Quivers. Basic Definitions and Path Algebra

Quivers 1 Introduction Throughout this paper, k will indicate an algebraically closed field of characteristic 0. 2 2.1 Quivers basic definitions and Path algebra Definition 2.1 A quiver is directed graph Q = (Q0 , Q1 ) where Q0 is the set of vertices and Q1 set of arrows with maps h, t : Q1 → Q0 which assign to each arrow its head and tail, respectively. Definition 2.2 A path is a sequence of arrows p = a1 a2 · · · an such that h(ak+1 ) = t(ak ) for every k = 1, . . . , n − 1. The head of the path is h(a1 ), and the tail of the path is t(an ). Definition 2.3 An oriented cycle is a path p such that h(p) = t(p), and h(ai ) = 6 t(aj ) for any other i 6= j + 1. Note that: this definitions implies that a quiver like an “eight” is not an oriented cycle. Proposition 2.1 A quiver with an oriented cycle has an infinite set of paths. Example 2.1 The Jordan quiver This quiver does not appear in this programm. Example 2.2 The 2-Kronecker a 1• b - •2 has no oriented cycle, but a 1•  b has an oriented cycle. - •2 2 2. Quivers Example 2.3 b a - •2 - c 1• has no oriented cycle, but - •3 b a - •2 - c 1•  •3 has an oriented cycle. Example 2.4 The 3-Kronecker quiver a - • - b • c Path algebra is an algebra build up from the paths. To form the path algebra, we must define a multiplication rule between the paths of the quiver: Definition 2.4 Given a quiver Q, the path algebra kQ is the k-vector space generated by all paths in Q with multiplication rule:  pq if h(q) = t(p) p∗q = 0 otherwise HAVE TO DEFINE THE IDENTITY ELEMENT END SAY THATS ASSOC, AND ... AS AN ALGEBRA. Now on we will denote the points in Q0 as its respective null path. Example 2.5 - • e2 f1 f2 - f3 e1 • - • e3 This quiver has path algebra with basis {e1 , e2 , e3 , f1 , f2 , f3 , f2 f1 }. Example 2.6 - • e2 f1 f2 f3 - e1 •  • e3 This quiver has an infinite dimensional path algebra with basis: {e1 , e2 , e3 , f1 , f2 , f3 , f2 f1 , f3 f2 , f1 f3 , f3 f2 f1 , f1 f3 f2 , f2 f1 f3 , . . .} since the quiver contains an oriented cycle. 3 2.2 quivers with relations We can impose further relations on the composition of the arrows to make paths. 3 Quiver Representations Definition 3.1 A quiver representation is a collection {Vi |i ∈ Q0 } of finite dimensional k-vector spaces together with a collection {φa : Vt(a) → Vh(a) |a ∈ Q1 } of k-linear maps. From now on, we will denote a representation R = (Vi , φa ). Definition 3.2 Suppose R = (Vi , φa ) and S = (Wi , ψa ) are representations of Q. The representation R0 is called a subrepresentation of R if • for every i ∈ Q0 , Wi is a subspace of Vi and • for every a ∈ Q1 , the restriction of φa : Vt(a) → Vh(a) to Wt(a) is equal to a : Wt(a) → Wh(a) . Definition 3.3 A non-zero representation V is called simple representation if the only subrepresentations of V are the zero representation and V itself. Definition 3.4 If R = (Vi , φa ) and S = (Wi , ψa ) are representations of a quiver Q then we can define the direct sum representation R ⊕ S = (Ui , ρa ) by: • Ui = Vi ⊕ Wi for every i ∈ Q0 , and  • ρa : Vt(a) ⊕ Wt(a) → Vh(a) ⊕ Wh(a) is defined by the matrix Va 0 0 Wa  Definition 3.5 If R and S are two representation of a quiver Q, then a representation morphism Φ : R → S is a collection k-linear maps {ϕi : Vi → Wi |i ∈ Q0 } such that the diagram Vt(a) ϕt(a) ? Wt(a) φa - Vh(a) ϕh(a) ? - Wh(a) a commutes for all a ∈ Q1 . If ϕi is invertible for every i ∈ Q0 , then the morphism Φ is called an isomorphism and R and S are called isomorphic representations. Definition 3.6 A representation R of a quiver Q is called decomposable if R∼ = S ⊕ T where S and T are nonzero representations of Q. A nonzero representation is called indecomposable if it is not decomposable. 