Quivers. Virginia, Lonardo, Tiago and Eloy UNICAMP. 28 July 2006

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1 Quivers Virginia, Lonardo, Tiago and Eloy UNICAMP 28 July Introduction This is our project Throughout this paper, k will indicate an algeraically closed field of characteristic 0 2 Quivers 21 asic definitions and Path algera Definition 21 A quiver is directed graph Q = (Q 0, Q 1 ) where Q 0 is the set of vertices and Q 1 set of arrows with maps h, t : Q 1 Q 0 which assign to each arrow its head and tail, respectively Definition 22 A path is a sequence of arrows p = a 1 a 2 a n such that h(a k+1 ) = t(a k ) for every k = 1,, n 1 The head of the path is h(a 1 ), and the tail of the path is t(a n ) Definition 23 An oriented cycle is a path p such that h(p) = t(p), and h(a i ) t(a j ) for any other i j + 1 Note that: this definitions implies that a quiver like an eight is not an oriented cycle Proposition 21 A quiver with an oriented cycle has an infinite set of paths Example 21 The Jordan quiver This quiver does not appear in this programm Example 22 The 2-Kronecker has no oriented cycle, ut 1 a 2 1 a 2 has an oriented cycle

2 2 2 Quivers Example 23 2 a 1 has no oriented cycle, ut c 3 2 a has an oriented cycle 1 c 3 Example 24 The 3-Kronecker quiver a c Path algera is an algera uild up from the paths To form the path algera, we must define a multiplication rule etween the paths of the quiver: Definition 24 Given a quiver Q, the path algera kq is the k-vector space generated y all paths in Q with multiplication rule: p q = { pq if h(q) = t(p) 0 otherwise HAVE TO DEFINE THE IDENTITY ELEMENT END SAY THATS ASSOC, AND AS AN ALGEBRA Now on we will denote the points in Q 0 Example 25 as its respective null path f 1 e 2 f 2 e 1 f 3 e 3 This quiver has path algera with asis {e 1, e 2, e 3, f 1, f 2, f 3, f 2 f 1 } Example 26 f 1 e 2 f 2 e 1 f 3 This quiver has an infinite dimensional path algera with asis: {e 1, e 2, e 3, f 1, f 2, f 3, f 2 f 1, f 3 f 2, f 1 f 3, f 3 f 2 f 1, f 1 f 3 f 2, f 2 f 1 f 3, } since the quiver contains an oriented cycle e 3

3 IRES Summer/Inverno quivers with relations We can impose further relations on the composition of the arrows to make paths 3 Quiver Representations Definition 31 A quiver representation is a collection {V i i Q 0 } of finite dimensional k-vector spaces together with a collection {φ a : V t(a) V h(a) a Q 1 } of k-linear maps From now on, we will denote a representation R = (V i, φ a ) Definition 32 Suppose R = (V i, φ a ) and S = (W i, ψ a ) are representations of Q The representation R is called a surepresentation of R if for every i Q 0, W i is a suspace of V i and for every a Q 1, the restriction of φ a : V t(a) V h(a) to W t(a) is equal to ψ a : W t(a) W h(a) Definition 33 A non-zero representation V is called simple representation if the only surepresentations of V are the zero representation and V itself Definition 34 If R = (V i, φ a ) and S = (W i, ψ a ) are representations of a quiver Q then we can define the direct sum representation R S = (U, ρ ) y: U i = V i W i for every i Q 0, and ρ a : V t(a) W t(a) V h(a) W h(a) is defined y the matrix ( ) Va 0 0 W a Definition 35 If R and S are two representation of a quiver Q, then a representation morphism Φ : R S is a collection k-linear maps {ϕ i : V i W i i Q 0 } such that the diagram V t(a) φ a Vh(a) ϕ t(a) ϕ h(a) W t(a) ψ a W h(a) commutes for all a Q 1 If ϕ i is invertile for every i Q 0, then the morphism Φ is called an isomorphism and R and S are called isomorphic representations Definition 36 A representation R of a quiver Q is called decomposale if R = S T where S and T are nonzero representations of Q A nonzero representation is called indecomposale if it is not decomposale

