6)\subset \mathrm{Spin}(12)^{\Pi }$. For this, we first look at the case of $n=4$. For

$$\begin{aligned} \Pi _0=\frac{1+e_1e_2}{\sqrt{2}}\frac{1+e_3e_4}{\sqrt{2}}\in \mathrm{Spin}(4), \end{aligned}$$

we have $\Pi _{0}^{2}=c_0=e_1e_2e_3e_4$. We have an isomorphism

$$\begin{aligned} \mathrm{Spin}(4)\cong \mathrm{Sp}(1)\times \mathrm{Sp}(1), \end{aligned}$$

which maps $-1\in \mathrm{Spin}(4)$ to $(-1,-1)\in \mathrm{Sp}(1)\times \mathrm{Sp}(1)$ and maps $c_0\in \mathrm{Spin}(4)$ to $(-1,1)\in \mathrm{Sp}(1)\times \mathrm{Sp}(1)$. Then $\Pi \in \mathrm{Spin}(4)$ is mapped to $(\mathbf i ,1)$ or $(\mathbf i ,-1)$ in $\mathrm{Sp}(1)\times \mathrm{Sp}(1)$. Since

$$\begin{aligned} (\mathrm{Sp}(1)\times \mathrm{Sp}(1))^{(\mathbf i ,\pm {1})}=\mathrm{U}(1)\times \mathrm{Sp}(1), \end{aligned}$$

so $(1,-1)$ is in the semisimple part $\mathrm{Sp}(1)$ of it. Then $-c_0\in \mathrm{Spin}(4)$ is in the $\mathrm{SU}(2)$ part of $\mathrm{Spin}(4)^{\Pi }$.

As $\Pi$ is in block form, so $-c\in \mathrm{SU}(6)\subset \mathrm{Spin}(12)^{\Pi }$ as well. Since $(-c)c= -1\ne 1\in \mathrm{Spin}(12), \, \pi (-1)=1$, and $\pi$ is a 2-fold covering, so

$$\begin{aligned} \mathrm{Spin}(12)^{\Pi }= (\mathrm{SU}(6)\times \mathrm{U}(1))/\langle (\omega I,\omega ^{-1})\rangle \end{aligned}$$

(here we identify $A$ and $\mathrm{U}(1)$). By Lemma 7.18 and Steinberg ’s theorem, we get that

$$\begin{aligned} \mathrm{Aut}({\mathfrak{e }}_7)^{\Gamma _1}= (\mathrm{Aut}({\mathfrak{e }}_7)^{\Gamma _1})_0\rtimes \langle (e_1\Pi ^{\prime }e_1,\mathbf{j})\rangle . \end{aligned}$$

The description of $(\mathrm{Aut}({\mathfrak{e }}_7)^{\Gamma _1})_0$ follows from the description of $\mathrm{Spin}(12)^{\Pi }$ as above. $\square$

In $G^{\sigma _1}\cong \mathrm{Spin}(12)\times \mathrm{Sp}(1)/\langle (c,1),(-c,-1)\rangle$, let

$$\begin{aligned} H_1&= \left\langle \sigma _1,\left(e_1\Pi e_1^{-1},\mathbf{i}\right),\left(e_1\Pi ^{\prime } e_1^{-1},\mathbf{j}\right)\right\rangle , \\ H_2&= \left\langle \sigma _1,(e_1e_2e_3e_4,1),(e_5e_6e_7e_8,1)\right\rangle \end{aligned}$$

and $H_3=\langle \sigma _1,(e_1\Pi e_1^{-1}, \mathbf{i}),(e_1e_2e_3e_4,1)\rangle$. Then any Klein four subgroup of $H_1$ is conjugate to $\Gamma _1$; any Klein four subgroup of $H_2$ is conjugate to $\Gamma _2$; a Klein four subgroup of $H_3$ is conjugate to $\Gamma _2$ if and only if it contains $(e_1e_2e_3e_4,1)$, otherwise it is conjugate to $\Gamma _1$.

Lemma 7.20

We have $G^{H_1}=(\mathrm{Sp}(3)/\langle -I\rangle )\times H_1$ and the involutions $I_{1,2}, \, \mathbf{i}I$ of $\mathrm{Sp}(3)/\langle -I\rangle$ are conjugate to $\sigma _1, \, \sigma _2$ in $\mathrm{Aut}({\mathfrak{e }}_7)$ respectively.

Proof

$G^{H_1}=(\mathrm{Sp}(3)/\langle -I\rangle )\times H_1$ follows from Lemma 7.19 and the fact

$$\begin{aligned} \mathfrak su (6)^{e_1\Pi ^{\prime }e_1^{-1}}=\mathfrak sp (3). \end{aligned}$$

A little more calculation by following the chain $\mathrm{Sp}(3)\subset \mathrm{SU}(6)\subset \mathrm{SO}(12)$ shows that $I_{1,2}, \, \mathbf{i}I\in \mathrm{Sp}(3)$ are conjugate to $e_1e_2e_3e_4, \, \Pi$ in $\mathrm{Spin}(12)$ respectively. Then they are conjugate to $\sigma _1, \, \sigma _2$ in $\mathrm{Aut}({\mathfrak{e }}_7)$ respectively. $\square$

Lemma 7.21

Any rank $3$ elementary abelian 2- pure $\sigma _1$ subgroup $F$ of $G$ is conjugate to one of $H_1,H_2,H_3$.

Proof

For a rank $3$ pure $\sigma _1$ elementary abelian $2$-subgroup $F$ of $G$, we may and do assume that $\sigma _1\in F$. Then

$$\begin{aligned} F\subset G^{\sigma _1}\cong \mathrm{Spin}(12)\times \mathrm{Sp}(1)/\langle (c,1),(-c,-1)\rangle \end{aligned}$$

and any element of $F-\{1,\sigma _1\}$ is conjugate to $(e_1\Pi e_1^{-1},\mathbf{i})$ or $(e_1e_2e_3e_4,1)$ in $G^{\sigma _1}$.

When any Klein four subgroup of $F$ is conjugate to $\Gamma _1$, we have $F\sim H_1$ by Lemma 7.17; when any Klein four subgroup of $F$ is conjugate to $\Gamma _2$, similarly we have $F\sim H_2$ by Lemma 7.17. For the remaining cases, it is clear that $F\sim H_3$. $\square$

We have defined the subgroups $\{F_{r,s}: r\le 2,s\le 3\}$ and $\{F^{\prime \prime }_{r,s}: r+s\le 3\}$ in the last two subsections. The subgroup $F_{r,s}$ contains a Klein four subgroup conjugate to $F_6$; $F^{\prime \prime }_{r,s}$ does not contain any element conjugate to $\sigma _2$ and we have $\mathrm{rank}(F^{\prime \prime }_{r,s}/H_{F^{\prime \prime }_{r,s}})=1$. For any $(r,s)$ with $r+s \le 3$, let

$$\begin{aligned} F^{\prime \prime \prime }_{r,s}=H_{F^{\prime \prime }_{r,s}} =\{1\}\cup \{x\in F^{\prime \prime }_{r,s}|x\sim \sigma _1\}; \end{aligned}$$

for any $r\le 2$, let

$$\begin{aligned} F^{\prime \prime }_{r}=H_{F_{r,3}}= \{1\}\cup \{x\in F_{r,3}|x\sim \sigma _1\}. \end{aligned}$$

Proposition 7.22

Any pure $\sigma _1$ elementary abelian 2-group $F\subset G$ is conjugate to $F^{\prime \prime }_{r+3}$ for some $r\le 2$ or $F^{\prime \prime \prime }_{r,s}$ for some $(r,s)$ with $r+s\le 3$.

