Regular Kac-Moody superalgebras and integrable highest weight modules
aa r X i v : . [ m a t h . R T ] N ov Regular Kac-Moody superalgebras andintegrable highest weight modules
Crystal Hoyt ∗ Abstract
We define regular Kac-Moody superalgebras and classify them usingintegrable modules. We give conditions for irreducible highest weightmodules of regular Kac-Moody superalgebras to be integrable. This pa-per is a major part of the proof for the classification of finite-growthcontragredient Lie superalgebras.
Keywords: Lie superalgebra; integrable modules; odd reflections.
The results of this paper are a crucial part of the proof for the classification ofcontragredient Lie superalgebras with finite growth, and in particular, for theclassification of finite-growth Kac-Moody superalgebras [2, 3]. Previously, sucha classification was known only for contragredient Lie superalgebras with eithersymmetrizable Cartan matrices [12, 13], or Cartan matrices with no zeros onthe main diagonal, i.e. contragredient Lie superalgebras without simple isotropicroots [5]. Several of the results of this paper are surveyed in [11].A contragredient Lie superalgebra g ( A ) is a Lie superalgebra defined by aCartan matrix A [4, 6]. A Lie superalgebra usually has more than one Cartanmatrix. However, an odd reflection at a regular simple isotropic root allowsone to move from one base to another (see Definitions 1.4, 1.5) [10]. An oddreflection yields a new Cartan matrix A ′ such that g ( A ′ ) and g ( A ) are isomorphicas Lie superalgebras.A matrix which satisfies certain numerical conditions is called a generalizedCartan matrix (see Definition 1.9). If A is a generalized Cartan matrix thenall simple isotropic roots are regular. A contragredient Lie superalgebra g ( A ) issaid to be regular Kac-Moody if A and any matrix A ′ , obtained by a sequenceof odd reflections of A , are generalized Cartan matrices. If A is a generalizedCartan matrix and g ( A ) has no simple isotropic roots, then g ( A ) is regular Kac-Moody by definition. Hence, we restrict our attention to regular Kac-Moody Supported in part by NSF grant DMS-0354321 at the University of California, Berkeleyand by ISF grant, no. 1142/07 at the Weizmann Institute of Science. ∗ Department of Mathematics, Weizmann Institute of Science, Rehovot 76100 Israel; Email:[email protected]. g ( A ) is a finite-growth contragredient Lie superal-gebra and the defining matrix A has no zero rows, then simple root vectors of g ( A ) act locally nilpotently on the adjoint module. This implies certain con-ditions on A which are only slightly weaker than the conditions for the matrixto be a generalized Cartan matrix. For a finite-growth Lie superalgebra, thesematrix conditions should still hold after odd reflections, which leads to the def-inition of a regular Kac-Moody superalgebra. Remarkably, these superalgebrasalmost always have finite growth. The exception is the family: Q ± ( m, n, t ) with m, n, t ∈ Z ≤− and not all equal to − Theorem 2.27. If A is a symmetrizable indecomposable matrix and the contra-gredient Lie superalgebra g ( A ) has a simple isotropic root, then g ( A ) is a regularKac-Moody superalgebra if and only if it has finite growth.If A is a non-symmetrizable indecomposable matrix and the contragredientLie superalgebra g ( A ) has a simple isotropic root, then g ( A ) is a regular Kac-Moody superalgebra if and only if it belongs to one of the following three families: q ( n ) (2) , S (1 , , α ) with α ∈ C \ Z , Q ± ( m, n, t ) with m, n, t ∈ Z ≤− . The non-symmetrizable contragredient Lie superalgebra S (1 , , α ) appearsin the list of conformal superalgebras [7]. It has finite growth, but is not regularKac-Moody when α ∈ Z . The non-symmetrizable regular Kac-Moody super-algebra Q ± ( m, n, t ) was discovered during this classification (see Section 3). If m, n, t = − Q ± ( m, n, t ) is just q (3) , which has finite growth. Otherwise, Q ± ( m, n, t ) has infinite growth, but is hyperbolic for small (absolute) values of m , n , and t . An explicit realization of Q ± ( m, n, t ) is still unknown and wouldbe interesting.For the proof of Theorem 2.27, we classify the corresponding connectedregular Kac-Moody diagrams (see Section 1.3) by using induction on the numberof vertices (i.e. simple roots). A subdiagram of a regular Kac-Moody diagramis regular Kac-Moody, however if the subdiagram does not have an isotropicvertex then it is not part of the classification. We work around this difficultyby using odd reflections.We say that a regular Kac-Moody diagram is subfinite if it is connected,has an isotropic vertex, and satisfies the condition that all connected propersubdiagrams which have an isotropic vertex are of finite type. In Section 2,we find all subfinite regular Kac-Moody diagrams by extending connected finitetype diagrams which have an isotropic vertex. A diagram of a finite-dimensionalor affine Kac-Moody superalgebra is subfinite regular Kac-Moody. In Section 4,we prove that every connected regular Kac-Moody diagram with an isotropicvertex is subfinite by using integrable modules and some explicit computations.2 highest weight module V of a regular Kac-Moody superalgebra g ( A ) iscalled integrable if for each real root α the element X α ∈ g ( A ) α is locallynilpotent on V . This is consistent with the original definition of integrablemodules for affine Lie superalgebras. It is shown in [9] that most non-twistedaffine Lie superalgebras do not have non-trivial irreducible integrable highestweight modules. The only exceptions are B (0 , n ) (1) , A (0 , m ) (1) and C ( n ) (1) .By the same method, one can show that the twisted affine Lie superalgebras,including q ( n ) (2) , but excluding A (0 , n − (2) , A (0 , n ) (4) and C ( n ) (2) , haveonly trivial irreducible integrable highest weight modules.A regular Kac-Moody superalgebra is integrable under the adjoint action,hence, so are its submodules. A non-trivial extension of a regular Kac-Moodydiagram Γ yields a non-trivial integrable highest weight module of g ( A Γ ). Thus,if the Kac-Moody superalgebra corresponding to Γ does not have non-trivialirreducible integrable highest weight modules, then Γ is not extendable. Thus,it remains to show that the diagrams for A (0 , m ) (1) , C ( n ) (1) , S (1 , , α ), and Q ± ( m, n, t ) are not extendable, which we do by direct computation.Integrable irreducible highest weight modules for affine Lie superalgebraswere described in [9]. If L ( λ ) is an irreducible highest weight module a regularKac-Moody superalgebra which does not have an isotropic root, then L ( λ ) isintegrable if and only if λ i ∈ p ( i ) Z ≥ . We describe the integrable irreduciblehighest weight modules for the remaining regular Kac-Moody superalgebraswhich have an isotropic root: S (1 , , α ) with α ∈ C \ Z , and Q ± ( m, n, t ) with m, n, t ∈ Z ≤− and not all equal to −
1. We show that these superalgebras havenon-trivial irreducible integrable highest weight modules, and give explicit con-ditions on the weights for an irreducible highest weight module to be integrable.
Acknowledgement.
The author would like to express her sincere grati-tude to Professor Vera Serganova for helpful discussions and suggestions whilesupervising this dissertation work.
Let A be a n × n matrix over C , I = { , . . . , n } and p : I → Z be a parityfunction. Fix a vector space h over C of dimension 2 n − rk( A ). Let α , . . . , α n ∈ h ∗ and h , . . . , h n ∈ h be linearly independent elements satisfying α j ( h i ) = a ij ,where a ij is the ij -th entry of A . Define a Lie superalgebra ¯ g ( A ) by generators X , . . . , X n , Y , . . . , Y n and h , and by relations[ h, X i ] = α i ( h ) X i , [ h, Y i ] = − α i ( h ) Y i , [ X i , Y j ] = δ ij h i , (1)where the parity of X i and Y i is p ( i ), and the elements of h are even.The contragredient Lie superalgebra given by the matrix A is defined to bethe quotient of ¯ g ( A ) by the unique maximal ideal that intersects h trivially, andit is denoted by g ( A ). We call A the Cartan matrix of g ( A ). If B = DA for3ome invertible diagonal matrix D , then g ( B ) ∼ = g ( A ). Hence, we may assumewithout loss of generality that a ii ∈ { , } for i ∈ I .The matrix A is said to be symmetrizable if there exists an invertible diagonalmatrix D such that B = DA is a symmetric matrix, i.e. b ij = b ji for all i, j ∈ I . In this case, we also say that g ( A ) is symmetrizable. A contragredientLie superalgebra is symmetrizable if and only if there exists a nondegenerateinvariant symmetric bilinear form. Hence, symmetrizability does not depend onthe choice of Cartan matrix.The matrix A is indecomposable if the the set I can not be decomposed intothe disjoint union of non-empty subsets J, K such that a j,k = a k,j = 0 whenever j ∈ J and k ∈ K . A proof of the following lemma can be found in [3]. Lemma 1.1.
For any subset J ⊂ I the subalgebra a J in g ( A ) generated by h , X i and Y i , with i ∈ J , is isomorphic to h ′ ⊕ g ( A J ) , where A J is the submatrixof A with coefficients ( a ij ) i,j ∈ J and h ′ is a subspace of h . More precisely, h ′ is amaximal subspace in ∩ i ∈ J Ker α i which trivially intersects the span of h i , i ∈ J . A superalgebra g = g ( A ) has a natural Z -grading g = ⊕ g m , called the principal grading , which is defined by g = h and g = g α ⊕ · · · ⊕ g α n . We saythat g is of finite growth if dim g n grows polynomially depending on n . Thismeans that the Gelfand-Kirillov dimension of g is finite.We recall a result from [3]. Theorem 1.2 (Hoyt, Serganova) . Suppose A is a matrix with no zero rows. If g ( A ) has finite growth, then ad X i are locally nilpotent for all i ∈ I . A proof of the following lemma can be found in [3].
Lemma 1.3.
Let g ( A ) be a contragredient Lie superalgebra. Then ad X i arelocally nilpotent for all i ∈ I if and only if A satisfies the following conditions,(after rescaling the rows of A such that a ii ∈ { , } for all i ∈ I ):1. if a ii = 0 and p ( i ) = 0 , then a ij = 0 for all j ∈ I ;2. if a ii = 2 , then a ij ∈ p ( i ) Z ≤ for all j ∈ I ;3. if a ij = 0 and a ji = 0 , then a ii = 0 . The Lie superalgebra g = g ( A ) has a root space decomposition g = h ⊕ M α ∈ ∆ g α . Every root is either a positive or a negative linear combination of the simpleroots, α , . . . , α n . Accordingly, we have the decomposition ∆ = ∆ + ∪ ∆ − , andwe call α ∈ ∆ + positive and α ∈ ∆ − negative. One can define p : ∆ → Z byletting p ( α ) = 0 or 1 whenever α is even or odd, respectively. By ∆ (resp. ∆ )we denote the set of even (resp. odd) roots.4et Π := { α , . . . , α n } be the set of simple roots of g ( A ). There are fourpossibilities for each simple root α i :1. if a ii = 2 and p ( α i ) = 0, then X i , Y i and h i generate a subalgebra isomor-phic to sl (2);2. if a ii = 0 and p ( α i ) = 0, then X i , Y i and h i generate a subalgebra isomor-phic to the Heisenberg algebra;3. if a ii = 2 and p ( α i ) = 1, then X i , Y i and h i generate a subalgebra isomor-phic to osp (1 | α i ∈ ∆;4. if a ii = 0 and p ( α i ) = 1, then X i , Y i and h i generate a subalgebra isomor-phic to sl (1 | Definition 1.4.
A simple root α i is isotropic if a ii = 0 and p ( α i ) = 1, andotherwise α i is non-isotropic . Definition 1.5.
A simple root α i is regular if for any other simple root α j , a ij = 0 implies a ji = 0.If α k is a simple root with a kk = 0, we define the (even) reflection r k at α k by r k ( α i ) = α i − α i ( h k ) α k , α i ∈ Π . If α k is a regular isotropic root, we define the odd reflection r k at α k asfollows: r k ( α i ) = − α k , if i = k ; α i , if a ik = a ki = 0, i = k ; α i + α k , if a ik = 0 or a ki = 0, i = k ; X ′ i := Y i , if i = k ; X i , if i = k , and a ik = a ki = 0;[ X i , X k ] , if i = k , and a ik = 0 or a ki = 0; Y ′ i := X i , if i = k ; Y i , if i = k , and a ik = a ki = 0;[ Y i , Y k ] , if i = k , and a ik = 0 or a ki = 0;and h ′ i := [ X ′ i , Y ′ i ] . Then h ′ i = ( − p ( α i ) ( a ik h k + a ki h i ) if i = k , and a ik or a ki = 0 ,h i if i = k , and a ik = a ki = 0 ,h k if i = k. Set α ′ i := r k ( α i ) for i ∈ I .A proof of the following lemma can be found in [3].5 emma 1.6. Let g ( A ) be a contragredient Lie superalgebra with base Π = { α , . . . , α n } . Suppose that Π ′ = { α ′ , . . . , α ′ n } is is obtained from Π by an oddreflection with respect to a regular isotropic root. Then α ′ , . . . , α ′ n are linearly in-dependent. The corresponding elements (defined above) X ′ , . . . , X ′ n , Y ′ , . . . , Y ′ n together with h ′ , . . . , h ′ n satisfy (1) . Moreover, h and X ′ , . . . , X ′ n , Y ′ , . . . , Y ′ n generate g ( A ) . It then follows that given a matrix A and a regular isotropic root α k , one canconstruct a new matrix A ′ such that g ( A ′ ) ∼ = g ( A ) as superalgebras. Explicitly,the entries of A ′ can be defined by A ′ ij = α ′ j ( h ′ i ). After possibly rescaling theelements h ′ i , we find that ( i = k and j = k ): a ′ kk := a kk ; a ′ kj := a kj ; a ′ ik := − a ki a ik ; a ′ ij := a ij , if a ik = a ki = 0; a ki a ij , if a ik or a ki = 0 , and a kj = a jk = 0; a ki a ij + a ik a kj + a ki a ik , if a ik or a ki = 0 , and a jk or a kj = 0 . Remark . We can rescale the rows of A ′ by rescaling the elements h ′ i . Soafter rescaling, we may assume that a ′ ii = α ′ i ( h ′ i ) = 0 or 2. In the case that a ′ ii = 0 for some i , it is our convention to rescale A ′ so that a ′ ij = 1 for some j .We say that A ′ is obtained from A (and Π ′ := { α ′ , . . . , α ′ n } is obtained fromΠ) by an odd reflection with respect to α k . If ∆ ′ + is the set of positive rootswith respect to Π ′ , then ∆ ′ + = (cid:0) ∆ + \ { α k } (cid:1) ∪ {− α k } . A root α is called real if α or α is simple in some base Π ′ , which is obtained fromΠ by a sequence of even and odd reflections. Otherwise, it is called imaginary otherwise. Remark . A reflection with respect to a regular isotropic simple root α k is indeed a reflection. If A ′′ is obtained from A by successively applying thereflection at α k twice, then there is an invertible diagonal matrix D such that A ′′ = DA and scalars b i , c i such that X ′′ i = b i X i and Y ′′ i = c i Y i . It is possibleto define an “odd reflection” at a simple isotropic root which is not regular,but in this case the subalgebra generated by X ′ , . . . , X ′ n , Y ′ , . . . , Y ′ n and h isnecessarily a proper subalgebra of g ( A ).The notion of finite growth does not depend on the choice of a base for g ( A )[3]. It thus follows from Theorem 1.2 that if g ( A ) has finite growth and A hasno zero rows, then A and any matrix A ′ obtained from A by a sequence of oddreflections satisfies the matrix conditions of Lemma 1.3. Definition 1.9.
