A Combinatorial Approach to Root Multiplicities of a Special Type Rank 3 Kac-Moody Algebras
aa r X i v : . [ m a t h . R T ] F e b A COMBINATORIAL APPROACH TO ROOT MULTIPLICITIES OF A SPECIALTYPE RANK 3 KAC-MOODY ALGEBRAS
BOWEN CHEN, HANYI LUO AND HAO SUN
Abstract.
In this paper, we calculate the dimension of root spaces g λ of a special type rank 3Kac-Moody algebras g . We first introduce a special type of elements in g , which we call elements instandard form. Then, we prove that any root space is spanned by these elements. By calculating thenumber of linearly independent elements in standard form, we obtain a formula for the dimension ofroot spaces g λ , which depends on the root λ . Introduction
Kac-Moody algebras are generalizations of finite-dimensional semisimple Lie algebras, of which thestructures can be defined from generalized Cartan matrices. The dimensions of root spaces (alsocalled root multiplicities ) are important data in understanding the structure of Kac-Moody algebras.Generally speaking, Kac-Moody algebras are divided into three groups: finite case, affine case andindefinite case (see [4]). Finite and affine Kac-Moody algebras were studied in many different ways inthe last decades, while the indefinite type is still mysterious as its name implies. As a special case ofindefinite Kac-Moody algebras, hyperbolic Kac-Moody algebras have some interesting properties, andpeople have made a full classification and studied the root multiplicities in the last several decades[1, 2, 5]. In this paper, we study the dimension of root spaces of a special type indefinite Kac-Moodyalgebras with rank 3.Let A be a generalized Cartan matrix, and denote by g ( A ) the corresponding Kac-Moody algebra.Let A be a matrix in the following type − a − a − a − a , where a , a are positive integers and at least one of them is larger than one. We will calculate thedimension of root spaces of the corresponding Kac-Moody algebra g := g ( A ) in this paper. The Kac-Moody algebra g has three simple roots α , α , α . Denote by e i the element spanning the root space g α i . From this special matrix A , we also know that [ e , e ] = 0, which is also an important property.In §
2, we first consider a special type of elements in g , which are called elements in standard form.These are elements that can be written in the following way[ e a n , [ e a n − , [ . . . , [ e a , e a ] . . . ]]] . Let λ = n α + n α + n α be a positive root, and n = n + n + n . We prove that anyroot space g λ can be spanned by elements in standard form (see Theorem 2.3 and Corollary 2.4).Note that given an n -tuple ( a n , a n − , . . . , a ), we can equate it with an element in standard form[ e a n , [ e a n − , [ . . . , [ e a , e a ] . . . ]]]. Therefore, we calculate the dimension of root spaces g λ by calculatingthe corresponding number of n -tuples. With respect to the basic property of Lie algebras (see [3])[ g α , g β ] ⊆ g α + β , ∗ MSC2010 Class: 17B67, 17B22 † Key words: Kac-Moody algebra, hyperbolic Kac-Moody algebra, root multiplicity where α, β are roots, we know that g λ is spanned by elements in standard form[ e a n , [ e a n − , [ . . . , [ e a , e a ] . . . ]]]such that the number of e i is n i . Although an n -tuple corresponds to an element in standard form,this element can be trivial, and several n -tuples may correspond to linearly dependent elements. Thus,we have to eliminate all of these bad cases.In §
3, we introduce a combinatorial problem, which will be used in calculating the dimension. In §
4, we make a careful discussion of the n -tuples together with the corresponding elements in g . Hereare several important properties.(1) If e and e are adjacent in a standard form g , then we will get a new element g ′ in standardform by switching the position of e and e . We find that g ′ = − g (see Lemma 4.3).(2) If e i and e are adjacent in a standard form g , where i = 1 ,
3, switching the position will givea new element g ′ , which is linearly independent to g (see Lemma 4.5).The properties above reduce the problem of calculating the dimension of g λ to a combinatorial problemsuch that we have n red balls, n blue balls, and we want to put them into n boxes (under someextra conditions in Lemma 4.5). With a careful discussion of the cases in the first box (see Lemma4.8), we have the following upper bound of dim g λ A := X ≤ i ≤ a , ≤ j ≤ a i,j ) =(0 , (cid:18) n + n − i − n − (cid:19)(cid:18) n + n − j − n − (cid:19) . However, there are some n -tuples left corresponding to trivial elements in g λ . With a careful discussionin § n -tuples C = C + C := (cid:18) n + n + n − a − n − a − (cid:19) + (cid:18) n + n + n − a − n − a − (cid:19) . After deleting all “trivial” elements, there are still some n -tuples linearly dependent to each other. Wediscuss this case in § B := (cid:18) n + n − n − (cid:19)(cid:18) n + n − n − (cid:19) + (cid:18) n + n − n − (cid:19)(cid:18) n + n − n − (cid:19) . With respect to the discussion above, we have our main theorem.
