Finite representation type
Stefan Wolf
1 Dynkin and Euclidean diagrams
Throughout the following section let Q be a quiver and Γ be its underlying graph.
1.1 Notations
Some Notation:
• (α, β) := hα, βi + hβ, αi
• q is positive definite if q(α) > 0 for all 0 6= α ∈ Zn
• q is positive semi-definite if q(α) ≥ 0 for all α ∈ Zn
• The radical of q is rad(q) = { α ∈ Zn | (α, −) = 0 }
• α, β ∈ Zn . α ≤ β if β − α ∈ Nn
• We say that α ∈ Zn is sincere if each component is non-zero.
q just depends on Γ and not on the orientation of Q.
Lemma 1.1. If Γ is connected and β ≥ 0 is a non-zero radical vector then β is sincere
and q is positive semi-definite. For α ∈ Zn we have
q(α) = 0 ⇔ α ∈ Qβ ⇔ α ∈ rad(q)
Proof. By assumption
0 = (εi , β) = (2 − 2nii )βi −
X
nij βj
j6=i
P
where nij is the number of edges between i and j in Γ. If βi = 0 then i6=j nij βj = 0,
j . Since Γ
and since each term is ≥ 0 we have βj = 0 whenever there is an edge i
1
is connected it follows that β = 0, a contradiction. Thus β is sincere. Now:
X
i<j
βi βj
nij
2
αi αj
−
βi
βj
2
=
X
=
X
=
X
nij
i<j
i<j
i6=j
X
βj 2 X
βi 2
αi −
nij αi αj +
nij
α =
2βi
2βj j
i<j
βj 2 X
nij
α −
nij αi αj =
2βi i
i<j
(2 − 2nii )βi
i
1 2 X
α −
nij αi αj
2βi i
i<j
= q(α)
Therefore q is positive semi-definite. If q(α) = 0 then
αi
βi
=
αj
βj
whenever there is an edge
j and since Γ is connected it follows that α ∈ Qβ. If α ∈ Qβ then α ∈ rad(q).
i
Finally if a ∈ rad(q) then certainly q(α) = 0.
1.2 Classification
Now we can classify the underlying graphs.
Theorem 1.2. Suppose Γ is connected.
1. If Γ is Dynkin then q is positive definite. By definition the (simply laced) Dynkin
diagrams are:
•
An : •
•
•
. . .
•
E6 : •
•
•
•
•
•
E7 : •
•
•
•
•
•
•
•
•
•?
??
Dn :
•


•
•
....
•
•
E8 : •
•
•
•
2. If Γ is Euclidean then q is positive semi-definite and rad(q) = Zδ. By definition
the Euclidean diagrams are as below. We have marked each vertex i with the value
2
of δi . Note that δ is sincere and δ ≥ 0.
1
em :
A

1 ??
?
.....
1
1
.....
1
1
2. . .

2 ??
?
1 ??
em
D
1

2
?