A matrix A is called a generalized Cartan matrix if it satisfiesthe matrix conditions of Lemma 1.3, where the third condition is strengthenedto the following: 3 ′ . if a ij = 0, then a ji = 0 .
6f a matrix satisfies the conditions of Lemma 1.3, then condition (3 ′ ) isequivalent to the condition that all simple isotropic roots are regular. We call g ( A ) a regular Kac-Moody superalgebra if A and any matrix obtained from A bya sequence of odd reflections is a generalized Cartan matrix.We call a root α ∈ ∆ principal if either α is even and belongs to some baseΠ ′ obtained from Π be a sequence of odd reflections, or if α = 2 β where β is oddand belongs to some base Π ′ obtained from Π be a sequence of odd reflections.For a principal root, the subalgebra generated by X α , Y α and h α := [ X α , Y α ] isisomorphic to sl , and we may choose X α , Y α such that α ( h α ) = 2. Note thatif α = 2 β , then X α = [ X β , X β ].Let Π ⊂ ∆ denote the set of principal roots. It is clear that Π ⊂ ∆ +0 . Ingeneral this set can be infinite, but this is not the case whenever g ( A ) has finitegrowth [3]. For any finite subset S ⊂ Π , we can define a matrix B by setting b ij = α j ( h i ).A proof of the following lemma can be found in [3]. Lemma 1.10.
If a Lie superalgebra g ( A ) has finite growth, then for any finitesubset S of Π the Lie algebra g ( B ) also has finite growth. In particular, B isa generalized Cartan matrix for a finite growth Kac-Moody algebra. Given a Cartan matrix A , we can associate a matrix diagram , denoted Γ A (orsimply Γ when A is fixed), as follows. Recall that we may assume that a i i = 0or 2. The vertices of Γ A correspond to the simple roots of g ( A ) and are givenby the following table. g ( A ) A p (1) Diagram A (2) 0 (cid:13) B (0 ,
1) (2) 1 • A (0 ,
0) (0) 1 N Heisenberg (0) 0 (cid:3)
We join vertex v i to vertex v j by an arrow if a ij = 0, and we label this arrowwith the number a ij . This correspondence between Cartan matrices and matrixdiagrams is a bijection. The condition that the matrix A is indecomposablecorresponds to the requirement that the diagram Γ A is connected. We say thata vertex is isotropic if it is of the type N . We use the notation of the followingtable to denote the possible vertex types.Notation Vertex Types • (cid:13) or NJ (cid:13) or • (cid:13) v (cid:13) or N or • Let A be a Cartan matrix with a ii = 0. Then rescaling the i th row of thematrix A (i.e. multiplying by a non-zero constant) corresponds to rescaling all7abels of arrows exiting the vertex v i of the diagram Γ A . This defines is anequivalence relation on matrix diagrams, where Γ A ′ ∼ Γ A if A ′ = DA for someinvertible diagonal matrix D , and in this case g ( A ′ ) ∼ = g ( A ). We consider matrixdiagrams modulo this equivalence.A diagram Γ A is called a regular Kac-Moody (or regular Kac-Moody di-agram) if the corresponding contragredient Lie superalgebra g ( A ) is regularKac-Moody. An odd reflection of a matrix diagram Γ A at an isotropic vertex v i is defined to be the diagram Γ A ′ , where A ′ is obtained from A by an oddreflection at the corresponding isotropic root α i of g ( A ). Note that the set of allregular Kac-Moody diagrams is closed under odd reflections. Denote by C (Γ)the collection of all diagrams obtained from sequences of odd reflections of aregular Kac-Moody diagram Γ. When α i is a simple odd isotropic root, wedenote by r i the odd reflection with respect to α i .By a subdiagram Γ ′ of the diagram Γ, we mean a full subdiagram, i.e. if thevertices v i and v j are in Γ ′ then the arrows with labels a ij and a ji also belong toΓ ′ . We say that a connected regular Kac-Moody diagram is extendable if it is aproper subdiagram of a connected regular Kac-Moody diagram. For a 3-vertexsubdiagram Γ ′ = { v i , v j , v k } , with v j isotropic, we refer to the fractions a ji a jk and a jk a ji as the ratios of the isotropic vertex v j in Γ ′ . In order to classify regular Kac-Moody superalgebras, we classify the corre-sponding regular Kac-Moody diagrams. A diagram for a finite-dimensional oraffine Kac-Moody superalgebra is regular Kac-Moody.
Definition 2.1.
We call a regular Kac-Moody diagram subfinite if it is con-nected, it contains an isotropic vertex, and it satisfies the following conditionfor all reflected diagrams: all connected proper subdiagrams which contain anisotropic vertex are of finite type.In Section 5, we will show that all regular Kac-Moody diagrams are subfinite.In this section, we classify subfinite regular Kac-Moody diagrams, by usinginduction on the number of vertices. A subdiagram of a subfinite regular Kac-Moody diagram is regular Kac-Moody, however if it does not contain an isotropicvertex then it is not part of our classification. We will work around this difficultyby using odd reflections.Note that we only need to find the extensions of one diagram Γ belonging toa collection of reflected diagrams C (Γ), and then include all reflections of eachextended diagram in our classification. In this section, we find all connected diagrams with two or three vertices whichare regular Kac-Moody and contain an isotropic vertex.8e say that an n -vertex diagram is a cycle if it is a connected diagram andeach vertex is connected to exactly two other vertices. We say that an n -vertexdiagram is a chain if the vertices can be enumerated by the set { , , ..., n } such that a ij = 0 if and only if j = i + 1 and i = j + 1. A proper connectedsubdiagram of a cycle is a chain. Lemma 2.2.
The connected regular Kac-Moody 2-vertex diagrams which con-tain an isotropic vertex are A (1 , and B (1 , .Proof. Recall that in the case that a ii = 0 for some i , it is our convention torescale so that a ij = 1 for some j . N / / o o a (cid:13) v ⋄ If a = −
1, then reflecting at v , we have a, − aa +1 ∈ Z < which implies a = − − aa +1 = − B (1 , . N / / o o a J −→ r N / / o o − a a J ⋄ If a = −
1, then by reflecting at v we have A (1 , N / / o o − (cid:13) −→ r N / / o o N We note that all 2-vertex diagrams of regular Kac-Moody superalgebras areof finite type.
Lemma 2.3.
The regular Kac-Moody extensions of A (1 , to three verticesare the following: A (0 , , A (1 , , B (1 , , B (2 , , C (3) , G (3) , D (2 , , α ) , A (1 , (2) , A (0 , (1) , B (1 , (1) , S (1 , , α ) , q (3) (2) , Q ± ( m, n, t ) .Proof. We consider each case for attaching a vertex to an A (1 ,
0) diagram.
Case 1: N / / o o N a / / o o b (cid:13) v a, b = 0 ⋄ If v is N , then by reflecting at v we have 1 + a, a ∈ Z ≤ , which implies a = −
1. This is A (1 , N / / o o N a / / o o N −→ r N (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E E − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a (cid:25) (cid:25) Y Y − (cid:13) a / / o o a (cid:13) a = − = ⇒ (cid:13) − / / o o N − / / o o − (cid:13) If v is J and b = −
2, then by reflecting at v we have 1 + a, a ∈ Z ≤ ,which implies a = −
1. This is B (1 , N / / o o N a / / o o − J −→ r N (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E E − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a (cid:25) (cid:25) Y Y − (cid:13) a / / o o a J a = − = ⇒ (cid:13) − / / o o N − / / o o − J ⋄ If v is (cid:13) and a, b = −
1, then this is A (0 , N / / o o N a / / o o − (cid:13) −→ r (cid:13) − / / o o N − / / o o − (cid:13)⋄ If v is (cid:13) , b = − a = −
1, then by reflecting at v we have 1 + a ∈{− , − } , which implies a ∈ {− , − } . If a = −
2, then 1 + a = − C (3). If a = −
3, then 1 + a = − G (3). N / / o o N a / / o o − (cid:13) −→ r N (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E E − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a (cid:25) (cid:25) Y Y a (cid:13) a / / o o − (1+ a ) N . Case 2: (cid:13) v b (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D a (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) d (cid:26) (cid:26) Z Z c N / / o o N a, b, c, d = 0 ⋄ If v is N , then after rescaling we have N (cid:5) (cid:5) (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) E E c (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) b (cid:25) (cid:25) Y Y N / / o o a N −→ r N (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E E − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) b (cid:25) (cid:25) Y Y − (cid:13) b + c / / o o a + b (cid:13) . a + b , b + c ∈ Z < or both zero. By symmetry it follows that1 + a + b = m b + c = n c + a = t ∈ Z < , or all equal to zero. (2)If they all equal zero, then this is D (2 , , α ). If they all equal −
1, then a, b, c = − q (3) (2) . If they are in Z < and not all equal to −
1, then this is Q ± ( m, n, t ). ⋄ If v is J and b, d = −
2, then by reflecting at v we have 4+ a , c , a + c ∈ Z < or all zero, which implies a, c = − . If v is (cid:13) then this is A (1 , (2) , andif v is • then this is B (1 , (1) . J − (cid:5) (cid:5) (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) E E a (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) − (cid:25) (cid:25) Y Y c N / / o o N −→ r J − (cid:5) (cid:5) (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) E E a (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) a (cid:25) (cid:25) Y Y Q N / / o o − (cid:13) Q = 1 + a + c. ⋄ If v is (cid:13) , b = − d = −
1, then by reflecting at v we have (cid:13) − (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E E a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − (cid:25) (cid:25) Y Y c N / / o o N −→ r • − (cid:5) (cid:5) (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) E E a (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) a (cid:25) (cid:25) Y Y Q N / / o o − (cid:13) Q = 1 + a + c. (3)Then either 1 + a + c ∈ Z < and 3 + a ∈ Z < , or both equal zero. If they areboth zero, then a = − , c = − and this is G(3).So now assume that 1 + a + c = 0. Then by reflecting at v we have (cid:13) − (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E E a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − (cid:25) (cid:25) Y Y c N / / o o N −→ r N − − c (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E E a + c (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c (cid:25) (cid:25) Y Y c (cid:13) − / / o o N . (4)This implies 1+ a + c ∈ {− , − } . Since 3+ a ≤ −
2, it follows that c ∈ ( − , − ]in the first case (1 + a + c = −
1) and c ∈ ( − , − ] in the second case.11f 1 + a + c = −
1, then by reflecting the last diagram of (4) at v we have: −→ r N P (cid:5) (cid:5) (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) E E P (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) c (cid:25) (cid:25) Y Y − N c / / o o − (cid:13) −→ r (cid:13) − (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E E − − c (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − (cid:25) (cid:25) Y Y R N c / / o o − • P = − − cR = c c . Then R = c c ∈ Z < , which contradicts c ∈ ( − , − ].If 1 + a + c = −
2, then by reflecting the last diagram of (4) at v we have N P (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E E − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c (cid:25) (cid:25) Y Y c (cid:13) − / / o o N −→ r N P (cid:5) (cid:5) (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) E E − (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) c (cid:25) (cid:25) Y Y − • S / / o o − (cid:13) P = − − cS = c c . Then S = c c ∈ Z < , which contradicts c ∈ ( − , − ]. ⋄ If v is (cid:13) and b, d = −
1, then by reflecting at v we have (cid:13) − (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E E a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − (cid:25) (cid:25) Y Y c N / / o o N −→ r N a (cid:5) (cid:5) (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) E E a (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) − − a (cid:25) (cid:25) Y Y Q N / / o o − (cid:13) Q = 1 + a + c. Then either 1 + a + c = 0 and 1 + 2 a = 0, or 1 + a + c ∈ {− , − } . If 1 + a + c = 0,then this is C (3). We have that 1 + a + c = − G (3).Suppose now that 1 + a + c = −
1. If c = −
1, then this is A (0 , (1) . For c = − c = − αα is reversible. After substituting and rescaling,we have (cid:13) − (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E E α − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − (cid:25) (cid:25) Y Y α +1 N − α / / o o − α N . This diagram is S (1 , , α ), and is regular Kac-Moody precisely when α is not12n integer. Indeed, by reflecting the following diagram at v we have: (cid:13) − (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E E Q − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Q +1 (cid:25) (cid:25) Y Y − N − Q / / o o − Q N Q = α + n −→ r N − R (cid:5) (cid:5) (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) E E − R (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) R − (cid:25) (cid:25) Y Y − N R +1 / / o o − (cid:13) R = Q − α + ( n − . Reflecting at v gives us (cid:13) − (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E E Q − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Q +1 (cid:25) (cid:25) Y Y − N − Q / / o o − Q N Q = α + n −→ r N R +1 (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E E − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − R (cid:25) (cid:25) Y Y − R (cid:13) − / / o o R − N R = Q + 1 = α + ( n + 1) . By induction, two diagrams given by labels α and α are connected by asequence of odd reflections precisely when α − α ∈ Z . Hence, S (1 , , α ) isregular Kac-Moody if and only if α is not an integer.We have found all regular Kac-Moody extensions of A (1 ,
0) by one vertex.