4 4 3 Quiver Representations 31 Isomorphism Classes of Representations The study of quiver representations will e simplified significantly if we can consider isomorphism classes of quiver representations, rather than specific representations To find a representative element oach isomorphism class, we apply representation isomorphisms to change the asis of the vector space at each vertex in order to simplify the matrices for the maps etween the vector spaces For square matrices and representations over C, this process is the same as the theory of the Jordan normal form Example 31 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 = ({V 1 }, {J 1 }) where J 1 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 invertile matrix, corresponding to the change of asis required to isolate the eigenspace oach eigenvalue {reference: Halmos} If we restrict ourselves to representations with invertile maps at each arrow, we may simultaneously descrie the isomorphism classes of representations of quivers which differ from each other only in the orientation of their arrows Note that the invertiility condition implies that the representation must have equidimensional vector spaces at all vertices The isomorphism classes of these representations can often e descried neatly, y analogy to the case of the Jordan quiver Example 32 For a quiver with two vertices and two arrows, given any representation R = ({V 1, V 2 }, {A, B}), where A and B are oth invertile, we can find an isomorphic representation of the form: V 1 A B V 2 = V 1 Id J V 2 To find this isomorphic representation, let B 1 and B 2 e ases for V 1 and V 2, respectively Take P 0 to e the change-of-asis matrix taking B 2 to AB 1 This is possile ecause invertiility implies equidimensionality Then the representation isomorphism Φ 0 = (Id, P 0 ) yields the isomorphic representation R = ({V 1, V 2 }, {Id, BA}) By Jordan s Theorem {Halmos}, we can find P 1 invertile such that BA = P1 1 JP 1, where J is a Jordan-form matrix Applying the representation isomorphism Φ 1 = (P 1, P 1 ) yields the desired isomorphic representation Example 33 Given a representation R = ({V 1, V 2, V 3 }, {A, B, C}) (with A, B, C all invertile) of the triangular quiver with three vertices and three arrows, we may find an isomorphic representation of the form: 2 2 A B = Id Id 1 C 3 1 J where J is a Jordan-form matrix The process for finding this representation is similar to the one descried in Example 32 aove 3

5 IRES Summer/Inverno Example 34 The case of the 3-Kronecker quiver is more complicated, ecause we may not e ale to simultaneously put the maps on the second and third arrows in Jordan normal form However, in the case dim(v 1 ) = dim(v 2 ) = 2, we will always e ale to conjugate ases and otain the form: V 1 A B C V 2 = V 1 Id J ( a 0 c ) V 2 where A = Id, B is in Jordan normal form, and C is a lower-triangular matrix 32 Simple Representations Definition 37 A canonical representation R for the quiver Q = (Q 0, Q 1 ) is a representation of the form { k for one i Q0 R = {V i =, φ {0} otherwise a = 0 for all a Q 1 } A canonical representation R must e simple, ecause the only suspace of it s V i is the null space at each vertex This implies that the only proper surepresentation is the trivial one, so R is simple Proposition 31 Let Q e 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 descried aove has a proper surepresentation of that form First, we egin with the following lemma Lemma 31 If Q = (Q 0, Q 1 ) is a quiver with no oriented cycles, then there is some vertex x Q 0 that is not the tail of any arrow Such an arrow is called a sink Proof: Proof y contrapositive Suppose for any v i Q 0, v i = t(a) for some a Q 1 Choose some v 1 Q 0 and form a path as follows: Choose a n such that t(a n ) = v n Write v n+1 = h(a n ), and repeat Since Q 0 is a finite set, eventually we will get v n + 1 = v i for some i n Then p = a i a n is an oriented cycle in Q Lemma 32 Let Q e a quiver with no oriented cycle, write x 1 Q 0 vertex such that t(a) x 1 for all a Q 1 is a Proof: Given an aritrary representation R = (V i, ρ a ), if V x1 {0}, then lael this vertex x (V h(a) = {0} for all a such that t(a) = x) If V x1 = {0}, define Q 1 = (Q 1 0 = Q 0 /{x 1 }, Q 1 1 = Q 1 /{a Q 1 h(a) = x 1 } Since Q contained no oriented cycles, and Q 0 Q 0, Q 1 Q 1, Q contains no oriented cycle, so we may apply the lemma Write x 2 Q 0 is a vertex such that t(a) x 2 for all a Q 1 Define the representation R of Q y restricting the representation R of Q If R is a non-trivial representation of Q, we will eventually find x n Q 0 such that V xn {0} ut V i = {0} for all i such that i = h(a) for some a Q 1 such that t(a) = x n Proposition { 32 Construct a representation k i = x S = {W i = {0} i x, φ a = 0 for all a Q 1 } for the quiver Q