4 3.1 Isomorphism Classes of Representations The study of quiver representations will be simplified significantly if we can consider isomorphism classes of quiver representations, rather than specific representations. To find a representative element of each isomorphism class, we apply representation isomorphisms to change the basis of the vector space at each vertex in order to simplify the matrices for the maps between the vector spaces. For square matrices and representations over C, this process is the same as the theory of the Jordan normal form. Example 3.1 For the so-called Jordan quiver with one vertex and one arrow, every isomorphism class of representations has a representative element of the form R = ({V1 }, {J1 }) where J1 is a matrix in Jordan normal form. This is a direct result of the theorem that every square matrix M = P −1 JP , where J is a Jordan-form matrix, and P is an invertible matrix, corresponding to the change of basis required to isolate the eigenspace of each eigenvalue. {reference: Halmos}. If we restrict ourselves to representations with invertible maps at each arrow, we may simultaneously describe the isomorphism classes of representations of quivers which differ from each other only in the orientation of their arrows. Note that the invertibility condition implies that the representation must have equidimensional vector spaces at all vertices. The isomorphism classes of these representations can often be described neatly, by analogy to the case of the Jordan quiver. Example 3.2 For a quiver with two vertices and two arrows, given any representation R = ({V1 , V2 }, {A, B}), where A and B are both invertible, we can find an isomorphic representation of the form: A V1 • Id  - • V2 B ∼ = V10 •  J 0 - • V2 To find this isomorphic representation, let B1 and B2 be bases for V1 and V2 , respectively. Take P0 to be the change-of-basis matrix taking B2 to AB1 . This is possible because invertibility implies equidimensionality. Then the representation isomorphism Φ0 = (Id, P0 ) yields the isomorphic representation R0 = ({V1 , V200 }, {Id, BA}). By Jordan’s Theorem {Halmos}, we can find P1 invertible such that BA = P1−1 JP1 , where J is a Jordan-form matrix. Applying the representation isomorphism Φ1 = (P1 , P1 ) yields the desired isomorphic representation. Example 3.3 Given a representation R = ({V1 , V2 , V3 }, {A, B, C}) (with A, B, C all invertible) of the triangular quiver with three vertices and three arrows, we may find an isomorphic representation of the form: •2 - - •2 •3 A Id 1•  J - - C Id B 1•  ∼ = •3 where J is a Jordan-form matrix. The process for finding this representation is similar to the one described in Example 3.2 above. 