Lemma 2.4.
The regular Kac-Moody extensions of B (1 , to three vertices thatare not extensions of A (1 , are the following: A (2 , (4) , D (2 , (2) .Proof. We consider each case for attaching a vertex to a B (1 ,
1) diagram.
Case 1: N / / o o − • a / / o o b (cid:13) v a, b = 0Reflecting at v we have N / / o o − • a / / o o b (cid:13) v −→ r N / / o o − (cid:13) − a / / o o b (cid:13) v . This implies a, − a ∈ Z < , which is a contradiction. Case 2: J − / / N a o o / / • − o o a = 013y reflecting at v we have J − / / o o a N / / o o − • −→ r N a (cid:5) (cid:5) (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) E E − (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) (cid:25) (cid:25) Y Y − J a / / o o a (cid:13) , which implies 2 + 2 a, a ∈ Z < . The unique solution is a = −
1. If v is (cid:13) then this is A (2 , (4) , and if v is • then this is D (2 , (2) . Case 3: J − (cid:4) (cid:4) (cid:10)(cid:10)(cid:10)(cid:10)(cid:10)(cid:10)(cid:10)(cid:10)(cid:10)(cid:10) D D a (cid:10)(cid:10)(cid:10)(cid:10)(cid:10)(cid:10)(cid:10)(cid:10)(cid:10)(cid:10) b (cid:26) (cid:26) Z Z c N / / o o − • a, b, c = 0Now a = 0, b ∈ Z < , and c ∈ Z < . By reflecting at v we have J − (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E E a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) b (cid:25) (cid:25) Y Y c N / / o o − • −→ r J − (cid:5) (cid:5) (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) E E a (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) P (cid:25) (cid:25) Y Y Q N / / o o − (cid:13) where P = 2 − b + a , Q = 2 − c + 2 a , and P, Q ∈ Z ≤ . But, the system ofequations b ≤ − , c ≤ − , − c + 2 a ≤ , − b + 2 a ≤ Q ± ( m, n, t ). Lemma 2.5.
Suppose that Γ is a subfinite regular Kac-Moody n -vertex chainwhere the vertex v n is isotropic and the vertices v i for < i < n are notisotropic. Then there is a sequence of odd reflections of the vertices v i with ≤ i ≤ n such that in the reflected diagram Γ ′ the vertex v is isotropic. Thevertex v is unchanged and a ′ = a . roof. If n = 2, then we are done. So suppose the lemma holds for such achain with n − n vertices satisfying thehypothesis of the lemma. Now a n,n = 0 because v n is isotropic, and a n − ,n − =2 because v n − is not isotropic. Since Γ is a chain, a n − ,n − = 0 and a n − ,n = a n,n − = 0. Since Γ is a subfinite regular Kac-Moody diagram, the subdiagramcontaining the vertices v n − , v n − , v n is of finite type, which by the three vertexclassification implies that a n − ,n = − v n − is even.By reflecting at v n we obtain a diagram Γ ′ where v ′ n − is isotropic. Weobserve that Γ ′ is again a chain. Indeed, a ′ in = a ′ ni = 0 and a ′ ij = a ij for i < n − j < n , because a in = a ni = 0 for i = n −
1. Also, a ′ n − ,i = a n,n − a n − ,i = 0 for i < n −
2. We can apply the induction hypothesis to Γ ′ − { v n } which gives usthe desired sequence of odd reflections of Γ ′ and hence of Γ. Lemma 2.6.
Let Γ be a subfinite regular Kac-Moody diagram. Suppose Γ ′ is aconnected m -vertex subdiagram of Γ such that Γ ′ does not contain an isotropicvertex. Then there is a sequence of odd reflections R of Γ such that R (Γ ′ ) = Γ ′ ,and there is a connected ( m + 1) -vertex subfinite regular Kac-Moody subdiagramof R (Γ) which contains Γ ′ = R (Γ ′ ) as a subdiagram and an isotropic vertex.Proof. First note that Γ contains an isotropic vertex because it is subfinite. Sowe may take a minimal subdiagram Γ ′′ containing Γ ′ and an isotropic vertex.Then by the minimality of Γ ′′ the set of vertices in Γ ′′ − Γ ′ form a chain with anisotropic vertex, and only one of these vertices is connected to the subdiagramΓ ′ . Denote this vertex v and the rest of the vertices in the chain Γ ′′ − Γ ′ by { v , ..., v n } so that v k is connected to v k − and v k +1 . Then v n is isotropic byminimality. Choose any vertex in Γ ′ that is connected to v and denote it v .Now apply the Lemma 2.5 to the subdiagram given by { v , v , ..., v n } to ob-tain a sequence of odd reflections R such that R ( v ) is isotropic. By minimalityof Γ ′′ each vertex v k for k ≥ ′ , so each reflection doesnot change the subdiagram Γ ′ , ie. a ′ ij = a ij if v i , v j ∈ Γ ′ . Then the subdiagram R (Γ ′ ∪ { v } ) of R (Γ) is a subfinite regular Kac-Moody m + 1-vertex diagramcontaining the diagram Γ ′ and an isotropic vertex. Lemma 2.7. If Γ is a subfinite regular Kac-Moody diagram with n ≥ vertices,then any odd non-isotropic vertex of Γ has degree one.Proof. First note that connected finite type 3-vertex regular Kac-Moody di-agrams satisfy the condition that odd non-isotropic vertices have degree one.Suppose that the vertex v is odd non-isotropic with degree greater than orequal to two. Let Γ ′ = { v , v , v } , where a , a , a , a = 0. By the 3-vertexclassification, Γ ′ does not contain an isotropic vertex since Γ is subfinite. ByLemma 2.6, we are reduced to the case when Γ has four vertices.If v is connected to an odd non-isotropic vertex v i , then v i has degree twoin a 3-vertex subdiagram { v , v i , v j } , which contradicts the assumption that Γis subfinite. Thus, a = a = 0. We may assume a , a = 0, which impliesthat v is even. By reflecting at v we obtain a diagram in which R ( v ) isisotropic, R ( v ) = v and a ′ i = a i for i = 1 , . . . ,
4. But then the subdiagram15 ( { v , v , v } ) is not of finite type. This contradicts the assumption that Γ issubfinite. Lemma 2.8. If Γ is a subfinite regular Kac-Moody diagram with n ≥ vertices,and Γ ′ is a 3-vertex subdiagram with an isotropic vertex of degree two in Γ ′ such that the ratio of the vertex is not a negative rational number, then Γ ′ is a D (2 , , α ) diagram.Proof. Since Γ ′ is finite type regular Kac-Moody, this follows from the 3-vertexclassification. Corollary 2.9. If Γ is a subfinite regular Kac-Moody diagram with n ≥ vertices and Γ ′ is a 4-vertex subdiagram containing an isotropic vertex of degreethree in Γ ′ , then Γ ′ contains a D (2 , , α ) diagram containing this vertex. We introduce the following notation in order to simplify the presentation. If Γ isa diagram and { v i } i ∈ I is a subset of the vertices of Γ, then we denote by Γ { i | i ∈ I } the subdiagram of Γ obtained by removing the vertices { v i } i ∈ I . For example,Γ , is the subdiagram of Γ obtained by removing the vertices v and v . Let F denote the set of all connected finite type regular Kac-Moody diagrams whichcontain an isotropic vertex.For each finite type 3-vertex diagram, we will consider each case for attachingan additional vertex to the diagram. Let Γ denote the corresponding extendeddiagram. Lemma 2.10.
The subfinite regular Kac-Moody extensions of D (2 , , α ) to fourvertices are D (3 , , D (2 , , F , B (2 , (2) , D (2 , , α ) (1) , G (3) (1) , A (1 , (2) ,and A (2 , (2) .Proof. We consider each case for attaching a vertex to a D (2 , , α ) diagram. Case 1: N (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D c (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) b (cid:26) (cid:26) Z Z (cid:13) v e / / o o d N / / o o a N a, b, c, d, e = 0 a + b = − b + c = − c + a = − ⋄ If v is N , then Γ ∈ F implies d = − ∈ F implies c = − d . Then by(5) we have a = − , b = − c = 1, and d = −
1. This is D (2 , ⋄ If v is J and e = −
2, then Γ ∈ F implies d = − ∈ F implies c = − d . Then by (5) we have a = − , b = − c = 1, and d = −
1. If v is (cid:13) ,then this is B (2 , (1) . If v is • , then this is A (2 , (2) . ⋄ If v is (cid:13) and e = −
1, then Γ ∈ F implies d ∈ {− , − , − } and Γ ∈ F implies c ∈ {− d, − d , − d } . If d ∈ {− , − } , then c ∈ {− d, − d } . This yields16hree distinct options, and we find that Γ is one of the following: D (3 , F , A (1 , (2) . If d = −
3, then c ∈ { , , } and by reflecting at v we obtain r (Γ) N − c / / o o − c (cid:15) (cid:15) O O − − BBBBBBBBBBBB ` ` − BBBBBBBBBBBB (cid:13) − (cid:15) (cid:15) O O c (cid:13) − / / o o N Then r (Γ) ∈ F implies − c ≤
0. Hence, c = 3. Then by (5) we have a = − , b = − . This is G (3) (1) . Case 2: N b / / o o (cid:15) (cid:15) O O c N f (cid:15) (cid:15) O O g N > > |||||||||||| ~ ~ a |||||||||||| d / / o o e (cid:13) v a, b, c, d, e, f, g = 0 a + b = − b + c = − c + a = − ⋄ If v is N , then Γ ∈ F implies d = − c and Γ ∈ F implies f = −
1. Thenby substitution and by (6), we have d + fa = − ( c + a ) = 1. Now Γ must be a D (2 , , α ) diagram, so (2) implies d + fa = −
1. This is a contradiction. ⋄ If v is (cid:13) , e = − g = −
2, then Γ ∈ F implies a = , d = − . NowΓ ∈ F implies dc <
0. Hence, c >
0. Then c + a >
0, which contradicts (6). ⋄ If v is (cid:13) and e = g = −
1, then Γ ∈ F implies d = fa = − , Γ ∈ F implies f <
0, and Γ ∈ F implies dc <
0. Since d, f < a, c >
0. But then c + a >
0, which contradicts (6).
Case 3: N a (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D a (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) b (cid:26) (cid:26) Z Z b (cid:13) v g O O (cid:15) (cid:15) dk x x pppppppppp f pppppppppp h ' ' NNNNNNNNNN g g e NNNNNNNNNN N c / / o o c N a, b, c, d, e, f, g, h, k = 0 a + b + c = 0 (7)Suppose v is (cid:13) . Then db , ad , ce , eb , fa , cf , ∈ Q < implies a + cb >
0, which contra-dicts (7). Hence, v is N . Thus all subdiagrams are D (2 , , α ) diagrams. So17e have g = d , h = e , and k = f . We also have a + b + c = 0, c + d + e = 0, a + e + f = 0, and b + d + f = 0. It follows that d = a , e = b and f = c . Hence,this is D (2 , , α ) (1) .Now that we have found all subfinite regular Kac-Moody extensions of D (2 , , α ) by one vertex, we may restrict our attention to diagrams that donot contain a D (2 , , α ) subdiagram. In particular, the ratio of an isotropic ver-tex must be a negative rational number by Lemma 2.8, and an isotropic vertexhas at most degree two by Corollary 2.9. Lemma 2.11.
The subfinite regular Kac-Moody 4-vertex extensions of C (3) that are not extensions of D (2 , , α ) are the following: C (4) , C (3) (1) , B (1 , (1) ,and A (1 , (2) .Proof. Since an isotropic vertex has degree at most two, we are reduced to thefollowing case. N (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z − (cid:13) v a / / o o b (cid:13) − / / o o N a, b = 0 ⋄ If v is N , then Γ ∈ F implies b = −
1. This is D (2 , ⋄ If v is • , then Γ ∈ F implies a = − b = −
1. This is B (1 , (1) . ⋄ If v is (cid:13) , then Γ ∈ F implies a = − b = −
1. This is A (1 , (2) . ⋄ If v is (cid:13) , then Γ ∈ F implies a = − b ∈ {− , − , − } . If b = −
1, then thisis C (4). If b = −
2, then this is C (3) (1) . If b = −
3, then by reflecting at v andthen at v we obtain −→ r (cid:13) − / / o o − N / / o o N − / / o o − (cid:13) −→ r N − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) (cid:26) (cid:26) Z Z − N / / o o − (cid:13) − / / o o − (cid:13) But, r ( r (Γ))
6∈ F . Hence, b = − Lemma 2.12.
All subfinite regular Kac-Moody extensions of G (3) are exten-sions of D (2 , , α ) .Proof. Since an isotropic vertex has degree at most two, we are reduced to the18ollowing case. N − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) (cid:26) (cid:26) Z Z (cid:13) v a / / o o b (cid:13) − / / o o − N a, b = 0But, Γ
6∈ F . Lemma 2.13.
The subfinite regular Kac-Moody extensions of B (1 , that arenot extensions of D (2 , , α ) , C (3) or G (3) are the following: B (1 , , B (2 , , D (2 , (2) , and A (2 , (4) .Proof. Since N has at most degree two and • has at most degree one, we arereduced to the following case. (cid:13) v a / / o o b N / / o o N − / / o o − • − o o a, b = 0 ⋄ If v is N , then Γ ∈ F implies b = −
1. This is B (2 , ⋄ If v is • , then Γ ∈ F implies a = − b = −
1. This is D (2 , (2) . ⋄ If v is (cid:13) , then Γ ∈ F implies a ∈ {− , − } and b ∈ {− , − , − } . Byassumption, Γ is not C (3) or G (3), which implies b = −
1. If a = −
1, then thisis B (1 , a = −
2, then this is A (2 , (4) . Lemma 2.14.
