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Apr 13, 2017 at 12:19 history edited CommunityBot
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Apr 24, 2015 at 18:18 vote accept jdc
Apr 24, 2015 at 17:41 answer added Jim Humphreys timeline score: 3
Apr 20, 2015 at 8:00 answer added Jay Taylor timeline score: 5
Apr 20, 2015 at 5:44 history edited jdc CC BY-SA 3.0
Improved statement by omitting mention of roots.
Apr 20, 2015 at 4:25 history edited jdc CC BY-SA 3.0
Fixed grammar, improved phrasing.
Apr 19, 2015 at 23:59 comment added jdc This suggests to me (I don't really understand diagram folding well, so I apologize if there's something obviously wrong in what follows) that the injection of Weyl groups should follow from folding. It also seems from your description that folding implies it should be a sufficient condition on $\mathfrak s$ that it lie within the tangent space to a maximal torus of an $F_4$ subgroup. I don't understand why necessity should follow from this line of reasoning, though, or indeed if you are claiming it should.
Apr 19, 2015 at 21:46 comment added Jay Taylor I would suggest this is very much related to the fact that the longest element of the Weyl group of type E6 is $-\sigma$ where $\sigma$ is the automorphism of the root system induced by the unique non-trivial automorphism of the Dynkin diagram. Thus the longest element acts as $-1$ on anything fixed by the diagram automorphism. The fixed point subgroup under $\sigma$ is the group obtained by the diagram fold, which is just the Weyl group of type F4.
Apr 19, 2015 at 21:12 history edited jdc
Added some tags in hopes of attracting readers able to answer; not having "reference request" was a clear mistake.
Apr 19, 2015 at 2:29 history asked jdc CC BY-SA 3.0