Proof

When $F$ contains a subgroup conjugate to $H_1$, we may and do assume that $H_1\subset F$, then

$$\begin{aligned} F\subset G^{H_1}=(G^{\sigma _1})^{(e_1\Pi e_1,\mathbf{i}),(e_1\Pi ^{\prime } e_1,\mathbf{j})} \cong (\mathrm{Sp}(3)/\langle -I\rangle )\times \langle \sigma _1,(e_1\Pi e_1,\mathbf{i}), (e_1\Pi ^{\prime } e_1,\mathbf{j})\rangle . \end{aligned}$$

Since $F$ is pure $\sigma _1$, by Lemma 7.20 we have any non-identity element of $F\cap (\mathrm{Sp}(3)/\langle -I\rangle )$ is conjugate to $I_{1,2}$ in $\mathrm{Sp}(3)/\langle -I\rangle$. Then $F\cap (\mathrm{Sp}(3)/\langle -I\rangle )$ is conjugate to a subgroup of $\langle I_{2,1},I_{1,2}\rangle$, which is a subgroup of $\langle \mathbf{i}I,\mathbf{j}I,I_{2,1},I_{1,2}\rangle$. We may and do assume that $\mathbf{i}I,\mathbf{j}I\in C_{G}(F)$. Since Non-identity elements of $\langle \mathbf{i}I,\mathbf{j}I\rangle$ are all conjugate to $\sigma _2$ in $G$, so $\langle F,\mathbf{i}I,\mathbf{j}I\rangle$ is conjugate to some $F_{r,s}$ (cf. Proposition 7.7). Then $F$ is conjugate to some $H_{F_{r,s}}= \{1\}\cup \{x\in F_{r,s}|x\sim \sigma _1\}$. Since we assume that $H_1\subset F$, so we have $s=3$. Then $F$ is conjugate to $F^{\prime \prime }_{r}$.

If $F$ does not contain any subgroup conjugate to $H_1$ but contains a subgroup conjugate to $\Gamma _1$, we may an do assume that $\sigma _1,(e_1 \Pi e_1^{-1},\mathbf{i})\in F$. Since $F$ does not contain any subgroup conjugate to $H_1$, so

$$\begin{aligned} F\subset \left((G^{\sigma _1})^{(e_1 \Pi e_1^{-1}, \mathbf{i})}\right)_0\cong (\mathrm{SU}(6)\times \mathrm{U}(1)\times \mathrm{U}(1))/\left\langle \left(e^{\frac{2\pi i}{3}},e^{\frac{2\pi i}{3}},1\right), (-I,1,1)\right\rangle . \end{aligned}$$

Since $F$ is pure $\sigma _1$, we have

$$\begin{aligned} F=(F\cap (\mathrm{SU}(6)/\langle -I\rangle )) \times \left\langle \sigma _1,\left(e_1 \Pi e_1^{-1}, \mathbf{i}\right)\right\rangle \end{aligned}$$

and any element in $F\cap (\mathrm{SU}(6)/\langle -I\rangle )$ is conjugate to $I_{2,4}$. Then $F$ is toral (cf. Proposition 2.4).

If $F$ does not contain any subgroup conjugate to $\Gamma _1$, then any Klein four subgroup of $F$ is conjugate to $\Gamma _2$. When $\mathrm{rank}(F)\ge 3$, we may an do assume that $H_2\subset F$. Since there are no elements $x\in (G^{\sigma _1})^{H_2}-H_2$ such that any Klein four subgroup of $\langle x,H_2\rangle$ is conjugate to $\Gamma _2$, so $\mathrm{rank}(F)\le 3$. Then $F$ is conjugate to one of $1,\langle \sigma _1\rangle ,\Gamma _2,H_2$, so $F$ is toral.

For a toral and pure $\sigma _1$ elementary abelian 2-subgroup $F$ of $G$, there exists a Cartan subalgebra $\mathfrak{h }_0$ such that $F\subset \exp (\mathfrak{h }_0)$. Choose a Chevelley involution $\theta$ of ${\mathfrak{e }}_7$ with respect to $\mathfrak{h }_0$. Then $F^{\prime }=\langle F,\theta \rangle$ satisfies $\mathrm{Res}(F^{\prime }/H_{F^{\prime }})=1$ and any involution in $F^{\prime }-H_{F^{\prime }}$ is conjugate to $\sigma _3$. Then $F^{\prime }$ is conjugate to $F^{\prime \prime }_{r,s}$ for some $(r,s)$ with $r+s\le 3$. Then $F$ is conjugate to $F^{\prime \prime \prime }_{r,s}$. $\square$

Proposition 7.23

For any $r+s\le 3$, we have $\mathrm{rank}A_{F^{\prime \prime \prime }_{r,s}}=r$; for any $r\le 2$, we have $\mathrm{rank}A_{F^{\prime \prime }_{r}}=r$.

Any two subgroups in $\{F^{\prime \prime \prime }_{r,s}: r+s\le 3\},\{F^{\prime \prime }_{r}: r\le 2\}$ are non-conjugate.

Proof

By Propositions 7.9 and 7.14, we get $\mathrm{rank}A_{F^{\prime \prime \prime }_{r,s}}=r$ and $\mathrm{rank}A_{F^{\prime \prime }_{r}}=r$. Then any two groups in $\{F^{\prime \prime \prime }_{r,s}: r+s\le 3\},\{F^{\prime \prime }_{r}: r\le 2\}$ are non-conjugate. $\square$

7.4 Automizer groups and inclusion relations

Corollary 7.24

$G$ has 78 conjugacy classes of elementary abelian 2-subgroups.

Proof

By Propositions 7.7, 7.10, 7.14, 7.15, 7.22 and7.23,we get that $G$ has

$$\begin{aligned} 3\times 4+3\times 4+3\times 6+3\times 3+10+4+10+3=78 \end{aligned}$$

conjugacy classes of elementary abelian 2-subgroups. $\square$

Proposition 7.25

For an isomorphism $f: F\longrightarrow F^{\prime }$ between two elementary abelian 2-subgroups of $G$, if $f(x)\sim x$ for any $x\in F$ and $m_{F^{\prime }}(f(x),f(y))=m_{F}(x,y)$ for any $x,y\in H_{F}$, then $f=\mathrm{Ad}(g)$ for some $g\in G$.

Proof

When $F$ contains an element conjugate to $\sigma _2$, we may and do assume that $\sigma _2\in F\cap F^{\prime }$ and $f(\sigma _2)=\sigma _2$, then

$$\begin{aligned} F,F^{\prime }\subset G^{\sigma _2}\cong \langle \omega \rangle \rtimes ((\mathrm{E}_{6}\times \mathrm{U}(1)) /\langle (c,e^{\frac{2\pi i}{3}})). \end{aligned}$$

From the description of conjugacy classes of elements in $G^{\sigma _2}$, we get that $f(x)\sim _{G^{\sigma _2}}x$ for any $x\in F$ by the assumption in the proposition. Then $f=\mathrm{Ad}(g)$ for some $g\in G^{\sigma _2}$ by Proposition 6.9.

