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Research article

Hilbert series of mixed braid monoid MB2,2

  • Received: 30 April 2022 Revised: 03 July 2022 Accepted: 12 July 2022 Published: 20 July 2022
  • MSC : 20F36, 20F05, 13D40

  • Hilbert series is a simplest way to calculate the dimension and the degree of an algebraic variety by an explicit polynomial equation. The mixed braid group Bm,n is a subgroup of the Artin braid group Bm+n. In this paper we find the ambiguity-free presentation and the Hilbert series of canonical words of mixed braid monoid MB2,2.

    Citation: Zaffar Iqbal, Xiujun Zhang, Mobeen Munir, Ghina Mubashar. Hilbert series of mixed braid monoid MB2,2[J]. AIMS Mathematics, 2022, 7(9): 17080-17090. doi: 10.3934/math.2022939

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  • Hilbert series is a simplest way to calculate the dimension and the degree of an algebraic variety by an explicit polynomial equation. The mixed braid group Bm,n is a subgroup of the Artin braid group Bm+n. In this paper we find the ambiguity-free presentation and the Hilbert series of canonical words of mixed braid monoid MB2,2.



    Hilbert series helps us in expressing the growth of dimension of homogenous components of a graded algebra. Presently, these ideas are extended to graded algebras, filtered algebras, and graded or filtered modules over these algebras, monoid in commutative algebra and coherent sheaves over projective schemes in algebraic geometry. The Hilbert series is a particular case of the Hilbert-Poincaré series of a graded vector space. It is also of a great significance in computational algebraic geometry due to the easiest way for computing the dimension and the degree of an algebraic variety defined by explicit polynomial equations. The Hilbert series is also helpful in providing us the important invariants of the algebraic varieties. This particular article focuses on the presentation and Hilbert series of the mixed braid monoid MB2,2.

    The classical braid group Bm+1, given by Artin [3], possesses the following presentation:

    Bm+1=a1,,am|aiaj=ajaiifij2ai+1aiai+1=aiai+1aiif1im1.

    The braid monoid MBm+1 has the same presentation as the braid group Bm+1. The braid group Bm+1 admits another presentation known as the band presentation given by Birman et al. in [5].

    Definition 1. [7] Let U be a set. A square matrix G=(guv)u,vU is calledCoxeter matrix over U such that muu=1 and muv=mvu{2,3,...,} for all u,vU,uv.

    Definition 2. [7] A Coxeter graph Γ is a labeled graph. It can be defined as

    i. A set of vertices of Γ is denoted by U.

    ii. Two vertices u,vU,uv are connected by an edge if muv3. The labeling of edge is muv if muv4.

    Definition 3. [7] Let M=(mst)s,tS be the Coxeter matrix of the Coxetergraph Γ. Then the group defined by

    W=sSs2=1,(st)mst=1foralls,tS,st,mst

    is called Coxeter group(of type Γ).

    Definition 4. [12] In mixed braid group Bm,n the first index m denotes trivialstrings in the braid group and the next n strings shows thebraiding by itself and with the m strings. The mixed braid group Bm,n generated by m+n strands is defined by

    Bm,n=α1,,αm,σ1,,σn1|σiσj=σjσiifij2σi+1σiσi+1=σiσi+1σiif1in1αiσj=σjαiifj2and1imαiσ1αiσ1=σ1αiσ1αiif1imαl(σ1αmσ11)=(σ1αmσ11)αlifl<m.

    If we remove the last n strands of Artin group Bm+n then wehave only identity braid with m strands. The collection of allsuch elements of Artin group Bm+n will be denoted by Bm,n.

    Definition 5. [12] The mixed braid monoid MBm,n has the following presentation:

    MBm,n=α1,,αm,σ1,,σn1|σiσj=σjσi,ij2σi+1σiσi+1=σiσi+1σiif1in1αiσj=σjαi,j2and1imαiσ1αiσ1=σ1αiσ1αi,1im.