6 6 3 Quiver Representations S is a surepresentation of R Proof: Clearly, for all i x {0} = w i v i Since v x is a non-zero k-vector space, k = w x v x Define p = {p i : i Q 0 } a representation morphism such that p i : w i v i is the inclusion map To check that all maps commute, first note that for a Q 1 such that t(a) x, W t(a) = {0} So ψ a : W t(a) W h(a) has {0} as its domain, ie ψ a = 0 Similarly P t(a) : W t(a) V t(a) is the inclusion of {0} implies P t(a) = 0 Hence, for all a Q 1 such that t(a) x we have: p h(a) ψ a = p h(a) 0 = 0 and ϕ a p t(a) = ϕ a 0 = 0 so the diagram commutes Now, for all a such that t(a) = x, we know that V h(a) = {0} So ϕ a : V t(a) V h(a) is ϕ a : V x {0}, ie ϕ a = {0} Similarly, ψ a = 0 and p h(a) : {0} {0} is also the zero map So for all a Q 1 such that t(a) = x, we have p h(a) = ψ a = 0 0 = 0 so the diagram commutes Therefore, S is a surepresentation of R of the desired form Example 35 1 a 2 2 a 1 c 3 These quivers have no oriented cycles, and only canonical simple representations Definition 38 A unique cycle is a cycle that has no other cycle on its vertices Proposition 33 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 e one parameter Example 36 1 a 2 2 a 1 c these quivers have unique oriented cycles, and 1-parameter families of simple representations 3

7 IRES Summer/Inverno Indecomposale Representations Here we will work y the examples we have shown aove The invertaility of maps and the dimensions vectors will play an important role in order to give all indecomposale representations for some given quiver representation Example 37 For the jordan quiver, if the map is invertile and there is an isomorphic representation with the canonical Jordan matrix, then it is indecomposale Example 38 For the example 23(?), we have the following classification: Proposition 34 A representation R = ({V 1 = C m, V 2 = C n }, {A, B}) (such that wlog m > n) of the quiver Q is indecomposale if and only if one of the following holds: R = R = ({W 1, W 2 }, {Id, J λ }) where J λ form with only one eigenlock is a matrix in Jordan normal (BA) k = 0 for some k Z and dim ker BA = 1 Corollary: The set of dimension vectors of indecomposale representations of Q is D Q = {(n, n), (n, n ± 1) n Z + } Proof of Proposition: We descrie the possile cases, and prove decomposaility or indecomposaility for each case 1 BA invertile R is indecomposale it is isomorphic to a representation with one Jordan lock 2 BA non-invertile (a) BA is nilpotent, ie (BA) k = 0 for some k Z + i nullba=1 R is indecomposale ii nullba 1 R is decomposale () BA is not nilpotent ie (BA) k 0 k Z + R is decomposale Proof of Claim: 1 BA invertile m=n, A, B are oth invertile as shown aove R is indecomposale it is isomorphic to a representation with one Jordan lock 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 = R R where R = ({W 1, W 2 }, {A W1, B W2 }), R = ({U 1, U 2 }, {A U1, B U2 }) are oth non-trivial wlog, x W 1 Suppose y U 1, y 0 y definition of decomposaility, (BA) i U 1 i Z +