Theorem 1.1 (Theorem 4.11) . Suppose that a ≤ a and n , n , n ≥ . Let λ = P i =1 n i α i . Denoteby g λ the corresponding root space. We have dim g λ = A − B, n < min { a , a } A − B − C , a ≤ n < a A − B − C − C , n ≥ max { a , a } . Note that in the theorem, we assume that n i ≥ i = 1 , ,
3. In Remark 4.12, we briefly discussthe general cases that n i ≥ × Standard Form
We first give the definition of Kac-Moody algebras. We refer the readers to [4] for more details.Let A be a r × r generalized Cartan matrix (GCM) of rank l . A realization of A is a triple ( h , Π , Π ∨ ),where h is a vector space, Π = { α , . . . , α r } ⊆ h ∗ and Π ∨ = { α ∨ , . . . , α ∨ r } ⊆ h , such that(1) both sets Π and Π ∨ are linearly independent.(2) ( α ∨ i , α j ) = a ij , 1 ≤ i, j ≤ r ,(3) r − l = dim h − r . OOT MULTIPLICITIES OF RANK 3 KAC-MOODY ALGEBRAS 3
We fix a generalized Cartan matrix A . Let ( h , Π , Π ∨ ) be a realization of it. The Kac-Moody algebra g := g ( A ) is a Lie algebra (with Lie bracket [ , ]) generated by h and e i , f i , where 1 ≤ i ≤ r , such that(1) [ e i , f j ] = δ ij α ∨ i , 1 ≤ i, j ≤ r .(2) [ h, h ′ ] = 0, where h, h ′ ∈ h .(3) [ h, e i ] = ( α i , h ) e i .(4) [ h, f i ] = − ( α i , h ) f i .(5) If i = j , we have ad( e i ) − a ij ( e j ) = 0 and ad( f i ) − a ij ( f j ) = 0, where ad( x )( y ) := [ x, y ] is theadjoint representation of g with respect to the Lie algebra structure.The elements α i are called simple roots . Let λ = r P i =1 n i α i be a root of g , and denote by g λ the rootspace of λ . In this section, we study some special elements in g , which we call elements in standardform . We prove that the root space g λ has a basis, of which the elements are in standard form.For simplicity, we only work on positive roots and the corresponding root space g λ is generated by { e i } ≤ i ≤ r . Definition 2.1.
An element g ∈ g is in standard form of length n if it can be written in the followingway [ e a n , [ e a n − , [ . . . [ e a , e a ] . . . ]]] , where 1 ≤ a i ≤ r for any integer r > e , [ e , e ]] is a standard form of length 3. [[ e , e ] , [ e , e ]] is anelement of length 4 but not in standard form. Lemma 2.2.
