Classification of finite-growth general Kac-Moody superalgebras
aa r X i v : . [ m a t h . R T ] O c t CLASSIFICATION OF FINITE-GROWTHGENERAL KAC-MOODY SUPERALGEBRAS
CRYSTAL HOYT AND VERA SERGANOVA
Abstract.
A contragredient Lie superalgebra is a superalgebra defined by a Car-tan matrix. A contragredient Lie superalgebra has finite-growth if the dimensionsof the graded components (in the natural grading) depend polynomially on thedegree. In this paper we classify finite-growth contragredient Lie superalgebras.Previously, such a classification was known only for the symmetrizable case. Introduction
Affine superalgebras have many interesting applications in combinatorics, numbertheory and physics (see [3,6,7,8]). Finite-growth contragredient Lie superalgebraswithout zeros on the diagonal were classified by V.G. Kac in [3]. Finite-growth sym-metrizable were classified by J.W. van de Leur in [10,11]. In the present paper wegive a complete classification of finite-growth contragredient Lie superalgebras with-out imposing any conditions on the Cartan matrices. This list includes examplespreviously known see [4,12] and some new superalgebras, however these new super-algebras are not simple.This also completes the classification of finite-growth Kac-Moody superalgebras,where we find that there are only two non-symmetrizable families, q ( n ) (2) and S (1 , , a ).In contrast to the Kac-Moody algebra situation, S (1 , , a ) is not a central extensionof a loop algebra. However, it appears in the list of conformal superalgebras [4].The main result is as follows. Theorem 1.1.
Let g ( A ) be a contragredient Lie superalgebra of finite growth andsuppose the matrix A is indecomposable with no zero rows. Then either A is sym-metrizable and g ( A ) is isomorphic to an affine or finite dimensional Lie superalgebraclassified in [10,11], or it is listed in Table 8.1. We also handle the case where the matrix A has a zero row. In this case, thealgebra g ( A ) is not simple and basically is obtained by extending a finite dimensionalalgebra by a Heisenberg algebra.The main idea of the classification is to use odd reflections (see [9]). Odd reflectionsare used to generalize the Weyl group for superalgebras, allowing one to move from Department of Mathematics, University of California, Berkeley, CA 94720, USA
Email: [email protected], [email protected]
Date : October 10, 2006 . one base to another. A Lie superalgebra usually has more than one Cartan matrix,unlike the Lie algebra case where the Cartan matrix is unique.Our proof goes in the following way. We show that in a finite-growth superalgebra,simple roots act locally nilpotently. This implies certain conditions on a Cartanmatrix (see Lemma 3.1). The crucial point of our classification is that these conditionsshould still hold after we apply an odd reflection. This gives rather strong conditionson a matrix, which allows us to list them all. These conditions are only slightlyweaker than the Kac-Moody conditions.The first author classifies in [1] all matrices with a zero on the main diagonal satis-fying the condition that the matrix and all its odd reflections are generalized Cartanmatrices (see conditions in definition 4.8). Remarkably, the superalgebras defined bysuch matrices almost always have finite-growth. The exception is a certain family ofmatrices of size 3, which is described in [1]. By comparing the classification in [1] tothe finite-growth symmetrizable classification in [10,11] we obtain the following
Theorem 1.2.
Let A be a symmetrizable generalized Cartan matrix with a zero onthe main diagonal. Suppose that any matrix A ′ obtained from A by a sequence ofodd reflections is again a generalized Cartan matrix. Then, g ( A ) has finite growth. Most of the calculations for our classification are in [1].2.
Definitions
Let I = { , . . . , n } , p : I → Z and A = ( a ij ) be a matrix. Fix a vector space h ofdimension n + cork ( A ), linearly independent α , . . . , α n ∈ h ∗ and h , . . . , h n ∈ h suchthat α j ( h i ) = a ij , define a Lie superalgebra ¯ g ( A ) by generators X , . . . , X n , Y , . . . , Y n and h with relations(2.1) [ h, X i ] = α i ( h ) X i , [ h, Y i ] = − α i ( h ) Y i , [ X i , Y j ] = δ ij h i . Here we assume that all elements of h are even and p ( X i ) = p ( Y i ) = p ( i ).By g ( A ) (or g when the Cartan matrix is fixed) denote the quotient of ¯ g ( A ) by theunique maximal ideal which intersects h trivially. It is clear that if a matrix B = DA for some invertible diagonal D then g ( B ) ∼ = g ( A ). Indeed an isomorphism can beobtained by mapping h i to d ii h i . Therefore without loss of generality we may assumethat a ii = 2 or 0.The Lie superalgebra g = g ( A ) has a root decomposition g = h ⊕ ⊕ α ∈ ∆ g α . Clearly, one can define p : ∆ → Z by putting p ( α ) = 0 or 1 whenever g α is evenor odd, respectively. By ∆ (∆ ) we denote the set of even (odd) roots. Every rootis a positive or negative linear combination of α , . . . , α n . According to this we calla root positive or negative and have the decomposition ∆ = ∆ + ∪ ∆ − . The roots α , . . . , α n are called simple roots . Sometimes instead of a ij we will write a αβ where α = α i , β = α j . LASSIFICATION OF FINITE-GROWTH GENERAL KAC-MOODY SUPERALGEBRAS 3
One sees easily that there are the following possibilities for each simple root α = α i :(1) if a αα = 2 and p ( α ) = 0, then X α , Y α and h α generate the subalgebraisomorphic to sl (2);(2) if a αα = 0 and p ( α ) = 0, then X α , Y α and h α generate the subalgebraisomorphic to the Heisenberg algebra;(3) if a αα = 2 and p ( α ) = 1, then [ X α , X α ] = [ Y α , Y α ] = 0 and X α , Y α and h α generate the subalgebra isomorphic to osp (1 | α ∈ ∆;(4) if a αα = 0 and p ( α ) = 1, then [ X α , X α ] = [ Y α , Y α ] = 0 and X α , Y α and h α generate the subalgebra isomorphic to sl (1 | α is isotropic and by definition any isotropic root isodd. In the other cases a root is called non-isotropic . A simple root α is regular if forany other simple root β , a αβ = 0 implies a βα = 0. Otherwise a simple root is called singular . Lemma 2.1.
For any subset J ⊂ I the subalgebra l in g ( A ) generated by h , X i and Y i , where i ∈ J , is isomorphic to h ′ ⊕ g ( A J ) , where A J is the submatrix of A with coefficients ( a ij ) i,j ∈ J and h ′ is a subspace of h . More precisely h ′ is the maximalsubspace in ∩ i ∈ J Ker α i which trivially intersects the span of h i , i ∈ J . Proof.
