Isotropic Schur roots Charles Paquette University of Connecticut May 1 st , 2016 joint with Jerzy Weyman Auslander Conference 2016, Woods Hole, MA
Outline Describe the perpendicular category of an isotropic Schur root. Describe the cone of dimension vectors of the above. Describe the ring of semi-invariants of an isotropic Schur root. Construct all isotropic Schur roots. Auslander Conference 2016, Woods Hole, MA
Quivers, dimension vectors k = ¯ k is an algebraically closed field. Q = ( Q 0 , Q 1 ) is an acyclic quiver with Q 0 = { 1 , 2 , . . . , n } . rep ( Q ) denotes the category of finite dimensional representations of Q over k . An element d ∈ ( Z ≥ 0 ) n is a dimension vector. Given M ∈ rep ( Q ), we denote by d M its dimension vector. Auslander Conference 2016, Woods Hole, MA
Geometry of quivers For d = ( d 1 , . . . , d n ) a dimension vector, denote by rep ( Q , d ) the set of representations M with M ( i ) = k d i . rep ( Q , d ) is an affine space. For such a d , we set GL ( d ) = � 1 ≤ i ≤ n GL d i ( k ). The group GL ( d ) acts on rep ( Q , d ) and for M ∈ rep ( Q , d ) a representation, GL ( d ) · M is its isomorphism class in rep ( Q , d ). Auslander Conference 2016, Woods Hole, MA
Bilinear form and roots We denote by �− , −� the Euler-Ringel form of Q . For M , N ∈ rep ( Q ), we have � d M , d N � = dim k Hom ( M , N ) − dim k Ext 1 ( M , N ) . Auslander Conference 2016, Woods Hole, MA
Roots and Schur roots d ∈ ( Z ≥ 0 ) n is a (positive) root if d = d M for some indecomposable M ∈ rep ( Q ). Then � d , d � ≤ 1 and we call d : real , if � d , d � = 1; isotropic , if � d , d � = 0; imaginary , if � d , d � < 0; A representation M is Schur if End ( M ) = k . If M is a Schur representation, then d M is a Schur root. We have real, isotropic and imaginary Schur roots. { iso. classes of excep. repr. } 1 − 1 ← → { real Schur roots } . Auslander Conference 2016, Woods Hole, MA
Perpendicular categories For d a dimension vector, we set A ( d ) the subcategory A ( d ) = { X ∈ rep ( Q ) | Hom ( X , N ) = 0 = Ext 1 ( X , N ) for some N ∈ rep ( Q , d ) } . A ( d ) is an exact extension-closed abelian subcategory of rep ( Q ). If V is rigid (in particular, exceptional), then A ( d V ) = ⊥ V . Proposition (-, Weyman) For a dimension vector d , A ( d ) is generated by an exceptional sequence ⇔ d is the dimension vector of a rigid representation. Auslander Conference 2016, Woods Hole, MA
Perpendicular category of an isotropic Schur root Let δ be an isotropic Schur root of Q (so � δ , δ � = 0). Proposition (-, Weyman) There is an exceptional sequence ( M n − 2 , . . . , M 1 ) in rep ( Q ) where all M i are simples in A ( δ ) . Complete this to a full exceptional sequence ( M n − 2 , . . . , M 1 , V , W ). Auslander Conference 2016, Woods Hole, MA
Perpendicular category of an isotropic Schur root · · We have { M n − 2 , . . . , M 2 , M 1 } = J ∪ K ∪ L . For I ⊆ { M n − 2 , . . . , M 2 , M 1 } , let E ( I ) denote the corr. exceptional sequence. We consider the following: ( E ( J ∪ K ) , V , W ) . Now, the M j and M k are pairwise orthogonal. Thus, we get the following: ( E ( J ) , E ( K ) , V , W ) . Reflecting V , W yields: ( E ( J ) , V ′ , W ′ , E ( K )) . Consider R ( Q , δ ) := Thick ( E ( J ) , V ′ , W ′ ). Auslander Conference 2016, Woods Hole, MA
Perpendicular category of an isotropic Schur root Theorem (-, Weyman) The category R ( Q , δ ) is tame connected with isotropic Schur root ¯ δ . It is uniquely determined by ( Q , δ ) . The simple objects in A ( δ ) are: The M i with M i ∈ K ∪ L, The quasi-simple objects of R ( Q , δ ) (which includes the M i with M i ∈ J). In particular, the dimension vectors of those simple objects are either ¯ δ or finitely many real Schur root. This gives the −�− , δ � -stable dimension vectors. Auslander Conference 2016, Woods Hole, MA
Cone of dimension vectors Take the cone in R n of all dimension vectors in A ( δ ). This cone lives in dimension n − 1 since it satisfies the equation �− , δ � = 0. Take an affine slice of it. This becomes an ( n − 2) dimensional polyhedron P . Auslander Conference 2016, Woods Hole, MA