Given two elements
X, Y ∈ g in standard form of length i and j respectively, [ X, Y ] canbe written as a linear sum of elements in standard form of length i + j .Proof. Denote by k i an element of length i in standard form. Without loss of generality, let X = k i and Y = k j .We first start from the base case [ k , k n ]. Let k be [ e a , e a ], then we have[ k , k n ] = − [ k n , [ e a , e a ]] . Based on Jacobi Identity, one can write=[ e a , [ e a , k n ]] + [ e a , [ k n , e a ]]=[ e a , [ e a , k n ]] − [ e a , [ e a , k n ]] . Since k n is in standard form, the two elements above [ e a , [ e a , k n ]] and [ e a , [ e a , k n ]] are also instandard form of length n + 2.Then, we keep computing [ k , k n ] to find the pattern. Here, we let k = [ e a , [ e a , e a ]].[ k , k n ] = − [ k n , [ e a , [ e a , e a ]]]=[ e a , [[ e a , e a ] , k n ]] + [[ e a , e a ] , [ k n , e a ]]=[ e a , ([ e a , [ e a , k n ] − [ e a , [ e a , k n ])] + [ e a , [ e a , [ e a , k n ]]] − [ e a , [ e a , [ e a , k n ]]]=[ e a , [ e a , [ e a , k n ]] − [ e a , [ e a , [ e a , k n ]]] + [ e a , [ e a , [ e a , k n ]]] − [ e a , [ e a , [ e a , k n ]]]Clearly, four elements in the last equation are in standard form of length n + 3.We now conduct induction to prove [ k m , k n ] can be written as the sum of standard forms of length m + n . Assume m < n and k m = [ e a , k m − ]. Therefore, we can write[ k m , k n ] =[[ e a , k m − ] , k n ]=[ e a , [ k m − , k n ]] + [ k m − , [ k n , e a ]=[ k , [ k m − , k n ]] + [ k m − , k n +1 ] BOWEN CHEN, HANYI LUO AND HAO SUN
By induction, [ k m − , k n ] can be written as a sum of elements in standard forms of length m + n − e a , [ k m − , k n ]] can be written as a sum of elements in standard forms of length m + n . The same argument holds for [ k m − , k n +1 ]. Therefore, [ k m , k n ] can be written as a sum ofelements in standard forms of length m + n . (cid:3) Lemma 2.2 proves two elements in standard forms in Lie bracket can be written as the linear sumof standard forms. Then, in the theorem below, we are going to prove a more general result.
Theorem 2.3.
For any element
X, Y of length m and n respectively ( m, n > , [ X, Y ] can be writtenas the sum of standard forms of length l = m + n .Proof. We try to prove this inductively on the total length l . Note that n = l − m since the totallength is l . When l = 2, we have the base case[ X, Y ] = [ e a , e a ] , where [ e a , e a ] is a standard form of length 2.Suppose when l < l ′ , [ X, Y ] can be converted to a linear sum of elements in standard form of length l . We want to prove the statement holds when l = l ′ . In this case, X is an element of length m and Y is an element of length l ′ − m . Both have length less than l ′ which implies they can be converted toa linear sum of standard forms from the induction hypothesis. Because of the bilinear property of liebracket, without loss of generality, suppose that X and Y are elements in standard forms of length m and l ′ − m respectively. Based on the results from Lemma 2.2, then [ X, Y ] can be written as a linearsum of length l ′ , which proves the theorem. (cid:3) With the theorem above, we can write, for example, [[ e , e ] , [[ e , e ] , [ e , e ]]], which is an elementof length 6, as a linear sum of elements in standard forms of length 6.[[ e , e ] , [[ e , e ] , [ e , e ]]]=[[ e , e ] , [ e , [ e , [ e , e ]]] + [[ e , e ] , [ e , [ e , [ e , e ]]]]=[ e , [ e , [ e , [ e , [ e , e ]]] + [ e , [ e , [ e , [ e , [ e , e ]]]]]+[ e , [ e , [ e , [ e , [ e , e ]]]]] + [ e , [ e , [ e , [ e , [ e , e ]]]]] . As an application of Theorem 2.3, we have the following corollary.
Corollary 2.4.
Let λ = r P i =1 n i α i be a root of g . The root space g λ is spanned by elements in standardform. A Combinatorial Problem
In this section, we are introducing a combinatorial problem that is related to our main problem.
Lemma 3.1.