When $\mathrm{rank}(F/H_{F})=1$ and $F$ contains no elements conjugate to $\sigma _2$, we may and do assume that $\sigma _3\in F\cap F^{\prime }$ and $f(\sigma _3)=\sigma _3$, then

$$\begin{aligned} F,F^{\prime }\subset G^{\sigma _3}\cong \langle \omega _0\rangle \rtimes (\mathrm{SU}(8)/\langle iI\rangle ) \end{aligned}$$

and any element in $(H_{F}\cup H_{F^{\prime }})-\{1\}$ is conjugate to $I_{4,4}$ in $\mathrm{SU}(8)/\langle iI\rangle$. Since the functions $m_{F}$ on $H_{F}\times H_{F}$ and $m_{F^{\prime }}$ on $H_{F^{\prime }}\times H_{F^{\prime }}$ are identical to the anti-symmetric bilinear form when $H_{F},H_{F^{\prime }}$ are regarded as subgroups of $\mathrm{PU}(8)$ (cf. Lemma 7.13). Then $f=\mathrm{Ad}(g)$ f or some $g\in G^{\sigma _3}$ by Proposition 2.24.

When $\mathrm{rank}(F/H_{F})=2$ and $F$ contains no elements conjugate to $\sigma _2$, we may and do assume that $\sigma _3,\omega _0\in F$, then

$$\begin{aligned} F,F^{\prime }\subset (G^{\sigma _3})^{\omega _0}\cong \mathrm{SO}(8)/\langle -I\rangle \end{aligned}$$

and any element in $(H_{F}\cup H_{F^{\prime }})-\{1\}$ is conjugate to $I_{4,4}$ in $\mathrm{SO}(8)/\langle -I\rangle$. Then $f=\mathrm{Ad}(g)$ for some $g\in (G^{\sigma _3})^{\omega _0}$ by Proposition 2.24.

When $F$ is pure $\sigma _1$, we get the conclusion by the considering the preserving of $m_{F},m_{F^{\prime }}$ under $f$. $\square$

Proposition 7.26

We have the following description for the automizer groups,  

• (1) for $r\le 2, \, s\le 3, \, W(F_{r,s})\cong \mathrm{Hom}(\mathbb{F }_2^{2},\mathbb{F }_2^{r})\rtimes (\mathrm{GL}(2,\mathbb{F }_2)\times P(r,s,\mathbb{F }_2))$;

• (2) for $r\le 2, \, s\le 3, \, W(F^{\prime }_{r,s})\cong \mathbb{F }_2^{r}\rtimes P(r,s,\mathbb{F }_2)$;

• (3) for $\epsilon +\delta \le 1, \, r+s\le 2$,

  $$\begin{aligned} W(F_{\epsilon ,\delta ,r,s})= (\mathbb{F }_2^{r+2s+\epsilon +2\delta +1}\rtimes \mathrm{Hom}(\mathbb{F }_2 ^{\epsilon +2\delta +2s+1},\mathbb{F }_2^{r})) \rtimes (\mathrm{GL}(r,\mathbb{F }_2)\times \mathrm{Sp}(\delta +s;\epsilon )). \end{aligned}$$

• (4) for $\epsilon +\delta \le 1, \, r+s\le 2$,

  $$\begin{aligned} W(F^{\prime }_{\epsilon ,\delta ,r,s})= \mathrm{Hom}(\mathbb{F }_2^{\epsilon +2\delta +2s+1},\mathbb{F }_2^{r})\rtimes (\mathrm{GL}(r,\mathbb{F }_2)\times \mathrm{Sp}(\delta +s;\epsilon )). \end{aligned}$$

• (5) for $r+s\le 3, \, W(F^{\prime \prime }_{r,s})\cong (\mathbb{F }_2^{r+2s}\rtimes \mathrm{Hom}(\mathbb{F }_2^{2s},\mathbb{F }_2^{r})) \rtimes (\mathrm{GL}(r,\mathbb{F }_2)\times \mathrm{Sp}(s))$;

• (6) for $r\le 3, \, W(F^{\prime }_{r})\cong \mathrm{Hom}(\mathbb{F }_2^{2},\mathbb{F }_2^{r})\rtimes (\mathrm{GL}(r,\mathbb{F }_2) \times \mathrm{GL}(2,\mathbb{F }_2))$;

• (7) for $r+s\le 3, \, W(F^{\prime \prime \prime }_{r,s})\cong \mathrm{Hom}(\mathbb{F }_2^{2s},\mathbb{F }_2^{r})\rtimes (\mathrm{GL}(r,\mathbb{F }_2)\times \mathrm{Sp}(s))$;

• (8) for $r\le 2, \, W(F^{\prime \prime }_{r})\cong P(r,3,\mathbb{F }_2)$.

Proof

By Proposition 7.25, we need to find all automorphisms of $F$ preserving the conjugacy classes of involutions and the form $m$ on $H_{F}$.

We prove $(4)$. Let $F=F^{\prime }_{\epsilon ,\delta ,r,s}$. Then $F$ has a decomposition $F=A_{F}\times F^{\prime }$ with $A_{F}=\mathbb{F }_2^{r}$ be the translation subgroup and $F^{\prime }\sim F^{\prime }_{\epsilon ,\delta ,0,s}$. By Proposition 7.25, we have

$$\begin{aligned} W(F)\cong \mathrm{Hom}(F^{\prime },A_{F})\rtimes (\mathrm{GL}(r,\mathbb{F }_2)\times W(F^{\prime })). \end{aligned}$$

So we only need to prove in the case of $r=0$. Assume that $r=0$ from now on.

Any element in $W(F)$ preserves the symplectic form $m$ on $H_{F}$. Since $\mathrm{rank}(\ker m)=\epsilon$, so we have a homomorphism

$$\begin{aligned} p: W(F)\longrightarrow \mathrm{Sp}(\delta +s;\epsilon ). \end{aligned}$$

We show that this homomorphism is an isomorphism, which finishes the proof.

For any $f: F\longrightarrow F$ with $f|_{H_{F}}=1$, since $F=H_{F}\rtimes \langle z\rangle$ with $z\sim \sigma _2$, let $f(z)=zx_0$ for some $x_0\in H_{F}$. The for any $x\in H_{F}, \, f(zx)=zxx_0$, so $zx\sim zxx_0$. This just said $x_0\in A_{F}$. Since we assume that $r=0$ (equivalent to $A_{F}=1$), so $x_0=1$. And so $f=\mathrm{id}$. Then $p$ is injective.