    In the below diagram of mixed braid monoid MBm,n, the single bonds shows relation of degree 3, the double bonds shows relation of degree 4 and two generators commute if they are not joined by a bond (see Figure 1).

    Figure 1.  Coxeter graph of MBm,m.

    Conveniently αi and σj are denoted by a single generator ak,1ki+j. Note that the mixed braid monoid MB1,2, which is isomorphic to the Artin monoid of type B2, is presented by

    MB1,2=a1,a2|a2a1a2a1=a1a2a1a2.

    The Coxeter diagram for MB1,2 takes the form in Figure 2.

    Figure 2.  Coxeter graph of MB1,2.

    The complete presentation and Hilbert series for MB1,2 are computed in [13]. This motivated us to compute the Hilbert series of MB2,2.

    Definition 6. The classical presentation of MB2,2 is given by

    MB2,2=a1,a2,a3|a2a1a2a1=a1a2a1a2,a3a2a3a2=a2a3a2a3,a3a1=a1a3.

    Let us denote the relations by

    R0:a3a1=a1a3,R1:a2a1a2a1=a1a2a1a2

    and

    R2:a3a2a3a2=a2a3a2a3.

    The Coxeter diagram of MB2,2 is given as Figure 3.

    Figure 3.  Coxeter graph of MB2,2.

    An example of a braid in B2,2 (see Figure 4).

    Figure 4.  braid in B2,2.

    In 2003, Bokut [6] gave the non-commutative complete presentation of the braid monoid MBn+1 with the length-lexicographic order induced by a1<<an. In [10] we computed the Hilbert series of braid monoid MB4 in band generators. In 1972, P. Deligne [8] proved that the Hilbert series of all the Artin monoids are rational functions. In [14] Saito computed the growth series of Artin monoids. In [13] we computed the Hilbert series of the Artin monoids M(I2(p)), where M(I2(4)) is isomorphic to MB1,2. In [7] we showed that the growth rate of all the Artin monoids is less than 4. In [11] we proved that the growth rates are exponential and the growth rates for MB3 and MB4 are approximately 1.618 and 2.0868. In this paper we construct a linear system for canonical words and then we find Hilbert series and the growth rate of the mixed braid monoid MB2,2. It is proved in [1] that the Hilbert series of the free G-graded F-algebra is a rational function. In [2] the authors proved that the Hilbert series of relatively free algebras is a rational function.

    To get a canonical form of a word in an algebra the diamond lemma by G. Bergman [4] is very useful. To understand the concepts of ambiguities and canonical words, we start with his terminology.

    Definition 7. Let X be a nonempty set and X be the free monoid on X. Let w1 and w2X, where w1=x1x2xr, w2=y1y2yr with xi,yiX. Then w1<w2 length-lexicographically if there is a kr such that xk<yk and xi=yi for all i<k.

    Definition 8. Let μ=ν be a relation in a monoid M and μ1=sw and μ2=wt be two words in M. Then the word of the form μ1×wμ2=swt is said to be anambiguity.

    If μ1t=sμ2 (in the length-lexicographic order) then we say that the ambiguity swt is solvable. A presentation of M is said to be complete if and only if all the ambiguities are solvable. Corresponding to the relation μ=ν, the changes αμβανβ give a rewriting system. A complete presentation is equivalent to a confluent rewriting system. In a complete presentation of a monoid a word containing μ will be called reducible word and a word that does not contain μ will be called an irreducible word or canonical word. For example a3a2a3=a2a3a2 is a basic relation in the braid monoid MB4. A word v=a23a2a3 contains α=a3a2a3. Hence v is a reducible word. Then a23a2a3=a2a3a22 and a2a3a22 is the canonical form of v. In a presentation of a monoid we fix a total order a1<a2<<an on the generators.