8 8 3 Quiver Representations But (BA) k = 0 pick the least j Z + such that (BA) j = 0 (BA) j 1 ker BA (BA) j 1 y = λx W 1 This yields a contradiction V 1 = W 1 Suppose y U 2, y 0 ecause of a dimension argument, either y = Ax for some x V 1 or By = x for some nonzero x V 1 in either case, y W 2 ecause of the invariance of surepresentations This yields a contradiction V 2 = W 2 R is the trivial surepresentation R is indecomposale 2(a)ii null(ba) 1 write ker BA = W 0 U 0 oth of which are non-zero for x V 1, write j x Z + is the minimal integer such that (BA) jx = 0, and define: W 1 = {x V 1 (BA)j x 1x W 0 }, U 1 = {x V 1 (BA) jx 1 U 0 )} These two sets define a decomposition of R R is decomposale 2() (BA) k 0 k Z + j Z + stv 1 = ker(ba) j W 1 and (BA) j W1 is invertile These sets define a decomposition of R R is decomposale 34 Representation of Path Algeras Given a representation R = (V i, φ a ) of the quiver Q = (Q 0, Q 1 ), we can construct a representation ρ : kq End( i Q 0 V i ) of the path algera kq It suffices to define the representation on the e i s and f j s, as these generate the asis as a ring ρ(e i ) := Id Vi, ρ(f j ) : V t(j) V h(j) x φ j (x) This extends to a representation on all elements of kq 4 Lie Algeras and Their Representations Definition 41 A Lie Algera g is an non-associative algera with the multiple rule given y the ilinear 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, ie [x, y] = [y, x] x, y g We can construct a Lie algera from any associative algera y defining the racket operation as the commutator [a, ] = a a Example 41 Let V e a vector space this is an associative algera we can define the commutator racket ecause V is aelian, [a,]=0 always

9 IRES Summer/Inverno Representation of sl 2 (k) The simple linear algera sl 2 (k)= {A M 2 (k) tra = 0)} of traceless 2 2 matrices is a Lie Algera with racket operation defined y the commutator [A, B] = AB BA and asis: ( ) ( ) ( ) e =, f =, h = We will descrie the equivalence classes of simple and indecomposale representations of sl 2 (k) and show that these correspond to simple and indecomposale representations of the quiver with relations descried in Example?? We will restrict ourselves to representations of sl 2 (k) such that V = k Z V k, where V k = {v V hv = kv} is the eigenspace with eigenvalue k for the action of h on V V k = 0 for k 0 Each V k is finite dimensional Given v V k, 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 V k with eigenvalue k to V k 2 with eigenvalue k 2, and the action o takes V k to V k+2 Since m Z is the maximal eigenvalue, we oserve that { k if k = m 2i for i Z+ V k = 0 otherwise Thus, if we take v 0 V m, we can define a asis B={v i i Z + } for V (m) = k m V k y the rule v i = f i (v 0 ) From the equations aove, we can calculate that e(v i ) = i(m i + 1)v i+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 = kv 0 f e V m 4 = kv 1 V(m) However, if the highest eigenvalue is positive, the action o will annihilate the eigenspace V m 2, so the representation will not e simple, in fact, it has M( m 2) for a surepresentation It will, however, e indecomposale, as the suspace m/leqk/leqm V k is not invariant under the action of f This representation is called V (m)