By Proposition 7.25, $W(F)$ permutes transitively elements of $F$ conjugate to $\sigma _2$. There are

$$\begin{aligned} \frac{2^{2\delta +2s+\epsilon }+(1-\epsilon ) (-1)^{\delta }2^{\epsilon +\delta +s}}{2}=2^{s+\delta -1} (2^{s+\delta +\epsilon }+ (1-\epsilon )(-1)^{\delta }2^{\epsilon }) \end{aligned}$$

such elements. It is clear that the stabilizer of $W(F)$ at $z$ is $\mathrm{Sp}(s;\epsilon ,\delta )$. So

$$\begin{aligned} |W(F)|=|\mathrm{Sp}(s;\epsilon ,\delta )| 2^{s+\delta -1}(2^{s+\delta +\epsilon }+(1-\epsilon )(-1) ^{\delta }2^{\epsilon }). \end{aligned}$$

By Propositions 2.32 and 2.33, this is also equal to $|\mathrm{Sp}(s+\delta ;\epsilon )|$. Then $p$ is surjective.

$(3)$ follows from $(4)$ immediately.

The proof for the other cases easier, we use the facts that $\mathrm{rank}A_{F}=r$ and the form $m$ on $H_{F}/A_{F}$ is non-degenerate. $\square$

Remark 7.27

We have the following containment relations,

$$\begin{aligned}&F^{\prime \prime \prime }_{\epsilon +r,\delta +s}\subset F^{\prime }_{\epsilon ,\delta ,r,s},\ F^{\prime \prime \prime }_{\epsilon +r,\delta +s}\subset F^{\prime \prime }_{\epsilon +r,\delta +s},\ F^{\prime \prime }_{\epsilon +r,\delta +s}\subset F_{\epsilon ,\delta ,r,s},\\&F^{\prime }_{r+s+\delta }\subset F_{\epsilon ,\delta ,r,s},\quad F^{\prime \prime }_{r+3}\subset F^{\prime }_{r,3}, \end{aligned}$$

together with those obvious relations, they consist in all containment relations (in the sense of conjugacy) between these subgroups

8 $\mathrm{E}_8$

Let $G=\mathrm{Aut}(\mathfrak e _8)$. By Table 2, $G$ has two conjugacy classes of involutions with representatives $\sigma _1,\sigma _2$ and we have

$$\begin{aligned}&G^{\sigma _1}\cong (\mathrm{E}_7\times \mathrm{Sp}(1))/\langle (c,-1)\rangle , \\&G^{\sigma _2}\cong \mathrm{Spin}(16)/\langle c^{\prime }\rangle , \end{aligned}$$

where $c$ is the unique non-trivial central element of $E_7$ and $c^{\prime }=e_1 e_2\ldots e_{16}\in \mathrm{Spin}(16)$.

In $G^{\sigma _1}\cong (\mathrm{E}_7\times \mathrm{Sp}(1))/\langle (c,-1)\rangle$, let $\eta _1,\eta _2\in \mathrm{E}_7$ be involutions such that there exists Klein four groups $F,F^{\prime }\subset \mathrm{E}_7$ with non-identity elements all conjugate to $\eta _1$ or $\eta _2$ respectively and

$$\begin{aligned} {\mathfrak{e }}_7^{F}\cong \mathfrak su (6)\oplus (i\mathbb{R })^2,\ {\mathfrak{e }}_7^{F^{\prime }}\cong \mathfrak so (8)\oplus (\mathfrak sp (1))^{3}. \end{aligned}$$

Then $c\eta _1\sim _{E_7}\eta _2, \, c\eta _2\sim _{E_7}\eta _1$. Let $\tau _1=(\eta _1,1), \tau _2=(\eta _2,1)\in G^{\sigma _1}$. Let $\eta _3,\eta _4\in \mathrm{E}_7$ be involutions with $\eta _3^{2}=\eta _4^{2}=c$ and

$$\begin{aligned} {\mathfrak{e }}_7^{\eta _3}\cong {\mathfrak{e }}_6\oplus i\mathbb{R },\ {\mathfrak{e }}_7^{\eta _4}\cong \mathfrak su (8). \end{aligned}$$

Then $c\eta _3\sim _{\mathrm{E}_7}\eta _3, \, c\eta _4\sim _{\mathrm{E}_7}\eta _4$. Let $\tau _3=(\eta _3,\mathbf{i}), \, \tau _4=(\eta _4,\mathbf{i})$. By [8, Page 17], we see that $\tau _1,\tau _2,\tau _3,\tau _4$ represent all conjugacy classes of involutions in $G^{\sigma _1}$ except $\sigma _1$ and we have the following conjugacy classes in $G$,

$$\begin{aligned}&\tau _1\sim \sigma _1,\ \tau _2\sim \sigma _2, \\&\tau _3\sim \sigma _1,\ \tau _4\sim \sigma _2 \end{aligned}$$

In $G^{\sigma _2}\cong \mathrm{Spin}(16)/\langle c\rangle$, let

$$\begin{aligned} \tau _1&= \ e_1e_2e_3e_4,\ \tau _2=e_1e_2e_3\ldots e_8, \\ \tau _3&= \Pi ,\ \tau _4=-\Pi , \end{aligned}$$

where

$$\begin{aligned} \Pi = \frac{1+e_1e_2}{\sqrt{2}}\frac{1+e_3e_4}{\sqrt{2}}\ldots \frac{1+e_{15} e_{16}}{\sqrt{2}}. \end{aligned}$$

By [8, Page 17], we see that $\tau _1,\tau _2,\tau _3,\tau _4$ represent all conjugacy classes of involutions in $G^{\sigma _2}$ except $\sigma _2$ and we have the following conjugacy classes in $G$,

$$\begin{aligned}&\tau _1\sim \tau _3\sim \sigma _1, \\&\tau _2\sim \tau _4\sim \sigma _2. \end{aligned}$$

Moreover in $G^{\sigma _2}$, we have

$$\begin{aligned}&\sigma _2\tau _1\sim _{G^{\sigma _2}}\tau _1,\ \sigma _2\tau _2\sim _{G^{\sigma _2}}\tau _2, \\&\sigma _2\tau _3\sim _{G^{\sigma _2}}\tau _4,\ \sigma _2\tau _4\sim _{G^{\sigma _2}}\tau _3. \end{aligned}$$

These are obtained from calcualtions in $\mathrm{Spin}(16)/\langle c\rangle$.

Definition 8.1

Let $F$ be an elementary abelian 2-subgroup of $G$. For any $x\in F$ with $x\sim \sigma _1$, let

$$\begin{aligned} H_{x}=\{y\in F|xy\not \sim y\}. \end{aligned}$$

Let

$$\begin{aligned} H_{F}:=\langle \{H_{x}: x\in F, x\sim \sigma _1\}\rangle = \langle \{x: x\in F,x\sim \sigma _1\}\rangle . \end{aligned}$$

Lemma 8.2

Let $F$ be an elementary abelian 2-subgroup of $G$. For any $x$ with $x\sim \sigma _1, \, H_{x}$ is a subgroup and $\mathrm{rank}(F/H_{x})\le 2$.