Suppose we have n non-distinguishable red balls and n non-distinguishable blue ballsputting into n boxes ( n , n , n ≥ , without exclusion (boxes can be empty). There are (cid:18) n + n − n − (cid:19)(cid:18) n + n − n − (cid:19) ways to arrange the balls.Proof. There are (cid:0) n + l − l − (cid:1) ways to put n non-distinguishable balls into l boxes without exclusion. Wecan prove the lemma based on this theorem above.(1) We first consider putting all red balls into the boxes regardless of the blue balls, which has (cid:0) n + n − n − (cid:1) ways.(2) Similarly, if we put all blue balls into the boxes, there are (cid:0) n + n − n − (cid:1) ways. OOT MULTIPLICITIES OF RANK 3 KAC-MOODY ALGEBRAS 5
Then we combine the situations together since putting red balls in the boxes and putting blue ones inthe boxes are two independent events. Hence, the total number of ways to put the balls are (cid:18) n + n − n − (cid:19)(cid:18) n + n − n − (cid:19) . (cid:3) Example 3.2.
We let n = 2 and l = 4 and see how we get (cid:0) n + l − l − (cid:1) as the number of ways to putwithout exclusion n balls into l boxes. We will explain how to derive the number.Let us first consider a sequence of numbers with the length of 5, which corresponds to n + l −
1, andwe have l − n = 2 balls to be put in all 5 positions. Clearly, we have (cid:0) (cid:1) = 10 waysin total. Put all the boards and balls into a sequence. Note that l − l parts. For example, ( a, b, b, a, b ) has four parts: with a in first part and third part (parts starting onthe left) and nothing in others. Therefore, each part corresponds to a box and the number of balls ineach part is the number of balls we will put into the corresponding box.This sequence corresponds to a possible case, several of which are listed as follow:(1) ( a, a, b, b, b )(2) ( a, b, a, b, b )(3) ( a, b, b, a, b )where a represents the ball, and b represents the board. We have several rules for these balls andboards.(1) Each interval between two b is a box, and the bracket and its closet b form a box.(2) The number of a inside each box is the number of balls that is put into the box.(3) The boxes can be empty.(4) The boxes are distinguishable, and the balls are non-distinguishable.When we look at our own example, we have 3 boards, which means we have 4 boxes, including theboxes bounded by the bracket. We have two balls to be put in four boxes. Thus, we transform thepermutation question into what we want to solve. Moreover, they are respectively corresponded to thesituations, several of which are listed as follow:(1) 2 0 0 0 corresponds to ( a, a, b, b, b )(2) 1 1 0 0 corresponds to ( a, b, a, b, b )(3) 1 0 1 0 corresponds to ( a, b, b, a, b )where the first number represents the number of balls in the first interval, and the second in the secondbox, and so on.Thus, the number of ways to put n balls into l boxes is equivalent to that to put l − n numbers, which is exactly (cid:0) n + l − l − (cid:1) .4. Dimension of Root Spaces
Properties of n-tuples.
Let A be a 3 × − a − a − a − a , where a and a are positive integers and at least one of them is larger than one. The correspondingKac-Moody algebra is denoted by A .For the special matrix A , we can write as follow four specific formulas from property (5) in thedefinition of Kac-Moody algebra (see § e ) a ( e ) = 0.(2) ad( e ) a ( e ) = 0.(3) ad( e ) a ( e ) = 0.(4) ad( e ) a ( e ) = 0. BOWEN CHEN, HANYI LUO AND HAO SUN (5) [ e , e ] = [ e , e ] = 0.We will use these properties frequently when proving our main result.Now we fix a root λ = P i =1 n i α i . Denote by g λ the root space. Given an element in standard form[ e a n , [ e a n − , [ . . . , [ e a , e a ] . . . ]]] ∈ g λ , it can be associated to an n -tuple ( a n , a n − , . . . , a ) naturally, where • ≤ a i ≤ ≤ i ≤ n ; • for each positive integer j , |{ a i | a i = j, ≤ i ≤ n }| = n j .Conversely, for each n -tuple ( a n , a n − , . . . , a ), the corresponding element in standard form is[ e a n , [ e a n − , [ . . . , [ e a , e a ] . . . ]]] . Therefore, any element in standard form corresponds to an n -tuple as the discussion above, and wewill use the n -tuples to represent the elements in standard form. If we want to find the dimension of g λ , we have to find a maximal linearly independent set in g λ . By Theorem 2.3, we can assume thatthe elements in this set are in standard form. In this case, the n -tuples is more convenient to workwith. However, if we work on n -tuples, there are some problems, when we do the calculations: • Several distinct n -tuples may correspond to linearly dependent elements in g λ . • A non-trivial n -tuple may correspond to a trivial element in g λ .In this section, we calculate the dimension of root spaces by solving the problems above. Definition 4.1. An n -tuple is nontrivial , if the corresponding element in standard form is nontrivial.A collection of n -tuples are linearly independent if their corresponding standard forms are linearlyindependent. Lemma 4.2.