10 10 4 Lie Algeras and Their Representations V m = kv 0 V m = kv m f V m 2 =kv m+1 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): V m = kv 0 f V m = kv m M (m) Because we have e(v m 2 ) = 0, we may define another linearly independent eigenvector w 0 with eigenvalue m 2, such that e(w 0 ) = v m This can e extended y the rule w i = f i (w 0 ) to find a second eigenvector for each eigenvalue λ = m 2(i + 1) Taking V k = span{v 1/2(m k) } for m > k > m and V k = span{w 1/2( m 2 k) } for k m 2, we find the representation M (m): e

11 IRES Summer/Inverno V m = kv 0 V m = kv m e V m 2 = kw 0 V m 4 = kw 1 P(-m-2) However, for k m 2, we may also allow V k to e two-dimensional, that is, V k = span{v 1/2(m k), w 1/2( m 2 k) } This descries the representation called P ( m 2) f e V m = kv 0 V m = kv m f e V m 2 =kv m+1 kw 0 V m 4 =kv m+2 e e f kw 1 In this case, note that e(w i ) = aw i 1 + vi m 2

12 12 4 Lie Algeras and Their Representations 5 Cartan Matrix and Kac-Moody Algeras Definition 51 A Cartan matrix is a matrix A = (a ij ) such that: a ii = 2 a ij 0 if i j and a ij Z a ij = 0 if and only if a ji = 0 Let Q e a quiver without loops, #Q 0 = n We can construct an n n Cartan matrix { aii = 2 (i = 1,, n) A = (a ij ) where a ij = #{arrows tw vertices i and j} (i 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 52 Given an n n Cartan matrix, we may construct a Lie algera, called the Kac-Moody algera, with the following generators and relations: generators: {e i, f i, h i i = 1,, n} relations: [h i, e j ] = a ij e j [h i, f j ] = a ij f j [e i, f j ] = δ i,j h i ad 1 aij (e i )e j = 0 ad 1 aij (f i )f j = 0 where ad(α) = ad α, α g(a) is the adjoint action ad α (β) = [α, β] Definition 53 The simple roots of a Kac-Moody algera g(a) are the functionals {α j },j = 1,, n, defined y: α j (h i ) = a ij Consider Γ = Zα i the set of all Z-linear cominations of simple roots Claim: If α is a root of g(a) then α Γ Definition 54 For (α i, α j ) = a ij the maps r i : h h, such that r i (λ) = λ (λ, α i ) α i are called simple reflections Remark: Since (r i ) = 1 h, r i is invertile for every i and the set of all compositions of r i is indeed a group, denoted y 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 55 Let α e an element of Γ, so α is a real root if w W such that w(α) = α i for some i {1, 2,, n} The othes are called imaginary roots Proposition 51 The real and imaginary roots may e calculate y the following formulas: R re = w W w(α i ) R im = w W w(m), where M = {α Γ + \0 (α, α i ) 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 aove

13 IRES Summer/Inverno sl 2 (k) The quiver Q A = gives Cartan matrix A = (2) The g(a) is the Lie algera generated y 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 R re = {α, α}, and since M = {α Γ + \0 (α, α) 0}, this implies that M =, which means that all the roots are real 52 sl 3 (k) The quiver Q B = ( ) 2 1 gives Cartan matrix B = 1 2 g(b) is the Lie Algera generated y {e 1, e 2, f 1, f 2, h 1, h 2 } with relations: [h 1, e 1 ] = 2e 1, [h 1, e 2 ] = e 2 [h 1, f 1 ] = 2f 1, [h 1, f 2 ] = f 2 [e 1, f 1 ] = h 1, [e 1, f 2 ] = 0, [e 2, f 1 ] = 0, [e 2, f 2 ] = h 2 [e 1, [e 1, e 2 ]] = 0, [e 2, [e 2, e 1 ]] = 0 [f 1, [f 1, f 2 ]] = 0, [f 2, [f 2, f 1 ]] = 0 [h 1, h 2 ] = 0 Indeed, g(b) sl 3 (k), the algera of 3 3 traceless matrices with racket operation defined y the commutator The simple roots are {α 1, α 2 } where α 1 (h 1 ) = 2, α 2 (h 1 ) = 1, α 1 (h 2 ) = 1, α 2 (h 2 ) = 2 and any root is of the form α = (z 1, z 2 ) Γ The simple reflections are r 1 (z 1, z 2 ) = (z 2 z 1, z 2 ) and r 2 (z 1, z 2 ) = (z 1, z 1 z 2 ) The Weyl Group is W = {I d, r 1, r 2, r 1 r 2, r 2 r 1, r 1 r 2 r 1 } where: r 1 r 2 (z 1, z 2 ) = ( z 2, z 1 z 2 ), r 2 r 1 (z 1, z 2 ) = (z 2 z 1, z 1 ) r 1 r 2 r 1 (z 1, z 2 ) = r 2 r 1 r 2 (z 1, z 2 ) = ( z 2, z 1 ) So, W is a finite group The caracterization of W makes possile to find all real and imaginary roots, as follows: R re = {α 1, α 1, α 2, α 2 α 1 + α 2, α 1 α 2 } since M = R im =

