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Unformatted text preview: ght zero part of W is W0 = F hα . So, W contains only one
irreducible V (even). Since we know that W contains Sα ∼ V (2), there cannot be any
other V (even). So, L2α = 0. In other words, twice a root cannot be a root. But then 1 α
also cannot be a root since twice of it is a root. And this implies that W1 = Lα/2 = 0.
So, W = V (2) is irreducible. This also implies that Lα is one dimensional.
Corollary 8.4.4. If α, β ∈ Φ and β is not ±α then the α-root string through β is irreducible and has the form:
V (m) ∼ Lβ +qα ⊕ Lβ +(q−1)α ⊕ · · · ⊕ Lβ −rα
and m = q + r. And β (hα ) = q − r is an integer. (β (hα ) are the Cartan integers.)
Proof. Let M be the α-root string through β . Then M0 = 0 since the root string does
not go through 0. (The proof of the last proposition showed that Sα = Lα ⊕ F hα ⊕ L−α
is the only α-root string through 0.) Therefore, M is a direct sum of V (odd)s. So, the
α-root string has only Modd but a diﬀerence of 2 in hα -weights corresponds to a diﬀerence
of α in roots since α(hα ) = 2. So, the root string contains only Lβ +kα for integer k . One
of these is M1 and thus is 1-dimensional. So, M is irreducible.
If M ∼ V (m) then Mm = Lβ +qα and M−m = Lβ −rα . The dimension of M is q + r + 1 =
m + 1. This implies that
q + r = m = (β + q α)(hα ) = β (hα ) + 2q
So, β (hα ) = r − q ∈ Z. 8.5. Inner product on H ∗ . We just proved that β (hα ) ∈ Z. We will rephrase this in
terms of the inner product on H ∗ .
Recall that hα = κ(t2αα α ) . And,...
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This document was uploaded on 03/09/2014 for the course MATH 223a at Brandeis.
- Fall '11