Proof

We may and do assume that $x=\sigma _1$, then

$$\begin{aligned} F\subset G^{\sigma _1}\cong \mathrm{E}_7\times \mathrm{Sp}(1)/\langle (c,-1)\rangle . \end{aligned}$$

For an element $y\in F\subset G^{\sigma _1}$ with $y^{2}=1, \, \sigma _1 y\not \sim y$ if and only if $y$ is conjugate to $1,\sigma _1,\tau _1,\tau _2$ in $G^{\sigma _1}$. Then it is also equivalent to $y\in \mathrm{E}_7\subset G^{\sigma _1}$. So $H_{x}=F \cap \mathrm{E}_7$. And so it is a subgroup. Then $F/H_{x}\subset G^{\sigma _1}/\mathrm{E}_7\cong \mathrm{Sp}(1)/ \langle -1\rangle$, so $\mathrm{rank}(F/H_{x})\le 2$. $\square$

Definition 8.3

Let $F$ an elementary abelian 2-subgroup of $G$, For any $x\in F$, define $\mu (x)=1$ if $x\sim \sigma _2$ or $x=1$; and $\mu (x)=-1$ if $x\sim \sigma _1$.

For any $x,y\in F$, define $m(x,y)=\mu (x)\mu (y)\mu (xy)$.

In general $m$ is not a bilinear form.

Definition 8.4

For an elementary abelian 2-subgroup $F$ of $G$, define the translation subgroup

$$\begin{aligned} A_{F}=\{x\in F|\mu (x)=1 \, and \, m(x,y)=1 \, for any \, y\in F\} \end{aligned}$$

and the defect index

$$\begin{aligned} \mathrm{defe}(F)=|\{x\in F:\mu (x)=1\}|-|\{x\in F:\mu (x)=-1\}|. \end{aligned}$$

Definition 8.5

For an elementary abelian 2-subgroup $F$ of $G$, we call $\mathrm{Res}(F):=\mathrm{rank}(F/H_{F})$ the residual rank of $F$, and

$$\begin{aligned} \mathrm{Res}^{\prime }(F)=max\{\mathrm{rank}(F/H_{x})|x\in F, x\sim \sigma _1\} \end{aligned}$$

the second residual rank of $F$.

Let $X=X_{F}=\{x\in F| x\sim \sigma _1\}$, define a graph with vertices set $X$ by drawing an edge connecting $x,y\in X$ if and only if $xy\sim \sigma _2$. It is clear that this graph $X$ is invariant under multiplication by elements in $A_{F}$. Let

$$\begin{aligned} \mathrm{Graph}(F)=X_{F}/A_{F} \end{aligned}$$

be the quotient graph of the graph $X_{F}$ modulo the action of $A_{F}$.

8.1 Subgroups from $\mathrm{E}_6$

For an elementary abelian 2-subgroup $F$ of $G$, if $F$ contains a Klein four subgroup conjugate to $\Gamma _1$, we may and do assume that $\Gamma _1=\langle \sigma _1,\tau _3\rangle \subset F$. Then

$$\begin{aligned} F\subset G^{\Gamma _1}=((\mathrm{E}_6\times \mathrm{U}(1)\times \mathrm{U}(1))/\langle (c,e^{\frac{2\pi i}{3}},1)\rangle ) \rtimes \langle \omega \rangle , \end{aligned}$$

where $\omega ^{2}=1, \, ({\mathfrak{e }}_6\oplus i\mathbb{R }\oplus i\mathbb{R })^{\omega }=\mathfrak{f }_4\oplus 0\oplus 0$ and $\Gamma _1=\langle (1,-1,1),(1,1,-1)\rangle$.

Let $G_{\Gamma _1}=\mathrm{E}_6\rtimes \langle \omega \rangle \subset G^{\Gamma _1}$. Let $\pi : G_{\Gamma _1}\longrightarrow \mathrm{Aut}({\mathfrak{e }}_6)$ be the adjoint homomorphism and $p: G_{\Gamma _1}\longrightarrow G^{\Gamma _1}$ be the inclusion. For an elementary abelian 2-subgroup $K$ of $\mathrm{Aut}({\mathfrak{e }}_6), \, p(\pi ^{-1}(K))\times \Gamma _1$ is the direct product of its (unique) Sylow 2-subgroup $F$ and $\langle (c,1,1)\rangle$. Let $\{F_{r,s}:r\le 2,s\le 3\}, \, \{F^{\prime }_{r,s}:r\le 2,s\le 3\}, \, \{F_{\epsilon ,\delta ,r,s}:\epsilon +\delta \le 1,r+s\le 2\}, \, \{F^{\prime }_{\epsilon ,\delta ,r,s}:\epsilon +\delta \le 1,r+s\le 2,s\ge 1\}$ be elementary abelian 2-subgroups of $\mathrm{E}_8$ obtained from elementary abelian 2-subgroups of $\mathrm{Aut}({\mathfrak{e }}_6)$ with the corresponding notation in this way.

Let $\theta _1, \, \theta _2\in \mathrm{E}_6$ be involutions with

$$\begin{aligned} {\mathfrak{e }}_6^{\theta _1}\cong \mathfrak su (6)\oplus \mathfrak sp (1),\ {\mathfrak{e }}_6^{\theta _2}\cong \mathfrak so (10)\oplus i\mathbb{R }. \end{aligned}$$

Let $\theta _3= \omega , \, \theta _4\in \omega \mathrm{E}_6$ be involutions with

$$\begin{aligned} {\mathfrak{e }}_6^{\theta _3}\cong \mathfrak{f }_4\oplus 0\oplus 0,\ {\mathfrak{e }}_6^{\theta _4}\cong \mathfrak sp (4)\oplus 0\oplus 0. \end{aligned}$$

From [8, Pages 16–18] (for Types $\mathbf E_6,\mathbf E_7,\mathbf E_8$), we have

$$\begin{aligned} \theta _1\sim \theta _3\sim \sigma _1 \end{aligned}$$

and

$$\begin{aligned} \theta _2\sim \theta _4\sim \sigma _2. \end{aligned}$$

More over we have

$$\begin{aligned} \theta _1\sigma \sim \theta _4\sigma \sim \sigma _2 \end{aligned}$$

and

$$\begin{aligned} \theta _2\sigma \sim \theta _3\sigma \sim \sigma _1, \end{aligned}$$

for any $\sigma \in \Gamma _1-\{1\}$.

Proposition 8.6

For an elementary abelian 2-subgroup $F$ of $G$, if $F$ contains a Klein four subgroup conjugate to $\Gamma _1$, then $F$ is conjugate to one of $\{F_{r,s}:r\le 2,s\le 3\}, \, \{F^{\prime }_{r,s}:r\le 2,s\le 2\}, \, \{F_{\epsilon ,\delta ,r,s}:\epsilon +\delta \le 1,r+s\le 2\}, \, \{F^{\prime }_{\epsilon ,\delta ,r,s}:\epsilon +\delta \le 1,r+s\le 2,s\ge 1\}$.

Proof

The proof is similar as that for Proposition 7.10. $\square$

Remark 8.7

Note that $F^{\prime }_{r,3}$ contains a rank 3 pure $\sigma _1$ subgroup. By Proposition 8.6, one can show that it is conjugate to $F_{r,2}$.