5 Example 3.4 The case of the 3-Kronecker quiver is more complicated, because we may not be able to simultaneously put the maps on the second and third arrows in Jordan normal form. However, in the case dim(V1 ) = dim(V2 ) = 2, we will always be able to conjugate bases and obtain the form: A V1 • B C Id - •V2 - ∼ = V10 • J  a 0 b c  - •V 0 - 2 where A = Id, B is in Jordan normal form, and C is a lower-triangular matrix. 3.2 Simple Representations Definition 3.7 A canonical representation R for the quiver Q = (Q0 , Q1 ) is a representation of the form  k for one i ∈ Q0 R = {Vi = , φa = 0 for all a ∈ Q1 }. {0} otherwise A canonical representation R must be simple, because the only subspace of it’s Vi is the null space at each vertex. This implies that the only proper subrepresentation is the trivial one, so R is simple Proposition 3.1 Let Q be a quiver with no oriented cycles. A representation R of Q is simple if and only if it is canonical. Proof (⇐) This follows from the definition of canonical representations. (⇒) We will show that every representation not of the form described above has a proper subrepresentation of that form. First, we begin with the following lemma. Lemma 3.1 If Q = (Q0 , Q1 ) is a quiver with no oriented cycles, then there is some vertex x ∈ Q0 that is not the tail of any arrow. Such an arrow is called a sink. Proof: Proof by contrapositive. Suppose for any vi ∈ Q0 , vi = t(a) for some a ∈ Q1 . Choose some v1 ∈ Q0 and form a path as follows: Choose an such that t(an ) = vn . Write vn+1 = h(an ), and repeat. Since Q0 is a finite set, eventually we will get vn + 1 = vi for some i ≤ n. Then p = ai · · · an is an oriented cycle in Q. Lemma 3.2 Let Q be a quiver with no oriented cycle, write x1 ∈ Q0 is a vertex such that t(a) 6= x1 for all a ∈ Q1 . Proof: Given an arbitrary representation R = (Vi , ρa ), if Vx1 6= {0}, then label this vertex x. (Vh(a) = {0} for all a such that t(a) = x). If Vx1 = {0}, define Q1 = (Q10 = Q0 /{x1 }, Q11 = Q1 /{a ∈ Q1 |h(a) = x1 }. Since Q contained no oriented cycles, and Q00 ⊂ Q0 , Q01 ⊂ Q1 , Q0 contains no oriented cycle, so we may apply the lemma. Write x2 ∈ Q00 is a vertex such that t(a) 6= x2 for all a ∈ Q01 . Define the representation R0 of Q0 by restricting the representation R of Q. If R is a non-trivial representation of Q, we will eventually find xn ∈ Q0 such that Vxn 6= {0} but Vi = {0} for all i such that i = h(a) for some a ∈ Q1 such that t(a) = xn . Proposition  3.2 Construct a representation k i=x S = {Wi = , φa = 0 for all a ∈ Q1 } for the quiver Q. {0} i 6= x 6 S is a subrepresentation of R. Proof: Clearly, for all i 6= x {0} = wi ≤ vi . Since vx is a non-zero k-vector space, k = wx ⊆ vx . Define p = {pi : i ∈ Q0 } a representation morphism such that pi : wi → vi is the inclusion map. To check that all maps commute, first note that for a ∈ Q1 such that t(a) 6= x, Wt(a) = {0}. So ψa : Wt(a) → Wh(a) has {0} as its domain, i.e. Similarly Pt(a) : Wt(a) → Vt(a) is the a = 0. inclusion of {0} implies Pt(a) = 0. Hence, for all a ∈ Q1 such that t(a) 6= x we have: ph(a) ◦ ψa = ph(a) ◦ 0 = 0 and ϕa ◦ pt(a) = ϕa ◦ 0 = 0 so the diagram commutes. Now, for all a such that t(a) = x, we know that Vh(a) = {0}. So ϕa : Vt(a) → Vh(a) is ϕa : Vx → {0}, i.e. ϕa = {0}. Similarly, ψa = 0 and ph(a) : {0} → {0} is also the zero map. So for all a ∈ Q1 such that t(a) = x, we have ph(a) = ψa = 0 ◦ 0 = 0 so the diagram commutes. Therefore, S is a subrepresentation of R of