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I want to know the number of components of the normalizer of an arbitrary circle subgroup $S$ of (the compact real form of) the exceptional Lie group $E_6$. This number will always be $1$ or $2$.

Write $T$ for a maximal torus of $E_6$. The number will be $2$ precisely when there is an element $w$ of the Weyl group $W = N(T)/T$ of $E_6$ such that for every $s \in S$, we have $\mathrm{Ad}(w)s = s^{-1}$.

On the Lie algebra level, if $X \in \mathfrak s < \mathfrak t$ is a nonzero tangent vector to $S$, this requires the existence of an element $w \in W$ such that $\mathrm{Ad}(w)X = -X$, or equivalently, that $\mathfrak s$ be a $(-1)$-eigenspace for some element under the adjoint action of $W$.

Now circles in $T$ correspond (up to multiplication by $\pm 1$) to primitive elements of $\pi_1(T) \cong \ker(\exp\colon \mathfrak t \to T) =: \Lambda$, so my project reduces to finding points in the integer lattice that meet $(-1)$-eigenspaces of $W$.

The most obvious brute-force way of accomplishing this goal seems to involve

  • finding a presentation somewhere of this $W$-action in terms of simple reflections through simple roots,
  • finding the lattice of fundamental weights in terms of the simple roots,
  • determining the action of $W$ on this lattice,
  • identifying the weight lattice with the coweight lattice through the nondegneracy of the Killing form, and recalling that the coweight lattice is the $\Lambda$ I'm actually interested in, in this case,
  • playing around until I find a way to convince myself I've found all the $(-1)$-eigenspaces. I know that all involutions in $W$ correspond to products of reflections through mutually orthogonal roots, so that should help a bit.

But I am not convinced that this process is the most efficient way to do what I want, nor that the answer I find in the last step, presumably in a rather ad hoc way, will be complete.

What else should I know before starting on this? Am I missing some better way to do it?

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    $\begingroup$ I have never thought about this, but feel impressed nevertheless. $\endgroup$ Jan 11, 2015 at 2:55
  • $\begingroup$ That's very kind of you. Thank you for saying so. $\endgroup$
    – jdc
    Jan 11, 2015 at 3:01
  • $\begingroup$ Just wanted to say hi, and sorry for being MIA lately. $\endgroup$ Jan 11, 2015 at 3:43
  • $\begingroup$ Hi there! No worries. I don't remember what last I sent you, but I'm sure it wasn't that critical. I'm glad to see you still patrol this place. $\endgroup$
    – jdc
    Jan 11, 2015 at 3:50

2 Answers 2

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I exchanged emails with the author of this question, so I thought I will post the answer here. I only write what I know :)

The normalizer of subgroup S in group E is {$x \in E$: $xsx^{-1}$ belongs to S for any $s \in S$}.

I propose following base vectors for $E_6$ lattice. It is located in $7$-dimensional space and sum of first three coordinates is $0$. $[0,1,-1, 1,1,0,0]/2, [0,0,0, -1,-1,1,1]/2, [0,0,0, -1,1,-1,-1]/2, [0,0,0, 1,-1,-1,1]/2, [-1,0,1, 1,0,1,0]/2, [1,0,-1, 0,-1,0,-1]/2;$ Remaining 30 vectors can be obtained by adding pairs of vectors having scalar product ${-^1/_2}$. I consider only one vector from pair {v,-v} belonging to E6 lattice.

As checked in GAP (see www.gap-system.org) the Weyl group of E6 - called further $W$ - has size 51840 and structure $U4(2).2$. The group $U4(2)$ can be seen on atlas of finite simple groups. The group $W$ contains 25 conjugacy classes, in this 4 conjugacy classes of involutions of sizes 270, 45, 540, 36. The involutions can be built as product of elementary reflections in 1,2,3,4 perpendicular vectors of the lattice.

The 4 perpendicular vectors in $E_6$ lattice span $SO_8$ lattice. So my guess is that the answer is 2 when circle S belongs to certain $SO_8$ subgroup of $E_6$ although I don't know how to prove all the steps.

Regards, Marek

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  • $\begingroup$ I may have asked this in the email, but do you know a reference for the involutions all being products of $\leq 4$ orthogonal simple reflections? Do you know why involutions suffice? I mean, it feels reasonable that they would be, and I'm in the process of trying to install GAP to manually check that the $(-1)$-eigenspaces of order-$2^n$ elements are all contained in those of involutions, but it'd be nice to have some theoretical justification. (math.stackexchange.com/questions/896398/…) $\endgroup$
    – jdc
    Aug 14, 2014 at 14:22
  • $\begingroup$ They do suffice. I've checked in GAP. Do you also use GAP to check that a quadruple of orthogonal root vectors span a four-dimensional vector space containing a D_4 lattice, or do you have some theoretical way of seeing it? $\endgroup$
    – jdc
    Aug 22, 2014 at 0:24
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Marek's answer can be expanded, very briefly, as follows. Replace $E_6$ with its universal cover $\tilde E_6$.

Every circle in $Spin(8)$ is reflected, so every circle in a $Spin(8)$ subgroup is reflected.

Within any given maximal torus of $\tilde E_6$, there are forty-five four-dimensional subtori that are contained in a $Spin(8)$. These span forty-five $D_4$ sublattices of the $E_6$ root lattice. These sublattices are reflected by products of four mutually orthogonal root reflections.

Because all roots in $E_6$ and $D_4$ are long, the Killing form takes these isometrically and $W(E_6)$-equivariantly to forty-five $D_4$ sublattices of the $\tilde E_6$ coroot lattice, which is the same as the integer lattice since $\tilde E_6$ is simply connected. The projectivization of this lattice is the set of circles in a maximal torus.

This construction identifies (the Lie algebras of maximal tori of $Spin(8)$ subgroups) that lie within a given $\mathfrak{t} \cong \mathbb{R}^6$ to the $D_4$ sublattices of the $E_6$ lattice, so it remains only to verify that all $(-1)$-eigenspaces lie within these 45 sublattices.

This can be done mechanically (by GAP, say).

To lessen the number of computations, note that it suffices to check for one representative of every conjugacy class within the Weyl group $W(\tilde E_6)$ of elements of even-power order that the $(-1)$-eigenspace lies within one of the 45. Because $w\cdot v = -v$ implies $w^k \cdot v = -v$ for $k$ odd, other orders are irrelevant.

A computer can then run the necessary checks in under two seconds.

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  • $\begingroup$ I was told much later a much simpler reason why this works: if anything reflects these elements, the longest word $w_0$ does. Much of this is set out here: mathoverflow.net/questions/203295/… $\endgroup$
    – jdc
    Feb 15, 2016 at 18:39

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