Let ( a n , . . . , a ) be a nontrivial n -tuple. Then, it is linearly equivalent to an n -tuple ( b n , . . . , b ) such that b = 2 .Proof. By property (5), we know that [ e , e ] = 0. Therefore, one of a and a should be “2” to makethe element non-trivial. Now we assume that a = 2. Since[ e , e a ] = − [ e a , e ] , the n -tuples ( a n , . . . , a , a ) and ( a n , . . . , a , a ) are linearly equivalent. This finishes the proof of thislemma. (cid:3) By this lemma, we can always assume that the first element in any nontrivial n -tuple is “2”. Lemma 4.3.
Let n ≥ be a positive integer. For any nontrivial n -tuple ( a n , . . . , p, q, . . . , a , a ) such that a i +1 = p , a i = q and p, q = 1 or , this element is linear equivalent to the element ( a n , . . . , q, p, . . . , a , a ) in standard form of the same length, where we swap the position of p and q .Proof. We will first prove elements in the following cases( p, q, a i − , . . . , a , a ) , where a i +1 = p and a i = q . It is easy to check that the argument holds when i = 2 ,
3. We nowfocus on elements in standard form of length greater than 4. We consider the [ e p , [ e q , A ]], where A = [ e a i − , [ . . . ., [ e a , e a ] . . . ] is an element in standard form of length i − e p , [ e q , A ]] = − [ e q , [ A, e p ]] − [ A, [ e p , e q ]]As we know [ e , e ] = 0 given by property (5), and [ e i , e i ] = 0 ( i = 1 , e p , [ e q , A ]] = [ e q , [ e p , A ]].The same proof holds for the general case ( a n , . . . , p, q, . . . , a ). (cid:3) OOT MULTIPLICITIES OF RANK 3 KAC-MOODY ALGEBRAS 7
Example 4.4.
Take [ e , [ e , [ e , [ e , e ]]] as an example. Let us find the number of elements that areequal to [ e , [ e , [ e , [ e , e ]]]]. By applying Jacobi Identity, we can get[ e , [ e , [ e , [ e , e ]]]]=[ e , [ e , [ e , [ e , e ]]]]=[ e , [ e , [ e , [ e , e ]]]] Lemma 4.5.
Let ( a n , . . . , a i , , . . . , a ) be a nontrivial n -tuple such that • a i is either “1” or “3” ; • there are at least three “2” on the right hand side of a i .Then, ( a n , . . . , a i , , . . . , a ) is linearly independent to ( a n , . . . , , a i , . . . , a ) .Proof. Similar to proof of Lemma 4.3, we reduce the n -tuple to the following one ( a i , , . . . , a ) andwe will prove that ( a i , , . . . , a ) is linearly independent to (2 , a i , . . . , a ).Let A = [ e a i − , [ . . . , [ e a , e a ] . . . ]]. Based on Jacobi identity, we have[ e a i , [ e , A ]] = [ e , [ e a i , A ] − [ A, [ e a i , e ]]From the equation above, we just need to check whether [ A, [ e a i , e ]] is zero. Clearly, it is zero onlyunder some specific situations listed as follows: • e a i = e • A = [ e a i , e ]Since the situations described above are not possible under the assumptions given by the lemma,the two elements, [ e a i , [ e , A ]] and [ e , [ e a i , A ]], are linearly independent. So are the two n -tuples( . . . , a i , , . . . ) and ( . . . , , a i , . . . ). (cid:3) Remark 4.6.