    Definition 9. [9] Let G be a finitely generated group and A be a finite set ofgenerators of G. The word length lA(g) of an element gG is the smallest integer n for which there exist a1,,anAA1 such that g=a1an.

    Definition 10. [9] The Hilbert function of a monoid M is given as H(M,n)=an, i.e., the number of elements of M of word length n. The Hilbertseries of the monoid MBn for arbitrary variable t isdenoted by HM(t) and is defined by HM(t)=n=0antn.

    Definition 11. For a sequence of positive numbers {bl}l1 the rate ofgrowth rR is given by

    ¯limlexp(logbll)=r.

    Lemma 1. [13]The following equations hold for the canonical words in MI2(4).

    (1) P(4)1=t(1t)+t(1t)P(4)2

    (2) P(4)2=t+tP(4)2+P(4)21

    (3) P(4)21=tP(4)1t2P(4)21t31tP(4)212

    (4) P(4)212=t2P(4)2t2P(4)21.

    The solution of the above equations is given in [13] which is as follows:

    P(3)1=t(1t)(1tt2t3),P(3)2=t(1+t+t2)1tt2t3,P(3)21=t2(1+t)1tt2t3,P(3)212=t31tt2t3.

    Lemma 2. [13]The Hilbert series of MI2(4) is given by

    P(3)M(t)=1(1t)(12t).

    While solving all the ambiguities we now give a complete presentation of MB2,2.

    Theorem 1. The braid monoid MB2,2 has a complete presentation

    MB2,2=a1,a2,a3|R0,R1,,R6,

    where

    R(4)0:a3a1=a1a3,R(4)1:a2a1a2a1=a1a2a1a2,R(4)2:a2an+11a2a1a2=a1a2a1a22an1,R(4)3:a3a2a3a2=a2a3a2a3,R(4)4:a3an+12a3a2a3=a2a3a2a23an2,R(4)5:a3a2an1a3a2a1a2=a2a3a2a1a3a2an1,R(4)6:a3an+12a3a2am1a3=a2a3a2a23an2am1,

    where n,m0.

    Proof. We denote the ambiguity formed by left sides of the relations Ri and Rj in MB2,2 by RiRj=swt (say). If in the ambiguity z=swt, L(z)=(sw)t and R(z)=s(wt) are different lexicographically, then we get a new relation in MB2,2 and if L(z)=(sw)t and R(z)=s(wt) are reduced to an identical word, then we say that ambiguity is solvable and no relation is formed. Here L(z) denotes the canonical form of (sw)t and R(z) denotes the conical form of s(wt). The above relations are obtained by solving the ambiguities involving the relations R(4)0, R(4)1 and R(4)2 and the new relations. In [13] we computed the relation (for p=4) R(4)3, which is given by

    R(4)3:a2an+11a2a1a2=a1a2a1a22an1.

    For an ambiguity R(4)2R(4)2=a3a2a3a2a3a2=w1(say), we have

    R(w1)=a3a2a3a2a3a2_=a3a22a3a2a3,L(w1)=a3a2a3a2_a3a2=a2a3a2a23a2.

    Hence we have a relation Rw1:a3a22a3a2a3=a2a3a2a23a2. Again by solving a new ambiguity Rw1R(4)2=a3a22a3a2a3a2=w2 we have

    R(w2)=a3a22a3a2a3a2_=a3a32a3a2a3,L(w2)=a3a22a3a2a3_a2=a2a3a2a23a22.

    which gives another relation Rw2:a3a32a3a2a3=a2a3a2a23a22. By continuing the same process we have the general relation

    R(4)4:a3an+12a3a2a3=a2a3a2a23an2,n1.

    In the ambiguity R(4)2R(4)1=a3a2a3a2a1a2a1=w3, we have

    R(w3)=a3a2a3a2a1a2a1_=a3a2a1a3a2a1a2,L(w3)=a3a2a3a2_a1a2a1=a2a3a2a1a3a2a1.