The subfinite regular Kac-Moody extensions of B (2 , that arenot extensions of D (2 , , α ) , C (3) or G (3) are the following: B (3 , , B (2 , , A (2 , (4) , and D (1 , (2) .Proof. We consider each case for attaching a vertex to a B (2 ,
1) diagram.
Case 1: N (cid:15) (cid:15) O O − / / o o − (cid:13) a (cid:15) (cid:15) O O b N d / / o o c (cid:13) v a, b = 0Then Γ
6∈ F even if c, d = 0.
Case 2: (cid:13) v a / / o o b N / / o o N − / / o o − (cid:13) a, b = 0 ⋄ If v is N , then Γ ∈ F implies b = −
1. This is B (2 , ⋄ If v is • , then Γ ∈ F implies a = − b = −
1. This is A (2 , (4) .19 If v is (cid:13) , then Γ ∈ F implies a ∈ {− , − } and b ∈ {− , − , − } . Byassumption, Γ is not C (3) or G (3), which implies b = −
1. If a = −
1, then thisis B (3 , a = −
2, then this is D (1 , (2) . Lemma 2.15.
The subfinite regular Kac-Moody extensions of A (0 , that arenot extensions of D (2 , , α ) , C (3) or G (3) are the following: A (0 , , A (1 , , B (1 , , B (3 , , A (0 , (1) , and q (4) (2) .Proof. We consider each case for attaching a vertex to an A (0 ,
2) diagram.
Case 1: (cid:13) v a / / o o b N / / o o N − / / o o − (cid:13) a, b = 0 ⋄ If v is N , then Γ ∈ F implies b = −
1. This is A (1 , ⋄ If v is • , then Γ ∈ F implies a = − b = −
1. This is B (1 , ⋄ If v is (cid:13) , then Γ ∈ F implies a ∈ {− , − } and b ∈ {− , − , − } . Byassumption, Γ is not C (3) or G (3), which implies b = −
1. If a = −
1, then thisis A (0 , a = −
2, then this is B (3 , Case 2: N / / o o N − / / o o − (cid:13) b / / o o a (cid:13) v a, b = 0 ⋄ If v is N , then Γ ∈ F implies b = −
1. This is A (1 , ⋄ If v is • , then Γ ∈ F implies a = − b = −
1. This is B (1 , ⋄ If v is (cid:13) , then Γ ∈ F implies a ∈ {− , − } and b ∈ {− , − , − } . Byassumption, Γ is not C (3) or G (3), which implies b = −
1. If a = −
1, then thisis A (0 , a = −
2, then this is B (3 , Case 3: N (cid:15) (cid:15) O O − / / o o − (cid:13) a (cid:15) (cid:15) O O b N d / / o o c (cid:13) v a, b, c, d = 0 ⋄ If v is (cid:13) , then Γ ∈ F implies c = − b ∈ {− , − , − } . By assumption,Γ is not C (3) or G (3), which implies b = −
1. Since we assume that Γ isnot C (3) or G (3), we have that Γ ∈ F implies d = −
1. Now Γ ∈ F implies a ∈ {− , − , − } and since we assume that it is not C (3) or G (3), we have a = −
1. This is A (0 , (1) . ⋄ If v is N , then Γ ∈ F implies a = −
1. Since we assume that Γ is not D (2 , , α ), we have that Γ ∈ F implies d = −
1. Finally, Γ ∈ F and not equalto C (3) or G (3) implies bc = −
1. This is q (4) (2) .20 emma 2.16. The subfinite regular Kac-Moody extensions of A (1 , that arenot extensions of D (2 , , α ) , C (3) or G (3) are the following: A (1 , , B (2 , , A (1 , (1) , and q (4) (2) .Proof. We consider each case for attaching a vertex to an A (1 ,
1) diagram.
Case 1: N / / o o N − / / o o N b / / o o a (cid:13) v a, b = 0 ⋄ If v is N , then Γ ∈ F implies b −
1. This is A (1 , ⋄ If v is • , then Γ ∈ F implies a = − b = −
1. This is B (2 , ⋄ If v is (cid:13) , then Γ ∈ F implies a ∈ {− , − } and b ∈ {− , − , − } . Byassumption, Γ is not C (3) or G (3), which implies b = −
1. If a = −
1, then thisis A (1 , a = −
2, then this is B (2 , Case 2: N (cid:15) (cid:15) O O − / / o o N a (cid:15) (cid:15) O O b N d / / o o c (cid:13) v a, b, c, d = 0 ⋄ If v is N , then Γ ∈ F implies a = −
1, Γ ∈ F implies d = −
1, and Γ ∈ F implies bc = −
1. This is A (1 , (1) . ⋄ If v is (cid:13) , then Γ ∈ F implies b = c = −
1. Then Γ , Γ ∈ F and not C (3) or G (3) imply a = d = −
1. This is q (4) (2) .This completes the classification of connected subfinite regular Kac-Moodydiagrams with four vertices. We observe that all are either of finite type or havefinite growth. For each finite type 4-vertex diagram, we will consider each case for attachinga vertex to the diagram. Let v denote the additional vertex, and let Γ denotethe corresponding extended diagram. Recall that a ij denotes the label of thearrow from the vertex v i to the vertex v j . Also, a ij = 0 if and only if a ji = 0. Lemma 2.17.
The subfinite regular Kac-Moody extensions of D (2 , are thefollowing: D (2 , , D (3 , , B (2 , (1) , A (4 , (2) , A (3 , (2) , D (2 , (2) .Proof. N (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z − N − / / o o − N / / o o N ∈ F implies a = a = 0. ⋄ If v is • , then Γ ∈ F implies a = 0, a = − a = −
1. This is B (2 , (1) . ⋄ If v is (cid:13) , then Γ ∈ F implies a = 0 and a , a ∈ {− , − } . If a = − a = −
1, then this is A (4 , (2) . If a = − a = −
2, then this is D (2 , (2) . If a = a = −
1, then this is D (2 , ⋄ If v is N , then Γ ∈ F implies that either a = a = 0, a = − a =1 and this is D (3 , a = − a = 1, a = a = − A (3 , (2) . Lemma 2.18.
The subfinite regular Kac-Moody extensions of D (3 , are thefollowing: D (2 , , D (4 , , B (3 , (1) , A (2 , (2) , A (3 , (2) , A (5 , (2) .Proof. N (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z − (cid:13) − / / o o − N / / o o N First observe that Γ ∈ F implies a = a = 0. ⋄ If v is • , then Γ ∈ F implies a = 0, a = − a = −
1. This is A (2 , (2) . ⋄ If v is (cid:13) , then Γ ∈ F implies that Γ satisfies the following conditions. If a = 0, then a = a = − a = a = 0, and this is A (3 , (2) . If a = 0,then a , a ∈ {− , − } . If a = 0, a = − a = −
1, then this is B (3 , (1) . If a = 0, a = − a = −
2, then this is A (5 , (2) . If a = 0, a = − a = −
1, then this is D (4 , ⋄ If v is N , then Γ ∈ F implies a = 0, a = − a = 1. This is D (2 , Lemma 2.19.
The subfinite regular Kac-Moody extensions of C (4) are thefollowing: C (5) , D (2 , , B (1 , (1) , D (1 , (1) , A (5 , (2) , A (6 , (2) .Proof. N (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z − (cid:13) − / / o o − (cid:13) − / / o o N First observe that Γ ∈ F implies a = a = 0. ⋄ If v is • , then Γ ∈ F implies a = 0, a = − a = −
1. This is B (1 , (1) . ⋄ If v is (cid:13) , then Γ ∈ F implies that Γ satisfies the following conditions. If a = 0, then a = a = − a = a = 0, and this is A (5 , (2) . If a = 0,22hen a , a ∈ {− , − } . If a = 0, a = − a = −
1, then this is A (6 , (2) . If a = a = 0, a = − a = −
2, then this is D (1 , (1) . If a = a = 0, a = − a = −
1, then this is C (5). ⋄ If v is N , then Γ ∈ F implies a = 0, a = 1, a = −
1. This is D (2 , Lemma 2.20.
The only subfinite regular Kac-Moody extensions of F (4) is F (4) (1) .Proof. N (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z − (cid:13) − / / o o − N / / o o N First observe that Γ ∈ F implies a = a = 0. Then Γ ∈ F implies a = 0, a = a = −
1. This is F (4) (1) .Let K = { Γ ′ ∈ F | R (Γ ′ ) is a chain for every sequence R of odd reflections } . Lemma 2.21.
The subfinite regular Kac-Moody extensions of A (0 , , A (1 , , B (1 , , B (3 , and B (2 , , that are not extensions of C (4) , D (3 , , D (2 , or F , are the following: A (0 , , A (1 , , A (2 , , B (1 , , B (2 , , B (3 , , B (4 , , A (0 , (1) , A (1 , (1) , A (4 , (2) , A (6 , (2) , D (2 , (2) , D (3 , (2) , D (4 , (2) , q (5) (2) .Proof. Note that every connected 4-vertex subdiagram of Γ containing an isotropicvertex is an element of K . We find the extensions for each of the diagrams: A (0 , A (1 , B (1 , B (3 ,
1) and B (2 , B (3 , N / / o o N − / / o o − (cid:13) − / / o o − (cid:13) First observe that since Γ , Γ ∈ K we have a = a = a = 0. ⋄ If v is N , then Γ ∈ K implies a = − a = 1. This is B (3 , ⋄ If v is • , then Γ ∈ K implies a = − a = −
2. This is A (6 , (2) . ⋄ If v is (cid:13) , then Γ ∈ K implies a = − a ∈ {− , − } . If a = −
1, thenthis is B (4 , a = −
2, then this is D (4 , (2) . B (1 , N / / o o N − / / o o − (cid:13) − / / o o − • First observe that since Γ , Γ ∈ K we have a = a = a = 0. ⋄ If v is N , then Γ ∈ K implies a = − a = 1. This is B (2 , ⋄ If v is • , then Γ ∈ K implies a = − a = −
2. This is D (2 , (2) . ⋄ If v is (cid:13) , then Γ ∈ K implies a = − a ∈ {− , − } . If a = −
1, then23his is B (1 , a = −
2, then this is A (6 , (2) . B (2 , N / / o o N − / / o o N − / / o o − • First observe that since Γ , Γ ∈ K we have a = a = a = 0. ⋄ If v is N , then Γ ∈ K implies a = − a = 1. This is B (2 , ⋄ If v is • , then Γ ∈ K implies a = − a = −
2. This is A (4 , (2) . ⋄ If v is (cid:13) , then Γ ∈ K implies a = − a ∈ {− , − } . If a = −
1, thenthis is B (3 , a = −
2, then this is D (3 , (2) .Finally, we restrict our attention to diagrams satisfying: a connected 4-vertexsubdiagram containing an isotropic vertex is a diagram for A (0 ,
3) or A (1 , A (1 , N / / o o N − / / o o N − / / o o N First observe that since Γ , Γ ∈ K we have a = a = 0. There are twodistinct cases: a = 0 and a , a = 0. ⋄ If v is N and a = 0, then a = 1, a = −
1. This is A (2 , ⋄ If v is (cid:13) and a = 0, then a , a = −
1. This is A (1 , ⋄ If v is N and a , a = 0, then a = 1, a , a , a = −
1. This is q (5) (2) . ⋄ If v is (cid:13) and a , a = 0, then a , a , a , a = −
1. This is A (1 , (1) . A (0 , (cid:13) − / / o o N − / / o o N − / / o o − (cid:13) Now we assume that a connected 4-vertex subdiagram is a diagram for A (0 , , Γ ∈ K we have a , a = 0. There are two distinctcases: a = 0 and a , a = 0. ⋄ If v is N and a = 0, then a = 1, a = −
1. This is A (1 , ⋄ If v is (cid:13) and a = 0, then a , a = −
1. This is A (0 , ⋄ If v is N and a , a = 0, then a = 1, a , a , a = −
1. This is q (5) (2) . ⋄ If v is (cid:13) and a , a = 0, then a , a , a , a = −
1. This is A (0 , (1) .This completes the classification of connected subfinite regular Kac-Moodydiagrams with five vertices. We observe that all are either of finite type or havefinite growth. n ≥ vertices Now we handle the general case. We find all connected subfinite regular Kac-Moody diagrams with six or more vertices. We will show in Theorem 5.5 thatthe subfinite regular Kac-Moody diagrams which are not of finite type are notextendable, completing the classification of regular Kac-Moody diagrams.24 emark . A finite type diagram Γ ′ with n ≥ • (cid:13) v • • n − _ _ • n − , (8)where Γ ′ is A ( k, l ) or A k , and the subdiagram { v , v , v } is one of the diagramsin the following table. Table 2.22 • NNNNNNNNN • • (cid:13) − ' ' NNNNNNNNN g g NNNNNNNNN − (cid:15) (cid:15) O O − • (cid:13) (cid:13) − ' ' NNNNNNNNN g g − NNNNNNNNN • (cid:13) − ppppppppp w w − ppppppppp N & & NNNNNNNNN f f NNNNNNNNN − (cid:15) (cid:15) O O − • (cid:13) • OOOOOOOOOO (cid:15) (cid:15) O O − • J N − & & NNNNNNNNN f f − NNNNNNNNN (cid:15) (cid:15) O O • N − ppppppppp x x − ppppppppp Lemma 2.23.