In this remark, we want to clarify the exceptions mentioned in Lemma 4.3. When e a i = e , it is obvious that ( . . . , a i , , . . . ) = ( . . . , , a i , . . . ) . When A = [ e a i , e ] , we have the followingexample [ e , [ e a i , [ e a i , e ]]]=[ e a i , [ e , [ e a i , e ]]] + [[ e a i , e ] , [ e , e a i ]]=[ e a i , [ e , [ e a i , e ]]] In this situation, we find that e a i and e are able to exchange their positions.Remark . Given an n -tuple ( a n , . . . , a ), we highlight the elements “2” in this n -tuple as follows( . . . , , . . . , , . . . , , . . . , , where “ . . . ” represents sequences of “1” and “3”. Lemma 4.5 tells us that the role of “2” is like aboard, and the elements “1” and “3” cannot go across the board. If we switch “1” (or “3”) with “2”(under the conditions in Lemma 4.5), then we will get a new linearly independent element. In themeantime, by Lemma 4.3, switching the positions of two adjacent “1” and “3” does not give us newlinearly independent tuples. More precisely, if e i and e j ( i, j = 1 ,
3) are between two “adjacent” e ,then exchanging the positions of e i and e j does not give us new element in standard form. Here, two“adjacent” e means that there is no e between them. This observation is very important for thecalculation of the dimension of root spaces.As we discussed above, to calculate the dimension of the root space g λ , it is equivalent to findthe largest set of linearly independent elements in standard form. Compared with the combinatorialproblem we discussed in §
3, we can take e and e as the red and blue balls respectively, and take e as the board. Since λ = n α + n α + n α , it is equivalent to consider the combinatorial problemthat there are n red balls, n blue balls and n − BOWEN CHEN, HANYI LUO AND HAO SUN
By Lemma 4.2, we can assume that the first element in an n -tuple is “2”, and by Remark 4.7, weknow that the elements “2” break the n -tuple into intervals. The i -th interval in a given n -tuple is theone between the i -th “2” and the ( i + 1)-th “2”. In Lemma 4.5, we made the assumption that thereare at least two “2” on the right hand side of a i in the following n -tuple( a n , . . . , , a i , . . . , a ) . Therefore, elements in the first interval cannot swap with the second “2” in the n -tuple. In the nextlemma, we make a careful discussion of all possible cases of the numbers of “1” and “3” in the firstinterval. Lemma 4.8.
Denote by n ′ and n ′ the numbers of “1” and “3” in the first interval of a given nontrivial n -tuple. Then, we have n ′ ≤ a and n ′ ≤ a .Proof. Based on the property (5), if n ′ ≤ a , n ′ ≤ a , we know the elements corresponding to thesetuples (1 , , . . . , | {z } n ′ ,
2) and (3 , , . . . , | {z } n ′ ,
2) are trivial elements, where n ′ ≥ a and n ′ ≥ a respectively. To avoid the trivial cases, we should have n ≤ a and n ≤ a . (cid:3) General Case.
We first give a general calculation for the case that the number of “1” is i andthe number of “3” is j in the first interval based on Lemma 4.2 and 4.5. Then, we sum over all possiblecases based on Lemma 4.8. Note that the elements in standard form corresponding to those n -tuplesspan the root space g λ . Thus, we give an upper bound of the dimension of g λ .By Lemma 4.8, we assume that the number of “1” is i and the number of “3” is j in the firstinterval, where i ≤ a and j ≤ a . Then, the rest number of “1” is n − i , and the rest number of “3”is n − j . All of these numbers will be arranged into n − (cid:18) n + n − i − n − (cid:19)(cid:18) n + n − j − n − (cid:19) ways in this situation. By summing over all possible situations in the first interval, we have X ≤ i ≤ a , ≤ j ≤ a i,j ) =(0 , (cid:18) n + n − i − n − (cid:19)(cid:18) n + n − j − n − (cid:19) ways in total.4.3. Special Cases. In § g λ by summing over allpossible n -tuples as discussed in § n -tuples and some n -tuples are linearly dependent. In this subsection, we make a careful discussion of these two cases.4.3.1. Trivial Tuples.