Cone of dimension vectors Let V be the set of vertices of P and for v ∈ V , let P v the convex hull of the points in V \{ v } . Only one dimension vector of simples is not real ⇒ we have | � v ∈ V P v | ≤ 1. Here are examples of such a cone in dimension 1 , 2 , 3: Auslander Conference 2016, Woods Hole, MA
Cone of dimension vectors Theorem (-, Weyman, H. Thomas ) Assume that P is an ( n − 2) -dimensional convex hull of points V with | � v ∈ V P v | ≤ 1 . Then R n − 2 = V 1 ⊕ · · · ⊕ V r where V i contains dim V i + 1 points in V ∪ { 0 } forming a dim V i -simplex containing the origin. Auslander Conference 2016, Woods Hole, MA
� � � � � An example Consider the quiver 2 1 4 3 We take δ = (3 , 2 , 3 , 1). We get an exceptional sequence whose dimension vectors are ((8 , 3 , 3 , 3) , (0 , 0 , 1 , 0) , (0 , 1 , 0 , 0) , (3 , 3 , 3 , 1)). We have δ = (3 , 3 , 3 , 1) − (0 , 1 , 0 , 0). δ does not lie in the τ -orbit of (1 , 1 , 0 , 1) or (1 , 0 , 1 , 1). ¯ δ = (3 , 2 , 1 , 1). Simple objects in A ( δ ) are of dimension vectors (0 , 0 , 1 , 0) , (8 , 3 , 3 , 3) or (3 , 2 , 1 , 1). We have R ( Q , δ ) of Kronecker type. Auslander Conference 2016, Woods Hole, MA
An example ¯ • δ • δ (0 , 0 , 1 , 0) • • (8 , 3 , 3 , 3) Figure : The cone of dimension vectors for δ = (3 , 2 , 3 , 1) Auslander Conference 2016, Woods Hole, MA
Semi-invariants Since Q is acyclic, the ring of invariants k [ rep ( Q , d )] GL ( d ) is trivial. Take SL ( d ) = � 1 ≤ i ≤ n SL d i ( k ) ⊂ GL ( d ). The ring SI ( Q , d ) := k [ rep ( Q , d )] SL ( d ) is the ring of semi-invariants of Q of dimension vector d . We have SI ( Q , d ) = � τ ∈ Γ SI ( Q , d ) τ where Γ = Hom Z ( Z n , Z ) . This ring is always finitely generated. Auslander Conference 2016, Woods Hole, MA
Semi-invariants f X For X ∈ rep ( Q ), let 0 → P 1 → P 0 → X → 0 be a projective resolution of X . For M ∈ rep ( Q ), the map Hom ( f X , M ) Hom ( P 1 , M ) → Ext 1 ( X , M ) → 0 0 → Hom ( X , M ) → Hom ( P 0 , M ) − → is given by a square matrix ⇔ � d X , d M � = 0 . We set C X ( M ) := detHom ( f X , M ). Proposition (Derksen-Weyman, Schofield-Van den Bergh) The function C X ( − ) is a non-zero semi-invariant of weight � d X , −� in SI ( Q , d ) provided � d X , d � = 0 . Moreover, these semi-invariants span SI ( Q , d ) over k. Auslander Conference 2016, Woods Hole, MA
Ring of semi-invariants of an isotropic Schur root The ring SI ( Q , d ) is generated by the C X ( − ) where X is simple in A ( d ). Theorem (-, Weyman) We have SI ( Q , δ ) ∼ = SI ( R , ¯ δ )[ { C M j ( − ) | j ∈ K ∪ L } ] . Corollary By a result of Skowro ´ nski - Weyman, SI ( Q , δ ) is a polynomial ring or a hypersurface. Auslander Conference 2016, Woods Hole, MA
Exceptional sequences of isotropic types A full exceptional sequence E = ( X 1 , . . . , X n ) is of isotropic type if there are X i , X i +1 such that Thick( X i , X i +1 ) is tame. Isotropic position is i and root type δ E is the unique iso. Schur root in Thick( X i , X i +1 ). The braid group B n acts on full exceptional sequences. This induces an action of B n − 1 on exceptional sequences of isotropic type. Auslander Conference 2016, Woods Hole, MA
An example Consider an exceptional sequence E = ( X , U , V , Y ) of isotropic type with position 2. The exceptional sequence E ′ = ( X ′ , Y ′ , U ′ , V ′ ) is of isotropic type with isotropic position 3. Auslander Conference 2016, Woods Hole, MA
Constructing isotropic Schur roots Theorem (-, Weyman) An orbit of exceptional sequences of isotropic type under B n − 1 always contains a sequence E with δ E an isotropic Schur root of an Euclidean full subquiver of Q. Corollary There are finitely many orbits under B n − 1 . We can construct all isotropic Schur roots starting from the easy ones . Auslander Conference 2016, Woods Hole, MA
THANK YOU Questions ? Auslander Conference 2016, Woods Hole, MA
Recommend
More recommend
