Why, conceptually, does the torus normalizer in $G_2$ split?

Question

Background: Let $G$ be a complex reductive group, $T$ a maximal torus, $N$ the normalizer of $T$ in $G$, and $W = N/T$ the Weyl group. All in all, we have a group extension $$ 1 \to T \to N \to W \to 1 $$ of a finite group by a torus. Say $N$ "splits" if this extension splits, that is, if $W$ lifts to a subgroup of $N$.

It is well known that $N$ splits when $G=\text{GL}_n$: here $T$ is the diagonal matrices, $N$ is the group of monomial matrices, and a splitting is given by the permutation matrices. On the other hand, it is not true for every $G$ that the torus normalizer splits, the simplest example being $\text{SL}_n$ ($n\geq2$), where one cannot use the permutation matrices because the odd ones have determinant $-1$. It's a fun exercise to work out the other classical groups.

The group $N$ comes up all over in Lie theory and it was first studied, as far as I know, by Tits in a 1966 paper (zbmath). More recently, Moshe Adrian (zbmath) has completely answered the question of when $N$ splits, for $G$ almost simple. In particular:

Theorem: If $G$ is almost simple and exceptional then $N$ splits if and only if $G$ is of type $G_2$.

This result is already suggested (and maybe even directly implied) by work of Gal't in exceptional finite reductive groups (mathscinet, in Russian). I'm interested in the "exceptional" case of $G_2$, where $W$ is the dihedral group of order $12$.

Question: For the group $G_2$, is there a "simple" or "conceptual" explanation why the torus normalizer splits?

For example, $G_2$ is the automorphism group of the octonions $\mathbb O$ and one should be able to describe everything in these terms. Already the book of Springer and Veldkamp (zbmath) describes $T$ (Lemma 2.3.1) as automorphisms of $\mathbb O$ and my hope is that someone on this site who has drunk deep from the spring of the octonions can also describe $N$ and the splitting.

Alternatively, Tits's group for $G_2$ is of order $48$ and one should be able to recognize it as a semidirect product, though the computations involved intimidate me.

Answer 1

Here's a description that doesn't use octonions; instead, it uses the definition of $\mathrm{G}_2$ as the stabilizer of a $3$-form on $\mathbb{R}^7$. For simplicity, I'll do this for the split-form, but a slight modification would work for the compact form or the complex form.

Let $\mathbb{R}^7$ have basis $e$, $f_1$, $f_2$, $f_3$, $g_1$, $g_2$, $g_3$ and consider the $3$-vector $$ E = e{\wedge}\bigl( f_1{\wedge}g_1 + f_2{\wedge} g_2 + f_3{\wedge} g_3\bigr) + f_1{\wedge} f_2{\wedge} f_3 + g_1{\wedge} g_2{\wedge} g_3\,. $$ It's classical (F. Engel, 1900), that the $\mathrm{GL}(7,\mathbb{R})$-stabilizer of $E\in\Lambda^3(\mathbb{R}^7)$ is isomorphic to (split) $\mathrm G_2$. (This $E$ looks special, but the $\mathrm{GL}(7,\mathbb{R})$-orbit of $E$ is an open set in $\Lambda^3(\mathbb{R}^7)$, as is not difficult to prove. Thus, fixing $E$ is $35$ equations on $g\in\mathrm{GL}(7,\mathbb{R})$, which is exactly the codimension of $\mathrm{G}_2$ in $\mathrm{GL}(7,\mathbb{R})$. In that sense, this is the most efficient definition of $\mathrm{G}_2$, as $\mathrm{G}_2 = \mathrm{Aut}(\mathbb{O})$ has codimension $50 = 64-14$ in $\mathrm{GL}(\mathbb{O},\mathbb{R})\simeq\mathrm{GL}(8,\mathbb{R})$. The realization that the compact form of $\mathrm{G}_2$ is the automorphism group of $\mathbb{O}$ dates from a 1908 article of Élie Cartan.)

