Lecture 6- Symplectic reflection algebras

Lecture 6- Symplectic reflection algebras - LECTURES ON...

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LECTURES ON SYMPLECTIC REFLECTION ALGEBRAS IVAN LOSEV 6. Symplectic reflection algebras 6.1. Definition of SRA. Let V be a finite dimensional complex vector space equipped with a non-degenerate skew-symmetric form ω . Then there is a distinguished filtered deformation of the symmetric algebra S ( V ). This is the Weyl algebra W ( V ) = T ( V ) / ( u v v u ω ( u, v )). Exercise 6.1. Show that W ( V ) is a filtered deformation of S ( V ) (the case dim V = 2 was considered above). Moreover, check that the Poisson bracket on S ( V ) induced from W ( V ) coincides with the initial bracket. Let Γ be a finite subgroup of Sp( V ). We want to study filtered deformations of S ( V ) Γ that are compatible with the Poisson bracket on S ( V ) Γ . For non-commutative deformations, this means that the bracket on S ( V ) Γ induced by the deformation has to coincide with (or be proportional to) the initial bracket on S ( V ) Γ . One can state a compatibility condition for commutative deformations as well (they itself have to be Poisson algebras) but we are not going to do this. As before, we are going to produce deformations of S ( V )#Γ first (and then pass to spher- ical subalgebras to get deformations of S ( V ) Γ , we will recall how this is done below). As in the case when dim V = 2, we have S ( V )#Γ = T ( V )#Γ / ( u v v u | u, v V ). So we can take a linear map κ : 2 V ( T ( V )#Γ) 6 1 and form the quotient H κ = T ( V )#Γ / ( u v v u κ ( u, v )). Of course, H 0 = S ( V )#Γ. Exercise 6.2. Show that if gr H κ = S ( V )#Γ , then κ is a Γ -equivariant map (where Γ acts on C Γ via the adjoint representation). Furthermore, show that if 1 V Γ , then the image of κ lies in C Γ . In general, it is still a good idea to consider only κ : 2 V C Γ. This is motivated, in part, by our compatibility condition of Poisson brackets: we want filtered deformations A of S ( V ) Γ with [ A 6 i , A 6 j ] ⊂ A i + j 2 . Of course, H κ deforms S ( V )#Γ, not S ( V ) Γ , but it is still reasonable to require that [ u, v ] H 6 0 κ (that should be equal to C Γ). So below we only consider Γ-equivariant κ with image in C Γ. We can write κ as γ Γ κ γ γ , where κ γ 2 V . It turns out that for gr H κ = S ( V )#Γ some κ γ must vanish. We map V to H κ via V , T ( V )#Γ H κ , this is an embedding whenever gr H κ = S ( V )#Γ. In H κ we must have [[ u, v ] , w ] + [[ v, w ] , u ] + [[ w, u ] , v ] = 0 for all u, v, w V , equivalently, (1) [ κ ( u, v ) , w ] + [ κ ( v, w ) , u )] + [ κ ( w, u ) , v ] = 0 . Exercise 6.3. We have [ κ ( u, v ) , w ] = γ Γ κ γ ( u, v )( γ ( w ) w ) γ . Since gr H κ = S ( V )#Γ, the map V C Γ H κ induced by the embedding V C Γ T ( V )#Γ is injective. It follows that the equalities (2) κ γ ( u, v )( γ ( w ) w ) + κ γ ( v, w )( γ ( u ) u ) + κ γ ( w, u )( γ ( v ) v ) = 0 1
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2 IVAN LOSEV hold for any γ Γ. Let us show that if rk( γ 1 V ) > 2, then κ γ = 0. Indeed, if u, v, w are such that γ ( u ) u, γ ( v ) v, γ ( w ) w are linearly independent, then we must have κ γ ( u, v ) = κ γ ( v, w ) = κ γ ( w, u ) = 0. But the linear independence definitely holds for vectors in general position provided rk( γ 1 V ) > 2, so κ is 0.
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