Let ¯ l be the subalgebra of ¯ g ( A ) generated by h , X i , Y i , i ∈ J . Then ¯ l isisomorphic to h ′ ⊕ ¯ g ( A J ), since by construction h ′ lies in the center of ¯ l , and (cid:2) ¯ l , ¯ l (cid:3) ∩ h ′ = { } . Let I be the maximal ideal in ¯ g ( A ) intersecting h trivially and J be the idealin ¯ l intersecting h trivially. We have to show that J = I ∩ ¯ l . Obviously, I ∩ ¯ l ⊂ J .On the other hand, write J = J + ⊕ J − , note that [ Y j , J + ] = [ X j , J − ] = 0 for any j ∈ I − J . Therefore, the ideal generated by J in ¯ g ( A ) intersects h trivially. Thatimplies J ⊂ I ∩ ¯ l , and lemma is proven. (cid:3) A superalgebra g = g ( A ) has a natural Z -grading g = ⊕ g n if we put g := h and g = g α ⊕ · · · ⊕ g α n . This grading is called principal . We say that g is of finite growth if dim g n grows polynomially depending on n . This means the Gelfand-Kirillov dimension of g is finite. The following Lemma is a straightforward corollaryof Lemma 2.1. Lemma 2.2. If g ( A ) is of finite growth, then for any subset J ⊆ I the Lie superal-gebra g ( A J ) is of finite growth. The matrix A is called indecomposable if the set I can not be decomposed intothe disjoint union of non-empty subsets J, K such that a ij = a ji = 0 whenever i ∈ I, j ∈ K . We say that A is elemental if it has no zero rows. Otherwise we call A non-elemental . 3. Integrability and finite growth
We say that g ( A ) is integrable if ad X i are locally nilpotent for all i ∈ I . In thiscase ad Y i are also locally nilpotent. CRYSTAL HOYT AND VERA SERGANOVA
Lemma 3.1.
Let g ( A ) be integrable and let A be elemental, indecomposable with n ≥ . Then after rescaling the rows A satisfies the following conditions (1) for any i ∈ I either a ii = 0 or a ii = 2 ; (2) if a ii = 0 then p ( i ) = 1 ; (3) if a ii = 2 then a ij ∈ p ( i ) Z − ; (4) if a ij = 0 and a ji = 0 , then a ii = 0 . Proof. If a ii = 0 and p ( i ) = 0, then h i , X i , Y i generate the Heisenberg subalgebra k . There exists j such that a ij = 0. Then Y j generates the irreducible infinitedimensional k -submodule with central charge − a ij . Hence (ad Y i ) m Y j = 0 for all m >
0. This proves the first statement.If a ii = 2, then h i , X i , Y i generate the subalgebra k isomorphic to sl for even i andosp (1 |
2) for odd i . Any Y j generates a k -submodule M with highest weight − a ij . Ifad Y i is locally nilpotent, this submodule must be finite-dimensional. From elementaryrepresentation theory we know that this implies − a ij ∈ Z ≥ for sl and − a ij ∈ Z ≥ for osp (1 | a ii = 2, a ij = 0, a ji = 0.Let k and M be as in the previous paragraph. Since − a ij = 0, M has the highestweight 0, therefore M is a trivial k -module or a Verma module. Since ad Y i is locallynilpotent, then M is trivial. Hence [ Y i , Y j ] = 0. But[ X j , [ Y i , Y j ]] = ± [ H j , Y i ] = ± a ji Y i = 0 . Contradiction. (cid:3)
Theorem 3.2.
Let A be elemental. If g ( A ) has finite growth, then g ( A ) is integrable. Proof.
We start with the following
Lemma 3.3.
Let n = 2 and a a = 0 . If ad X does not act nilpotently on X , then g ( A ) has infinite growth. Proof.
We use the fact that A is symmetrizable, therefore g ( A ) admits an invariantsymmetric form ( , ). By the condition of Lemma (ad X ) m X = 0 for any m . Let E k = (ad X ) k X , F k = (ad Y ) k Y , H k = [ E k , F k ] , β k = 3 kα + α , k ∈ Z ≥ . Obviously, β k − β l / ∈ ∆ if k = l . Hence[ E k , F l ] = δ kl H k . Let l be the Lie subalgebra generated by H k , E k , F k with k ∈ Z + . We claim that l has infinite growth with respect to the grading induced by the principal grading of g ( A ), and therefore g ( A ) has infinite growth.To prove our claim we will show first that there exists k such that(3.1) ( β k , β k + · · · + β k r ) = 0for any k ≤ k < · · · < k r . The claim follows immediately from the following LASSIFICATION OF FINITE-GROWTH GENERAL KAC-MOODY SUPERALGEBRAS 5
Lemma 3.4.
Let V be a two-dimensional space with non-zero symmetric bilinearproduct ( · , · ) . For any two vectors v, w ∈ V with ( v, w ) = 0 , there exist k > and c ∈ C such that Re c ( v + pw, v + qw ) > for all p, q ≥ k . Proof.
Assume first that ( w, w ) = 0. Put c = w,w ) . Thenlim p,q →∞ c ( v + pw, v + qw ) pq = 1 . Now let ( w, w ) = 0. Put c = v,w ) . Then againlim p,q →∞ c ( v + pw, v + qw ) pq = 1 , and Lemma 3.4 is proven. (cid:3) Recall that ( β k + · · · + β k r ) ( H k ) is proportional to ( β k , β k + · · · + β k r ) with non-zero coefficient and therefore it is not zero. Hence[ F k , [ E k [ E k . . . E k r ]]] = [ H k , [ E k . . . E k r ]] = ( β k + · · · + β k r ) ( H k ) [ E k . . . E k r ] = 0if k ≤ k < · · · < k r .The last calculation shows that the commutator [ E k , . . . , E k r ] is not zero. More-over, the same calculation shows that all such commutators are linearly independent. Lemma 3.5.
Let m be a positive integer, and let f m ( n ) be the number of ways towrite n as the sum m + · · · + m r with m r > m r − > · · · > m ≥ m . Then f m ( n ) grows exponentially on n . Finally, introduce a new Z -grading on g ( A ), by putting deg ′ X = 1, deg ′ X = 3.One can see that deg ′ [ E k , . . . , E k r ] = 3 ( k + 1) + · · · + 3 ( k r + 1) . Therefore dim g ( A ) ′ n ≥ f k +1 ( n ). Hence g ( A ) has infinite growth in this new grading.However | deg ′ ( X ) | ≤ | deg ( X ) | for any X ∈ g ( A ), therefore g ( A ) has infinitegrowth in the principal grading. Hence, Lemma 3.3 is proven. (cid:3) Lemma 3.6.
Let n = 2 , a = 2 , a = 0 , a = 0 . Then g ( A ) is of infinite growth. Proof.
Let X = [ X , X ], Y = ( − p (1) p (2) a [ Y , Y ]. Then one can check easily thefollowing relations[ X , Y ] = [ X , Y ] = H , [ X , Y ] = [ Y , X ] = 0 . Then X , Y , X , Y and h generate a subalgebra isomorphic to a quotient of ¯ g ( B )for b = b = b = b = 2. In particular, ad X is not nilpotent. By Lemma 3.3such an algebra has infinite growth, therefore g ( A ) has infinite growth. (cid:3) CRYSTAL HOYT AND VERA SERGANOVA
Lemma 3.7.
Let n = 3 , a = a = 0 and a = 0 . If ad X does not act nilpotentlyon X , then g ( A ) has infinite growth. Proof.