1
?
2
1 ??
(m ≥ 0)
e6 : 1
E
2
3
2
1
2
(m ≥ 4)
e7 : 1
E
2
3
4
3
2
1
1
3
e8 : 2
E
4
6
5
4
3
2
1
(n = m + 1) vertices
e0 has one vertex and one loop, and A
e1 has two vertices joined by two
Note that A
edges.
3. Otherwise, there is a vector α ≥ 0 with q(α) < 0 and (α, εi ) ≤ 0 for all i.
Proof.
2. To check that δ is radical we have to verify
X
2δi =
δj
neighbours j of i
Now q is positive semi-definite by lemma (1.1). Since some δi is 1, we obtain
rad(q) = Qδ ∩ Zn = Zδ.
1. Any Dynkin graph is contained in its corresponding Euclidean graph. Since q is
positive semi-definite and zero iff α = Zδ, the restriction of q to a proper connected
subgraph is positive definite.
3. By inspection, any non-Dynkin graph Γ contains an Euclidean subgraph Γ0 , say
with radical vector δ. If all vertices of Γ are in Γ0 then q(δ) < 0. If i is a vertex
not in Γ0 , connected to Γ0 by an edge, q(2δ + εi ) < 0.
A vertex i of an Euclidean graph Γ with δi = 1 is called an extending vertex. Removing
i from Γ gives the corresponding Dynkin diagram.
1.3 Roots
Suppose Γ is Dynkin or Euclidean.
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Definition 1.3. The set of roots is defined as
∆ := { α ∈ Zn | α 6= 0, q(α) ≤ 1 } .
A root α is real, if q(α) = 1 and imaginary if q(α) = 0.
For the wild case one also can define a root system, but in a slightly different way.
Lemma 1.4. The root system has the following properties:
1. Each i is a root.
2. If α ∈ ∆ ∪ {0}, so are −α and α + β with β ∈ rad(q).
(
∅
{imaginary roots} =
{ rδ | 0 6= r ∈ Z }
3.
(Dynkin)
(Euclidean)
4. Every root α is positive or negative.
5. If Γ is Euclidean then ∆ ∪ {0}/Zδ is finite.
6. If Γ is Dynkin then ∆ is finite.
Proof.
f0 this is clear. In the other cases q(i ) = 1 since there are no loops.
1. For A
2. q(β ± α) = q(β) + q(α) ± (β, α) = q(α)
3. Use lemma (1.1)
4. Let α = α+ −α− where α+ , α− ≥ 0 are non-zero and have disjoint support. Clearly
we have (α+ , α− ) ≤ 0, so that
1 ≥ q(α) = q(α+ ) + q(α− ) − (α+ , α− ) ≥ q(α+ ) + q(α− ).
Thus one of α+ ,α− is an imaginary root, and hence sincere. This means that the
other is zero, a contradiction.
5. Let e be an extending vertex. If α is a root with αe = 0, then δ − α and δ + α are
roots which are positive at the vertex e, and hence are positive roots. Thus
{ α ∈ ∆ ∪ {0} | αe = 0 } ⊆ { α ∈ Zn | −δ ≤ α ≤ δ }
which is finite. Now if β ∈ ∆ ∪ {0} then β − βe δ belongs to the finite set
{ α ∈ ∆ ∪ {0} | αe = 0 }.
e with extending vertex e. We can
6. Embed Γ in the corresponding Euclidean graph Γ
e with αe = 0, so the result follows from 5.
now view a root α for Γ as a root for Γ
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2 Finite Representation Type
We now use the work done before to prove Gabriel’s theorem.
Theorem 2.1. Suppose Q is a quiver with underlying graph Γ Dynkin. The assignment
X 7→ dimX induces a bijection between the iso-classes of indecomposable modules and
the positive roots of q.
Proof. If X is indecomposable then X is a brick, for otherwise there is a Y < X which
is a brick and has self-extensions. Hence
0 < q(dimY ) = dim End(Y ) − dim Ext(Y, Y ) ≤ 0,
a contradiction.
Now for every brick X we still have 0 < q(dimX) = 1 − dim Ext(X, X) ≤ 1, hence X
has no self-extensions and is a positive root. If X, Y are two indecomposables bricks with
no self-extensions of the same dimension vector then OX and OY are open, therefore
intersect and so are the same. Hence X ∼
=Y.
The last thing to show is that for every positive root α there is an indecomposable
representation X with dimX = α. Now pick an orbit OX of maximal dimension in
Rep(α). If X decomposes, say X = U ⊕ V , then Ext(U, V ) = Ext(V, U ) = 0. Thus
1 = q(α) = q(dimU + dimV ) =
q(dimU ) + q(dimV ) + hdimU, dimV i + hdimU, dimV i =
= q(dimU ) + q(dimV ) + dim Hom(U, V ) + dim Hom(V, U ) ≥ 2,
a contradiction.
Theorem 2.2. If Q is a connected quiver with graph Γ then there are only finitely many
indecomposable representations of Q iff Γ is Dynkin.
Proof. If Γ is Dynkin then the indecomposables correspond to the positive roots, and
there are only a finite number of them.
Conversely, suppose there are only a finite number of indecomposables. Any module is
a direct sum of indecomposables, so it follows that there are only finitely many iso-classes
of modules of dimension α for all α ∈ Nn . Thus there are only finitely many orbits in
Rep(α). By a previous lemma we have q(α) > 0 for all 0 6= α ∈ Nn . Now theorem (1.2)
shows that Γ is Dynkin.
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