    Hence we have a relation Rw3:a3a2a1a3a2a1a2=a2a3a2a1a3a2a1. Again by solving a new ambiguity Rw3R(4)1=a3a2a1a3a2a1a2a1=w4 we have

    R(w4)=a3a2a1a3a2a1a2a1_=a3a2a21a3a2a1a2,
    L(w4)=a3a2a1a3a2a1a2_a1=a2a3a2a1a3a2a21,

    which gives another relation Rw4:a3a2a21a3a2a1a2=a2a3a2a1a3a2a21. By continuing the same process we have the general relation

    R(4)5:a3a2an1a3a2a1a2=a2a3a2a1a3a2an1,n1.

    In the ambiguity R(4)4R(4)0=a3an+12a3a2a3a1=w5, we have

    R(w5)=a3an+12a3a2a3a1_=a3an+12a3a2a1a3,L(w5)=a3an+12a3a2a3_a1=a2a3a2a23an2a1.

    Hence we have a relation Rw5:a3an+12a3a2a1a3=a2a3a2a23an2a1. Again by solving a new ambiguity Rw5R(4)0=a3an+12a3a2a3a1a3a1=w6 we have

    R(w6)=a3an+12a3a2a3a1a3a1_=a3an+12a3a2a21a3,
    L(w6)=a3an+12a3a2a3a1a3_a1=a2a3a2a23an2a21,

    which gives another relation Rw6:a3an+12a3a2a21a3=a2a3a2a23an2a21. By continuing the same process we have the general relation

    R(4)6:a3an+12a3a2am1a3=a2a3a2a23an2am1,n,m1.

    All other ambiguities all solvable. Hence we have the complete set of relations.

    For our convenience we use these notations for canonical and reducible words in MB2,2. In general B(4) denotes the set of reducible words and A(4) denotes the set of canonical words in MB2,2. Particularly, B(4)i denotes the reducible word of R(4)i in MB2,2 and A(n+m)j(j1)k denotes collection of all canonical words in MBm,n starting with ajaj1ak. These sets are graded by length-lexicographic order, so we can compute the Hilbert series of these sets. The words Xa3×3a3Y and Xa3a2×32a3a2Y are equivalent to products Xa3Y and Xa3a2Y, respectively. Let Q(4) denote the Hilbert series for B(4) and P(4) denote the Hilbert series for A(4). Let H(m+n)M(t) denote the Hilbert series for the set

    A(m+n)=eA(m+n)1A(m+n)2A(m+n)k

    (for some kZ) for a monoid M. Then we have

    H(m+n)M(t)=1+P(m+n)1+P(m+n)2++P(m+n)k.

    Theorem 2. The following relations hold for the Hilbert series of reducible words in MB2,2.

    a.Q(4)1=t4,b.Q(4)2=t61t,c.Q(4)3=t4,d.Q(4)4=t61t,e.Q(4)5=t71t,f.Q(4)6=t7(1t)2.

    Proof. Let α(i,,j) be a wordin ai,ai+1,,aj in MB2,2, then Σα(i,,j) be a word in ai+1,ai+2,,aj+1 in MB2,2. B(4) denotes the reducible wordsin MB2,2 corresponding to relation R(4). Q(4) denotes the Hilbert series of B(4). Hilbertseries of A(2)1={a1,a21,} is P(2)1=t+t2+=t1t. Therefore we have

    a. B(4)1={a2a1a2a1}. This implies Q(4)1=t4.

    b. Since B(4)2={a2an+11a2a1a2}={a2a1}×A(2)1×{a2a1a2}. Hence Q(4)2=t61t.

    c. Similarly B(4)3={a3a2a3a2} implies Q(4)3=t4.

    d. The decomposition B(4)4={a3an+12a2a2a3}={a3a2}×ΣA(2)1×{a3a2a3} gives Q(4)4=t61t.

    e. B(4)5={a3a2am1a3a2a1a2}={a3a2}×A(2)1×{a3a2a1a2} gives Q(4)5=t71t.

    f. B(4)6={a3an+12a2a2am1a3}={a3a2}×ΣA(2)1×{a3a2}×A(2)1×a3 gives Q(4)6=t7(1t)2.