In this special case, the first interval only has one element, and then there areat least a i − a i depends on the number “ i ” in thefirst interval. More precisely, the sequences are given as follows,( . . . , , . . . , | {z } a , ,
2) or ( . . . , , . . . , | {z } a , , , where the number of “2” are at least a and a respectively. Recall the property (5): ad( e ) a ( e )= ad( e ) a ( e ) = 0. Thus, the two sequences we wrote above, namely ( . . . , , . . . , | {z } a , ,
2) and( . . . , , . . . , | {z } a , , Lemma 4.9.
Suppose that n ≥ max { a , a } . There are (cid:18) n + n + n − a − n − a − (cid:19) + (cid:18) n + n + n − a − n − a − (cid:19) OOT MULTIPLICITIES OF RANK 3 KAC-MOODY ALGEBRAS 9 n -tuples counted in § n -tuples are in the followingform ( . . . , , . . . , | {z } a , , , ( . . . , , . . . , | {z } a , , . Proof.
Since the number of “2” in the front should be least a i , there will have a i intervals reducedfrom the right hand side, where i depends on the number in the first interval. Then the total numberof the rest of intervals is n − a i . Eliminating the first interval, the number of e is n − n .By Lemma 3.1, there will be (cid:18) ( n − n ) + ( n − a − n − a − (cid:19) + (cid:18) ( n − n ) + ( n − a − n − a − (cid:19) = (cid:18) n + n + n − a − n − a − (cid:19) + (cid:18) n + n + n − a − n − a − (cid:19) possible cases in the first special case. (cid:3) Linearly Dependent Tuples.
In Remark 4.6, we give an example to explain why we have theextra conditions in Lemma 4.5. This example tells us that the n -tuples we counted in § e , [ e a i , [ e a i , e ]]] = [ e a i , [ e , [ e a i , e ]]] . Therefore, in the situation above, e or e can exchange their position with e , leading to the repetitionof many different permutations of elements. For example, (1 , , , ,
2) and (1 , , , ,
2) are linearlyequivalent. It is easy to check that the n -tuples ( . . . , , i, i,
2) and ( . . . , i, , i,
2) are linearly equivalent,where i = 1 or 3. However, we count both of them in § n -tuples in the form of ( . . . , , i, i, Lemma 4.10.
Suppose that n , n , n ≥ and a i > , there are (cid:18) n + n − n − (cid:19)(cid:18) n + n − n − (cid:19) + (cid:18) n + n − n − (cid:19)(cid:18) n + n − n − (cid:19) situations with the form ( . . . , , i, i, , where i is either or .Proof. In the proof, we will only discuss the e case, as the e case shares the same reasoning. Notethat there is always an n -tuple of the form ( ..., , ,
2) where there is at least one “1” in the secondinterval such that it corresponds to a linearly dependent n -tuple of the form ( ..., , , , , , , , ,
2) and (3 , , , , ,
2) are linearly dependent. Therefore, we can easily conclude that each n -tuple of the form ( ..., , , ,
2) can be linearly dependent to an n -tuple of the form ( ..., , , n -tuple isof the form ( ..., , , , e case can derived in the same way, the total numberto be reduced from second special case is (cid:18) n + n − n − (cid:19)(cid:18) n + n − n − (cid:19) + (cid:18) n + n − n − (cid:19)(cid:18) n + n − n − (cid:19) . (cid:3) Main Result.
We define the following numbers: A = X ≤ i ≤ a , ≤ j ≤ a i,j ) =(0 , (cid:18) n + n − i − n − (cid:19)(cid:18) n + n − j − n − (cid:19) B = (cid:18) n + n − n − (cid:19)(cid:18) n + n − n − (cid:19) + (cid:18) n + n − n − (cid:19)(cid:18) n + n − n − (cid:19) C = (cid:18) n + n + n − a − n − a − (cid:19) , C = (cid:18) n + n + n − a − n − a − (cid:19) With the discussion in § Theorem 4.11.
Suppose that a ≤ a and n , n , n ≥ . Let λ = P i =1 n i α i . Denote by g λ thecorresponding root space. We have dim g λ = A − B, n < min { a , a } A − B − C , a ≤ n < a A − B − C − C , n ≥ max { a , a } Proof.
