Finite representation type

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1 Finite representation type Stefan Wolf Dynkin and Euclidean diagrams Throughout the following section let Q be a quiver and Γ be its underlying graph.. Notations Some Notation: (α, β) := α, β + β, α q is positive definite if q(α) > 0 for all 0 α Z n q is positive semi-definite if q(α) 0 for all α Z n The radical of q is rad(q) = { α Z n (α, ) = 0 } α, β Z n. α β if β α N n We say that α Z n is sincere if each component is non-zero. q just depends on Γ and not on the orientation of Q. Lemma.. 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 Z n we have q(α) = 0 α Qβ α rad(q) Proof. By assumption 0 = (ε i, β) = ( n ii )β i j i n ij β j where n ij is the number of edges between i and j in Γ. If β i = 0 then i j n ijβ j = 0, and since each term is 0 we have β j = 0 whenever there is an edge i j. Since Γ

2 is connected it follows that β = 0, a contradiction. Thus β is sincere. Now: n ij β i β j ( αi α ) j = β j n ij αi n ij α i α j + β i n ij αj = β i β j β i β j = i j = i = q(α) β j n ij αi n ij α i α j = β i ( n ii )β i αi n ij α i α j β i Therefore q is positive semi-definite. If q(α) = 0 then α i β i = α j β j whenever there is an edge i j and since Γ is connected it follows that α Qβ. If α Qβ then α rad(q). Finally if a rad(q) then certainly q(α) = 0.. Classification Now we can classify the underlying graphs. Theorem.. Suppose Γ is connected.. If Γ is Dynkin then q is positive definite. By definition the (simply laced) Dynkin diagrams are: A n :... E 6 : D n :.... E 7 : E 8 :. 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

3 of δ i. Note that δ is sincere and δ 0. à m : D m... 3 (m 0) Ẽ 6 : 3 (m 4) Ẽ 7 : Ẽ 8 : (n = m + ) vertices Note that Ã0 has one vertex and one loop, and à has two vertices joined by two edges. 3. Otherwise, there is a vector α 0 with q(α) < 0 and (α, ε i ) 0 for all i. Proof.. To check that δ is radical we have to verify δ i = neighbours j of i δ j Now q is positive semi-definite by lemma (.). Since some δ i is, we obtain rad(q) = Qδ Z n = Zδ.. 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 Γ, say with radical vector δ. If all vertices of Γ are in Γ then q(δ) < 0. If i is a vertex not in Γ, connected to Γ by an edge, q(δ + ε i ) < 0. A vertex i of an Euclidean graph Γ with δ i = is called an extending vertex. Removing i from Γ gives the corresponding Dynkin diagram..3 Roots Suppose Γ is Dynkin or Euclidean. 3

4 Definition.3. The set of roots is defined as := { α Z n α 0, q(α) }. A root α is real, if q(α) = and imaginary if q(α) = 0. For the wild case one also can define a root system, but in a slightly different way. Lemma.4. The root system has the following properties:. Each ɛ i is a root.. If α {0}, so are α and α + β with β rad(q). 3. {imaginary roots} = { (Dynkin) { rδ 0 r Z } (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.. For Ã0 this is clear. In the other cases q(ɛ i ) = since there are no loops.. q(β ± α) = q(β) + q(α) ± (β, α) = q(α) 3. Use lemma (.) 4. Let α = α + α where α +, α 0 are non-zero and have disjoint support. Clearly we have (α +, α ) 0, so that 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 α e = 0, then δ α and δ + α are roots which are positive at the vertex e, and hence are positive roots. Thus { α {0} α e = 0 } { α Z n δ α δ } which is finite. Now if β {0} then β β e δ belongs to the finite set { α {0} α e = 0 }. 6. Embed Γ in the corresponding Euclidean graph Γ with extending vertex e. We can now view a root α for Γ as a root for Γ with α e = 0, so the result follows from 5. 4

5 Finite Representation Type We now use the work done before to prove Gabriel s theorem. Theorem.. Suppose Q is a quiver with underlying graph Γ Dynkin. The assignment X 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) = dim Ext(X, X), 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 O X and O Y 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 O X of maximal dimension in Rep(α). If X decomposes, say X = U V, then Ext(U, V ) = Ext(V, U) = 0. Thus a contradiction. = q(α) = q(dimu + dimv ) = q(dimu) + q(dimv ) + dimu, dimv + dimu, dimv = = q(dimu) + q(dimv ) + dim Hom(U, V ) + dim Hom(V, U), Theorem.. 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 N n. Thus there are only finitely many orbits in Rep(α). By a previous lemma we have q(α) > 0 for all 0 α N n. Now theorem (.) shows that Γ is Dynkin. 5