An obvious abelian subgroup of $\mathrm G_2$ is the subgroup $T\subset (\mathbb{R}^\times)^3$ consisting of triples $\lambda = (\lambda_1,\lambda_2,\lambda_3)$ (where $\lambda_i\not=0$) satisfying $\lambda_1\lambda_2\lambda_3=1$, where $\lambda$ acts on $\mathbb{R}^7$ as $$ \lambda\cdot(e\ \ f_1\ \ f_2\ \ f_3\ \ g_1\ \ g_2\ \ g_3) = (e\ \ \lambda_1 f_1\ \ \lambda_2 f_2\ \ \lambda_3f_3 \ \ \lambda_1^{-1}g_1\ \ \lambda_2^{-1}g_2\ \ \lambda_3^{-1}g_3). $$ (Thus, this basis provides a diagonal representation of the maximal torus in $\mathrm{G}_2$.) There is an obvious element $\iota\in \mathrm G_2$ that acts as $$ \iota\cdot(e\ \ f_1\ \ f_2\ \ f_3\ \ g_1\ \ g_2\ \ g_3) = (-e\ \ g_1\ \ g_2\ \ g_3\ \ f_1\ \ f_2\ \ f_3). $$ (Note that $\iota$ conjugates $\lambda$ to $\lambda^{-1}$.) There is also a homomorphism of $S_3$ into $\mathrm G_2$ given by $$ \pi\cdot(e\ \ f_1\ \ f_2\ \ f_3\ \ g_1\ \ g_2\ \ g_3) = (e\ \ {\pm f_{\pi(1)}}\ \ {\pm f_{\pi(2)}}\ \ {\pm f_{\pi(3)}}\ \ {\pm g_{\pi(1)}}\ \ {\pm g_{\pi(2)}}\ \ {\pm g_{\pi(3)}}) $$ with the $+$ if $\pi\in S_3$ is an even permutation and the $-$ sign when $\pi\in S_3$ is an odd permutation.

Then $S_3$ and $\iota$ clearly generate a 12-element group $W$ that normalizes $T\subset\mathrm{G}_2$ and $W\cap T$ consists of the identity transformation (since no non-identity element of $W$ preserves the $7$ lines spanned by the given basis). This is the desired splitting.

Answer 2

Fix an $\mathbb{H} \subset \mathbb{O}$. This choice selects an involution of $\mathbb{O}$ which is $+1$ on $\mathbb{H}$ and $-1$ on $\mathbb{H}^\perp$. The centralizer of this involution is an $SO(4)$, acting on $\mathbb{H}^\perp$ as the vector representation and on $\mathbb{H}$ as (trivial plus) the action on self-dual 2-forms. Any inclusion of a centralizer of a diagonalizable element is automatically of full rank. So this lets you "see" the maximal torus of $G_2$ as the maximal torus of $SO(4)$.

How much of its normalizer can you see in this description? The Weyl of $SO(4)$ is a Klein-4 group, mapping into the Weyl of $G_2$. If you know already that the Weyl of $G_2$ is dihedral of order $12$, then you know that on Tits groups, this map $SO(4) \to G_2$ is a Sylow 2-subgroup. How do you know that the Weyl of $G_2$ is (at most) of order $12$? You can do it by hand: decompose $\mathbb{O}$ over $Spin(2) \times Spin(2) \to Spin(3) \times Spin(3) = Spin(4) \to G_2$, and directly see a regular hexagon.

But splitability of a group of shape $(\text{2-group}).(\text{quotient})$ is detected on Sylow 2-subgroups. So to see that the Tits group of $G_2$ splits (equivalently, by Tits' observation, splitability of the full torus normalizer) it is equivalent to see that the Tits group of $SO(4)$ splits.