Note that X , X , Y , Y generate the subalgebra isomorphic to g (cid:0) A { , } (cid:1) .However A { , } is non-elemental, and by Lemma 10.1, g (cid:0) A { , } (cid:1) contains an infiniteHeisenberg Lie superalgebra l = l − ⊕ C h ⊕ l . Note that ad l Y = 0 and ad h ( Y ) = a Y . Therefore the l -submodule M generated by Y is isomorphic to U ( l ) ⊗ U ( C h ⊕ l ) C Y . It is not difficult to see that M has infinite growth in the principal grading.Hence g ( A ) has infinite growth. (cid:3) Now we can prove the theorem. Assume that ad X i is not locally nilpotent. Thenad X i does not act nilpotently on some X j . If a ij = a ji = 0 then [ X i , X j ] = 0.Therefore either a ij = 0 or a ji = 0. Consider the following cases(1) If a ij a ji = 0, then A { i,j } satisfies the conditions of Lemma 3.3, and therefore g (cid:0) A { i,j } (cid:1) has infinite growth;(2) If a ii = 2 , a ij = 0, then A { i,j } satisfies the conditions of Lemma 3.6, andtherefore g (cid:0) A { i,j } (cid:1) has infinite growth;(3) The case a ii = a ij = 0 is impossible since Lemma 10.1 implies ad X i X j = 0;(4) If a ii = 2 , a ij = 0, a ji = a jj = 0 then a jk = 0 for some k since A is non-elemental. Then by Lemma 3.7 g (cid:0) A { i,j,k } (cid:1) has infinite growth;(5) If a ii = a jj = 2 , a ij = 0, a ji = 0, then A { i,j } satisfies the conditions ofLemma 3.6, and therefore g (cid:0) A { i,j } (cid:1) has infinite growth;(6) If a ii = 0 , a ij = 0, a ji = a jj = 0 then by the same argument as in the fourthcase g (cid:0) A { i,j,k } (cid:1) has infinite growth for some k ;(7) If a ii = a ji = 0 , a ij = 0, a jj = 2 then A { i,j } satisfies the conditions ofLemma 3.6, and therefore g (cid:0) A { i,j } (cid:1) has infinite growth.Thus, we have shown that g ( A ) contains a subalgebra of infinite growth. Therefore g ( A ) itself has infinite growth. Contradiction. (cid:3) Odd reflections
Suppose a kk = 0 and p ( α k ) = 1. Define r k ( α i ) by the following formula r k ( α k ) = − α k ,r k ( α i ) = α i if a ik = a ki = 0 , i = k,r k ( α i ) = α i + α k if a ik = 0 or a ki = 0 , i = k. LASSIFICATION OF FINITE-GROWTH GENERAL KAC-MOODY SUPERALGEBRAS 7
Define X ′ i = [ X i , X k ] if i = k and a ik = 0 or a ki = 0 X i if i = k and a ik = a ki = 0 Y i if i = kY ′ i = [ Y i , Y k ] if i = k and a ik = 0 or a ki = 0 Y i if i = k and a ik = a ki = 0 X i if i = kh ′ i = [ X ′ i , Y ′ i ] . Then we have h ′ i = ( − p ( α i ) ( a ik h k + a ki h i ) if i = k and a ik = 0 or a ki = 0 h i if i = k and a ik = a ki = 0 h k if i = k Lemma 4.1.
The roots r i ( α ) , . . . , r i ( α n ) are linearly independent. The elements X ′ , . . . , X ′ n , Y ′ , . . . , Y ′ n together with h ′ , . . . , h ′ n satisfy (2.1) . Moreover if α i is regular,then h and X ′ , . . . , X ′ n , Y ′ , . . . , Y ′ n generate g . Proof.
We have to check that (cid:2) X ′ j , Y ′ k (cid:3) = 0 for any j = k . If j, k = i , then r i ( α j ) − r i ( α k ) / ∈ ∆, that forces (cid:2) X ′ j , Y ′ k (cid:3) = 0. So assume that j = i . Then for some constant c we have [ X ′ i , Y ′ k ] = c [ Y i , [ Y i , Y k ]] = 0 , as ad Y i = 0. Similarly one deals with the case k = i .To prove the second statement note that if r i ( α j ) = α j + α i , then X j = 1 a ij [ Y i , [ X i , X j ]] = d (cid:2) X ′ i , X ′ j (cid:3) for some non-zero constant d . Similarly, Y j = d ′ (cid:2) Y ′ i , Y ′ j (cid:3) . If r i ( α j ) = α j , then X j = X ′ j , Y j = Y ′ j . Finally, X i = Y ′ i and Y i = X ′ i . Hence every generator X j , Y j canbe expressed in terms of X ′ , . . . , X ′ n , Y ′ , . . . , Y ′ n . (cid:3) We see from Lemma 4.1 that if one has a matrix A and some regular isotropic simpleroot α i for g ( A ), then one can construct another matrix A ′ such that g ( A ′ ) ∼ = g ( A ).In this case we say that A ′ is obtained from A and a base Π ′ = { α ′ , . . . , α ′ n } isobtained from Π = { α , . . . , α n } by the regular odd reflection with respect to α i . If(∆ + ) ′ denotes the set of positive roots with respect to Π ′ , then one can check easilythat(4.1) (cid:0) ∆ + (cid:1) ′ = ∆ + ∪ {− α i } − { α i } . Thus, any two sets of positive roots differ only by finite set of isotropic roots. Inparticular, the set of even positive roots ∆ +0 is defined independently of a choice of abase Π. CRYSTAL HOYT AND VERA SERGANOVA
After rescaling the rows of the matrix A ′ we have that it is defined by the following,(where we assume 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 = 0 or a ki = 0 , and a kj = a jk = 0 a ki a ij + a kj a ik + a ki a ik if a ik = 0 or a ki = 0 , and a jk = 0 or a kj = 0 . Remark . In practice, we rescale the rows of the matrix A ′ so that the nonzerodiagonal entries are equal to 2. Remark . If α i is a singular isotropic root, then the corresponding odd reflectionis called singular . In this case, we can only show that g ( A ′ ) is a subalgebra in g ( A ).Though, the roots r i ( α ) , . . . , r i ( α n ) still form a base in the geometric sense, i.e.every α ∈ ∆ is a linear combination P nj =1 m j r i ( α j ), where all m j are non-negativeintegers or all m j are non-positive integers. The set Π ′ obtained from Π by a sequenceof odd reflections, singular or regular, is called a geometric base. Lemma 4.4.
Suppose β is a regular isotropic root, and let A ′′ and X ′′ α , Y ′′ α , H ′′ α denotethe matrix and generators obtained by reflecting twice at β . Then by rescaling therows of A ′′ and the generators X ′′ α accordingly we obtain the original matrix andgenerators. Thus, r β ◦ r β ∼ id . Proof.