Proposition 8.8

We have the following formulas for $\mathrm{Res}F, \, \mathrm{Res}^{\prime } F, \, \mathrm{rank}A_{F}$ and $\mathrm{defe}F$,  

• (1) for $F=F_{r,s}, \, r\le 2, \, s\le 3, \, (\mathrm{Res}F,\mathrm{Res}^{\prime } F)=(0,2), \, \mathrm{rank}A_{F}=r, \, \mathrm{defe}F=3\cdot 2^{r+1}(2^{s}-2)$;

• (2) for $F=F^{\prime }_{r,s}, \, r\le 2, \, s\le 2, \, (\mathrm{Res}F,\mathrm{Res}^{\prime } F)=(0,1), \, \mathrm{rank}A_{F}=r, \, \mathrm{defe}F=2^{r+1}(2^{s}-2)$;

• (3) for $F=F_{\epsilon ,\delta ,r,s}, \, \epsilon +\delta \le 1, \, r+s\le 2, \, (\mathrm{Res}F,\mathrm{Res}^{\prime }F)=(1,2), \, \mathrm{rank}A_{F}=r, \, \mathrm{defe}F=(1-\epsilon )(-1)^{\delta +1}2^{r+s+\delta +1}+2^{\epsilon +r+2\delta +2s}$;

• (4) for $F=F^{\prime }_{\epsilon ,\delta ,r,s}, \, \epsilon +\delta \le 1, \, r+s\le 2, \, s\ge 1, \, (\mathrm{Res}F,\mathrm{Res}^{\prime }F)=(0,1), \, \mathrm{rank}A_{F}=r, \, \mathrm{defe}F= (1-\epsilon )(-1)^{\delta +1} 2^{r+s+\delta +1}$.

Proof

These formulas follow from the construction of these subgroups and the comparison of the conjugacy classes of involutions in $G^{\Gamma _1}$ and in $G$. $\square$

Proposition 8.9

The subgroups $\{F_{r,s}:r\le 2,s\le 3\}, \, \{F^{\prime }_{r,s}:r\le 2,s\le 2\}, \, \{F_{\epsilon ,\delta ,r,s}:\epsilon +\delta \le 1, r+s\le 2\}, \, \{F^{\prime }_{\epsilon ,\delta ,r,s}:\epsilon +\delta \le 1, r+s\le 2,s\ge 1\}$ are not conjugate to each other.

Proof

The numbers $(\mathrm{Res}F,\mathrm{Res}^{\prime } F,\mathrm{rank}A_{F},\mathrm{rank}F,\mathrm{defe}F)$ clearly distinguish most of these c onjugacy classes except for some possible pairs $(F^{\prime }_{r^{\prime },s^{\prime }},F^{\prime }_{\epsilon ,\delta ,r,s})$. Suppose that some $(F^{\prime }_{r^{\prime },s^{\prime }}$ is conjugate to some $F^{\prime }_{\epsilon ,\delta ,r,s})$. By the formulas in Proposition 8.8, we have $r^{\prime }=r$ (by $A_{F}$), $s^{\prime }-1=(1-\epsilon )(-1)^{\delta }$ (by the sign of $\mathrm{defe}F$) and $s^{\prime }=2s+2\delta +\epsilon$ (by $\mathrm{rank}F/A_{F}$). Since $s^{\prime }\le 2$ and $s\ge 1$, the last equality implies that $\epsilon =\delta =0, \, s=1$ and $s^{\prime }=2$. Then the second equality implies that $s^{\prime }=1$. So we get a contradiction. $\square$

8.2 Other subgroups

In $G^{\sigma _1}\cong (\mathrm{E}_7\times \mathrm{Sp}(1))/\langle (c,-1)\rangle$, choose $x_1,x_2\in \mathrm{E}_7$ with $x_1\sim x_2\sim x_1x_2\sim \tau _4$, then

$$\begin{aligned} (G^{\sigma _1})^{x_1,x_2}=\mathrm{SO}(8)/\langle -I\rangle \times \langle \sigma _1,x_1,x_2\rangle . \end{aligned}$$

Let $z_1=\mathrm{diag}\{-I_4,I_4\}$,

$$\begin{aligned} z_2&= \mathrm{diag}\{-I_2,I_2,-I_2,I_2\}, \\ z_3&= \mathrm{diag}\{-1,1,-1,1,-1,1,-1,1\}. \end{aligned}$$

Define

$$\begin{aligned} F^{\prime \prime }_{r,s}=\langle z_1,\ldots ,z_{r},\sigma _1,x_1,\ldots ,x_{s}\rangle \end{aligned}$$

for any $r\le 3, \, s\le 2$,

In $G^{\sigma _2}\cong \mathrm{Spin}(16)/\langle c\rangle , \, c=e_1e_2\ldots e_{16}$, let $y_1=\sigma _1=-1$,

$$\begin{aligned} y_2&= e_1e_2e_3e_4e_5e_6e_7e_8, \\ y_3&= e_1e_2e_3e_4e_{9}e_{10}e_{11}e_{12}, \\ y_4&= e_1e_2e_5e_6e_9e_{10}e_{13}e_{14}, \\ y_5&= e_1e_3e_5e_7e_9e_{11}e_{13}e_{15}. \end{aligned}$$

Define $F^{\prime }_{r}=\langle y_1,\ldots ,y_{r}\rangle$ for any $r\le 5$.

Lemma 8.10

For an elementary abelian 2-subgroup $F$ of $G$, if $F$ contains no Klein four subgroup conjugate to $\Gamma _1$, but contains an element conjugate to $\sigma _1$, then

$$\begin{aligned} \mathrm{rank}H_{F}/A_{F}=1. \end{aligned}$$

Proof

Recall that, $H_{F}$ is a subgroup of $F$ generated by elemnts conjugate to $\sigma _1$. Let

$$\begin{aligned} Y_{F}=\{x\in H_{F}:x\sim \sigma _2\}\cup \{1\}. \end{aligned}$$

We show that $A_{F}=Y_{F}$ under the assumption of the lemma.

Choose any $x_0\in F$ with $x_0\sim \sigma _1$. For any other $x\in F$ with $x\sim \sigma _1$, since $F$ contains no Klein four subgroup conjugate to $\Gamma _1$, so $xx_0\sim \sigma _2$. Then $x\in H_{x_0}$. By this, we get that $H_{F}\subset H_{x_0}$. So $H_{x_{0}}=H_{F}$ as the containment relation in the converse direction is obvious. Similarly we have $H_{x}=H_{F}$ for any $x\in F$ with $x\sim \sigma _1$. Then for any two distinct $y_1,y_2\in Y_{F}$ with $y_1\sim y_2\sim \sigma _2$, we have $y_1y_2\sim \sigma _2$. So $Y_{F}$ is a subgroup of $H_{F}$.

Then it is clear that $Y_{F}=A_{F}$. So $\mathrm{rank}H_{F}/A_{F}=\mathrm{rank}H_{F}/Y_{F}=1$. $\square$

Proposition 8.11

For an elementary abelian 2-subgroup $F$ of $G$, if $F$ contains no Klein four subgroup conjugate to $\Gamma _1$, then $F$ is conjugate to one of $\{F^{\prime \prime }_{r,s}:r\le 3,s\le 2\},\{F^{\prime }_{r}:r\le 5\}$.