    Next we construct a linear system for canonical forms in MB2,2.

    Theorem 3. The following relations hold for the Hilbert series of canonicalwords in MB2,2.

    (1) P(4)1=P(3)1+P(3)1P(4)3.

    (2) P(4)2=P(3)2+P(3)2P(4)3.

    (3) P(4)3=t+tP(4)3+P(4)32.

    (4) P(4)21=P(3)21+P(3)21P(4)3.

    (5) P(4)32=tP(4)2t2P(4)32t31tP(4)323t31tP(4)3212t6(1t)2P(4)3.

    (6) P(4)212=P(3)212+P(3)212P(4)3

    (7) P(4)323=t2P(4)3t2P(4)32.

    (8) P(4)3212=tP(4)212.

    Proof. As we have already defined A(4) denotes the set of canonical words in MB2,2 and B(4) denotes the set of reducible words in MB2,2. Now we define the corresponding Hilbert series. Let Q(4) denotes the Hilbert series for B(4) and P(4) denotes the Hilbert series for A(4). Then using the set decomposition we have

    (1) The decomposition of A(4)1 can be given as A(4)1=A(3)1(A(3)1×A(4)3). This implies Relation 1.

    (2) It follows directly from A(4)2=A(3)2(A(3)2×A(4)3).

    (3) The set A(4)3 can be written as A(4)3={a3}(a3×A(4)3)A(4)32. This results the Relation 2.

    (4) It follows from A(4)21=A(3)21(A(3)21×A(4)3).

    (5) The set A(4)32 is decomposed as

    A(4)32={a3}×A(4)2(B(4)3×32A(4)32)(B(4)323×323A(4)323)(B(4)5×3212A(4)3212)(B(4)6×3A(4)3)

    which gives the required equation.

    (6) The decomposition of A(4)212 can be written as A(4)212=A(3)212(A(3)212×A(4)3). This implies Relation 6.

    (7) It follows from A(4)323={a3a2}×A(4)3(B(4)3×32A(4)32).

    (8) The decomposition of A(4)3212={a3}×A(4)212 immediately gives us Relation 8.

    Theorem 4. Hilbert Series for the canonical words in MB2,2 is given as

    HMB2,2(t)=12t+2t22t3+t4+t6(1t)(14t+5t25t3+6t43t5+3t64t7t8t9+t10).

    Proof. As we have already computed the Hilbert series of canonical words and reducible words in Theorem 3 and in Theorem 2, respectively. Now we construct a linear system forcanonical words in MB2,2. For this we put the values of P(3)1,P(3)2,P(3)21 and P(3)212 form Lemma 1 in Theorem 3, we get following system of equations:

    (1) P(4)1=t(1t)(1tt2t3)+t(1t)(1tt2t3)P(4)3

    (2) P(4)2=t(1+t+t2)(1tt2t3)+t(1+t+t2)(1tt2t3)P(4)3

    (3) P(4)3=t+tP(4)3+P(4)32

    (4) P(4)21=t2(1+t)(1tt2t3)+t2(1+t)(1tt2t3)P(4)3

    (5) P(4)32=tP(4)2t2P(4)32t31tP(4)323t31tP(4)3212t6(1t)2P(4)3

    (6) P(4)212=t31tt2t3+t31tt2t3P(4)3

    (7) P(4)323=t2P(4)3t2P(4)32

    (8) P(4)3212=tP(4)212.

