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 = k̄ 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) = k di .
rep(Q, d) is an affine space.
For such a d, we set GL(d) =
Q
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 h−, −i the Euler-Ringel form of Q.
For M, N ∈ rep(Q), we have
hdM , dN i = dimk Hom(M, N) − dimk Ext1 (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 hd, di ≤ 1 and we call d:

if hd, di = 1;
 real,
isotropic, if hd, di = 0;

imaginary, if hd, di < 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.
1−1
{iso. classes of excep. repr.} ←→ {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 hδ, δi = 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 0 , W 0 , E (K )).
Consider R(Q, δ) := Thick(E (J), V 0 , W 0 ).
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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
δ̄. 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 δ̄ or finitely many real Schur root.
This gives the −h−, δi-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 h−, δi = 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
T 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
T 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
xo
2f
1f
4
3
x
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).
δ̄ = (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
δ̄ •
δ •
(0, 0, 1, 0)
•
•
(8, 3, 3, 3)
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) =
Q
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.
L
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
f
X
For X ∈ rep(Q), let 0 → P1 →
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 ⇔ hdX , dM i = 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 hdX , −i
in SI(Q, d) provided hdX , di = 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, δ̄)[{C Mj (−) | j ∈ K ∪ L}].
Corollary
By a result of Skowroński - Weyman, SI(Q, δ) is a polynomial ring
or 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 δ E 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 0 = (X 0 , Y 0 , U 0 , V 0 ) 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 δ E 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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