Proof

When $F$ contains no Klein four subgroup conjugate to $\Gamma _1$, but contains an element conjugate to $\sigma _1$, we may and do assume that $\sigma _1\in F$. Then

$$\begin{aligned} F\subset G^{\sigma _1}\cong (\mathrm{E}_7\times \mathrm{Sp}(1))/\langle (c,-1)\rangle . \end{aligned}$$

Modulo $\mathrm{Sp}(1)$, we get a homomorphism

$$\begin{aligned} \pi : F\longrightarrow \mathrm{E}_7/\langle c\rangle \cong \mathrm{Aut}({\mathfrak{e }}_7). \end{aligned}$$

Since we assume that $F$ contains no Klein four subgroup conjugate to $\Gamma _1$, so any element in $F-\langle \sigma _1\rangle$ is conjugate to $\tau _1=(\eta _1,1), \, \tau _2=(\eta _2,1)$ or $\tau _4=(\eta _4,\mathbf{i})$ in $(\mathrm{E}_7\times \mathrm{Sp}(1))/\langle (c,-1)\rangle$; and any Klein four subgroup of $F\cap \mathrm{E}_7$ has at least one element conjugate to $\eta _2$. Then $F^{\prime }=\pi (F)\subset \mathrm{Aut}({\mathfrak{e }}_7)$ contains no elements conjugate to $\eta _3$, and no Klein four subgroups whose fixed point subalgebra is isomorphic to $\mathfrak su (6)\oplus (i\mathbb{R })^2$. In the case of $\mathrm{E}_7$ (Sect. 7), it corresponds to the elementary abelian 2-subgroup $F^{\prime }$ with no elements conjugate to $\sigma _2$ and the map $m$ on $H_{F^{\prime }}$ is trivial. By Propositions 7.14 and 7.22, we get that $F\sim F^{\prime \prime }_{r,s}$ for some $(r,s)$ with $r\le 3,s\le 2$.

When $F$ is pure $\sigma _2$, we may and do assume that $\sigma _2\in F$. Then

$$\begin{aligned} F\subset G^{\sigma _2}\cong \mathrm{Spin}(16)/\langle c\rangle \end{aligned}$$

and any element in $F-\langle \sigma _2\rangle$ is conjugate to $e_1e_2e_3e_4e_5e_6e_7e_8$ in $\mathrm{Spin}(16)/\langle c\rangle$. Then $F\sim F^{\prime }_{r}$ for some $r\le 5$. $\square$

Proposition 8.12

For any $(r,s)$ with $r\le 3$ and $s\le 2$, we have $\mathrm{rank}A_{F^{\prime \prime }_{r,s}}=r$; for any $r\le 5$, we have $\mathrm{rank}A_{F^{\prime }_{r}}=r$.

Any two subgroups in $\{F^{\prime \prime }_{r,s}: r\le 3,s\le 2\},\{F^{\prime }_{r}: r\le 5\}$ are non-conjugate.

Proof

The equalities $\mathrm{rank}A_{F^{\prime \prime \prime }_{r,s}}=r$ and $\mathrm{rank}A_{F^{\prime \prime }_{r}}=r$ are clear. By them, we get that any two subgroups in $\{F^{\prime \prime \prime }_{r,s}: r+s\le 3\},\{F^{\prime \prime }_{r}: r\le 2\}$ are non-conjugate. $\square$

8.3 Involution types and Automizer groups

Corollary 8.13

$G$ has 66 conjugacy classes of elementary abelian 2-subgroups.

Proof

By Propositions 8.6, 8.9, 8.11, 8.12, we get that $G$ has

$$\begin{aligned} 3\times 4+3\times 3+3\times 6+3\times 3+4\times 3+6=66 \end{aligned}$$

conjugacy classes of elementary abelian 2-subgroups. $\square$

Proposition 8.14

For an isomorphism $f:F\longrightarrow F^{\prime }$ between two elementary abelian 2-subgroups of $G$, if $f(x)\sim x$ for any $x\in F$, then $f=\mathrm{Ad}(g)$ for some $g\in G$.

Proof

When $F$ contains a Klein four subgroup conjugate to $\Gamma _1$, this reduces to the similar statement in $\mathrm{Aut}({\mathfrak{e }}_6)$ case.

When $F$ does not contain any Klein four subgroup conjugate to $\Gamma _1$, this is already showed in the proof of Proposition 8.11. $\square$

Definition 8.15

For an elementary abelian 2-subgroup $F$ of $G$, we say that $F$ is the orthogonal direct product of other subgroups $K_1,\ldots ,K_{t}$ if there exists an isomorphism $f: K_1\times \cdots \times K_{t}\longrightarrow F$ with

$$\begin{aligned} \mu (f(x_1,\ldots ,x_{t}))=\mu (x_1)\ldots \mu (x_{t}) \end{aligned}$$

for any $(x_1,\ldots ,x_{t})\in K_1\times \cdots \times K_{t}$.

Let $A=\langle \sigma _2\rangle$. Let $B_{s}$($s\le 3$) be a rank $s$ pure $\sigma _1$ subgroup. Let $B=B_1, \, C=F_3$ and $D$ be a rank 3 subgroup with only one element conjugate to $\sigma _1$. Then the involution types of some elementary abelian 2-subgroups of $\mathrm{E}_8$ have the following description

$$\begin{aligned}&F_{r,s}=A^{r}\times B_{s}\times B_{3};\ F^{\prime }_{r,s}=A^{r}\times B_{s}\times B_2;\\&F^{\prime }_{\epsilon ,\delta ,r,s}=A^{r}\times C^{s}\times B^{\epsilon }\times B_2^{1+\delta };\\&F^{\prime \prime }_{r,1}=A^{r}\times B,\ F^{\prime \prime }_{r,2}=A_{r}\times C,\\&F^{\prime \prime }_{r,3}=A^{r}\times D;\ F^{\prime }_{r}=A^{r} \end{aligned}$$

$F_{\epsilon ,\delta ,r,s}$ ($s\ge 1$) does not have a similar decomposition since elements in $F_{\epsilon ,\delta ,r,s}-F^{\prime }_{\epsilon ,\delta ,r,s}$ are all conjugate to $\sigma _2$.

With the involution types available, we can describe the graphs $\mathrm{Graph}(F)$. The graphs of $F_{r,s}$ is a complete bipartite graph with $s,3$ vertices in two parts; that of $F^{\prime }_{r,s}$ is a complete bipartite graph with $s,2$ in two parts; that of $F^{\prime \prime }_{r,s}$ ($s\ge 1$) is a single vertex graph; that of $F^{\prime }_{r}$ is an empty graph. The graphs of $F_{\epsilon ,\delta ,r,s},F^{\prime }_{\epsilon ,\delta ,r,s}$ are not of bipartite form and a little more complicated.

In summary, we have the following statement

“the conjugacy class of an elementary abelian 2-subgroup $F\subset G$ is determined by the datum $(\mathrm{rank}F, \mathrm{rank}A_{F}, \mathrm{Graph}(F))$”.

Proposition 8.16

For an elementary abelian 2-subgroup $F\subset \mathrm{E}_8, \, m$ is a bilinear form on $F$ if and only if $F$ is not conjugate to any of $\{F_{r,s}:r\le 2,s\le 3\}\cup \{F_{\epsilon ,\delta ,r,s}:\epsilon +\delta \le 1,r+s\le 2\}\cup \{F^{\prime \prime }_{r,3}:r\le 2\}$.