In $SO(4)$, the maximal torus is $$ \begin{pmatrix} \cos \alpha & \sin \alpha && \\ -\sin \alpha & \cos \alpha && \\ && \cos \beta & \sin \beta \\ && -\sin \beta & \cos \beta \end{pmatrix}.$$ The Weyl is a Klein-4, as I said. It lifts to $SO(4)$ by $$ \begin{pmatrix} & 1 && \\ 1 & && \\ && & 1 \\ && 1 & \end{pmatrix}, \quad \begin{pmatrix} && 1 & \\ && & 1 \\ 1 & && \\ & 1 && \end{pmatrix} $$ These act by sending $(\alpha,\beta)$ to $(-\alpha,-\beta)$ and to $(\beta,\alpha)$, respectively.

Answer 3

Let $\mathbb{O}$ denote the octonions, in the so-called Zorn model. One writes elements of $\mathbb{O}$ as $2 \times 2$ matrices $x = \left(\begin{array}{cc} a & v \\ \phi & d \end{array}\right)$ where $v$ is in the standard representation of $\mathrm{SL}_3$ and $\phi$ is in the dual representation. Let $b_1, b_2, b_3$ be the standard basis of the standard representation of $\mathrm{SL}_3$ and $b_1', b_2', b_3'$ the dual basis. For ease of notation, I identify these elements with elements of the octonions $\mathbb{O}$.

Consider the map $w: \mathbb{O} \rightarrow \mathbb{O}$ that switches the components of the diagaonal matrices, i.e., $w(\mathrm{diag}(a,d)) = \mathrm{diag}(d,a)$, and on the off-diagonal matrices takes

$$ b_1 \mapsto -b_2' \mapsto b_3 \mapsto -b_1' \mapsto b_2 \mapsto -b_3' \mapsto b_1.$$

I haven't checked it, but it may be that $w$ preserves the multiplication on $\mathbb{O}$, and thus is an element of $G_2$. It has order $6$.

For an element of order $2$, let $\tau$ be the image in $G_2$ of the matrix

$$\left(\begin{array}{ccc} -1 & & \\ & & -1 \\ & -1 & \end{array}\right)$$

under the map $\mathrm{SL}_3 \rightarrow G_2$.

The group generated by $w$ and $\tau$ may give you the dihedral group of order $12$ you are looking for.

Word of caution: I remember that there might be a typo in the definition of the multiplication on $\mathbb{O}$ in Springer-Veldkamp, so one would need to proceed carefully.

Answer 4

While the following might not be conceptual, but at least it's simple if one knows group cohomology. The obstruction to splitting lives in $H^2(W;T)$, and we have $H^n(W;T)=0$ for all $n\geq 0$!

One way to see this is the following: Let $S$ be the Sylow $3$-subgroup of $W$. As $S$ is normal in $W$, the Lyndon-Hochschild-Serre spectral sequence converging to $H^*(W;T)$ is given by $E_2^{p,q} = H^p(W/S;H^q(S;T))$. Since $S\cong C_3$ a computation shows that $H^q(S;T)\cong T^S=\mathbb{Z}/3$ for $q$ even and $H^q(S;T)=0$ for $q$ odd. Since $W/S\cong (C_2)^2$ is a $2$-group we get $E_2^{p,q}=0$ for $p>0$. For $p=0$, $q$ even, we have $E_2^{0,q}=(T^S)^{W/S} = T^W = 0$. Thus $E_2^{p,q}=0$ for all $p,q\geq 0$ and we are done.

Answer 5

$\newcommand\ldef{\mathrel{:=}}$Let $\alpha$ be the short, and $\beta$ the long, simple root of $G_2$. Thus $\langle\alpha^\vee, \beta\rangle$ equals $-3$ and $\langle\beta^\vee, \alpha\rangle$ equals $-1$.