First observe that a ′ ki = a ki and a ki = 0 ⇔ a ik = 0. Then there are three casesto check. First, r β ( r β ( β )) = r β ( − β ) = β and r β ( r β ( X β )) = r β ( Y β ) = X β . Second,if a ki = 0 then r β ( r β ( α )) = r β ( α ) = α and r β ( r β ( X α )) = r β ( X α ) = X α . Finally, if a ki = 0 then a ′ ki = 0, and we have r β ( r β ( α )) = r β ( α + β ) = ( α + β ) − α = α and r β ( r β ( X α ) = r β ([ X α , X β ]) = [[ X α , X β ] , Y β ] = − a ki X α . (cid:3) Lemma 4.5.
Let A ′ be obtained from A by an odd reflection. If A is elemental(non-elemental), then A ′ is elemental (non-elemental). Proof.
If the i -th row of A is zero, then an odd reflection does not change α i and h i and the i -th row of A ′ is also zero. Suppose the i -th row of A ′ is zero. Then α ′ j ( h ′ i ) = 0 for all j . Let A ′ be obtained from A by an odd reflection r j . Then if a ji = 0 h ′ i = c ( a ij h j + a ji h i ) = ca ji h i for some non-zero c . If a ji = 0, then h ′ i = h i . Any way, the i -th row of A is zero. (cid:3) Note that the notion of finite growth does not depend on choice of a base as followsfrom
LASSIFICATION OF FINITE-GROWTH GENERAL KAC-MOODY SUPERALGEBRAS 9
Lemma 4.6.
Let Π ′ be obtained from Π by an odd reflection. Then dim g ′ n ≤ dim g − n ⊕ · · · ⊕ g n . In particular, if g ( A ) is of finite growth, then g ( A ′ ) is of finitegrowth. Theorem 4.7. If g ( A ) has finite growth and A is elemental, then matrix A andany matrix A ′ obtained from A by a sequence of odd reflections (singular or regular)satisfies the matrix conditions of Lemma 3.1. Proof.
This follows immediately from Lemma 4.6, Lemma 3.1 and Theorem 3.2. (cid:3)
Definition 4.8.
The matrix A is called a generalized Cartan matrix if the followingconditions hold(1) for any i ∈ I either a ii = 0 or a ii = 2;(2) if a ii = 0 then p ( i ) = 1;(3) if a ii = 2 then a ij ∈ p ( i ) Z − ;(4) if a ij = 0 then a ji = 0.Note that the above conditions are just a little bit stronger than conditions ofLemma 3.1. More precisely Lemma 4.9.
A matrix satisfying the conditions of Lemma 3.1 is a generalized Cartanmatrix if and only if Π does not contain a singular root. We call g ( A ) a regular Kac-Moody superalgebra if A itself and any matrix obtainedfrom A by a sequence of odd reflections is a generalized Cartan matrix. Corollary 4.10. If A is symmetrizable with a zero on the diagonal and g ( A ) hasfinite growth, then g ( A ) is regular Kac-Moody. Proof.
This is an immediate consequence of Theorem 4.7 and Lemma 4.9. (cid:3) principal roots We call α ∈ ∆ a principal root if α belongs to some base Π ′ obtained from aninitial base Π by a sequence of regular odd reflections. A principal root defines asubalgebra of g generated by X α , Y α and h α = [ X α , Y α ]. We choose X α , Y α and h α insuch a way that α ( h α ) = 0 or 2. So the choice of these generators is unique up toproportionality. For two roots α and β put a αβ = β ( h α ).We define the subset Π ⊆ ∆ as follows. Let α ∈ Π if α is even and belongs tosome geometric base, or α = 2 β , where β is odd, non-isotropic and belongs to somegeometric base. Obviously, Π ⊂ ∆ +0 . For a Lie superalgebra l we denote by l ′ itscommutator [ l, l ] Lemma 5.1.
Let S be a subset of Π , and let g S be the subalgebra of g generated by X β , Y β for all β ∈ S and h β = [ X β , Y β ] . These elements satisfy the following relationsfor a certain matrix B : [ h β , X γ ] = b β,γ X β , [ h β , Y γ ] = − b β,γ Y β , [ X β , Y γ ] = δ β,γ h β , for β, γ ∈ S. There exists a surjective homomorphism g S → g ′ ( B ) /c where c is some central sub-algebra of the Cartan subalgebra in g ′ ( B ) = [ g ( B ) , g ( B )] . Proof.
In order to check the relations it suffices to show that β − γ / ∈ ∆ for any β, γ ∈ Π . Indeed, assume that α = β − γ . Then α ∈ ∆ and without loss ofgenerality we may assume that α is positive. Thus, we have β = α + γ . On the otherhand, there exists a geometric base Π ′ such that either β ∈ Π ′ or β ∈ Π ′ . Anyway, β can not be decomposed into a sum of some positive roots not proportional to itself.Contradiction.Now let us prove the second statement. There exists a surjective homomorphism j : ¯ g ′ ( B ) → g S . Let i : ¯ g ′ ( B ) → g ′ ( B ) be the standard projection. Let I , J bethe kernels of i, j respectively. Then (see [5]) I = I + ⊕ I − , where I ± is the uniquemaximal ideal in ¯ n ± . We claim that J = J + ⊕ J ⊕ J − , where J ± is an ideal in ¯ n ± and J is some ideal in the Cartan subalgebra of ¯ g ′ ( B ). Indeed, let us choose h ∈ h such that β ( h ) > β ∈ S and ¯ h in the Cartan subalgebra of ¯ g ( B ) such that β (¯ h ) = β ( h ) for all β ∈ S . Then one can extend j to the surjection C ¯ h + ¯ g ′ ( B ) → C h + g S by putting j (¯ h ) = h . If h ∈ g S we take ¯ h ∈ j − ( h ), and then we do not need to extend j since it is already defined on ¯ h . Since J is also an ideal in C ¯ h + ¯ g ′ ( B ), we have thedecomposition J = J + ⊕ J ⊕ J − , where J ± is the sum of the eigenspaces of ad(¯ h ) with positive (negative) eigenvalues and J is the zero eigenspace, hence J isan ideal in the Cartan subalgebra. Now the maximality of I ± implies J ± ⊂ I ± . Let c = i ( J ). obviously, c commutes with all X β , Y β , hence c is in the center of g ′ ( B ).Then clearly the surjective homomorphism g S → g ′ ( B ) /c exists. (cid:3) Corollary 5.2.
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 is aneven generalized Cartan matrix. Remark . We denote by g ( B ) the algebra given by all roots in Π .Generally speaking it is possible that Π is infinite, but this is certainly not possibleif g ( A ) is regular and has finite growth. Lemma 5.4. If g ( A ) is a regular Kac-Moody superalgebra of finite growth, then Π is finite. Moreover, | Π | ≤ n . Proof.
Since all odd reflections involved are regular, a αα = 2 for any α ∈ Π . Forany finite subset S ⊂ Π the corresponding matrix B has finite growth, therefore itconsists of affine and finite blocks. The rank of B is not bigger than n , on the otherhand it is bounded by the size of B minus the number of affine blocks. That implies | Π | ≤ n . (cid:3) LASSIFICATION OF FINITE-GROWTH GENERAL KAC-MOODY SUPERALGEBRAS 11 Regular Case
We have the following results from [1].
Lemma 6.1.
