Degree of Parabolic Quantum Groups - Dipartimento di Matematica ...

4. General Theory 49
Thus we introduce the skew symmetric matrix A = (aij) with aij = (βi|βj)
if i < j.
Let ki the class of Ki, relations 4.9 we obtain a n × N matrix
B = ((ωi|βj)) 1≤i≤n,1≤j≤N
Let t = 2 unless the Cartan matrix is of type G2 in which case t = 6, and
let Z ′ = Z
1 . We wish to think the matrix A as the matrix of a skew form
t
on a free Z ′ module V with basis u1, . . .,uN. Identifying V with its dual
V ∗ using the given basis, we may also think A as linear operator from V to
itself.While we may think of the matrix B as a linear map from the module
V to the module Q∗ ⊗Z Z ′ , where Q∗ = HomZ(Λ, Z) be the dual lattice.
Construct the matrix T:
T =
A − t B
B 0
T is the matrix associated to the twisted polynomial algebra Uǫ(b). To study
this the matrix we need the following
Lemma 4.5.4. Let w ∈ W and fix a reduced expression w = si1 · · ·sik .
Given ω = n i=1 δiωi, with δi = 0 or 1. Set
Then
,
Iω(w) := {t ∈ {1, . . .,k} : sit(ω) = ω} .
ω − w(ω) =
Proof. We proceed by induction on the length of w. The hypothesis made
implies si(ω) = ω or si(ω) = ω − αi. Write w = w ′ sik . If k /∈ Iω, then
w(ω) = w ′ (ω) and we are done by induction. Otherwise
and again we are done by induction.
t∈Iω
βt
w(ω) = w ′ (ω − αik ) = w′ (ω) − βk
Consider the operator T1 = A − t B and N = B 0 so that
T = T1 ⊕ N.
Lemma 4.5.5. (i) The operator T1 is surjective
(ii) The vector vω :=
t∈Iω ut
− ω − w0(ω), as ω run thought the fundamental
weights, form a basis of the kernel of T1.
(iii) N(vω) = ω − w0(ω) =
t∈Iω βt.
Proof. See [DCKP95] or [DCP93] §10.