Let Γ be a subfinite regular Kac-Moody cycle with n ≥ verticesand let v j be a vertex of Γ connected to v i and v k . Then either v j is (cid:13) with a ji = a jk = − ; or v j is N with a ji a jk = − . (9) If Γ has an even number of odd roots, then g ( A ) is an A ( k − , l ) (1) diagram. If Γ has an odd number of odd roots, then g ( A ) is a q ( n ) (2) diagram.Proof. By the above classification, all subfinite regular Kac-Moody 4-vertexand 5-vertex cycles are of type A ( k, l ) (1) or q ( n ) (2) , and satisfy (9). Let Γ bea subfinite regular Kac-Moody cycle with n ≥ v j be a vertexwhich is connected to v i and v k . Since n ≥ v j is contained in a proper 5-vertex subdiagram where v j is the middle vertex of the chain. By the 5-vertexclassification, we see that for every finite type chain diagram with five vertices,the middle vertex satisfies (9). Hence, every vertex of Γ satisfies (9). If Γ has aneven number of odd roots, then the corresponding matrix A is symmetrizable25nd g ( A ) is an A ( k, l ) (1) diagram. If Γ has an odd number of odd roots, then A is non-symmetrizable and g ( A ) is a q ( n ) (2) diagram. Lemma 2.24. If Γ contains a proper subdiagram Γ ′ which is a cycle with n ≥ vertices then Γ is not subfinite regular Kac-Moody.Proof. Suppose Γ is a subfinite regular Kac-Moody diagram which contains acycle with n ≥ ′ ∪ { v n +1 } is subfinite regular Kac-Moody. By the above classification, asubfinite regular Kac-Moody 5-vertex diagram does not contain a 4-vertex cyclesubdiagram. So now suppose Γ ′ is a cycle with n vertices where n ≥ ′ ∪ { v n +1 } is subfinite regular Kac-Moody. If Γ ′ contains an isotropicvertex, then by Lemma 2.23 the diagram Γ ′ is either A ( m, n ) (1) or q ( n ) (2) , whichis not of finite type.Suppose Γ ′ does not contain an isotropic vertex. Then the additional vertex v n +1 of Γ is isotropic. Since Γ is connected we have the following subdiagram: J / / o o J (cid:15) (cid:15) O O / / o o J / / o o JN n +1 , where the double lines are necessarily connected, and the dotted lines are possi-bly connected. But by the 5-vertex classification, we see that this is not a finitetype subdiagram. Lemma 2.25.
Let Γ be the following diagram • • • • • (cid:13) v O O (cid:15) (cid:15) where Γ contains an isotropic vertex and Γ is a diagram for A ( k, l ) or A k .Then Γ is not a subfinite regular Kac-Moody diagram.Proof. If v is isotropic, then Γ
6∈ F , which is a contradiction. We may assumethat v is isotropic, by using odd reflections in the subdiagram Γ , . ThenΓ , Γ ∈ F implies that the vertices v , v , v are even. Moreover, Γ ∈ F implies that the ratio of v in the subdiagram { v , v , v } is 1, while Γ ∈ F implies that the ratio is −
1, which is a contradiction.
Proposition 2.26. If Γ is a subfinite regular Kac-Moody diagram with n ≥ vertices and Γ is not a cycle, then Γ is of the form • OOOOOOOOOO • n − mmmmmmmmm • • · · · • • n − (cid:13) v (cid:13) v n , (10)26 here the subdiagram Γ \{ v , v n } is A ( k, l ) or A k , and the subdiagrams { v , v , v } and { v n , v n − , v n − } are diagrams from Table 2.22.Proof. First note that diagrams satisfying (10) reflect by odd reflections to di-agrams that again satisfy (10). So it suffices to prove the proposition for adiagram obtained by odd reflections from the original diagram.Let Γ be a subfinite regular Kac-Moody diagram with n ≥ v n such that Γ ′ := Γ \ { v n } is connected andcontains an isotropic vertex. Then Γ ′ satisfies (8). We may assume that thesubdiagram { v , v , v } contains an isotropic vertex, by using odd reflections inthe subdiagram { v , . . . , v n − } . If Γ ′ \ { v } does not contain an isotropic vertex,then v is isotropic. In this case, by Table 2.22, Γ ′ is the standard diagram for A (1 , n −
2) and an odd reflection at v results in a diagram with v isotropic.So we may assume that v or v is isotropic.Suppose that v n is connected only to the vertex v . If a , a = 0, then a , a = 0 and { v n , v , v , v , v } ∈ F implies that the subdiagram { v n , v , v } is A ( k, l ) or A k . Then by Lemma 2.25, the diagram { v n , v , v , v , v , v } is nota subfinite regular Kac-Moody diagram. If a , a = 0, then Γ n − ∈ F implies a , a = 0. The condition { v n , v , v , v , v } ∈ F yields two possibilities: either { v n , v , v } is a diagram from Table 2.22 and { v , v , v , v } is A ( k, l ) or A k , or { v , v , v } is a diagram from Table 2.22 and { v n , v , v , v } is A ( k, l ) or A k . Inthe first case, Γ n is A ( k, l ) or A k , and we are done. In the second case, Γ is A ( k, l ) or A k , and we are done.Now we may assume that Γ \ { v } is connected, and hence satisfies (8).Thus, the vertex v n is not connected to vertices v j with 3 < j < n −
2. Firstsuppose that v n is connected to v n − or v n − . Then Γ ∈ F implies that v n is not connected to v or v and Γ satisfies (8). By Lemma 2.24, v n is notconnected to v . Thus, Γ satisfies (10). Now suppose that v n is not connectedto v n − or v n − . If v n is connected to v then { v n , v , v , v , v } satisfies (8),implying v n is not connected to v or v , a , a = 0, and the subdiagram { v , v , v } is A ( k, l ) or A k . But then by Lemma 2.25, { v n , v , v , v , v , v } isnot a subfinite regular Kac-Moody diagram. Hence, v n is not connected to v .Finally, since subdiagram Γ \ { v n − } satisfies (8) with v n connected only to v or v we conclude that Γ satisfies (10).27 .6 Classification theorem Theorem 2.27. If A is a symmetrizable matrix and g ( A ) has a simple isotropicroot, then g ( A ) is regular Kac-Moody if and only if it has finite growth. If A isa non-symmetrizable matrix and g ( A ) has a simple isotropic root, then g ( A ) isregular Kac-Moody if and only if it is one of the following three classes: Algebra Dynkin diagrams q ( n ) (2) • a v v nnnnnnnnnnn b ( ( PPPPPPPPPPP • nnnnnnnnnnn / / • o o / / · · · o o / / • o o / / • o o h h PPPPPPPPPPP
There are n • .Each • is either (cid:13) or N .An odd number of them are N .If • is (cid:13) , then a = b = − .If • is N , then ab = − . S (1 , , α ) N − α (cid:25) (cid:25) (cid:5) (cid:5) (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) N E E (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) − − α / / (cid:13) − o o − Y Y α = 0 , α ∈ C \ Z N b (cid:25) (cid:25) (cid:5) (cid:5) (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) N c E E (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) / / N a o o Y Y N b (cid:25) (cid:25) (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:13) − E E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) b + c / / (cid:13) a + b o o − Y Y Q ± ( m, n, t ) N c (cid:25) (cid:25) (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:13) − E E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c + a / / (cid:13) b + c o o − Y Y N a (cid:25) (cid:25) (cid:5) (cid:5) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:13) − E E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a + b / / (cid:13) c + a o o − Y Y a + b = m b + c = n c + a = t m, n, t ∈ Z ≤− andnot all equal to − ,a, b, c ∈ R \ Q . Remark . This theorem follows from the classification of connected subfiniteregular Kac-Moody diagrams in Section 2 and Theorem 5.5. The fact that Q ± ( m, n, t ) is not symmetrizable and does not have finite growth will be provenin Section 3. Note that Q ± ( m, n, t ) ∼ = Q ± ( n, t, m ) as algebras (see Remark 3.3).28 The Lie superalgebra Q ± ( m, n, t ) Q ± ( m, n, t ) N (cid:5) (cid:5) (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) E E c (cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11)(cid:11) b (cid:25) (cid:25) Y Y N / / o o a N a + b = m b + c = n c + a = tm, n, t ∈ Z ≤− , not all equal to -1.In this section, we describe the parameters of the defining matrices for Q ± ( m, n, t ), and then we show that Q ± ( m, n, t ) is not symmetrizable and doesnot have finite growth. Lemma 3.1.
For each diagram Q ± ( m, n, t ) , it follows that a, b, c ∈ R \ Q ,and there are two solutions of the above equations, namely Q − ( m, n, t ) with a, b, c < − and Q + ( m, n, t ) with − < a, b, c < .Proof. First, we show that a solution exists. To simplify the calculations we let M = 1 − m, N = 1 − n, T = 1 − t. Then
M, N, T ∈ Z ≥ . Solving Equation 1 for b and Equation 3 for c , and thensubstituting into Equation 2 yields − N = − a + M + − aaT + 1 . By clearing denominators and regrouping, we have f ( a ) = 0 where f ( a ) = ( N T − a + ( M N T − M + N − T ) a + ( M N − . Since
M, N, T ≥
2, one has
N T − >
0. The discriminant of f ( a ) is D = ( M N T − M − N − T ) −
4. Now since M − , N − , T − ∈ Z ≥ and notall equal to 1, we have3 ≤ ( M − N − T −
1) + 1=
M N T − ( N − M − ( T − N − ( M − T< M N T − M − N − T, which implies D >
0. Hence, f ( a ) has two real roots. Moreover, these roots arenot rational, since by taking k = M N T − M − N − T and y = D , we obtain theequation y = k −
4, which has only two solutions with integral k and rational y , namely k = ± y = 0. Since D = 0, we conclude that y is not rational.Hence, a ∈ R \ Q , and it follows from the defining equations that b, c ∈ R \ Q .Let a > a be the roots of the quadratic equation f ( a ) = 0. Since N T − , M N − >
0, the function f ( a ) is concave up with f (0) >
0. We can express f ( −
1) as f ( −
1) = − ( M − N − T − − ( M − T −
2) + 1 , f ( − < M, N, T ∈ Z ≥ , not all equal to 2.Hence, a < − < a < . Denote by b , b , c , c the corresponding values of b, c . From a i + b − i ≤ − − < a < ⇒ − < b < ⇒ − < c < a < − ⇒ c < − ⇒ b < − . Corollary 3.2.
The determinant of the Cartan matrix equals abc and isnonzero. Hence, the dimension of the Cartan subalgebra is .Remark . It is clear that Q ± ( m, n, t ) ∼ = Q ± ( n, t, m ) as algebras by cyclicpermutation of the variables a, b, c . We also have Q ± ( m, n, t ) ∼ = Q ∓ ( m, t, n ) bytransforming the equations: a → b , b → a , c → c . Lemma 3.4. Q ± ( m, n, t ) is not symmetrizable.Proof. Suppose that we have a symmetrizable solution a, b, c . We show that thisimplies a ∈ Q , which is a contradiction. If the matrix is symmetrizable then abc = 1. So, we substitute c = ab into the defining equations. This yields1 + a + 1 b = m a + 1) b = n ab + 1 a = t From the first equation we have b = m − a − . Substituting this into the secondequation and solving for a , we have a = mn − m − nn ∈ Q . Hence, there is nosymmetrizable solution. Lemma 3.5. Q ± ( m, n, t ) does not have finite growth.Proof. The set of principal roots of Q ± ( m, n, t ) is Π = { α + α , α + α , α + α } . The Cartan matrix B of the subalgebra of Q ± ( m, n, t ) generated by Π is B = m mn nt t . All off diagonal entries of B are negative integers. Since they are not all equalto −
1, this is not a Cartan matrix of a finite-growth Kac-Moody algebra. ByLemma 1.10, Q ± ( m, n, t ) does not have finite growth.30 Kac-Moody superalgebra is called hyperbolic if after the removal of anysimple root the superalgebra is either of finite or affine type. Lemma 3.6.
