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

3. Twisted polynomial algebras 29
(a) xixj = bijxjxi + Pij if i > j, where bij ∈ k, Pij ∈ A i−1 .
(b) A i is a finite module over Z i 0 .
(c) Formulas σi(xj) = bijxj for j < i define an automorphism of A i−1
which is the identity on Z i−1
0 .
Note that letting Di(xj) = Pij for j < i, we obtain A i = A i−1
σ,D [xi],
so that A is an iterated twisted polynomial algebra. We may consider the
twisted polynomial algebra A i with zero derivations, so that the relations
are xixj = bijxjxi for j < i. we call this the associated twisted polynomial
algebra.
We can state the main theorem of this section.
Theorem 3.3.8. Under the above assumptions, the degree of A is equal to
the degree of the associated quasi polynomial algebra A.
Proof. By theorem 3.2.9 A is obtained from A with a sequence of simple
degenerations, hence by corollary 3.3.7, it follows that they have the same
degree.
3.4 Representation theory of twisted polynomial algebras
Let k a field and 0 = q ∈ k a given element. Given n × n skew symmetric
matrix H = (hij) over Z, we construct the twisted polynomial algebra
kH[x1, . . .,xn]. This is the algebra on generators x1, . . .,xn and the following
defining relations:
(3.2)
xixj = q hij xjxi
for i, j = 1, . . .,n. It can be viewed as an iterated twisted polynomial algebra
with respect to any order of the xi’s. Similarly we can define the twisted Lau-
rent polynomial algebra kH[x1, x −1
1 , . . .,xn, x −1
n ]. Note that both algebras
have no zero divisors.
To study its spectrum we start with a simple general lemma.
Lemma 3.4.1. If M is an irreducible Aσ[x] module, then there are two
possibilities:
(i) x = 0, hence M is actually an A module.
(ii) x is invertible, hence M is actually an Aσ[x, x −1 ] module.
Proof. It is clear that im x and ker x are submodules of M.
Corollary 3.4.2. In any irreducible kH[x1, . . .,xn] module, each element xi
is either 0 or invertible.