If the matrix A is symmetrizable, irreducible and has a zero on thediagonal, then g ( A ) is regular Kac-Moody if and only if g ( A ) has finite growth. Lemma 6.2.
Let A be nonsymmetrizable with a zero on the diagonal and g ( A ) beregular Kac-Moody. Then g ( A ) has finite growth if and only if it is either q ( n ) (2) or S (1 , , a ) for non-integer a . S (1 , , a ) is isomorphic to S (1 , , b ) iff a ± b ∈ Z . Seediagrams in table 8.1. Non-regular elemental case
Here we classify all finite growth Lie superalgebras g ( A ) where A is an irreducible,elemental and g ( A ) is not regular Kac-Moody. It is useful to describe A by thecorresponding Dynkin diagram, see [2,10]. Here we use matrix diagrams, but stillfollow the same labeling conventions for the vertices.(7.1) g ( A ) A p (1) Dynkin diagram A (2) 0 (cid:13) B (0 ,
1) (2) 1 • A (0 ,
0) (0) 1 N We assume the the matrix has only 0 or 2 on the diagonal and call A admissible ifany matrix obtained from A by a sequence of odd reflections satisfies the conditionsof Lemma 3.1. We join vertex i to vertex j be an arrow if a ij = 0 and we write thenumber a ij on this arrow.We proceed by induction on the number of vertices. We say that Γ ′ is a subdiagramof Γ if A ′ is a principal submatrix of A . We observe that if a diagram has only twovertices and one of them is singular, then the diagram is non-elemental. So ourclassification begins with n = 3 vertices. Lemma 7.1.
Let A be an admissible × -matrix such that a = 0 , a = 2 , and a = 0 . If p (2) = 0 , then a = − or − . If p (2) = 1 , then a = − . Proof.
We may assume a = 1 without loss of generality. Also, we know a = 0.Now by reflecting at the isotropic vertex v , we have (cid:18) a (cid:19) −→ r (cid:18) − a a + 1) (cid:19) . If a = −
1, then after rescaling the new matrix we have (cid:18) − a a +1 (cid:19) . Then a , − a a +1 ∈ Z − implies that a = −
2. Hence, a ∈ {− , − } . Since a ∈ p ( i ) Z − , the result follows. (cid:3) Lemma 7.2. If a = 0 , then the following matrix is not admissible: A = a a a − . Proof.
By reflecting at the isotropic vertex v and then rescaling, we have A ′ = a a − a − . For matrix A and A ′ to be admissible, we must have that a < − a < (cid:3) Theorem 7.3.
Every admissible diagram with 3 vertices which contains a singularvertex and defines an algebra of finite growth is isomorphic to S (1 , , or D (2 , , . Proof.
Let Π = { α , α , α } and assume that α is singular. Then p ( α ) = 1. Withoutloss of generality we have a = a = 0, a = 0 , a = 1. We have these assumptionsfor A in the following 3 lemmas. Lemma 7.4. p ( α ) = 0 , and therefore a = 2 . Proof.
Consider the singular reflection r , and let A ′ be the new Cartan matrix, α ′ i = r ( α i ). Then α ′ = α + α , h ′ = a h and a ′ = a a = 0, a ′ = 0. Thisimplies p ( α ′ ) = 1. Since p ( α ′ ) = p ( α ) + p ( α )(mod 2), we conclude p ( α ) = 0 andhence a = 2. (cid:3) Lemma 7.5. If a = 0 , then A equals N − (cid:24) (cid:24) (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:12)(cid:12)(cid:12) N E E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / / (cid:13) − o o − X X , which is S (1 , , . Proof.
Note that by Lemma 7.4, a = 0 and therefore without loss of generality weassume that a = 1. By reflecting at v we obtain the matrix A ′ − a Since g ( A ′ ) must have finite growth a ′ = − −
2. By Lemma 7.2, a ′ = −
2. Thus,1 + a = − a = −
2. This implies a = 0. By considering the odd reflection r we have the even roots α + α , α ∈ Π . The subalgebra with the corresponding LASSIFICATION OF FINITE-GROWTH GENERAL KAC-MOODY SUPERALGEBRAS 13 positive generators X ′ = [ X , X ] and X ′ = X must have finite growth. So byCorollary 5.2, the Lie algebra g ( B ) must have finite growth, where we calculate that B = (cid:18) − a + a (cid:19) . Recall that a , a ∈ Z < , so a + a ≤ −
2. Thus, a + a = − a = a = − (cid:3) Lemma 7.6. If a = 2 , then A equals (cid:13) − / / N o o / / (cid:13) − o o , which is D (2 , , . Proof.
Then a , a ∈ Z − . Now a = 0 and a = 2 imply a = 0. Then byLemma 7.1 a = − −
2. Suppose a = −
2. Then after an application of r weobtain the matrix − − a This implies 2 − a ∈ Z < , but this is impossible since a ∈ Z < . Hence a = − v and then at v we obtain −→ r a − −→ r a − a a − − a − . Then a , a a − ∈ Z − implies a = 0. Thus a = 0.Now we need to find a . Observe that α , α + α + α ∈ Π , and and thesubalgebra generated by X ′ = [[ X , X ] , [ X , X ]] and X ′ = X must have finitegrowth. By Corollary 5.2, the Lie algebra g ( B ) must have finite growth, where wecalculate that B = (cid:18) a + 2 2 (cid:19) . We can conclude that a = − (cid:3)(cid:3) Lemma 7.7.
Suppose that A is a × generalized Cartan matrix, but g ( A ) isnot regular Kac-Moody. If g ( A ) has finite growth, then A is one of the followingdiagrams: (cid:13) − (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:12)(cid:12)(cid:12) N a − E E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − a / / N − a o o a +1 Y Y a ∈ Z \ { } . Proof. If A does not have a singular root and g ( A ) is not regular Kac-Moody, then A must reflect by a sequence of regular reflections to some A ′ which contains a singularroot. Then A ′ must contain at least two isotropic roots, one singular and one regular.Hence A ′ is isomorphic to S (1 , , S (1 , , a ) for integer a is closed under regular odd reflections andcontains the S (1 , ,
0) diagram with a singular vertex. (cid:3)
Theorem 7.8.
Every admissible diagram with 4 vertices which contains a singularvertex and defines an algebra of finite growth is isomorphic to b D (2 , , . An admis-sible diagram with n ≥ vertices which contains a singular vertex does not define analgebra of finite growth. Proof.
Every admissible non-regular contragredient Lie superalgebra of finite growthwith n > S (1 , ,
0) and D (2 , , Lemma 7.9. S (1 , , is not extendable. Proof.