the desired form. Example 3.5 a - •2 1• b b a - •2 These quivers representations. have - c 1• no oriented - •3 cycles, and only canonical simple Definition 3.8 A unique cycle is a cycle that has no other cycle on its vertices. Proposition 3.3 Every unique cycle generates a 1-parameter family of simple representations. Sketch of proof: Form the identity around the quiver. Get the first column of the quiver and it will be one parameter. Example 3.6 a 1•  - •2 b c - 1•  b a - •2 •3 these quivers have unique oriented cycles, and 1-parameter families of simple representations. 7 3.3 Indecomposable Representations Here we will work by the examples we have shown above. The invertability of maps and the dimensions vectors will play an important role in order to give all indecomposable representations for some given quiver representation. Example 3.7 For the jordan quiver, if the map is invertible and there is an isomorphic representation with the canonical Jordan matrix, then it is indecomposable. Example 3.8 For the example 2.3(?), we have the following classification: Proposition 3.4 A representation R = ({V1 = Cm , V2 = Cn }, {A, B}) (such that wlog m > n) of the quiver Q is indecomposable if and only if one of the following holds: • R∼ = R0 = ({W1 , W2 }, {Id, Jλ }) where Jλ is a matrix in Jordan normal form with only one eigenblock. • (BA)k = 0 for some k ∈ Z and dim ker BA = 1. Corollary: The set of dimension vectors of indecomposable representations of Q is DQ = {(n, n), (n, n ± 1)|n ∈ Z+ }. Proof of Proposition: We describe the possible decomposability or indecomposability for each case. cases, and prove 1. BA invertible → R is indecomposable ↔ it is isomorphic to a representation with one Jordan block. 2. BA non-invertible (a) BA is nilpotent, i.e. (BA)k = 0 for some k ∈ Z+ i. nullBA=1 → R is indecomposable. ii. nullBA¿1 → R is decomposable. (b) BA is not nilpotent. i.e. (BA)k 6= 0 ∀ k ∈ Z+ → R is decomposable. Proof of Claim: 1. BA invertible → m=n, A, B are both invertible → as shown above R is indecomposable representation with one Jordan block. ⇐⇒ it is isomorphic to a 2(a)i. (BA)k = 0 for some k ∈ Z+ , null(BA)=1. Let x ∈ ker BA. Then BAy = 0 → y = λx for some λ ∈ Z+ Suppose R = R0 ⊕ R00 where R0 = ({W1 , W2 }, {A|W1 , B|W2 }) , R00 = ({U1 , U2 }, {A|U1 , B|U2 }) are both non-trivial. wlog, x ∈ W1 . Suppose y ∈ U1 , y 6= 0. → by definition of decomposability, (BA)i ∈ U1 ∀i ∈ Z+ 8 But (BA)k = 0 → pick the least j ∈ Z+ such that (BA)j = 0. → (BA)j−1 ∈ ker BA → (BA)j−1 y = λx ∈ W1 . This yields a contradiction → V1 = W1 . Suppose y ∈ U2 , y 6= 0 → because of a dimension argument, either y = Ax for some x ∈ V1 or By = x for some nonzero x ∈ V1 . → in either case, y ∈ W2 because of the invariance of subrepresentations. This yields a contradiction → V2 = W2 → R00 is the trivial subrepresentation. → R is indecomposable. 2(a)ii. null(BA) ¿ 1 write ker BA = W0 ⊕ U0 both of which are non-zero. for x ∈ V1 , write jx ∈ Z+ is the minimal integer such that (BA)jx = 0, and define: W1 = {x ∈ V1 |(BA)jx − 1x ∈ W0 } , U1 = {x ∈ V1 |(BA)jx −1 ∈ U0 )} These two sets define a decomposition of R. → R is decomposable. 2(b) (BA)k 6= 0 ∀ k ∈ Z+ → ∃j ∈ Z+ s.t.V1 = ker(BA)j ⊕ W1 and (BA)j |W1 is invertible. These sets define a decomposition of R → R is decomposable. 