The regular Kac-Moody superalgebra Q ± ( m, n, t ) is hyperbolic forthe following m ≥ n ≥ t . m n t-1 -1 -2-1 -1 -3-1 -1 -4-1 -2 -2-2 -2 -2 Let g ( A ) be a regular Kac-Moody superalgebra, and let g ( A ) = h ⊕ ( ⊕ α ∈ ∆ g ( A ) α )be the root space decomposition. Recall that we may assume that a ii ∈ { , } without loss of generality. An element ρ ∈ h ∗ such that ρ ( h i ) = α i ( h i ) = a ii for all i ∈ I is called a Weyl vector .A root α is called real if α or α is simple in some base obtained by a sequenceof even and odd reflections, and it is called imaginary otherwise. For each realroot α , the vector space [ g α , g − α ] is a one dimensional subspace of the Cartansubalgebra h . Moreover, if α is nonisotropic, then for each nonzero h ∈ [ g α , g − α ]we have that α ( h ) = 0. In this case, we may choose h α to be the unique vectorin [ g α , g − α ] which satisfies α ( h α ) = 2 (see Remark 1.7). Then take x α ∈ g α , y α ∈ g − α such that [ x α , y α ] = h α . If α is even, then { x α , y α , h α } forms an sl -triple. If α is odd nonisotropic, then { x α , y α , h α , [ x α , x α ] , [ y α , y α ] } is a basis fora subalgebra which is isomorphic to osp (1 , {− [ x α , x α ] , [ y α , y α ] , h α } forms an sl -triple.Let U ( g ) denote the universal enveloping algebra of a Lie superalgebra g .Let n + (resp. n − ) denote the subalgebra of g ( A ) generated by the elements X i (resp. Y i ), i ∈ I . Then one has the triangular decomposition g ( A ) = n − + h + n + .A g ( A )-module V is called a weight module if V = ⊕ µ ∈ h ∗ V µ , where V µ = { v ∈ V | hv = µ ( h ) v, for all h ∈ h } . If V µ is non-zero, then µ is called a weight .A g ( A )-module V is called a highest weight module with highest weight λ ∈ h ∗ if there exists a vector v λ ∈ V such that n + v λ = 0 , hv = λ ( h ) v λ for h ∈ h , U ( g ( A )) v λ = V. A highest weight module is a weight module. We let L ( λ ) denote the irreduciblehighest weight module with highest weight λ . To simplify notation we define λ i := λ ( h i ), for i ∈ I .A subalgebra s of g ( A ) is locally finite on a module V if dim U ( s ) v < ∞ forany v ∈ V . An element x ∈ g is locally nilpotent on V if for any v ∈ V thereexists a positive integer N such that x N v = 0.31 .1 Integrable modules over affine Lie superalgebras Let g be a basic Lie superalgebra, that is, a finite-dimensional simple Lie su-peralgebra over C with a nondegenerate even symmetric invariant bilinear form( · , · ), such that g ¯0 is reductive [4]. The associated (non-twisted) affine Lie su-peralgebra is b g = (cid:0) C [ t, t − ] ⊗ C g (cid:1) ⊕ C K ⊕ C d with commutation relations[ a ( n ) , b ( l )] = [ a, b ]( n + l ) + nδ n, − l ( a | b ) K, [ d, a ( n )] = − na ( n ) , [ K, b g ] = 0where a, b ∈ g ; n, l ∈ Z and a ( n ) = t n ⊗ a . By identifying g with 1 ⊗ g , we havethat the Cartan subalgebra of b g is h = ◦ h ⊕ C K ⊕ C d where ◦ h is the Cartan subalgebra of g . Let δ be the linear function defined on h by δ | ◦ h ⊕ C K = 0 and δ ( d ) = 1. Let θ be the unique highest weight of the g . Then α = δ − θ is the additional simple root which extends g to b g .Now suppose σ is an automorphism of g with order m = 1. Then g decom-poses into eigenspaces: g = M j ∈ Z /m Z g ( σ ) j , g ( σ ) j = { x ∈ g | σ ( x ) = e j im x } . The associated twisted affine Lie superalgebra is g ( m ) = M j ∈ Z t j ⊗ g ( σ ) j mod m ⊕ C K ⊕ C d with the same commutation relations as given above. By identifying g ( σ ) with1 ⊗ g ( σ ) , we have that the Cartan subalgebra of g ( m ) is h = ◦ h ⊕ C K ⊕ C d where ◦ h is the Cartan subalgebra of g ( σ ) . Let δ be the linear function definedon h by δ | h ⊕ C K = 0 and δ ( d ) = 1. Let θ be the unique highest weight ofthe g ( σ ) -module g ( σ ) . Then α = δ − θ is the additional simple root whichextends g ( σ ) to g ( m ) .The bilinear form ( · , · ) on g gives rise to a nondegenerate symmetric invariantbilinear form on b g and on g ( m ) by:( a ( n ) , b ( l )) = δ n, − l ( a, b ) ( C [ t, t − ] ⊗ g , C K ⊕ C d ) = 0( K, K ) = ( d, d ) = 0 , ( K, d ) = 1 ( M j ∈ Z t j ⊗ g ( σ ) j mod m , C K ⊕ C d ) = 0 . The restriction of this form to h is nondegenerate and is also denoted by ( · , · ).We use this form to identify h with h ∗ , which induces a nondegenerate bilinear32orm on h ∗ . Let V ( λ ) be a highest module over b g . Then k = λ ( K ) is the level of L ( λ ).A highest weight module V over an affine Lie superalgebra b g (resp. g ( m ) )is called integrable if it is integrable over the affine Lie algebra b g ¯0 (resp. g ( m )¯0 ),that is, if g α ( n ) is locally finite on V for every root α of g ¯0 and n ∈ Z .Since g ¯0 = ⊕ j ∈ Z /m Z ( g ( σ ) j ) ¯0 is a graded reductive Lie algebra, it followsthat ( g ( σ ) ) ¯0 is reductive. Hence we may write g ¯0 (resp. ( g ( σ ) ) ¯0 ) as a sum ⊕ Ni =0 g ¯0 i , where g ¯00 is abelian and g ¯0 i for i = 1 , . . . , N are simple Lie algebras.Then for each i , the superalgebra b g (resp. g ( m ) ) contains an affine Lie algebra b g ¯0 i associated to g ¯0 i . Explicitly, we have b g ¯0 i = (cid:0) C [ s, s − ] ⊗ C g ¯0 i (cid:1) ⊕ C K ′ ⊕ C d, where s = t , K ′ = K (resp. s = t m , K ′ = mK ). The Cartan subalgebra of b g ¯0 i is h i = ◦ h i ⊕ C K ⊕ C d , where ◦ h i = ◦ h ∩ g ¯0 i . If V is integrable over b g (resp. g ( m ) ),then it follows from the definition that V is integrable over b g ¯0 i , i = 1 , . . . , N .It was shown in [9], that most non-twisted affine Lie superalgebras have onlytrivial irreducible integrable highest weight modules, which led to the consider-ation of weaker notions of integrability. Proposition 4.1 (Kac, Wakimoto) . The only non-twisted affine Lie superalge-bras with non-trivial irreducible integrable highest weight modules are B (0 , n ) (1) , C ( n ) (1) and A (0 , m ) (1) . In [1], S. Eswara Rao and V. Futorny show that over non-twisted affineLie superalgebras all irreducible integrable highest weight modules with non-zero level are highest weight modules. The proof of the following statement issimilar to the non-twisted case.
Proposition 4.2.
Let g ( m ) be a twisted affine Lie superalgebra which is not oneof the algebras: A (0 , n − (2) , A (0 , n ) (4) and C ( n ) (2) . Then an irreducibleintegrable highest weight module over g ( m ) is trivial.Proof. First we consider the non-symmetrizable twisted affine Lie superalgebra, q ( n ) (2) . The Lie superalgebra q ( n ) (2) is not covered by the construction givenabove, so we must handle this case separately. If n is odd, then we have aCartan matrix defined by a i,i +1 = − a i +1 ,i = 1 for i = 1 , . . . , n ( mod n ), andall other entries zero. All simple roots in this case are odd isotropic. By an oddreflection at α i , we obtain an even root α i + α i +1 with h α i + α i +1 = h i +1 − h i .By Lemma 4.6, the conditions for integrability are λ i +1 − λ i ∈ Z ≥ , ( mod n ).This implies λ i = 0 for all i = 1 , . . . , n .If n is even, the we have a Cartan matrix defined by a i,i +1 = − a i +1 ,i = 1for i = 1 , . . . , n − a , = 2, a ,n = a n, = −
1, and all other entries zero. In thiscase, the simple root α is even, and all other simple roots are odd isotropic.Using odd reflections, we obtain the set of principal roots { α , α + α , α + α , . . . , α n − + α n , α n + α + α } . For i = 2 , . . . , n − h α i + α i +1 = h i +1 − h i , and h α n + α + α = h − h n − h . This yields integrability conditions33 ∈ Z ≥ , λ i +1 − λ i ∈ Z ≥ for i = 2 , . . . , n −
1, and λ − λ n − λ ∈ Z ≥ . Togetherthese imply λ i = 0 for i = 1 , . . . , n .For a symmetrizable twisted affine Lie superalgebra g ( m ) , we have the stan-dard nondegenerate symmetric invariant bilinear form. The structure of g ( m )0 ,the even part of g ( m ) , is given by van de Leur in [12]. Let g ( m ) be a sym-metrizable twisted affine Lie superalgebra, which is not one of the algebras: A (0 , n − (2) , A (0 , n ) (4) and C ( n ) (2) , and choose a base Π = { α , . . . , α l } with a unique simple isotropic root, α d . Using the bilinear form to identify h with h ∗ , we have α i = α i ∨ for i ≤ d and α i = − α i ∨ for i > d . Note that d = 0and d = l by choice of g ( m ) .Then g ( m )0 has as subalgebras, both an affine subalgebra g ′ on which this formis positive definite, and an affine subalgebra g ′′ on which this form is negativedefinite. Denote by ◦ g ′ (resp. ◦ g ′′ ) the finite part of g ′ (resp. g ′′ ). Let θ ′ (resp. θ ′′ ) denote the highest weight of the module ◦ g ′ (resp. ◦ g ′′ ) over ◦ g ′ (resp. ◦ g ′′ ).Then the simple root α ′ = δ − θ ′ (resp. α ′′ = δ − θ ′′ ) extends ◦ g ′ (resp. ◦ g ′′ ) to g ′ (resp. g ′′ ). Also, θ ′ = P d − i =0 c i α i and θ ′′ = P li = d +1 c i α i with c i ∈ Z > .Let L ( λ ) be an irreducible integrable highest weight module over g ( m ) . Let k = λ ( K ), k i = λ ( α i ∨ ), k ′ = λ ( α ′ ∨ ) and k ′′ = λ ( α ′′ ∨ ). Then by Lemma 4.6, k i ∈ Z ≥ for i ∈ I \ { d } and k ′ , k ′′ ∈ Z ≥ . Now using the bilinear form toidentify h with h ∗ , we have α ′ ∨ = α ′ = δ − θ ′ = K − θ ′∨ α ′′ ∨ = − α ′′ = − ( δ − θ ′′ ) = − ( K − θ ′′∨ ) . Hence, k = λ ( K ) = λ ( α ′ ∨ ) + λ ( θ ′∨ ) = k ′ + λ ( θ ′∨ ) − k = − λ ( K ) = λ ( α ′′ ∨ ) − λ ( θ ′′∨ ) = k ′′ − λ ( θ ′′∨ ) . Since θ ′∨ (resp. θ ′′∨ ) is a positive (resp. negative) combination of α i ∨ with i ∈ I \ { d } and λ ( α i ∨ ) ≥ i ∈ I \ { d } , we have that λ ( θ ′∨ ) ≥ λ ( θ ′′∨ ) ≤
0. Hence, k = 0 and the level of L ( λ ) is zero. The above equations thenimply that k ′ , k ′′ , λ ( θ ′∨ ) , λ ( θ ′′∨ ) = 0. Since λ ( θ ′∨ ) = P d − i =0 c i k i and λ ( θ ′′∨ ) = − P li = d +1 c i k i with c i ∈ Z > , we have that k i = 0 for i ∈ I \ { d } . Finally, k ′ = 0then implies k d = 0. Hence, the weight λ is zero, and so the module L ( λ ) istrivial. Remark . We can summarize this as follows. If ◦ g ¯0 is a direct sum of two ormore simple Lie algebras, then the corresponding affine (or twisted affine) Liesuperalgebra does not have any nontrivial irreducible integrable highest weightmodules. Let g ( A ) be a regular Kac-Moody superalgebra, and let V be a weight moduleover g ( A ). We call V integrable if for every real root α the element X α ∈ g ( A ) α is34ocally nilpotent on V . If g ( A ) is an affine Lie superalgebra then this definitioncoincides with the definition given in Section 4.1. This follows from the factthat every even root of an affine Lie superalgebra with non-zero length is real,as was shown in the dissertation of V. Serganova.The following lemma follows from Lemma 1.3. Lemma 4.4.
The adjoint module of a regular Kac-Moody superalgebra is anintegrable module. In particular, ad X α is locally nilpotent for every real root α ,where X α ∈ g ( A ) α . The following lemma follows from Proposition 4.1 and Proposition 4.2.
Lemma 4.5.
Suppose g ( A ) is a regular Kac-Moody superalgebra with a subfiniteregular Kac-Moody diagram which is not of finite type and it is not one of thealgebras: A (0 , m ) (1) , C ( n ) (1) , S (1 , , α ) , and Q ± ( m, n, t ) . Then all irreducibleintegrable highest weight modules are trivial. The conditions for an irreducible highest weight module over A (0 , m ) (1) or C ( n ) (1) to be integrable were given in [9]. We are interested in the irreducibleintegrable highest weight modules for the regular Kac-Moody superalgebras: S (1 , , α ) and Q ± ( m, n, t ). We will see that they have non-trivial irreducibleintegrable highest weight modules, and we will describe the weights. First weneed the following lemma. Lemma 4.6.
Let g ( A ) be a Kac-Moody superalgebra, and let L ( λ ) be an irre-ducible highest weight module. If α i is a simple non-isotropic root of g ( A ) , then Y i ∈ g ( A ) − α i is locally nilpotent on L ( λ ) if and only if λ i ∈ p ( i ) Z ≥ . If α i is asimple isotropic root, then Y i ∈ g ( A ) − α i is locally nilpotent on L ( λ ) .Proof. If α i is a simple even root, let e = X i , f = Y i and h = h i . Then { e, f, h } is an sl -triple. If α i is a simple odd non-isotropic root, let e = − [ X i , X i ], f = [ Y i , Y i ], and h = h i . Then { e, f, h } is an sl -triple. Since g ( A ) is aKac-Moody superalgebra, it follows that f is locally nilpotent on V if and onlyif f is nilpotent on the highest weight vector v λ . Now f is nilpotent on v λ ifand only if λ ( h ) ∈ Z ≥ . If α i is odd, then [ Y i , Y i ] v = 2( Y i ) v for v ∈ V . Thus, f is nilpotent on v λ if and only if Y i is nilpotent on v λ . Hence, Y i ∈ g ( A ) − α i islocally nilpotent on L ( λ ) if and only if λ i ∈ p ( i ) Z ≥ . If α i is a simple isotropicroot, then ( Y i ) v = 0 for all v ∈ L ( λ ). Lemma 4.7.
Let L ( λ ) be an irreducible integrable highest weight module of aregular Kac-Moody superalgebra g ( A ) , and let α s be a simple root. Let n ′ + = r s ( n + ) and A ′ = r s ( A ) . Then after a simple even or odd reflection r s themodule L ( λ ) is an irreducible integrable highest weight module of g ( A ′ ) withhighest weight given below.1. If λ ( h s ) = 0 , then v ′ λ = v λ is a highest weight vector with respect to n ′ + and the highest weight is λ ′ = λ .2. If λ ( h s ) = 0 and α s is isotropic, then v ′ λ = Y s v λ is a highest weight vectorwith respect to n ′ + and the highest weight is λ ′ = λ − α s . . If λ ( h s ) = k = 0 and α s is even, then v ′ λ = ( Y s ) k v λ is a highest weightvector with respect to n ′ + and the highest weight is λ ′ = λ − kα s .4. If λ ( h s ) = 2 n = 0 and α s is odd non-isotropic, then v ′ λ = ( Y s ) n v λ isa highest weight vector with respect to n ′ + and the highest weight is λ ′ = λ − nα s .Proof. (1) and (2) follow immediately from the facts that [ Y s , Y s ] v λ = 2( Y s ) v λ ,and that Y s v λ = 0 if and only if λ ( h s ) = 0. For (3) and (4) suppose now that α s is a non-isotropic root. Then by Lemma 4.6, we have k = λ ( h s ) ∈ p ( s ) Z ≥ sincethe module L ( λ ) is assumed to be integrable. If α s is a simple even root, then { X s , Y s , h s } is an sl -triple. Set v j = j ! ( Y s ) ( j ) v λ . Then h s v j = ( k − j ) v j and X s v j = ( k + 1 − j ) v j − . Thus X s v k +1 = 0, and X i v k +1 = 0 for all i ∈ I \ { s } .Since the module L ( λ ) is irreducible this implies that v k +1 = 0. Thus Y s v k = 0with v k = 0. Hence, L ( λ ) is a highest weight module with respect to n ′ + , withhighest weight vector v k = k ! ( Y s ) k v λ and the highest weight is λ − kα s .Finally, suppose that α s is an odd non-isotropic root. Set v j = ( Y s ) ( j ) v λ .Then h s v j = (2 n − j ) v j , X s v i = (2 i ) v i − and X s v i − = (2 n + 2 − i ) v i − .Thus, X s ( Y s ) n +1 v λ = 0, and X i ( Y s ) n +1 v λ = 0 for i ∈ I \ { s } . Since themodule L ( λ ) is irreducible, we conclude that ( Y s ) n +1 v λ = 0. Hence, Y s v n = 0with v n = 0. Therefore, L ( λ ) is a highest weight module with respect to n ′ + ,with highest weight vector ( Y s ) n v λ and highest weight λ − (2 n ) α s . The factthat the module L ( λ ) is integrable as a g ( A ′ ) module follows from the fact thatthe real roots of g ( A ) and g ( A ′ ) coincide. Lemma 4.8.
