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How many Weyl Chambers/bases does $ B_2$ have?

I thought it was 8, but if instead of for bases using obtuse root pairs you use orthogonal pairs, you get 8 different chambers intersect partially with the chambers associated with obtuse pairs.

This also happens with $ G_2$.

Thanks for the help

dylan7
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If I am not mistaken, the type $B_{2}$ corresponds to Lie algebra $\mathcal{so}(3)$. Its Weyl group should have order $$2\times 6=12$$ as it should be isomorphic to $\mathbb{Z}_{2}\rtimes S_{3}$. So the Weyl chamber should have equally many chambers.

Bombyx mori
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  • But can Weyl chambers intersect? – dylan7 Aug 15 '14 at 00:15
  • I do not think they would intersect. – Bombyx mori Aug 15 '14 at 00:17
  • There's something strange I found. Say for $ B_2$ you choose an orthogonal pair of simple roots: find the chamber their hyperplanes enclose, which is eachother. Then say take one of those roots and create a new base with that root and one obtuse to it. Find their chamber enclosed by their hyperplanes and notice that these two chambers intersect. – dylan7 Aug 15 '14 at 00:21
  • To be honest, I do not really know what you are talking about. You should be able to construct the Weyl chambers yourself. In my past experience I never had this happened. And I do not think this would happen to $\mathcal{so}(3)$. – Bombyx mori Aug 15 '14 at 00:26
  • Ah, I got. I just realized I wasn't considering one of the axioms for simple roots to be considered a base. The all coefficients must be the same sign axiom. – dylan7 Aug 15 '14 at 01:14
  • I think the order of the Weyl group of type $B_2$ is $8$, and it is isomorphic to the dihedral group $D_4$...... – Z. Liu Apr 11 '23 at 06:46