14 14 5 Cartan Matrix and Kac-Moody Algeras 53 ˆ sl2 The quiver Q C = ( ) 2 2 gives Cartan matrix C = 2 2 g(c) is the Lie Algera generated y {e 1, e 2, f 1, f 2, h 1, h 2 } with relations: [h 1, e 1 ] = 2e 1, [h 1, e 2 ] = 2e 2 [h 1, f 1 ] = 2f 1, [h 1, f 2 ] = 2f 2 [e 1, f 1 ] = h 1, [e 1, f 2 ] = 0, [e 2, f 1 ] = 0, [e 2, f 2 ] = h 2 [e 1, [e 1 [e 1, e 2 ]]] = 0, [e 2, [e 2, [e 2, e 1 ]]] = 0, [f 1, [f 1 [f 1, f 2 ]]] = 0, [f 2, [f 2, [f 2, f 1 ]]] = 0 [h 1, h 2 ] = 0 Indeed, g(c) sl 2 [t, t 1 ] kc for some constant c The simple roots are {α 1, α 2 } where α 1 (h 1 ) = 2 = α 2 (h 2 ), α 2 (h 1 ) = 2 = α 1 (h 2 ) and any root is of the form α = (z 1, z 2 ) Γ The simple reflections are r 1 (z 1, z 2 ) = ( z 1 + 2z 2, z 2 ) and r 2 (z 1, z 2 ) = (z 1, 2z 1 z 2 ) The Weyl Group W = r 1, r 2 / r 1 2 = 1, r 2 2 = 1 is an infinite dimensional group with the following properties, for r W: ( 2j + 1, 2j) if r = (r 2 r 1 ) j (2j + 1, 2j) if r = (r r(α 1 ) = 1 r 2 ) j ( 2j 1, 2j) if r = r 1 (r 2 r 1 ) j (2j + 1, 2(j + 1)) if r = r 2 (r 1 r 2 ) j (2j, 2j + 1) if r = (r 2 r 1 ) j ( 2j), 2j + 1) if r = (r r(α 2 ) = 1 r 2 ) j (2(j + 1), 2j + 1) if r = r 1 (r 2 r 1 ) j ( 2j, 2j 1)) if r = r 2 (r 1 r 2 ) j where j N So y choosing any positive integer j, we find that the real roots are: R re = {(m, n) m, n Z, m n = 1} To calculate the imaginary roots, we note that: M = {(z 1, z 2 ) z 1, z 2 Z +, 2z 1 2z 2 0, 2z 1 + 2z 2 0} = {(z 1, z 2 ) z 1, z 2 Z +, z 1 = z 2 } = Z + (1, 1) These elements are preserved y the action of W, so we have: R im = Z + (α 1 + α 2 ) = {(n, n) n Z}