Tits's elements $a_r$ satisfy $a_r^2 = r^\vee(-1)$. In place of $a_\alpha$, take $a_\alpha' \ldef \beta^\vee(-1)a_\alpha$, so that $a_\alpha^{\prime\,2} $ equals $(w_\alpha\beta^\vee + \beta^\vee)(-1)a_\alpha^2 = (\alpha^\vee + 2\beta^\vee)(-1)\alpha^\vee(-1) = 1$; and in place of $a_\beta$, take $a_\beta' \ldef \alpha^\vee(-1)a_\beta$, so that $a_\beta^{\prime\,2} = (w_\beta\alpha^\vee + \alpha^\vee)(-1)a_\beta^2 = (2\alpha^\vee + 3\beta^\vee)(-1)\beta^\vee(-1) = 1$.

We have that $(a_\alpha' a_\beta')^3$ equals $$(\beta^\vee + w_\alpha\alpha^\vee + (w_\alpha w_\beta)\beta^\vee + (w_\alpha w_\beta w_\alpha)\alpha^\vee + (w_\alpha w_\beta)^2\beta^\vee + (w_\alpha w_\beta)^2 w_\alpha\alpha^\vee)(-1)(a_\alpha a_\beta)^3 = (-2(3\alpha^\vee + 4\beta^\vee))(-1)(a_\alpha a_\beta)^3$$ and $(a_\beta' a_\alpha')^3$ equals $$(\alpha^\vee + w_\beta\beta^\vee + (w_\beta w_\alpha)\alpha^\vee + (w_\beta w_\alpha w_\beta)\beta^\vee + (w_\beta w_\alpha)^2\alpha^\vee + (w_\beta w_\alpha)^2 w_\beta\beta^\vee)(-1)(a_\beta a_\alpha)^3 = (-2(2\alpha^\vee + 5\beta^\vee))(-1)(a_\beta a_\alpha)^3,$$ so $(a_\alpha' a_\beta')^3$ equals $(a_\beta' a_\alpha')^3$.

Thus, for the split form of $G_2$, the map $w_\alpha \mapsto a_\alpha'$, $w_\beta \mapsto a_\beta'$ is a splitting of the projection from the Tits group to the Weyl group.

Actually, this is just the first splitting I tried, and it worked. Another amusing choice might be to take $a_\alpha'' = \rho^\vee(-1)a_\alpha$ and $a_\beta'' = \rho^\vee(-1)b_\beta$, where $\rho^\vee$ is the half-sum of the positive coroots. Then again it's relatively easy to check that these are involutions, and now $(a_\alpha''a_\beta'')^3(a_\alpha a_\beta)^{-3}$ equals $$ ((1 + w_\alpha + w_\alpha w_\beta + w_\alpha w_\beta w_\alpha + (w_\alpha w_\beta)^2 + (w_\alpha w_\beta)^2 w_\beta)\rho^\vee)(-1) = (10\beta^\vee)(-1) $$ and $(a_\beta''a_\alpha'')^3(a_\beta a_\alpha)^{-3}$ equals $$ ((1 + w_\beta + w_\beta w_\alpha + w_\beta w_\alpha w_\beta + (w_\beta w_\alpha)^2 + (w_\beta w_\alpha)^2 w_\beta)\rho^\vee)(-1) = (6\alpha^\vee)(-1). $$ This has the advantage that it works also for the compact form of $G_2$.

Answer 6

There are already many lovely answers, but since the question mused whether one can resolve this by using the size 48 "extended Weyl group" defined in Tits' 1966 paper "Normalisateurs de tores. I. Groupes de Coxeter étendus", I thought I'd add that as yet another option, working solely with its presentation in terms of generators and relations.

