Lecture 5- Deformed preprojective algebras cont'd

# Lecture 5- Deformed preprojective algebras cont'd -...

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• jweng1292
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LECTURES ON SYMPLECTIC REFLECTION ALGEBRAS IVAN LOSEV 4. Deformed preprojective algebras, cont’d 4.1. Recap. Recall that in the previous lecture we have identified C x, y #Γ with T C Γ ( C 2 C Γ), and C Q with T ( C Q ) 0 ( C Q ) 1 . Also recall that f C x, y f = C Q , where f = r i =0 f i C Γ = r i =0 End( N i ) with f i being a primitive idempotent in End( N i ) . Under this identi- fication, f i C Q becomes the path ϵ i . Further, to i Q 0 we have assigned an element [ a , a ] i ϵ i ( C Q ) 2 ϵ i by the formula [ a , a ] i = a Q 1 ,t ( a )= i a a a Q 1 ,h ( a )= i aa . Also to c ( C Γ) Γ we assign λ = ( λ i ) i Q 0 by λ i = tr N i c . The main result we are going to prove is a theorem of Crawley-Boevey and Holland. Theorem 4.1. The ideal f ( xy yx c ) C x,y f is generated by the elements [ a , a ] i λ i ϵ i , i Q 0 . A key step in the proof is the following lemma again due to Crawley-Boevey and Holland. Lemma 4.2. To each a Q 1 one can associate η a Hom Γ ( N t ( a ) , C 2 N h ( a ) ) , θ a Hom Γ ( N h ( a ) , C 2 N t ( a ) ) that combine to form bases in the spaces Hom Γ ( N i , C 2 N j ) are all i, j and satisfy (1) a Q 1 ,t ( a )= i (1 C 2 θ a ) η a a Q 1 ,h ( a )= i (1 C 2 η a ) θ a = δ i ( ζ 1 N i ) , (the equality of maps N i C 2 C 2 N i ) for all i . To prove the lemma we have introduced explicit mutually inverse isomorphisms of the spaces Hom Γ ( M, C 2 M ) , Hom Γ ( C 2 M, M ). Namely, we map ψ Hom Γ ( M, C 2 M ) to ψ := ( ω 1 M ) (1 C 2 ψ ) Hom Γ ( C 2 M, M ), and we map φ Hom Γ ( C 2 M, M ) to (1 C 2 φ ) ( ζ 1 M ) Hom( M, C 2 M ). Here ω is the skew-symmetric form on C 2 given by ω ( y, x ) = 1 = ω ( x, y ) (and viewed as a map C 2 C 2 C ) and ζ = x y y x (viewed as a map C C 2 C 2 ). 4.2. Proof of the CBH lemma. From now on we concentrate on the non-cyclic case. A special feature of this case is that Γ is a tree. The spaces Hom Γ ( N i , C 2 N j ) when i = t ( a ) , j = h ( a ) or vice versa are 1-dimensional. For a moment, choose arbitrary nonzero η a , θ a , they are defined up to a nonzero scalar multiple. Then θ a η a is a nonzero endomorphism of N t ( a ) , while η a θ a is a nonzero endomorphism of N h ( a ) . Multiplying θ a by a nonzero scalar k , we also multiply those two endomorphisms by k . We claim that there are nonzero scalars d i , i Q 0 , with the property that (after rescaling the θ i ’s) we get (2) θ a η a = d h ( a ) 1 N t ( a ) , η a θ a = d t ( a ) 1 N h ( a ) . 1

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2 IVAN LOSEV This is a consequence of Q being a tree. Namely, we fix all η a and some vertex i . Pick d i . This fixes θ a for all a with h ( a ) = i (from θ a η a = d h ( a ) 1 N t ( a ) ) or t ( a ) = i (from η a θ a = d t ( a ) 1 N h ( a ) ) and so also d j for all vertices j connected to i (for example, if h ( a ) = i , then d t ( a ) is determined from η a θ a = d t ( a ) 1 N h ( a ) , where we now know the left hand side). Then we proceed with i replaced by one of these j ’s. Since our graph is a tree, we see that every vertex appears only
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• Fall '12
• IvanLosev
• Algebra, Hamiltonian mechanics, Symplectic manifold, Symplectic geometry, Cγ, xy − yx

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