Suppose A is an admissible matrix, such that the submatrix given by excludingthe vertex v is S (1 , , a − a − − a a a a a . Case 1: p (4) = 0Then a = 2, and a , a , a ∈ Z − . The subalgebra corresponding to the evenroots α + α , α , α ∈ Π , with positive generators [ X , X ] , X , X must have finitegrowth. So by Corollary 5.2, the Lie algebra g ( B ) must have finite growth, wherewe calculate that B = − a + a − a a + a a . Hence, a = a = 0, a + a = 0, and a + a = 0. Since a , a ≤ a = a = 0. It follows that a = a = 0. Hence, case 1 is not admissible.Case 2: p (4) = 1 and a = 2Then a , a , a ∈ Z − . The subalgebra corresponding to the even roots α + α , α , α ∈ Π , with positive generators [ X , X ], X , [ X , X ]. must have finitegrowth. So by Corollary 5.2, the Lie algebra g ( B ) must have finite growth, wherewe calculate that B = − a + a ) − a ( a + a ) a . LASSIFICATION OF FINITE-GROWTH GENERAL KAC-MOODY SUPERALGEBRAS 15
By the same argument as in case 1, we conclude that case 2 is not admissible.Case 3: p (4) = 1 and a = 0, with a = 0Then a = 0 by Lemma 7.4, and WLOG a = 1. The subalgebra correspondingto the even roots α + α , α , α + α ∈ Π , with positive generators [ X , X ], X ,[ X , X ] must have finite growth. So by Corollary 5.2, the Lie algebra g ( B ) musthave finite growth, where we calculate that B = − a + a − a −
11 + a (1 + a ) − a a a . Hence, a − a = 1. But, a ∈ Z − , which is a contradiction. Hence,case 3 is not admissible.Case 4: p (4) = 1 and a = 0, with a = 0Then a = 0 by Lemma 7.4. For the submatrix excluding v to be admissible, either a = 0 or a = −
1. Suppose a = 0. Then a = 0. Since A is elemental, a = 0.Then a = 0. For the submatrix excluding v to be admissible a ∈ {− , − } byLemma 7.1. By Lemma 7.2, a = −
2. Hence, a = − a = 1. Thenby reflecting at v we have A ′ is − − − . But, the submatrix excluding v of A ′ is not admissible. Hence, a = −
1. Thenthe submatrix excluding v of A must be S (1 , ,
0) by Lemma 7.5, which implies a = − a = 1 and a = −
2. Then by reflecting at v we have −
11 0 − − − −
11 0 − −→ r − − −
11 0 − − − . The submatrix excluding v of A ′ is not admissible by the three vertex classification.Hence, case 4 is not admissible. Therefore, S (1 , ,
0) is not extendable. (cid:3)
Lemma 7.10.
The only admissible extension of D (2 , , is the diagram b D (2 , , (cid:13) − (cid:127) (cid:127) ~~~~~~~ (cid:13) − / / N o o ? ? ~~~~~~~ − (cid:31) (cid:31) @@@@@@@ (cid:13) − _ _ @@@@@@@ . The diagram b D (2 , , is not extendable. Proof.
Suppose A is an admissible matrix, such that the submatrix given by excludingthe vertex v is D (2 , , − a a − a a a a a . Case 1: a = 0, a = 0Then the for the submatrix given by excluding the vertex v to be admissible it mustbe D (2 , , a = a = 0, a = − p (4) = 0. By reflecting at v weobtain the matrix A ′ − −
11 0 0 01 0 0 01 0 0 0 . But, the submatrix given by excluding v of A ′ is not admissible.Case 2: a = 0, a = 0Then by reflecting at v we obtain A ′ − a a a a . Consider the submatrix of A ′ given by excluding the vertex v . Since S (1 , ,
0) is notextendable, we have that a = 0. Now consider the original matrix A . If a = 0then a = 0 since a = 2. Since A is elemental this implies a = 2. Then a = 0and so A is decomposable. So we have that a = 0. Since a ∈ Z − , the submatrixof A ′ given by excluding the vertex v is not admissible.Case 3: a = 0Then the for the submatrix given by excluding the vertex v to be admissible it mustbe D (2 , ,
0) since S (1 , ,
0) is not extendable. Then a = a = 0, a = − a = 2and p (4) = 0. For the submatrix given by excluding v to have finite growth, either a = a = a = −
1; or a = a = 0. So first suppose a = a = a = −
1. Thenby reflecting at v we obtain the matrix A ′ − −
11 0 0 − − − . LASSIFICATION OF FINITE-GROWTH GENERAL KAC-MOODY SUPERALGEBRAS 17
But, the submatrix given by excluding v is not admissible. Thus a = a = 0.Then by reflecting at v we obtain the matrix A ′ − − − a a − − a − − a a − a . For the submatrix excluding v of A ′ to have finite growth, − − a = 0. So a = −
1. This case is admissible, and we call it b D (2 , , b D (2 , ,
0) does not extend to an admissible 5 vertex diagram. So supposethat A is a Cartan matrix of an admissible 5 vertex extension of b D (2 , , D (2 , ,
0) must be b D (2 , , (cid:13) − (cid:15) (cid:15) (cid:13) − / / N a O O b o o c (cid:15) (cid:15) / / (cid:13) − o o (cid:13) − O O , where a = − b , b = − c , and c = − a . But, this is a contradiction since a, b, c = 0.Hence, b D (2 , ,
0) is not extendable. (cid:3)
For n ≥
6, an admissible diagram with a singular vertex must contain an admissible5 vertex subdiagram with a singular vertex. Since there are none with 5 vertices, wehave found all admissible diagrams containing a singular vertex. (cid:3)
Corollary 7.11.
Suppose the matrix A is a generalized Cartan matrix, but g ( A ) isnot regular Kac-Moody. If g ( A ) has finite growth, then A is one of the diagrams inLemma 7.7. Proof.