3.4 Representation of Path Algebras Given a representation R = (Vi , φa ) ofLthe quiver Q = (Q0 , Q1 ), we can construct a representation ρ : kQ → End( i∈Q0 Vi ) of the path algebra kQ. It suffices to define the representation on the ei ’s and fj ’s, as these generate the basis as a ring. ρ(ei ) := Id |Vi , ρ(fj ) : Vt(j) → Vh(j) x 7→ φj (x) This extends to a representation on all elements of kQ. 4 Lie Algebras and Their Representations Definition 4.1 A Lie Algebra g is an non-associative algebra with the multiple rule given by the bilinear map [, ] which satisfies • [x, x] = 0 for all x ∈ g, • [x, [y, x]] + [y, [z, x]] + [z, [x, y]] = 0 ∀x, y, z ∈ g. These two properties imply that the [, ] operation is anti-symmetric, i.e. [x, y] = −[y, x] ∀x, y ∈ g. We can construct a Lie algebra from any associative algebra by defining the bracket operation as the commutator [a, b] = ab − ba. Example 4.1 Let V be a vector space. this is an associative algebra. we can define the commutator bracket. because V is abelian, [a,b]=0 always. 9 4.1 Representation of sl2 (k) The simple linear algebra sl2 (k)= {A ∈ M2 (k)|trA = 0)} of traceless 2 × 2 matrices is a Lie Algebra with bracket operation defined by the commutator [A, B] = AB − BA and basis:       0 1 0 0 1 0 e= , f= , h= . 0 0 1 0 0 −1 We will describe the equivalence classes of simple and indecomposable representations of sl2 (k) and show that these correspond to simple and indecomposable representations of the quiver with relations described in Example ?.?. We will restrict ourselves to representations of sl2 (k) such that L • V = k∈Z Vk , where Vk = {v ∈ V |hv = kv} is the eigenspace with eigenvalue k for the action of h on V . • Vk = 0 for k  0. • Each Vk is finite dimensional. Given v ∈ Vk , we calculate to find the following properties: • h(f (v)) = (k − 2)f (v) • h(e(v)) = (k + 2)e(v) In other words, the action of f takes the eigenspace Vk with eigenvalue k to Vk−2 with eigenvalue k − 2, and the action of e takes Vk to Vk+2 . Since m ∈ Z is the maximal eigenvalue, we observe that  k if k = m − 2i for i ∈ Z+ Vk = 0 otherwise Thus, if we take v0 ∈ Vm , we can define a basis B={vi |i ∈ Z+ } for V (m) = ⊕k≤m Vk by the rule vi = f i (v0 ). From the equations above, we can calculate that e(vi ) = i(m − i + 1)vi+1 . M(-m-2) If the highest eigenvalue of a representation is negative, then the map e does not annihilate any of the eigenspaces, and we have an infinite-dimensional, simple representation called M (−m − 2). In this notation, m is positive, so −m − 2 is the desired negative integer. V−m−2 = kv0 6 e f ? V−m−4 = kv1 6 f e ? .. . V(m) However, if the highest eigenvalue is positive, the action of e will annihilate the eigenspace V−m−2 , so the representation will not be simple, in fact, it has M (−m − 2) for a subrepresentation. It will, however, be indecomposable, as the subspace ⊕−m/leqk/leqm Vk is not invariant under the action of f . This representation is called V (m). 10 Vm = kv0 6 f e ? .. . 6 f e ? V−m = kvm f ? V−m−2 =kvm+1 6 f e ? .. . M(m) Taking the quotient representation V (m)/M (−m − 2) gives the second simple representation, and the only one with positive maximal eigenvalue, M(m): Vm = kv0 6 f e ? .. . 