Re: Sphere packing in six dimensions, root system for E6, Wikipedia article


On 4-Oct-2009, David Bernier <david250@xxxxxxxxxxxx>
wrote in message <haaud101ik9@xxxxxxxxxxxxxxxxx>:

[...]

In books such as "Lie Groups and Algebras, beyond an Introduction",
it's explained that simple roots generate the root lattice
(as an abelian group with vector addition),

Yes, as long as you scale the roots appropriately to satisfy the
crystallographic condition.

and
inclusion of Dynkin diagrams induces an inclusion of
root systems (something like that).

Yes, that's an almost trivial result of Coxeter group theory.

By deleting the alpha_8 and alpha_7 simple roots, the furthest
from the root connected to three others in the E8 Dynkin diagram,
I believe we get roots that generate the E6 root lattice ...

Right.

Out of the Z-span W of {alpha_i}_{ i =1, ... 6},
I think alpha_8 will be orthogonal to everything in W,
but alpha_7 won't be orthogonal to everything in W, e.g.
<alpha_6, alpha_7 > =/= 0 . On Wikipedia, someone
may have made a slip, thinking that alpha_7 is orthogonal
to all of W ...

Yes, you're correct for alpha_8 and alpha_7. I've looked at the
Wikipedia article, and it's definitely wrong. I should edit it when
I get a chance.

[...]

I wrote my program(s) about a week or 10 days ago. Getting the E8 root
system Phi_E8 (actually 2Phi_E8, changing +/- 1/2 to +/- 1, etc.) was
straightforward, just following instructions from sources that agreed.

For the E7 root system, I had the program try each of the 240 roots
in the E8 root system and compute how many vectors in Phi_E8 were
orthogonal to the chosen root. It always came out to 126.

Yes. E8 acts transitively on its root system, and the stabilizer of
a root is E7.

Then I tried pairs of roots from Phi_E8. With {alpha, -alpha},
I got 126 orthogonal roots.

I should hope so. Clearly a root beta is orthogonal to a root
alpha if and only if beta is also orthogonal to -alpha.

Then I believe I tried all pairs
{alpha, beta} in Phi_E8 with alpha and beta orthogonal,
counting orthogonal roots as usual. That didn't seem to work,
but I wasn't sure why. Then I tried {alpha, beta} with
2 < alpha, beta >/<alpha, alpha> = -1 (pi/3 angle), and that
gave 72 vectors, then I gave an answer to the kissing balls
question. [ But after reading your first reply,
I thought I might have made mistakes somewhere, or
forgotten some detail.]

Taking alpha and beta orthogonal didn't work because it's simply a
fact that the 2-dimensional plane P orthogonal to the
6-dimensional subspace spanned by the roots of an E6 has exactly
six roots of E8, arranged in the A2 configuration. That is, any two
linearly independent roots in P subtend an angle of pi/3 or 2pi/3.

To see this, you can take the vector-space basis
{omega_1, ..., omega_8} dual to the basis of the simple roots
{alpha_1, ..., alpha_8}, and look for all roots in the span of
{omega_7, omega_8}. See a previous post of mine where I calculated
omega_7 and omega_8 (I called them s_7 and s_8) for a particular
choice of {alpha_1, ..., alpha_8} (I called them {r_1, ..., r_8}).

When I replied to you last time, I took this:
"The root system E6 is the set of vectors in E7 that are perpendicular
to a fixed root in E7."

That's the line in the Wikipedia article that's wrong. As I
explained previously, the roots of E7 that are perpendicular to any
given root form the root system of D6, not E6.

to mean: ...perpendicular to some (particular) root of E7.

Doesn't matter. All roots are the same in this case, because E7,
like E8, acts transitively on its root system.

Then I looked for more references, and then I put these things off
till later.

--
Jim Heckman
.

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