Let g ( A ) be a Kac-Moody superalgebra, and let L ( λ ) be a highestweight module. Let α s and α i be simple non-isotropic roots. Let r s be the evenreflection with respect to α s (or α s if α s is odd). Suppose Y i ∈ g ( A ) − α i islocally nilpotent on L ( λ ) , then Y ′ i ∈ g ( A ) − r s ( α i ) is locally nilpotent on L ( λ ) .Proof. The reflection r s does not change the Cartan matrix, so g ( A ′ ) is again aKac-Moody superalgebra. Thus, it suffices to show that Y ′ i is nilpotent on thehighest weight vector v λ ′ of L ( λ ′ ). It is sufficient to consider the case when both α s and α i are even roots. Then this is equivalent to the condition λ ′ ( h ′ i ) ∈ Z ≥ .Now λ ′ = λ − λ ( h s ) α s . We have α ′ s = − α s and α ′ i = α i − a si α s . Also, h ′ s = − h s and h ′ i = h i − a is h s . Then λ ′ ( h ′ i ) = ( λ − λ ( h s ) α s )( h i − a is h s ) = λ ( h i ) . Since Y i is locally nilpotent on L ( λ ) we have λ ( h i ) ∈ Z ≥ . Hence Y ′ i is locallynilpotent. Corollary 4.9.
Let g ( A ) be a regular Kac-Moody superalgebra. The irreduciblehighest weight module L ( λ ) is an integrable module if and only if the element Y α ∈ g ( A ) − α is locally nilpotent on L ( λ ) for each principal root α . We call a weight λ typical if for any real isotropic root α we have ( λ + ρ )( h α ) =0. 36 .3 Integrable modules of the Lie superalgebra S (1 , , α ) Now we describe the weights for integrable highest weight modules of the Liesuperalgebra S (1 , , α ). The Cartan matrix for the superalgebra S (1 , , α ) is B = − − − α − − α with α = 0 and α Z . Lemma 4.10.
Let L ( λ ) be an irreducible highest weight module for S (1 , , α ) .If λ = 0 or λ = 0 , then L ( λ ) is integrable if and only if λ ∈ Z ≥ λ + λ − ∈ Z ≥ . If λ = λ = 0 , then L ( λ ) is integrable if and only if λ ∈ Z ≥ .Proof. By Corollary 4.9, it suffices to find the conditions for the principal rootsto be locally nilpotent on L ( λ ). The principal roots of S (1 , , α ) are α and α + α . We have that h α + α = h + h by Lemma 1.6 and the formula aboveit, (after rescaling h α + α so that α α + α ( h α + α ) = 2; see Remark 1.7). ByLemma 4.6, Y ∈ g ( A ) − α is locally nilpotent on L ( λ ) if and only if λ ∈ Z ≥ .First suppose λ = 0 and consider the odd reflection r . We have that r ( α ) = α + α and by Lemma 4.7 we have that λ ′ = λ − α . Then by Lemma 4.6, Y ′ ∈ g ( A ) − ( α + α ) is locally nilpotent on L ( λ ) if and only if λ ′ ( h ′ ) ∈ Z ≥ where λ ′ ( h ′ ) = ( λ − α )( h + h ) = λ + λ − . The argument for λ = 0 is similar and yields the same condition. Finally, if λ = λ = 0 then the additional integrability condition is λ ( h + h ) ∈ Z ≥ ,which is vacuously satisfied. Q ± ( m, n, t ) Now we describe the weights for integrable highest weight modules of the Liesuperalgebra Q ± ( m, n, t ). The Cartan matrix for Q ± ( m, n, t ) is ab c a + b = m b + c = n c + a = t with m, n, t ∈ Z ≤− , not all equal to -1, and the simple roots are odd isotropic.The principal even roots are { α + α , α + α , α + α } . One can check thatfor i = j , h α i + α j = h j a ji + h i a ij using the formula appearing before Lemma 1.6 and rescaling h α i + α j so that( α i + α j )( h α i + α j ) = 2. 37 emma 4.11. A highest weight module V ( λ ) for the algebra Q ± ( m, n, t ) istypical if and only if λ , λ , λ = 0 .Proof. Since odd reflections of the diagram Γ do not yield new simple odd roots,the only conditions for the module to be typical are λ ( h ) , λ ( h ) , λ ( h ) = 0. Lemma 4.12.
An irreducible highest weight module for Q ± ( m, n, t ) , with typicalweight, is integrable if and only if λ + 1 b λ − λ + 1 c λ − ∈ Z ≥ .λ + 1 a λ − Proof.
By Corollary 4.9, it suffices to find the conditions for Y α ∈ g ( A ) − α to belocally nilpotent on L ( λ ) when α is a principal root. Let α i be a simple isotropicroot and let r i be the odd reflection with respect to α i . Since the weight λ istypical, λ i = 0. Then by Lemma 4.7, λ ′ = λ − α i . Since the simple even rootsof g ( A ′ ) are α i + α j for i = j , we have by Lemma 4.6 that the conditions ofintegrability are λ ′ ( h ′ j ) ∈ Z ≥ , where λ ′ ( h ′ j ) = λ ′ ( h α i + α j ) = ( λ − α i ) (cid:18) h j a ji + h i a ij (cid:19) = λ j a ji + λ i a ij − . Proposition 4.13.
The non-trivial irreducible integrable highest weight modulesof Q ± ( m, n, t ) are L ( λ ) such that λ λ λ = (cid:18)
11 + abc (cid:19) − b bc ac − c − a ba xyz , with x, y, z ∈ Z > . These weights are typical.Proof. The dimension of the Cartan subalgebra is 3 by Corollary 3.2, and hence λ is determined by its values on h , h and h . First we consider the case when λ is typical. We can rewrite the conditions of Lemma 4.12 using matrices: b
00 1 c a λ λ λ = xyz with x, y, z ∈ Z > . The determinant of the left most matrix is abcabc , whichis nonzero by Corollary 3.2. Hence, the matrix is invertible, and is calculated38bove. Finally, suppose that λ is not typical. Then without loss of general-ity suppose λ = 0. Consider the odd reflection r with respect to α . ByLemma 4.7, λ ′ = λ . Then by Lemma 4.6, we have the integrability conditions x, z ∈ Z ≥ with x = λ ′ ( h α + α ) = λ + 1 b λ = 1 b λ z = λ ′ ( h α + α ) = λ + 1 a λ = λ . If λ or λ is nonzero, then by a reflection at the corresponding simple root weobtain the integrability condition y ∈ Z ≥ with y = λ + 1 c λ − . By Lemma 3.1, b, c <
0. But this together with b λ , λ ≥ λ + c λ − <
0, which is a contradiction. Hence, if λ is not typical then λ = 0. In this section, we prove that a subfinite regular Kac-Moody diagram with anisotropic vertex, which is not of finite type, is not extendable. Hence, a regularKac-Moody diagram with an isotropic vertex is subfinite.
Lemma 5.1.
Suppose Γ and Γ ′ = Γ ∪ { v n +1 } are connected regular Kac-Moodydiagrams. Let g ( A ) (resp. g ( A ′ ) ) be the Kac-Moody superalgebra with diagram Γ (resp. Γ ′ ), and let Y n +1 ∈ g ( A ′ ) − α n +1 be the generator corresponding to thevertex v n +1 . Then the submodule M of g ( A ′ ) generated by g ( A ) acting on Y n +1 is an integrable highest weight module over the subalgebra g ( A ) .Proof. The fact that M is a highest weight module follows immediately from[ X i , Y n +1 ] = 0 for all i = 1 , . . . , n . The module M has highest weight − α n +1 . Areal root α of the subalgebra g ( A ) is also a real root of g ( A ′ ). By Lemma 4.4, theadjoint module of g ( A ′ ) is integrable. Thus for each real root α of g ( A ) we havethat Y α ∈ g ( A ) − α acts locally nilpotently on the submodule M of g ( A ′ ). Hencethe submodule M is an integrable highest weight module over the subalgebra g ( A ). Corollary 5.2. If Γ is a diagram for a regular Kac-Moody superalgebra g ( A ) that does not have non-trivial irreducible integrable highest weight modules, then Γ is not extendable.Proof. If g ( A ) has only trivial irreducible integrable highest weight modules,then the highest weight of the module M is 0. Hence − α n +1 = 0, which implies a i,n +1 = 0 for i = 1 , . . . , n . Since we assumed that the matrix A is a generalizedCartan matrix, this implies a n +1 ,j = 0 for j = 1 , . . . , n . This is not possiblewith Γ ′ being a connected diagram. 39 orollary 5.3. Suppose that Γ is a subfinite regular Kac-Moody diagram for g ( A ) which is not of finite type and not one of the algebras: A (0 , m ) (1) , C ( n ) (1) , S (1 , , α ) , and Q ± ( m, n, t ) . Then the diagram Γ is not extendable.Proof. This follows immediately from Lemma 4.5.From the classification of 3-vertex regular Kac-Moody diagrams we obtain:
Lemma 5.4. If Γ is a 3-vertex regular Kac-Moody diagram which is not a dia-gram for Q ± ( m, n, t ) , S (1 , , α ) or D (2 , , α ) , then the ratio of an isotropic ver-tex in Γ is rational and negative. The ratio of an isotropic vertex for Q ± ( m, n, t ) is real, irrational and negative. For S (1 , , α ) and D (2 , , α ) , if the ratios arerational then at most one of them is positive. Theorem 5.5.
A connected regular Kac-Moody diagram containing an isotropicvertex is subfinite. In particular, if Γ is a subfinite regular Kac-Moody diagramwhich is not of finite type, then Γ is not extendable.Proof. We prove this by induction on the number of vertices. Suppose that theclaim is true for all diagrams with less than n vertices, and let Γ be a connected n -vertex diagram which is regular Kac-Moody and contains an isotropic vertex.If Γ ′ is a connected proper subdiagram of Γ containing an isotropic vertex, thenΓ ′ is regular Kac-Moody and so by the induction hypothesis Γ ′ is subfinite.Let S denote the set of connected subfinite regular Kac-Moody which areextendable. By Corollary 5.3, Γ ′ ∈ S is either of finite type or can be a diagramfor A (0 , m ) (1) , C ( n ) (1) , S (1 , , α ), Q ± ( m, n, t ). In the following lemmas, weprove that if Γ ′ is a diagram for: A (0 , m ) (1) , C ( n ) (1) , S (1 , , α ), or Q ± ( m, n, t ),then Γ ′ is not a proper subdiagram of a connected regular Kac-Moody diagramΓ which satisfies the condition for all reflected diagrams: all proper connectedregular Kac-Moody diagrams containing an isotropic vertex are in S . Lemma 5.6. Q ± ( m, n, t ) is not extendable and hence Q ± ( m, n, t )
6∈ S .Proof.
We consider each case for attaching a vertex to a Q ± ( m, n, t ) diagram.Let Γ denote the extended diagram. Recall that a, b, c ∈ ( R \ Q ) < satisfy:1 + a + b = m b + c = n c + a = t ∈ Z < , or all equal to zero. (11) Case 1: (cid:13) v e / / o o d N (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D c (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) b (cid:26) (cid:26) Z Z N / / o o a N d, e = 0Now Γ , Γ ∈ S implies d, bd ∈ Q . But then b ∈ Q , which is a contradiction.40 ase 2: N (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D c (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) b (cid:26) (cid:26) Z Z (cid:13) v e x x pppppppppp d pppppppppp f ' ' NNNNNNNNNN g g g NNNNNNNNNN N / / o o a N d, e, f, g = 0Now Γ ∈ S implies g <
0, and Γ ∈ S implies cd <
0. Since a, b, c <
0, thisimplies d, ga >
0. By Lemma 5.4, this implies Γ
6∈ S , which is a contradiction.
Case 3: N (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D c (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) b (cid:26) (cid:26) Z Z (cid:13) v d O O (cid:15) (cid:15) ek x x pppppppppp h pppppppppp f ' ' NNNNNNNNNN g g g NNNNNNNNNN N / / o o a N d, e, f, g, h, k = 0The ratios of v , v , v in Γ are a, b, c ∈ ( R \ Q ) < . If Γ is a subdiagram suchthat the ratios of v and v in Γ are real negative numbers, then the ratio of v in Γ and of v in Γ are real positive numbers. But then by Lemma 5.4 allof the ratios of v are real negative numbers, which is a contradiction. Hence,Γ , Γ , Γ are diagrams for D (2 , , α ) or S (1 , , α ). ⋄ If v is N and Γ , Γ , Γ are diagrams for D (2 , , α ), then by (2) the we have: h + ga = − e + hc = − g + eb = − ag + kf = − kd + 1 e = − be + fd = − fk + 1 h = − ch + dk = − df + 1 g = − . (14)By solving (12) for h we find that h = bc − c − abc abc ∈ R , since abc = − d, e, f, g, k ∈ R and all vertex ratios are real. Now at leastone ratio at each vertex v , v , v must be positive, and the diagrams Γ , Γ , Γ are D (2 , , α ) diagrams and so they have at most one positive vertex ratio. Thisimplies that all of the ratios at v are negative, which is a contradiction. ⋄ If v is (cid:13) and Γ , Γ , Γ are diagrams for S (1 , , α ), then d, f, k = − h + ga = − e + hc = − g + eb = −
2. Solving for e we find that e = ab − b − abc )1+ abc ∈ R since abc = − h, g ∈ R . Now by reflecting at v we41ave that Γ ′ is N (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) b (cid:26) (cid:26) Z Z − N e O O (cid:15) (cid:15) eS x x ppppppppp − ppppppppp R & & NNNNNNNNN f f − NNNNNNNNN (cid:13) P / / o o Q (cid:13) , P = b + c + 1 Q = a + b + 1 R = − b − eS = − − e which implies that Γ ′ is a diagram Q ± ( m, n, t ), and the ratio of v in Γ ′ is − b − e − − e ∈ ( R \ Q ) < . But by substituting e = ab − b − abc )1+ abc we have − b − e − − e = 4 abc + 3 b − ab − ab c abc + 4 b − ab − > , which is a contradiction.We conclude that Q ± ( m, n, t ) is not extendable. Lemma 5.7. S (1 , , α ) is not extendable and hence S (1 , , α )
6∈ S .Proof.