6 f e ? V−m = kvm M∗ (m) Because we have e(v−m−2 ) = 0, we may define another linearly independent eigenvector w0 with eigenvalue −m − 2, such that e(w0 ) = v−m . This can be extended by the rule wi = f i (w0 ) to find a second eigenvector for each eigenvalue λ = −m − 2(i + 1). Taking Vk = span{v1/2(m−k) } for m > k > −m and Vk = span{w1/2(−m−2−k) } for k ≤ −m − 2, we find the representation M∗ (m): 11 Vm = kv0 6 f e ? .. . 6 f e ? V−m = kvm  e V−m−2 = kw0 6 f e ? kw1 V−m−4 = .. . P(-m-2) However, for k ≤ −m − 2, we may also allow Vk to be two-dimensional, that is, Vk = span{v1/2(m−k) , w1/2(−m−2−k) }. This describes the representation called P (−m − 2). Vm = kv0 6 e f ? .. . 6 f e ? V−m = kvm  e f ⊕  ? V−m−2 =kvm+1 6 e ? V−m−4 =kvm+2 .. . e f ⊕ kw0 6 e f ? kw1 .. . In this case, note that e(wi ) = awi−1 + bvi − m − 2 12 5 Cartan Matrix and Kac-Moody Algebras Definition 5.1 A Cartan matrix is a matrix A = (aij ) such that: • aii = 2 • aij ≤ 0 if i 6= j and aij ∈ Z • aij = 0 if and only if aji = 0. . Let Q be a quiver without loops, #Q0 = n. We can construct an n × n Cartan matrix  aii = 2 (i = 1, . . . , n) A = (aij ) where −aij = #{arrows btw vertices i and j} (i 6= j) The orientation of the arrows in Q does not matter in the construction of the Cartan matrix, so we may disregard the orientation. Definition 5.2 Given an n × n Cartan matrix, we may construct a Lie algebra, called the Kac-Moody algebra, with the following generators and relations: generators: {ei , fi , hi | i = 1, . . . , n} relations: • [hi , ej ] = aij ej • [hi , fj ] = −aij fj • [ei , fj ] = δi,j hi • ad1−aij (ei )ej = 0 • ad1−aij (fi )fj = 0. where ad(α) = adα , α ∈ g(A) is the adjoint action adα (β) = [α, β]. Definition 5.3 The simple roots of a Kac-Moody algebra g(A) are the functionals {αj } ,j = 1, . . . , n, defined by: αj (hi ) = aij . Consider Γ = ⊕Zαi the set of all Z-linear combinations of simple roots. Claim: If α is a root of g(A) then α ∈ Γ. Definition 5.4 For (αi , αj ) = aij the maps ri : h∗ → h∗ , ri (λ) = λ − (λ, αi ) αi are called simple reflections. such that Remark: Since (ri ) = 1h∗ , ri is invertible for every i and the set of all compositions of ri is indeed a group, denoted by W, called the Weyl Group. The Weyl group is a very important group as we will see in the next proposition when we want to find the real and imaginary roots, as the following definitions: Definition 5.5 Let α be an element of Γ, so α is a real root if ∃w ∈ W such that w(α) = αi for some i ∈ {1, 2, . . . , n}. The othes are called imaginary roots. Proposition 5.1 The real and imaginary roots may be calculate by the following formulas: • Rre = ∪w∈W w(αi ) • Rim = ∪w∈W w(M ), where M = {α ∈ Γ+ \0|(α, αi ) ≤ 0, ∀i} remark Indeed, the definition of M contains one more property which does not make difference in the examples we are working with. From now on we will find the simple roots, simple reflections, real end immaginary roots of the five exemples above. 