15 IRES Summer/Inverno ˆ sl 3 k The quiver Q D = gives Cartan matrix D = g(d) is the Lie Algera generated y {e 1, e 2, e 3, f 1, f 2, f 3, h 1, h 2, h 3 } with relations: [h 1, e 1 ] = 2e 1, [h 1, e 2 ] = e 2, [h 1, e 3 ] = e 3, [h 2, e 1 ] = e 1, [h 2, e 2 ] = 2e 2, [h 2, e 3 ] = e 3, [h 3, e 1 ] = e 1, [h 3, e 2 ] = e 2, [h 3, e 3 ] = 2e 3, [h 1, f 1 ] = 2f 1, [h 1, f 2 ] = f 2, [h 1, f 3 ] = f 3, [h 2, f 1 ] = f 1, [h 2, f 2 ] = 2f 2, [h 2, f 3 ] = f 3, [h 3, f 1 ] = f 1, [h 3, f 2 ] = f 2, [h 3, f 3 ] = 2f 3, [e 1, f 1 ] = h 1, [e 1, f 2 ] = 0, [e 1, f 3 ] = 0, [e 2, f 1 ] = 0, [e 2, f 2 ] = h 2, [e 2, f 3 ] = 0, [e 3, f 1 ] = 0, [e 3, f 2 ] = 0, [e 3, f 3 ] = h 3, [e 1, [e 1, e 2 ]] = 0, [e 1, [e 1, e 3 ]] = 0, [e 2, [e 2, e 1 ]] = 0, [e 2, [e 2, e 3 ]] = 0, [e 3, [e 3, e 1 ]] = 0, [e 3, [e 3, e 2 ]] = 0, [f 1, [f 1, f 2 ]] = 0, [f 1, [f 1, f 3 ]] = 0, [f 2, [f 2, f 1 ]] = 0, [f 2, [f 2, f 3 ]] = 0, [f 3, [f 3, f 1 ]] = 0, [f 3, [f 3, f 2 ]] = 0 [h 1, h 2 ] = 0, [h 2, h 3 ] = 0 Indeed, g(d) sl 3 [t, t 1 ] Cc for some constant c The simple roots are {α 1, α 2, α 3 } where: α 1 (h 1 ) = α 2 (h 2 ) = α 3 (h 3 ) = 2 α 2 (h 1 ) = α 3 (h 1 ) = α 1 (h 2 ) = α 3 (h 2 ) = α 1 (h 3 ) = α 2 (h 3 ) = 1 and any root is of the form α = α = (z 1, z 2, z 3 ) Γ The simple reflections are r 1 (z 1, z 2, z 3 ) = ( z 1 + z 2 + z 3, z 2, z 3 ), r 1 (z 1, z 2, z 3 ) = (z 1, z 1 z 2 + z 3, z 3 ), r 1 (z 1, z 2, z 3 ) = (z 1, z 2, z 1 + z 3 z 3 ) The Weyl Group W = r 1, r 2 / r 1 2 = 1, r 2 2 = 1 is an infinite dimensional group with the following properties: r 1 (k ± 1, k, k) = (k 1, k, k)

