Isotropic Schur roots Charles Paquette University of Connecticut - - PowerPoint PPT Presentation

Isotropic Schur roots

Charles Paquette

University of Connecticut

May 1st, 2016

joint with Jerzy Weyman

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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.

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Quivers, dimension vectors

k = ¯ k is an algebraically closed field. Q = (Q0, Q1) is an acyclic quiver with Q0 = {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 dM its dimension vector.

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Geometry of quivers

For d = (d1, . . . , dn) a dimension vector, denote by rep(Q, d) the set of representations M with M(i) = kdi. rep(Q, d) is an affine space. For such a d, we set GL(d) =

1≤i≤n GLdi(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).

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Bilinear form and roots

We denote by −, − the Euler-Ringel form of Q. For M, N ∈ rep(Q), we have dM, dN = dimkHom(M, N) − dimkExt1(M, N).

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Roots and Schur roots

d ∈ (Z≥0)n is a (positive) root if d = dM 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 dM is a Schur root. We have real, isotropic and imaginary Schur roots. {iso. classes of excep. repr.} 1−1 ← → {real Schur roots}.

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Perpendicular categories

For d a dimension vector, we set A(d) the subcategory A(d) = {X ∈ rep(Q) | Hom(X, N) = 0 = Ext1(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(dV) =⊥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.

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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 (Mn−2, . . . , M1) in rep(Q) where all Mi are simples in A(δ). Complete this to a full exceptional sequence (Mn−2, . . . , M1, V , W ).

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Perpendicular category of an isotropic Schur root

We have {Mn−2, . . . , M2, M1} = J

·

∪ K

·

∪ L. For I ⊆ {Mn−2, . . . , M2, M1}, let E(I) denote the corr. exceptional sequence. We consider the following: (E(J ∪ K), V , W ). Now, the Mj and Mk 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 ′).

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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 Mi with Mi ∈ K ∪ L, The quasi-simple objects of R(Q, δ) (which includes the Mi with Mi ∈ 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.

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Cone of dimension vectors

Take the cone in Rn 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.

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Cone of dimension vectors

Let V be the set of vertices of P and for v ∈ V , let Pv the convex hull of the points in V \{v}. Only one dimension vector of simples is not real ⇒ we have |

v∈V Pv| ≤ 1.

Here are examples of such a cone in dimension 1, 2, 3:

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Cone of dimension vectors

Theorem (-, Weyman, H. Thomas) Assume that P is an (n − 2)-dimensional convex hull of points V with |

v∈V Pv| ≤ 1. Then

Rn−2 = V1 ⊕ · · · ⊕ Vr where Vi contains dimVi + 1 points in V ∪ {0} forming a dimVi-simplex containing the origin.

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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.

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An example

• (0, 0, 1, 0)
• (8, 3, 3, 3)
• δ
• ¯

δ

Figure : The cone of dimension vectors for δ = (3, 2, 3, 1)

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Semi-invariants

Since Q is acyclic, the ring of invariants k[rep(Q, d)]GL(d) is trivial. Take SL(d) =

1≤i≤n SLdi(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 = HomZ(Zn, Z). This ring is always finitely generated.

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Semi-invariants

For X ∈ rep(Q), let 0 → P1

fX

→ P0 → X → 0 be a projective resolution of X. For M ∈ rep(Q), the map

0 → Hom(X, M) → Hom(P0, M)

Hom(fX ,M)

− → Hom(P1, M) → Ext1(X, M) → 0

is given by a square matrix ⇔ dX, dM = 0. We set C X(M) := detHom(fX, M). Proposition (Derksen-Weyman, Schofield-Van den Bergh) The function C X(−) is a non-zero semi-invariant of weight dX, − in SI(Q, d) provided dX, d = 0. Moreover, these semi-invariants span SI(Q, d) over k.

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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 Mj(−) | j ∈ K ∪ L}]. Corollary By a result of Skowro´ nski - Weyman, SI(Q, δ) is a polynomial ring

• r a hypersurface.

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Exceptional sequences of isotropic types

A full exceptional sequence E = (X1, . . . , Xn) is of isotropic type if there are Xi, Xi+1 such that Thick(Xi, Xi+1) is tame. Isotropic position is i and root type δE is the unique iso. Schur root in Thick(Xi, Xi+1). The braid group Bn acts on full exceptional sequences. This induces an action of Bn−1 on exceptional sequences of isotropic type.

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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.

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Constructing isotropic Schur roots

Theorem (-, Weyman) An orbit of exceptional sequences of isotropic type under Bn−1 always contains a sequence E with δE an isotropic Schur root of an Euclidean full subquiver of Q. Corollary There are finitely many orbits under Bn−1. We can construct all isotropic Schur roots starting from the easy ones.

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THANK YOU Questions ?

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