Note that an odd reflection of a S (1 , , α ) diagram is again a S (1 , , α )diagram, but with a different α . Let a = α . Then a Z . Case 1: (cid:13) − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D a − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z a +1 (cid:13) v c / / o o b N − a / / o o − a N b, c = 0 ⋄ If v is N , then Γ ∈ S implies c = 1 and b = a . By reflecting at v we have (cid:13) − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D − a (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z a +1 N / / o o − (cid:13) − / / o o − a N . But Γ ′
6∈ S . ⋄ If v is J and c = −
2, then Γ ∈ S implies b = a . Then Γ ∈ S implies a − a ∈ {− , − } , and so a ∈ { , } . By reflecting at v and then at v , we42btain Γ ′′ := r ( r (Γ)) −→ r N − a − (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) a +1 (cid:26) (cid:26) Z Z a +1 J − / / o o − (cid:13) − / / o o − a N −→ r N − a − (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D − a − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) a +1 (cid:26) (cid:26) Z Z − J − / / o o a +2 N a +3 / / o o − (cid:13) . Then Γ ′′ ∈ S implies a +3 a +2 ∈ {− , − } , which is a contradiction. ⋄ If v is (cid:13) and c = −
1, then Γ ∈ S implies b = ka with k ∈ { , , } .Substituting b = ka and reflecting at v yields Γ ′ N a − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D a − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − a +2 (cid:26) (cid:26) Z Z − N P O (cid:15)
Pka x x ppppppppp ka ppppppppp R & MMMMMMMMMM f Q MMMMMMMMMM N − a / / o o − (cid:13) P = 1 − ( k + 1) aQ = 1 − kR = (1 − k ) a. First suppose that P = 1 − ( k + 1) a = 0. Then Γ ′ ∈ S implies Q ∈ { , − } andso k ∈ { , } . If k = 1, then we are reduced to a previous case. If k = 2, then P = 0 implies that a = . Reflecting at v of Γ ′ then yields Γ ′′ N − x x qqqqqqqqqq − qqqqqqqqqq & & NNNNNNNNN f f NNNNNNNNN (cid:13) N N f f MMMMMMMMMM & & − MMMMMMMMMM − ppppppppp x x − ppppppppp . But, Γ ′′
6∈ S .Now we assume that P = 0. Then reflecting at v of Γ ′ yields Γ ′′ N a − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) T (cid:26) (cid:26) Z Z T (cid:13) − O (cid:15)
P V ' NNNNNNNNNN g U NNNNNNNNNN (cid:13) − / / o o a − N P = 1 − ( k + 1) aT = 2 − aU = (2 k + 1)( a − V = − (2 k +1) a − ( k +1) a . ′′ ∈ S implies V ∈ { , − , − } . If V = 0, then U = (2 k + 1)( a −
1) = 0and so a = 1, which contradicts a Z . If V = −
2, then Γ ′′ ∈ S implies P/T = − /
3, and so k = 1, a = . But then Γ ′′
6∈ S . If V = −
1, then a = k +2 . Now Γ ′′ ∈ S implies that either Γ ′′ is S (1 , , β ) and P + U = − T ,or Γ ′′ is C (3) and P/T = − /
2. If P + U = − T , then a = 2 which contradicts a Z . If P/T = − /
2, then k = 1, a = . Reflecting at v of Γ ′′ yields Γ ′′′ (cid:13) − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z N x ppppppppp ppppppppp − & NNNNNNNNN f − NNNNNNNNN N − / / o o − N . But then Γ ′′′
6∈ S . Case 2: (cid:13) v c / / o o b (cid:13) − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D a − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z a +1 N − a / / o o − a N b, c = 0By reflecting at v , we are reduced to Case 1. Case 3: (cid:13) − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D a − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z a +1 (cid:13) v b x x ppppppppp c ppppppppp d & & NNNNNNNNN f f e NNNNNNNNN N − a / / o o − a N b, c, d, e = 0Now Γ ∈ S implies v is either N or (cid:13) . ⋄ If v is N , then Γ ∈ S implies b = c , d = e , and e = a − c . Also, Γ , Γ ∈ S implies a − c , a +1 a − c ∈ {− , − , − } . If a − c = a +1 a − c , then c = a − or a − c = 2,which is a contradiction. If either fraction equals −
1, then a reflection at thecorresponding vertex returns us to Case 1. So without loss of generality bysymmetry we have that a − c = − a +1 a − c = −
3. Then a = and c = . By44ubstituting and then reflecting at v , we obtain Γ ′ N − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) (cid:26) (cid:26) Z Z (cid:13) − O (cid:15) − − ' ' NNNNNNNNNN g g − NNNNNNNNNN (cid:13) − / / o o − N . But, Γ ′
6∈ S . ⋄ If v is (cid:13) , then Γ ∈ S implies that either b, d = − b = − d = − b = − d = −
2, then Γ is a G (3) diagram and c = a , e = a . ThenΓ ∈ S implies e = − a −
1. Thus a = − . By substituting and then reflectingat v , we obtain Γ ′ N − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) (cid:26) (cid:26) Z Z − N O (cid:15) − x x ppppppppp − ppppppppp N / / o o − (cid:13) . But then Γ ′
6∈ S .If b, d = −
1, then Γ ∈ S implies that either Γ is a C (3) diagram and − ca = − ea = − , or Γ is a S (1 , , β ) diagram and c + e = 2 a . Now by reflectingΓ at v we obtain Γ ′ N a − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D a − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) N (cid:26) (cid:26) Z Z − N P O (cid:15) Pc x x ppppppppp c ppppppppp R & & MMMMMMMMMM f f Q MMMMMMMMMM N − a / / o o − (cid:13) N = 2 − aP = 1 − a − cQ = a − e − ca R = a − c. If Γ is a C (3) diagram with − ca = − ea = − , then R, Q = 0 and Γ ′ reducesto the previous subcase. Thus Γ is a S (1 , , β ) diagram with e = 2 a − c and45 = −
1. Then by reflecting Γ at v we obtain Γ ′′ N U (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) a +1 (cid:26) (cid:26) Z Z a +1 N T O (cid:15) TW x x qqqqqqqqqq − qqqqqqqqqq V & & NNNNNNNNN f f V NNNNNNNNN (cid:13) − / / o o − a N T = c − − aU = − a − V = 2 a − cW = 2 c − a. If P = 0, then Γ ′ ∈ S implies that either N = R = − P/ N + R = − P .But if N = R = − P/
2, then by reflecting Γ ′ at v we reduce to Case 1. Similarly,if T = 0, then Γ ′′ ∈ S implies that either U = W = − T / U + W = − T .But if U = W = − T /
2, then by reflecting Γ ′′ at v we reduce to Case 1. • If P, T = 0, then 1 − a = c = 1 + 3 a implies a = 0, contradicting a Z . • If P = 0, T = 0, then c = 1 − a . Then by substitution Γ ′′ ∈ S impliesthat − a = 8 a , which contradicts a Z . • If P = 0, T = 0, then c = 1 + 3 a . Then by substitution Γ ′ ∈ S impliesthat − a = 8, which contradicts a Z . • If P, T = 0, then a = 2 + 2 c and 5 a = 2 + 2 c , contradicting a Z . Case 4: (cid:13) − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D a − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z a +1 (cid:13) v b O O (cid:15) (cid:15) c d & & NNNNNNNNN f f e NNNNNNNNN N − a / / o o − a N b, c, d, e = 0Now Γ ∈ S implies v is N and Γ ∈ S implies e = a . By reflecting at v , weare reduced to Case 3. Case 5: (cid:13) − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D a − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z a +1 (cid:13) v b O O (cid:15) (cid:15) cd x x ppppppppp e ppppppppp f & & NNNNNNNNN f f g NNNNNNNNN N − a / / o o − a N b, c, d, e, f, g = 046f v is (cid:13) , then Γ
6∈ S . If v is N then Γ is D (2 , , α ). But then a reflectionat v reduces to a previous case.We conclude that S (1 , , α ) is not extendable. Lemma 5.8. C ( n ) (1) is not extendable and hence C ( n ) (1)
6∈ S . N n +11 (cid:2) (cid:2) (cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6) B B − (cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6)(cid:6) − (cid:26) (cid:26) Z Z − (cid:13) − / / o o − (cid:13) __ (cid:13) n − − / / o o N n Proof.
Let v n +2 denote the additional vertex. Now Γ ∈ S implies a n +2 ,n +1 , a n +2 ,n =0, and Γ n +1 ∈ S implies a n +2 ,j = 0 for j = 1 , . . . , n −
1. Hence, C ( n ) (1) is notextendable. Lemma 5.9. A (0 , m ) (1) is not extendable and hence A (0 , m ) (1)
6∈ S .Proof.
First we show that A (0 , (1) is not extendable. Case 1: (cid:13) − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z − (cid:13) v b x x ppppppppp c ppppppppp d & & NNNNNNNNN f f e NNNNNNNNN N / / o o N b, c = 0 ⋄ If v is N , then Γ ∈ S implies c >
0. But then the ratio of v in Γ is positive,so Γ is a D (2 , , α ) diagram and d, e = 0. Then Γ ∈ S implies e >
0. Butthen the ratio of v in Γ is also positive, so Γ
6∈ S by Lemma 5.4. ⋄ If v is (cid:13) , then Γ ∈ S implies c <
0. Then Γ ∈ S implies b = −
1, since theratio at v is positive. By reflecting at v we obtain Γ ′ N − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) (cid:26) (cid:26) Z Z − N P O O (cid:15) (cid:15) Pc x x ppppppppp c ppppppppp Q & MMMMMMMMMM f R MMMMMMMMMM N / / o o − (cid:13) P = 1 − cQ = cd − c − R = c + e + 1 . But Γ ′
6∈ S since the ratio at v is positive, namely P = 1 − c > ase 2: (cid:13) − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z − (cid:13) v b O O (cid:15) (cid:15) c N / / o o N b, c = 0By reflecting at v we return to Case 1. Case 3: (cid:13) − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z − (cid:13) v b O O (cid:15) (cid:15) cd x x ppppppppp e ppppppppp N / / o o N b, c, d, e = 0Now Γ ∈ S implies e <
0. Then Γ
6∈ S since the ratio at v is positive. Case 4: (cid:13) − (cid:4) (cid:4) (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) D D − (cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9)(cid:9) − (cid:26) (cid:26) Z Z − (cid:13) v f O O (cid:15) (cid:15) gb x x ppppppppp c ppppppppp d & & NNNNNNNNN f f e NNNNNNNNN N / / o o N b, c, d, e, f, g = 0Now Γ , Γ ∈ S implies v is N and c, e >
0. But then the vertex ratios of v and v in Γ are positive. Hence Γ
6∈ S by Lemma 5.4. Therefore, A (0 , (1) isnot extendable.Now we show that A (0 , (1) is not extendable. N (cid:15) (cid:15) O O − / / o o − (cid:13) − (cid:15) (cid:15) O O − (cid:13) v z z zD D D {{{ CCC N − / / o o − (cid:13) a = 0. ⋄ If v is N then Γ , ∈ S implies a >
0. Then Γ , is D (2 , , α ), and so a , = 0. Then Γ , ∈ S implies a >
0. So Γ , has two isotropic vertices witha positive ratio, which contradicts Lemma 5.4. ⋄ If v is (cid:13) then Γ , ∈ S implies a <
0. Then then Γ , is D (2 , , α ), and so a = −
1. Then by reflecting at v we return a previous case.Therefore, a = 0 and by symmetry a = 0. Now without loss of generalitysuppose that a = 0. Then by reflecting at v we have v ′ is N and a ′ = 0,but this is the previous case. Hence, A (0 , (1) is not extendable.Finally we show that A (0 , m ) (1) is not extendable, for m ≥
3. Let Γ bea diagram for A (0 , m ) (1) . Then Γ has m + 2 ≥ v denote thevertex being added to the diagram. First suppose v is connected to an isotropicvertex, which we denote v . Let v and v denote the vertices adjacent to v in Γ ′ . Since v is isotropic with degree 3, it must be contained in a D (2 , , α )subdiagram. Since the subdiagram { v , v , v } is not D (2 , , α ), we have withoutloss of generality that the subdiagram { v , v , v } is D (2 , , α ). Then v is either N or (cid:13) . If v is N then we have the following subdiagram: (cid:13) v / / o o N / / o o ! ! CCCCCC a a CCCCCC N } } {{{{{{ = = {{{{{{ / / o o (cid:13) v N , where the double lines are necessarily connected, and the dotted lines are pos-sibly connected. By the 5-vertex classification, this diagram is not subfiniteregular Kac-Moody. If v is (cid:13) , then by reflecting at v we return to the thiscase.Next suppose v is not connected to an isotropic vertex. Then let Γ ′′ be aminimal subdiagram containing v and an isotropic vertex, which we denote v n .By minimality of Γ ′′ it is a chain such that v i is not isotropic for 1 < i < n .Thus by Lemma 2.5, there is a sequence of odd reflections R such that R ( v ) isisotropic and connected to v . This reduces us to the previous case.This concludes the proof of Theorem 5.5. References [1] S. Eswara Rao and V. Futorny,
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