16 16 5 Cartan Matrix and Kac-Moody Algeras r 1 (k, k ± 1, k) = (k ± 1, k ± 1, k) r 1 (k, k, k ± 1) = (k ± 1, k, k ± 1) r 2 (k ± 1, k, k) = (k ± 1, k ± 1, k) r 2 (k, k ± 1, k) = (k, k 1, k) r 2 (k, k, k ± 1) = (k, k 1, k ± 1) r 3 (k ± 1, k, k) = (k ± 1, k, k ± 1) r 3 (k, k ± 1, k) = (k, k ± 1, k ± 1) r 3 (k, k, k ± 1) = (k, k, k 1) With these properties, we can compute the real and imaginary roots which leads to: R re = {(k ± 1, k, k), (k, k ± 1, k), (k, k, k ± 1)} k Z To calculate the imaginary roots, we note that: M = {(z 1, z 2, z 3 ) z 1, z 2, z 3 Z +, 2z 1 z 2 z 3 0, z 1 + 2z 2 z 3 0 z 1 z 2 + 2z 3 0} = {(z 1, z 2, z 3 ) z 1, z 2, z 3 Z +, z 1 = z 2 = z 3 } = Z + (1, 1, 1) This set is preserved y W, so we have: R im = Z + (α 1 + α 2 + α 3 ) 55 3-Kronecker The quiver Q E = ( ) 2 3 gives Cartan matrix E = 3 2 g(e) is the Lie Algera generated y {e 1, e 2, f 1, f 2, h 1, h 2 } given y the relations: [h 1, e 1 ] = 2e 1, [h 1, e 2 ] = 3e 2, [h 1, f 1 ] = 2f 1, [h 1, f 2 ] = 3f 2, [e 1, f 1 ] = h 1, [e 1, f 2 ] = 0, [e 2, f 1 ] = 0, [e 2, f 2 ] = h 2, [e 1, [e 1, [e 1, [e 1, e 2 ]]]] = 0, [e 2, [e 2, [e 2, [e 2, e 1 ]]]] = 0, [f 1, [f 1, [f 1, [f 1, f 2 ]]]] = 0, [f 2, [f 2, [f 2, [f 2, f 1 ]]]] = 0, [h 1, h 2 ] = 0

17 IRES Summer/Inverno The simple roots are {α 1, α 2 } where: α 1 (h 1 ) = α 2 (h 2 ) = 2 α 2 (h 1 ) = α 1 (h 2 ) = 3 and any root is of the form α = α = (z 1, z 2 ) Γ The simple reflections are r 1 (z 1, z 2 ) = ( z 1 + 3z 2, z 2 ) and r 1 (z 1, z 2 ) = (z 1, 3z 1 z 2 ) The Weyl Group W = r 1, r 2 / r 2 1 = 1, r 2 2 = 1 is an infinite dimensional group Its action on the set of simple roots giver rise to a sequence {a n } satisfying a n = 3a n 1 a n 2 After solving the equation for this Fionacci sequence, we find the sequence of roots given y ( 3 ) ( 5 a n = ) n ( ) ( 3 ) n 5 2 This sequence allows us to explicitlygive the properties of the elements of the Weyl Group ( ) (r 2 r 1 ) j aj a = j+1 a j+1 a j+2 ( ) r 1 (r 2 r 1 ) j aj+2 a = j+3 a j+1 a j+2 ( ) (r 2 r 1 ) j aj+1 a r 2 = j a j+2 a j+1 ( ) (r 1 r 2 ) j aj+2 a = j+1 a j+1 v a j Finally, we have that R re = {(±a k, ±a k+1 ), (±a k+1, ±a k ), where a k Z and a n satisfies a n = 3a n 1+a n 2}!!!!!!!!! We need to formalize what we will say aout imaginary roots of the 3-kronecker M = {(m, n) m, n Z; 2 3 m n 3 2 } References [1] Fuchs,J and Schweigert, C ; Symmetries, Lie Algeras and Representations - A Graduate Course for Physics Camridge University Press, 30-36, 1997 [2] Kac, V G ; Infinite Dimensional Lie Algeras Camridge University Press, third edition; chapter five, 1990 [3] Humphreys, JE ; Introduction to Lie Algeras and Representation Theory Springer, chapters one, two and three, 1987 [4] Fulton, W and Harris, J; Representation Theory : a First Course Springer - Graduate texts in mathematics, 1991 [5] Lecture notes from: hderksen/math711w01/math711html [6] http : //xxxlanlgov/p S c ach/math/pdf/0505/ pdf [7] Halmos, P R ; Finite-Dimensional Vector Spaces, 2nd ed Van Nostrand, 112, 1958