This follows immediately from Lemma 7.7 and the fact that b D (2 , ,
0) doesnot contain a regular isotropic vertex. (cid:3) Non-symmetrizable Contragredient Lie Superalgebras of FiniteGrowth (with elemental matrices)
Algebra Dynkin diagrams Table 8.1 D (2 , , (cid:13) − / / N o o / / (cid:13) − o o b D (2 , , (cid:13) − (cid:127) (cid:127) ~~~~~~~ (cid:13) − / / N o o ? ? ~~~~~~~ − (cid:31) (cid:31) @@@@@@@ (cid:13) − _ _ @@@@@@@ S (1 , , (cid:13) − (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:12)(cid:12)(cid:12) N a − E E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − a / / N − a o o a +1 Y Y a ∈ Z S (1 , , a ) (cid:13) − (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:12)(cid:12)(cid:12) N a − E E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − a / / N − a o o a +1 Y Y a Z q ( n ) (2) • a } {{{{{{{{ = {{{{{{{{ b ! CCCCCCCC a CCCCCCCC • / / · · · o o / / • o o There are n • .Each • is either (cid:13) or N .An odd number of them are N .If • is (cid:13) , then a = b = − • is N , then ab = − Description of non-symmetrizable contragredient superalgebras
As follows from our classification there are the following non-symmetrizable super-algebras: D (2 ,
1; 0), b D (2 ,
1; 0), q ( n ) (2) and S (1 , a ).The Lie superalgebra D (2 ,
1; 0) is the degenerate member of the family D (2 , α )when α = 0, described in [2]. D (2 ,
1; 0) is not simple, but has the ideal isomorphic to psl (2 | D (2 ,
1; 0) ∼ = Der ( psl (2 | D (2 ,
1; 0) is isomorphic to a semidirect sum of sl (2) and psl (2 | H, E and F denote the standard basis of sl (2), l denote the Lie superalgebra psl (2 | Z -grading l = l − ⊕ l ⊕ l , where l = sl (2) ⊕ sl (2), l and l − are isomorphic l -modules. For every g ∈ l i define [ H, g ] = ig , furthermorelet [ E, l ] = [ F, l ] = [ E, l ] = [ F, l − ] = 0 , [ E, x ] = γ ( x ) , x ∈ l − , [ F, y ] = γ − ( y ) , y ∈ l , where γ : l − → l is an isomorphism of l -modules. To identify the Chevalleygenerators let X = F, Y = E and let X , X be the simple roots of l . Description of b D (2 , , b l isomorphic to l ⊗ C (cid:2) t, t − (cid:3) ⊕ C c ⊕ C d, where c belongs to the center and the commutator is determined by the formulas[ x ⊗ t m , y ⊗ t n ] = [ x, y ] ⊗ t m + n + ( x, y ) mδ m, − n c, [ d, x ⊗ t m ] = mx ⊗ t m , where x, y ∈ l . Define the semidirect product of sl (2) and b l by the formula[ g, x ⊗ t m + ac + bd ] = [ g, x ] ⊗ t m for all g ∈ sl (2), x ∈ l , a, b ∈ C . The Lie superalgebra b D (2 ,
1; 0) is isomorphic tothis semidirect product. To identify Chevalley generators let again X = F, Y = E and X , X , X be the simple roots b l . Description of q ( n ) (2) . The Lie superalgebra q ( n ) (2) is an extension of the loopalgebra of the simple Lie superalgebra t = q ( n ) twisted by the involutive automor-phism φ such that φ | t = id, φ | t = − id. The Lie superalgebra q ( n ) (2) is isomorphicto t ⊗ C (cid:2) t , t − (cid:3) ⊕ t ⊗ t C (cid:2) t , t − (cid:3) ⊕ C c ⊕ C d with d = t ∂∂t and c is the central element. For any x, y ∈ t , the commutator is definedby the formula[ x ⊗ t m , y ⊗ t n ] = [ x, y ] ⊗ t m + n + δ m, − n (1 − ( − m ) tr ( xy ) c. Description of S (1 , , a ). The Lie superalgebra S (1 , , a ) appears in the list ofconformal Lie superalgebras by Kac and Van de Leur [4]. Consider commutativeassociative Lie superalgebra R = C [ t, t − , ξ , ξ ] with even generator t and two odd LASSIFICATION OF FINITE-GROWTH GENERAL KAC-MOODY SUPERALGEBRAS 21 generators ξ , ξ . By W (1 ,
2) we denote the Lie superalgebra of derivations of R , i.e. all linear maps d : R → R such that d ( f g ) = d ( f ) g + ( − p ( d ) p ( f ) f d ( g ) . An element d ∈ W (1 ,
2) can be written as d = f ∂∂t + f ∂∂ξ + f ∂∂ξ for some f, f , f ∈ R . It is easy to see that the subset of all d ∈ W (1 ,
2) satisfyingthe condition at − f + ∂f∂t = ( − p ( d ) (cid:18) ∂f ∂ξ + ∂f ∂ξ (cid:19) form a subalgebra of W (1 , S a . One can check that S a issimple if a / ∈ Z . If a ∈ Z , then the ideal S ′ a = [ S a , S a ] is simple and has codimension1, more precisely S a = C ξ ξ t − a ∂∂t ⊕ S ′ a . Set X = − ∂∂ξ , X = − aξ ξ t − ∂∂ξ + ξ ∂∂t , X = ξ ∂∂ξ ,Y = ( a + 1) ξ ξ ∂∂ξ + ξ t ∂∂t , Y = t ∂∂ξ , Y = ξ ∂∂ξ ,h = − ( a + 1) ξ ∂∂ξ − t ∂∂t , h = aξ ∂∂ξ + ξ ∂∂ξ + t ∂∂t , h = ξ ∂∂ξ − ξ ∂∂ξ . Then X i , Y i , h i , i = 1 , ,
3, generate S a or S ′ a if a ∈ Z . They satisfy the rela-tions (2.1) with Cartan matrix S (1 , , a ). Hence the contragredient Lie superalgebra S (1 , , a ) can be obtained from S a ( S ′ a ) by adding the element d = t ∂∂t and taking acentral extension. The formula for this central extension can be found in [4].The fact that the algebras in the table 8.1 have finite growth follows directly fromtheir descriptions. 10. Non-elemental case
The description of superalgebras with non-elemental Cartan matrices can be easilyobtained from the following
Lemma 10.1.
Let A be indecomposable with a i = 0 for all i ∈ I , and let J = I \ { } .Then g ( A ) has a Z -grading g = g − ⊕ g ⊕ g such that g ∼ = g ( A J ) ⊕ C h ⊕ C d , where d is a non-zero vector in ∩ i ∈ J \{ } Ker α i not proportional to h , g − is an irreducible g -module with highest weight − α , g is the Z -graded dual of g − (which is a lowestweight module with lowest weight α ) and the commutator between g − and g isdefined by the formula [ x, y ] = h x, y i h , here h x, y i denotes the natural pairing between x ∈ g − and y ∈ g . Proof.
First, we have to show that the algebra g is a quotient of ¯ g ( A ). Indeed, let X i , Y i with i ≥ g , X be the lowest vector of g and, Y be thehighest vector of g − , h be the direct sum of the Cartan subalgebra of g , C h and C d .It is easy to see that X , . . . , X n , Y , . . . , Y n and h , . . . , h n satisfy the relations (2.1).Now we have to check that g does not have non-zero ideals, which intersect h trivially.If I is such ideal, then I ∩ g = { } , and therefore I = ( I ∩ g ) ⊕ ( I ∩ g − ). Since g is an irreducible g -module, I ∩ g = { } or I = g . If I = g , then X ∈ I , and h ∈ I . This is impossible, hence I ∩ g = { } . In the same way I ∩ g − = { } . Thelemma is proven. (cid:3) Corollary 10.2.
Let A be an indecomposable Cartan matrix, I ′ be the subset ofall i ∈ I such that a ij = 0 for all j ∈ I , A ′ = A I − I ′ , g ′ = g ( A ′ ) . Then g ( A ) is asemi-direct sum of g ′ and the ideal H = g − ⊕ g + ⊕ ∩ i ∈ I − I ′ Ker α i , where g − is a directsum of | I ′ | irreducible highest weight g ′ -modules, g + is a direct sum of | I ′ | irreduciblelowest weight g ′ -modules, dual to g − , [ g + , g + ] = [ g − , g − ] = 0 , g + and g − generatea direct sum of Heisenberg Lie superalgebras ⊕ i ∈ I ′ H i . Furthermore, g ( A ) has finitegrowth if and only if g ′ has finite growth and g ± have finite growth in the natural Z -grading induced by the principal grading of g ′ . Corollary 10.3.