Proof

When $F$ is conjugate to one of $\{F_{r,s}:r\le 2,s\le 3\}\cup \{F_{\epsilon ,\delta ,r,s}:\epsilon +\delta \le 1,r+s\le 2\}$, it contains a subgroup conjugate to $B_3,F_{0,0,0,0}$ or $D$. The subgroups $B_3,F_{0,0,0,0},D$ contains $7,3,1$ elements with $\mu$-value -1 respectively, so $m$ is not bilinear on them by Proposition 2.30.

When $F$ is conjugate to a subgroup in the other four families, $m$ is bilinear on $F$ follows from the orthogonal decomposition of it. $\square$

We can write the decomposition of involution types for some subgroups in a simpler way,

$$\begin{aligned}&F^{\prime }_{r,0}=A^{r}\times B_2,\\&F^{\prime }_{r,1}=A^{r}\times B\times B_2=A^{r}\times B\times C,\\&F^{\prime }_{r,2}=A^{r}\times B_2\times B_2=A^{r}\times C\times C,\\&F^{\prime }_{1,0,r,s}=A^{r}\times C^{s}\times B\times B_2^{1} =A^{r}\times B\times C^{s+1},\\&F^{\prime }_{0,\delta ,r,s}=A^{r}\times C^{s}\times B_2^{1+\delta } =A^{r}\times B_2^{1-\delta }\times C^{s+2\delta }. \end{aligned}$$

Proposition 8.17

 

• (1) $r\le 2, \, s\le 2, \, W(F_{r,s})\cong \mathrm{Hom}(\mathbb{F }_2^{3+s},\mathbb{F }_2^{r})\rtimes \big (\mathrm{GL}(r,F_2)\times (\mathrm{GL}(s,\mathbb{F }_2)\times \mathrm{GL}(3,\mathbb{F }_2))\big )$;

• (2) $r\le 2, \, W(F_{r,3})\cong \mathrm{Hom}(\mathbb{F }_2^{6},\mathbb{F }_2^{r})\rtimes \big (\mathrm{GL}(r, \mathbb F _2)\times ((\mathrm{GL}(3,\mathbb{F }_2)\times \mathrm{GL}(3,\mathbb{F }_2))\rtimes S_2)\big )$;

• (3) $r\le 2, \, s\le 2, \, W(F^{\prime }_{r,s})\cong \mathrm{Sp}(r,s;2s-s^{2},\frac{(s-1)(s-2)}{2})$;

• (4) $\epsilon +\delta \le 1, \, r+s\le 2, \, W(F_{\epsilon ,\delta ,r,s})=\mathbb{F }_2^{r+2s+\epsilon +2\delta +2} \rtimes \mathrm{Sp}(r,s+\epsilon +2\delta ;\epsilon ,(1-\epsilon )(1-\delta ))$;

• (5) $\epsilon +\delta \le 1, \, r+s\le 2, \, W(F^{\prime }_{\epsilon ,\delta ,r,s})=\mathrm{Sp}(r,s+\epsilon +2\delta ;\epsilon , (1-\epsilon )(1-\delta ))$.

• (6) $r\le 3, \, s\le 2, \, W(F^{\prime \prime }_{r,s})\cong \mathrm{Hom}(\mathbb{F }_2^{s},\mathbb{F }_2^{r+1})\rtimes \big ( (\mathbb F _2^{r}\rtimes \mathrm{GL}(r,\mathbb{F }_2))\times \mathrm{GL}(s))\big )$;

• (7) $r\le 5, \, W(F^{\prime }_{r})\cong \mathrm{GL}(r,\mathbb F _2)$.

Proof

By Proposition 8.14, we need to calculate automorphisms of $F$ preserving the function $\mu$ on $F$.

$W(F_{r,s})=\mathrm{Hom}(\mathbb{F }_2^{3+s},\mathbb{F }_2^{r})\rtimes W(F_{0,s})$ and $W(F_{0,s})$ stabilizes $B_{s}\cup B_3\subset F_{0,s}$. By this we get $(1)$ and $(2)$.

When $m$ is bilinear, $(F,m,\mu )$ is a symplectic metric space, then we can identify $W(F)$ with the automorphism group of $(F,m,\mu )$. By this we get (3) and (5).

$(4)$ follows from $(5)$ immediately.

For (6), we have $A_{F}\subset H_{F}\subset F$ and $A_{F},H_{F}$ are preserved by $W(F)$. By Lemma 8.10, we have $\mathrm{rank}A_{F}=r, \, \mathrm{rank}H_{F}/A_{F}=1, \, \mathrm{rank}F/H_{F}=1$, then we get the conclusion.

(7) is clear. $\square$

We have an inclusion $p:\mathrm{E}_7\subset \mathrm{E}_8$ since $\mathrm{E}_8^{\sigma _1}\cong (\mathrm{E}_7\times \mathrm{Sp}(1))/ \langle (c,-1)\rangle$. Let $\pi :\mathrm{E}_7\longrightarrow \mathrm{Aut}(\mathfrak e _7)$ be the adjoint homomorphism, which is a 2-fold covering. For a pure $\sigma _1$ (that for $\mathrm{E}_7$ case) elementary abelian 2-subgroup $F$ of $\mathrm{Aut}(\mathfrak e _7), \, p(\pi ^{-1}F)$ is an elementary abelian 2-subgroup of $\mathrm{E}_8$.

Proposition 8.18

An elementary abelin 2-subgroup $F$ of $\mathrm{E}_8$ is conjugate to the subgroup $p(\pi ^{-1}(K))$ for some pure $\sigma _1$ subgroup $K$ of $\mathrm{Aut}(\mathfrak e _7)$ if and only if $F$ contains an elementary $x$ such that $x\sim \sigma _1$ and $H_{x}=F$.

Proof

It follows from the description of the conjugacy classes of involutions in $\mathrm{E}_8^{\sigma _1}\cong (\mathrm{E}_7\times \mathrm{Sp}(1))/\langle (c,-1)\rangle$. $\square$

Remark 8.19

Any subgroup of $\mathrm{E}_8$ satisfying the condition in Proposition 8.18 is conjugate to one of

$$\begin{aligned} \{F_{r,1}: r\le 2\},\ \{F^{\prime }_{r,1}: r\le 2\},\ \{F^{\prime }_{1,0,r,s}: r+s\le 2, s\ge 1\},\ \{F^{\prime \prime }_{r,1}: r\le 3\}. \end{aligned}$$

There are 13 such conjugacy classes in total. On the other hand, there are 13 classes of pure $\sigma _1$ subgroups of $\mathrm{Aut}(\mathfrak e _7)$, so for any two elementary abelian 2-subgroups $K_1, \, K_2$ of $\mathrm{Aut}(\mathfrak e _7)$, we have

$$\begin{aligned} p(\pi ^{-1} K_1)\sim _{\,\mathrm{E}_8} p(\pi ^{-1} K_2) \Leftrightarrow K_1\sim _{\mathrm{Aut}(\mathfrak e _7)} K_2. \end{aligned}$$