13 5.1 sl2 (k) The quiver QA = • gives Cartan matrix A = (2) The g(A) is the Lie algebra generated by e, f, h such that: • [h, e] = 2e • [h, f ] = −2f • [e, f ] = h So g(A) has one simple root, namely α(h) = 2. The simple reflection is r(λα) = λα − (λα, α) α = −λα. Then the Weyl group is W = {e, r}. Now the real root are Rre = {α, −α}, and since M = {α ∈ Γ+ \0 | (α, α) ≤ 0}, this implies that M = ∅, which means that all the roots are real. 5.2 sl3 (k) The quiver QB = •  gives Cartan matrix B = - • 2 −1 −1 2  g(B) is the Lie Algebra generated by {e1 , e2 , f1 , f2 , h1 , h2 } with relations: • [h1 , e1 ] = 2e1 , [h1 , e2 ] = −e2 • [h1 , f1 ] = −2f1 , [h1 , f2 ] = f2 • [e1 , f1 ] = h1 , [e1 , f2 ] = 0, [e2 , f1 ] = 0, [e2 , f2 ] = h2 • [e1 , [e1 , e2 ]] = 0, [e2 , [e2 , e1 ]] = 0 • [f1 , [f1 , f2 ]] = 0, [f2 , [f2 , f1 ]] = 0 • [h1 , h2 ] = 0. Indeed, g(B) ' sl3 (k), the algebra of 3 × 3 traceless matrices with bracket operation defined by the commutator. The simple roots are {α1 , α2 } where α1 (h1 ) = 2, α2 (h1 ) = −1, α1 (h2 ) = −1, α2 (h2 ) = 2 and any root is of the form α = (z1 , z2 ) ∈ Γ. The simple reflections are r1 (z1 , z2 ) = (z2 −z1 , z2 ) and r2 (z1 , z2 ) = (z1 , z1 −z2 ). The Weyl Group is W = {Id , r1 , r2 , r1 r2 , r2 r1 , r1 r2 r1 } where: • r1 r2 (z1 , z2 ) = (−z2 , z1 − z2 ), • r2 r1 (z1 , z2 ) = (z2 − z1 , −z1 ) • r1 r2 r1 (z1 , z2 ) = r2 r1 r2 (z1 , z2 ) = (−z2 , −z1 ) So, W is a finite group. The caracterization of W makes possible to find all real and imaginary roots, as follows: Rre = {α1 , −α1 , α2 , −α2 α1 + α2 , −α1 − α2 } . Rim = ∅ since M = ∅. 14 5.3 slˆ2 The quiver QC = • - •   2 −2 gives Cartan matrix C = −2 2 g(C) is the Lie Algebra generated by {e1 , e2 , f1 , f2 , h1 , h2 } with relations: • [h1 , e1 ] = 2e1 , [h1 , e2 ] = −2e2 • [h1 , f1 ] = −2f1 , [h1 , f2 ] = 2f2 • [e1 , f1 ] = h1 , [e1 , f2 ] = 0, [e2 , f1 ] = 0, [e2 , f2 ] = h2 • [e1 , [e1 [e1 , e2 ]]] = 0, [e2 , [e2 , [e2 , e1 ]]] = 0, • [f1 , [f1 [f1 , f2 ]]] = 0, [f2 , [f2 , [f2 , f1 ]]] = 0 • [h1 , h2 ] = 0. Indeed, g(C) ' sl2 [t, t−1 ] ⊕ kc for some constant c. The simple roots are {α1 , α2 } where α1 (h1 ) = 2 = α2 (h2 ), α2 (h1 ) = −2 = α1 (h2 ) and any root is of the form α = (z1 , z2 ) ∈ Γ. The simple reflections are r1 (z1 , z2 ) = (−z1 + 2z2 , z2 ) and r2 (z1 , z2 ) = (z1 , 2z1 − z2 ). The Weyl Group W = hr1 , r2 i/hr1 2 = 1, r2 2 = 1i is an infinite dimensional group with the following properties, for r ∈ W:  (−2j + 1, −2j) if r = (r2 r1 )j    (2j + 1, 2j) if r = (r1 r2 )j r(α1 ) = (−2j − 1, −2j) if r = r1 (r2 r1 )j    (2j + 1, 2(j + 1)) if r = r2 (r1 r2 )j  (2j, 2j + 1) if r = (r2 r1 )j    (−2j), −2j + 1) if r = (r1 r2 )j r(α2 ) = (2(j + 1), 2j + 1) if r = r1 (r2 r1 )j    (−2j, −2j − 1)) if r = r2 (r1 r2 )j where j ∈ N. So by choosing any positive integer j, we find that the real roots are: Rre = {(m, n) | m, n ∈ Z, |m − n| = 1} . To calculate the imaginary roots, we note that: M = {(z1 , z2 ) | z1 , z2 ∈ Z+ , 2z1 − 2z2 ≤ 0, −2z1 + 2z2 ≤ 0} = {(z1 , z2 ) | z1 , z2 ∈ Z+ , z1 = z2 } = Z+ (1, 1) These elements are preserved by the action of W, so we have: Rim = Z+ (α1 + α2 ) = {(n, n) | n ∈ Z} 15 5.4 slˆ3 k The quiver QD = - • - •  •  2 −1 −1 2 −1  . gives Cartan matrix D =  −1 −1 −1 2 g(D) is the Lie Algebra generated by {e1 , e2 , e3 , f1 , f2 , f3 , h1 , h2 , h3 } with relations: • [h1 , e1 ] = 2e1 , [h1 , e2 ] = −e2 , [h1 , e3 ] = −e3 , • [h2 , e1 ] = −e1 , [h2 , e2 ] = 2e2 , [h2 , e3 ] = −e3 , • [h3 , e1 ] = −e1 , [h3 , e2 ] = −e2 , [h3 , e3 ] = 2e3 , • [h1 , f1 ] = −2f1 , [h1 , f2 ] = f2 , [h1 , f3 ] = f3 , • [h2 , f1 ] = f1 , [h2 , f2 ] = −2f2 , [h2 , f3 ] = f3 , • [h3 , f1 ] = f1 , [h3 , f2 ] = f2 , [h3 , f3 ] = −2f3 , • [e1 , f1 ] = h1 , [e1 , f2 ] = 0, [e1 , f3 ] = 0, • [e2 , f1 ] = 0, [e2 , f2 ] = h2 , [e2 , f3 ] = 0, • [e3 , f1 ] = 0, [e3 , f2 ] = 0, [e3 , f3 ] = h3 , • [e1 , [e1 , e2 ]] = 0, [e1 , [e1 , e3 ]] = 0, • [e2 , [e2 , e1 ]] = 0, [e2 , [e2 , e3 ]] = 0, • [e3 , [e3 , e1 ]] = 0, [e3 , [e3 , e2 ]] = 0, • [f1 , [f1 , f2 ]] = 0, [f1 , [f1 , f3 ]] = 0, • [f2 , [f2 , f1 ]] = 0, [f2 , [f2 , f3 ]] = 0, • [f3 , [f3 , f1 ]] = 0, [f3 , [f3 , f2 ]] = 0 • [h1 , h2 ] = 0, [h2 , h3 ] = 0. Indeed, g(D) ' sl3 [t, t−1 ] ⊕ Cc for some constant c. The simple roots are {α1 , α2 , α3 } where: α1 (h1 ) = α2 (h2 ) = α3 (h3 ) = 2 α2 (h1 ) = α3 (h1 ) = α1 (h2 ) = α3 (h2 ) = α1 (h3 ) = α2 (h3 ) = −1 and any root is of the form α = α = (z1 , z2 , z3 ) ∈ Γ. The simple reflections are r1 (z1 , z2 , z3 ) = (−z1 + z2 + z3 , z2 , z3 ), r1 (z1 , z2 , z3 ) = (z1 , z1 − z2 + z3 , z3 ), r1 (z1 , z2 , z3 ) = (z1 , z2 , z1 + z3 − z3 ). The Weyl Group W = hr1 , r2 i/hr1 2 = 1, r2 2 = 1i is an infinite dimensional group with the following properties: • r1 (k ± 1, k, k) = (k ∓ 1, k, k) 16 • r1 (k, k ± 1, k) = (k ± 1, k ± 1, k) • r1 (k, k, k ± 1) = (k ± 1, k, k ± 1) • r2 (k ± 1, k, k) = (k ± 1, k ± 1, k) • r2 (k, k ± 1, k) = (k, k ∓ 1, k) • r2 (k, k, k ± 1) = (k, k ∓ 1, k ± 1) • r3 (k ± 1, k, k) = (k ± 1, k, k ± 1) • r3 (k, k ± 1, k) = (k, k ± 1, k ± 1) • r3 (k, k, k ± 1) = (k, k, k ∓ 1) With these properties, we can compute the real and imaginary roots which leads to: Rre = {(k ± 1, k, k), (k, k ± 1, k), (k, k, k ± 1)}k∈Z . To calculate the imaginary roots, we note that: M = {(z1 , z2 , z3 ) | z1 , z2 , z3 ∈ Z+ , 2z1 − z2 − z3 ≤ 0, −z1 + 2z2 − z3 ≤ 0 − z1 − z2 + 2z3 ≤ 0} = {(z1 , z2 , z3 ) | z1 , z2 , z3 ∈ Z+ , z1 = z2 = z3 } = Z+ (1, 1, 1) This set is preserved by W, so we have: Rim = Z+ (α1 + α2 + α3 ) 5.5 3-Kronecker The quiver QE = • - •   2 −3 gives Cartan matrix E = −3 2 g(E) is the Lie Algebra generated by {e1 , e2 , f1 , f2 , h1 , h2 } given by the relations: • [h1 , e1 ] = 2e1 , [h1 , e2 ] = −3e2 , • [h1 , f1 ] = −2f1 , [h1 , f2 ] = 3f2 , • [e1 , f1 ] = h1 , [e1 , f2 ] = 0, • [e2 , f1 ] = 0, [e2 , f2 ] = h2 , • [e1 , [e1 , [e1 , [e1 , e2 ]]]] = 0, [e2 , [e2 , [e2 , [e2 , e1 ]]]] = 0, • [f1 , [f1 , [f1 , [f1 , f2 ]]]] = 0, [f2 , [f2 , [f2 , [f2 , f1 ]]]] = 0, • [h1 , h2 ] = 0. 17 The simple roots are {α1 , α2 } where: α1 (h1 ) = α2 (h2 ) = 2 α2 (h1 ) = α1 (h2 ) = −3 and any root is of the form α = α = (z1 , z2 ) ∈ Γ. The simple reflections are r1 (z1 , z2 ) = (−z1 + 3z2 , z2 ) and r1 (z1 , z2 ) = (z1 , 3z1 − z2 ). The Weyl Group W = hr1 , r2 i/hr1 2 = 1, r2 2 = 1i is an infinite dimensional group. Its action on the set of simple roots giver rise to a sequence {an } satisfying an = 3an−1 − an−2 . After solving the equation for this ”Fibonacci” sequence, we find the sequence of roots given by ! ! √ !n √ √ !n √ 3+ 5 3 5 1 3− 5 3 5 1 + + − + an = 10 2 2 10 2 2 This sequence allows us to explicitlygive the properties of the elements of the Weyl Group.   −aj aj+1 • (r2 r1 )j = aj+1 aj+2   −aj+2 aj+3 • r1 (r2 r1 )j = −aj+1 aj+2   −aj+1 −aj j • (r2 r1 ) r2 = aj+2 −aj+1   aj+2 −aj+1 j • (r1 r2 ) = aj+1 v −aj Finally, we have that Rre = {(±ak , ±ak+1 ), (±ak+1 , ±ak ), where ak ∈ Z and an satisfies an = 3an −1+an −2} . !!!!!!!!! We need to formalize what we will say about imaginary roots of the 3-kronecker.... M = {(m, n) | m, n ∈ Z; 23 m ≤ n ≤ 32 }