The number A is the total number in the general case as we discussed in § B corresponds to the special case we discussed in § C and C comes from § (cid:3) Remark 4.12.
In this remark, we consider the general case that n i ≥ . Note that if n i = 0 , then wereduce to a Kac-Moody algebra g ′ , of which the associated generalized Cartan matrix is a × matrix.Therefore, we assume that n i ≥ for i = 1 , , .Let n = 1 . by Lemma 4.2, we know that we can place the only e at the beginning of the n -tuple.Then, by Lemma 4.5, we know that if the root space g λ is nonempty, then dim g λ = 1 .Now we consider n = 1 and n = 1 . The case of n = 1 can be discussed similarly. Let g ′ := g ( A ′ ) be the Kac-Moody algebra associated to the matrix A ′ = (cid:18) − a − a (cid:19) , and g ′ is a hyperbolic Kac-Moody algebra. Let λ ′ = n α + n α . If we forget the only e , then thedimension of g λ can be calculated from the dimension of g ′ λ ′ (see [6] ). Therefore, the only thing we haveto consider is the position of the “1” in the n -tuple. With a similar discussion as we did in Lemma4.8, we can get a similar formula for dim g λ in this case. A Special Case of Hyperbolic Kac-Moody Algebras.
People have made a full classificationof hyperbolic Kac-Moody algebras. Some rank-3 Hyperbolic Kac-Moody algebras are defined by thegeneralized Cartan matrix we consider in this paper (see § a = 1 and a = 2. The corresponding Kac-Moody algebra is a hyperbolic Kac-Moody algebra(see [1, 7]). With the same discussion as in § A := X ≤ i ≤ , ≤ j ≤ i,j ) =(0 , (cid:18) n + n − i − n − (cid:19)(cid:18) n + n − j − n − (cid:19) . For the special case of trivial tuples, we have C + C := (cid:18) n + n + n − n − (cid:19) + (cid:18) n + n + n − n − (cid:19) many situations, in which n ≥ max { a , a } and n , n ≥ OOT MULTIPLICITIES OF RANK 3 KAC-MOODY ALGEBRAS 11
For the special case about linearly dependent tuples, since a = 1, we will only consider e in thisexample. Therefore, the total number of elements for the special case is B := (cid:18) n + n − n − (cid:19)(cid:18) n + n − n − (cid:19) , where n , n , n ≥ Corollary 4.13.
Suppose that n , n , n ≥ . Let λ = P i =1 n i α i . Denote by g λ the correspondingroot space. We have dim g λ = A − B, n < A − B − C , n = 3 A − B − C − C , n ≥ . Acknowledgements
We are thankful to great opportunity provided by Pioneer China Program (PCP).
References [1] Carbone, L., Chung, S., Cobbs, L., McRae, R., Nandi, D., Naqvi, Y., Penta, D. (2010). Classification of hyperbolicDynkin diagrams, root lengths and Weyl group orbits. Journal of Physics A: Mathematical and Theoretical, 43(15),155-209.[2] Frenkel, I. B. (1985). Representations of Kac–Moody algebras and dual resonance models, Applications of group the-ory in physics and mathematical physics (Chicago, 1982), Lectures in Appl. Math., vol. 21, American MathematicalSociety, Providence, RI, 325–353.[3] Humphreys, J. E. (2012). Introduction to Lie algebras and representation theory (Vol. 9). Springer Science &Business Media.[4] Kac, V. G. (1990). Infinite-dimensional Lie algebras. Cambridge University Press.[5] Kang, S. J. (1994). Root multiplicities of Kac-Moody algebras. Duke Mathematical Journal, 74(3), 635-666.[6] Kang, S. J., Lee, K. H., Lee, K. (2017). A combinatorial approach to root multiplicities of rank 2 hyperbolicKac–Moody algebras. Communications in Algebra, 45(11), 4785-4800.[7] Sa¸clioglu, C. (1989). Dynkin diagrams for hyperbolic Kac-Moody algebras. Journal of Physics A: Mathematical andGeneral, 22(18), 3753.
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