Specifically, we can use the presentation from section 4.6 of that paper. Specialized to $G_2$, it is as follows: $$ W^{ext}(G_2) := \langle q_1, q_2 \mid q_1^4=q_2^4=1,\ [q_1^2,q_2^2]=1,\ (q_1q_2)^3 = (q_2q_1)^3,\ q_1q_2^2q_1^{-1} = q_2^2 q_1^2,\ q_2q_1^2q_2^{-1} = q_1^2 q_2^6 \rangle $$ Things become a bit clearer if we introduce two auxiliary generators $t_1,t_2$ that correspond to torus elements, and simplify the result a bit (using Tietze transformations): $$ W^{ext}(G_2) \cong N := \langle q_1, q_2, t_1, t_2 \mid q_1^2=t_1,\ q_2^2=t_2,\ (q_1q_2)^3 = (q_2q_1)^3,\ t_1^2=t_2^2=1,\ [t_1,t_2]=1,\ t_2^{q_1} = t_1^{q_2} = t_1 t_2 \rangle $$ (Note that I used among other things that ${}^{q_1}t_2 = t_2^{q_1}$ is a consequence of the relations).

Then the subgroup $T:=\{1,t_1,t_2,t_1t_2\}$ is normal in $N$ and isomorphic to $C_2^2$. Now set $s_1:=q_1t_2$ and $s_2:=q_2 t_1$. Then I claim $U:=\langle s_1, s_2\rangle$ is isomorphic to $W(G_2)$, a dihedral group of order 12, and we have $N=T\rtimes U$.

Indeed $s_1^2 = q_1t_2 q_1t_2 = q_1^2 t_2^{q_1} t_2 = t_1 (t_1 t_2) t_2 = 1$ and likewise $s_2^2=1$. Moreover, using that $t_i$ and $q_i$ commute, $$ s_1s_2s_1 = q_1t_2 \cdot q_2 t_1 \cdot q_1t_2 = q_1 q_2 t_2 q_1 t_1 t_2 = q_1 q_2 q_1 t_2^{q_1} (t_1 t_2) = q_1 q_2 q_1 .$$ By symmetry we also have $s_2s_1s_2= q_1 q_2 q_1$, therefore $$ (s_1s_2)^3 = (s_1s_2s_1)(s_2s_1s_2) = (q_1q_2)^3 = (q_2 q_1)^3 = \dots = (s_2s_1)^3. $$ Hence $U$ satisfies the relations of the Coxeter presentation of $W(G_2)$ and so is a quotient of that.

It is not hard to see that it is equal if one is willing to believe Tit's that $W^{ext}(G_2)\cong N$ has 48 elements: then it is routine to verify that $N/T$ consists of 12 cosets with representatives $$1, q_1, q_2,\ q_1q_2, q_2q_1,\ q_1q_2q_1, q_2q_1q_2, \dots, (q_1q_2)^3=(q_2q_1)^3 $$ and that $U$ contains at least (and hence exactly) one element of each coset.

If one is not willing to believe Tits at this point, one can do all kinds of thing, and e.g. ask GAP, OSCAR, Magma, ... to verify it. Here's some GAP code to do just that:

gap> F:=FreeGroup("q1", "q2");
<free group on the generators [ q1, q2 ]>
gap> AssignGeneratorVariables(F);
#I  Assigned the global variables [ q1, q2 ]
gap> W := F / [q1^4, q2^4, Comm(q1^2,q2^2), (q1*q2)^3/(q2*q1)^3, q1*q2^2*q1^-1/(q2^2*q1^2), q2*q1^2*q2^-1/(q1^2*q2^6) ];
<fp group on the generators [ q1, q2 ]>
gap> Size(W);
48
gap> T := Subgroup(W, [W.1^2, W.2^2]);
Group([ q1^2, q2^2 ])
gap> StructureDescription(T);
"C2 x C2"
gap> U := Subgroup(W, [W.1*W.2^2, W.2*W.1^2]);
Group([ q1*q2^2, q2*q1^2 ])
gap> StructureDescription(U);
"D12"
gap> Size(Intersection(T,U));
1

Naturally we could also just have asked GAP to find a complement to T:

gap> ComplementClassesRepresentatives(W, T);
[ Group([ (q2^-1*q1^-1)^2*q2^-3, q1^-1*q2^2*q1^2, (q1^-1*q2^-1)^2 ]) ]