Let g ′ be a contragredient Lie superalgebra with Cartan matrix A ′ of size n , V , . . . , V r be irreducible highest weight g ′ -modules with highest weight λ , . . . , λ r . Let V ∗ , .., V ∗ r denote the graded duals of V , . . . , V r . Let H denote thevector space H = V ⊕ ... ⊕ V r ⊕ V ∗ ⊕ ... ⊕ V ∗ r ⊕ C r . Choose a basis c , . . . , c r , d , . . . , d r in C r and define the Lie superalgebra structureon H assuming that c i lie in the center of H and the following relations hold [ x, y ] = δ ij h x, y i c i , [ d k , x ] = δ ik x , [ d k , y ] = − δ jk y for all x ∈ V ∗ i , y ∈ V j . Let g be the semi-direct product of g ′ and the ideal H , wherethe action of g ′ on V i , V ∗ i is already defined and the action of g ′ on C r is trivial.Then g is a contragredient Lie superalgebra, its Cartan matrix A has size n + r andcan be obtained from A ′ in the following manner a ij = a ′ ij , if i, j ≤ n , a ij = λ i − n (cid:0) h ′ j (cid:1) if i > n, j ≤ n , a ij = 0 if i ≤ n, j > n. Let g be a contragredient Lie superalgebra, by Corollary 10.2 g = g ′ ⋊ H . One canapply Corollary 10.2 to g ′ again and obtain g ′ = g ′′ ⋊ H ′ , then repeat for g ′′ and get g ′′ = g ′′′ ⋊ H ′′ e.t.c. Lemma 10.4. If g is a contragredient algebra of finite growth, then H ′ is finite-dimensional, ( g ′′ ) ± are purely odd, H ′′′ = 0 and hence g ′′′′ = g ′′′ . LASSIFICATION OF FINITE-GROWTH GENERAL KAC-MOODY SUPERALGEBRAS 23
Proof.
Let i ∈ I ′′ , then h i is a central element of H ′ . There exists j ∈ I ′ such that α j ( h i ) = 0. The corresponding irreducible component g − j has a non-zero centralcharge. A highest weight module over a Heisenberg Lie superalgebra with a non-zerocentral charge has finite growth only if the algebra is finite dimensional. Thus, H ′ i isfinite-dimensional for each i ∈ I ′′ , which implies that H ′ is finite-dimensional.Repeat now this argument for g ′′ . Let i ∈ I ′′′ , then there exists j ∈ I ′′ suchthat α j ( h i ) = 0. Then the corresponding irreducible component (cid:0) g ′ j (cid:1) − is finite-dimensional. A highest weight module with non-zero central charge over a Heisenbergalgebra is finite-dimensional if and only if the quotient of the algebra by the centeris purely odd. Hence H ′′ i / C h i is purely odd, which implies ( g ′′ ) ± are purely odd.Finally, applying this argument to g ′′′ and using the fact that there is no purelyodd highest weight modules with non-zero central charge over a Heisenberg algebra,we obtain the last statement of Lemma. (cid:3) Theorem 10.5.
Let g be an affine Lie algebra, V be a non-trivial irreducible highestweight g -module. Then V has infinite growth. Proof.
Let λ be a highest weight of V , c be the central element of g . Assume first,that λ ( c ) = 0. Let H be an infinite Heisenberg subalgebra of g generated by allimaginary roots. If v is a highest vector of V and M be the H -submodule of V generated by v . Then M is the irreducible highest weight H -module with non-zerocentral charge. If H − is the subalgebra of H generated by all negative imaginaryroots, then M is a free U ( H − )-module, and therefore M has infinite growth. Hence V has infinite growth.Now assume that λ ( c ) = 0. Then there exists a simple root α such that λ ( h α ) / ∈ Z ≥ . Then X n − α v = 0 for any v . Let δ be a positive imaginary root. One can seeeasily from the the general description of affine algebras, that there exist t i ∈ g iδ ,and f i ∈ g − α + iδ . such that [ t i , f j ] = f i + j and [ f i , f j ] = 0. Since f i ∈ ∆ + for i > f i v = 0 for all i >
0. Let a be the abelian subalgebra of g spanned by f i for all i ≤
0. We claim that U ( a ) v is free over U ( a ). Indeed, assume that X i
1; 0).
Lemma 10.6. If g ( A ) is a contragredient Lie superalgebra of finite growth withindecomposable non-elemental A , then A ′′′ is of quasi-finite type. The proof of this Lemma follows immediately from the following
Lemma 10.7.
Let g be a contragredient Lie superalgebra of finite-growth with a ele-mental indecomposable Cartan matrix and suppose there exists a non-trivial highestweight g -module of finite growth. Then g is either finite-dimensional or b D (2 , , . Proof.
Assume that g is infinite-dimensional. If its Cartan matrix is symmetrizable,then g is affine superalgebra and its even part g is a subquotient of a direct sum ofaffine Lie algebras. If V is a non-trivial g -module of finite-growth, then it contains anon-trivial irreducible g -subquotient V . However, it is impossible by Theorem 10.5.In the same manner, one can argue that g is not q ( n ) (2) , since its even part g containsthe loop algebra of sl ( n ). Finally, consider the case when g = S (1 , , a ). Let V λ bean irreducible highest weight module of finite growth with highest weight λ . Notethat g contains a subalgebra isomorphic to sl (2) (1) whose simple roots are α + α and α . By Theorem 10.5, λ ( h ) = 0 and(10.1) λ ( h α + α ) = λ ( h ) − λ ( h ) = 0 . Assume that λ ( h ) = 0. Then λ − α be the highest weight of V λ with respect to thebase − α , α + α , α + α , obtained from initial base by the odd reflection r . Thenwe have λ − α ( h − h ) = λ ( h ) − λ ( h ) + 1 = 0 , which obviously contradicts (10.1).Thus, we showed that g is finite-dimensional or b D (2 ,
1; 0). The latter case ispossible, since the superalgebra has a finite-dimensional quotient isomorphic to sl (2),hence one can consider any highest weight sl (2)-module and induce b D (2 , α ) actionon it assuming that the ideal acts trivially. (cid:3) Theorem 10.8.
The contragredient Lie superalgebra g ( A ) with a non-elementalindecomposable Cartan matrix has finite growth if and only if the following conditionshold (1) A ′′′ is of quasi-finite type; (2) p ( i ) = 1 for every i ∈ I ′′′ ; (3) for every i ∈ I ′′′ , j ∈ I − ( I ′ ∪ I ′′ ∪ I ′′′ ) , a ji = 0 implies that α j is either asimple root of a purely even block or is the left simple root of the diagram intable 8.1 in D (2 ,
1; 0) or b D (2 ,
1; 0) block;
LASSIFICATION OF FINITE-GROWTH GENERAL KAC-MOODY SUPERALGEBRAS 25 (4) for any i ∈ I ′′ , the restriction of − α i on the Cartan subalgebra of g ′′′ is anintegral dominant weight (a highest weight of finite-dimensional irreducible g ′′′ -module); (5) if i ∈ I , j is an index from b D (2 ,
1; 0) -block of A ′′′ and a ji = 0 , then α j is theleft simple root in the diagram in table 8.1. The proof of this theorem follows from Lemmas 10.4, 10.6 and Corollary 10.2 andwe will leave it to the reader.
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