    Now we use Matrix Inversion Method to solve the system of these equations. Writing above equations in matrix form AX=B we have

    A=(10t(1t)(1tt2t3)0000001t(1+t+t2)(1tt2t3)00000001t0100000t2(1+t)(1tt2t3)100000tt6(1t)201+t20t31tt31t00t31tt2t30010000t20t201000000t01),X=(P(4)1P(4)2P(4)3P(4)21P(4)32P(4)212P(4)323P(4)3212),B=(t(1t)(1tt2t3)t(1+t+t2)(1tt2t3)tt2(1+t)(1tt2t3)0t3(1tt2t3)00).

    By solving the above system of equations, we get

    P(4)1=t2t2+2t32t4+t5+t7(1t)(14t+5t25t3+6t43t5+3t64t7t8t9+t10),P(4)2=tt2+t32t4+t5t6+2t7+t8+t914t+5t25t3+6t43t5+3t64t7t8t9+t10,P(4)3=t2t2+2t33t4+2t5t6+2t7t1014t+5t25t3+6t43t5+3t64t7t8t9+t10,P(4)21=t2t3t6+t7+t8+t914t+5t25t3+6t43t5+3t64t7t8t9+t10,P(4)32=t32t4+2t52t6+t7+t914t+5t25t3+6t43t5+3t64t7t8t9+t10,P(4)212=t9+2t8t5t3+t214t+5t25t3+6t43t5+3t64t7t8t9+t10,P(4)323=t33t4+3t53t6+3t7t8+2t92t10t11t1214t+5t25t3+6t43t5+3t64t7t8t9+t10,P(4)3212=t42t5+2t62t7+t8+t1014t+5t25t3+6t43t5+3t64t7t8t9+t10.

    Hence the Hilbert series of MB2,2 is computed as

    HMB2,2(t)=1+P(4)1+P(4)2+P(4)3=1+t2t2+2t32t4+t5+t7(1t)(14t+5t25t3+6t43t5+3t64t7t8t9+t10)+tt2+t32t4+t5t6+2t7+t8+t914t+5t25t3+6t43t5+3t64t7t8t9+t10+t2t2+2t33t4+2t5t6+2t7t1014t+5t25t3+6t43t5+3t64t7t8t9+t10=12t+2t22t3+t4+t6(1t)(14t+5t25t3+6t43t5+3t64t7t8t9+t10)=1+3t+8t2+21t3+53t4+132t5+327t6+807t7+1988t8+.

    In the following remark we find the growth rate of MB2,2. Using Maple we convert the Hilbert series into partial fraction and by expanding its each term.

    Remark 1. The partial fraction of above Hilbert series is calculated as

    12t+2t22t3+t4+t6(1t)(14t+5t25t3+6t43t5+3t64t7t8t9+t10)=0.05443+0.01323t0.560471.44679t+t2+0.00582+0.00567t0.767900.22627t+t2+0.501.00+t+0.07088+0.10684t1.26443+1.00955t+t2+0.03341+0.04181t2.14793+2.17401t+t2+0.062382.10385t+0.605190.40664t.

    As the last term

    0.605190.40664t=1.48827(1+2.45917t+(2.45917)2t2+(2.45917)3t3+)

    gives us the approximation of the series and the rest of the termshave negligible effect on it. Hence a(4)l1.48827(2.45917)l. Thus, in this case the growth function a(4)l of MB2,2 is also exponential and growth rate isapproximately equal to 2.45917.

    The mixed braid group Bm,n is the subgroup of the Artinian braid group Bm+n. In this article we compute the canonical forms(ambiguity free presentation) of the words in the braid monoid and the corresponding Hilbert series. This work can also be utilized to compute Hilbert series of braid monoid MBm,n.

    The authors thank the National Key Research and Development Program under Grant 2018YFB0904205, Science and Technology Bureau of Chengdu (2020-YF09-00005-SN), Sichuan Science and Technology program (2021YFH0107), and Erasmus+SHYFTE Project (598649-EPP-1-2018-1-FR-EPPKA2-CBHE-JP).

    Authors declare that there are no conflicts of intereset.



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