26
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- skein relation
- equation satisfied
- relation corresponding
- polynomial invariants
- ?2?2 ?
- group representations
- algebra associated
- satisfying cubic
- hecke algebra

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PAOLO BELLINGERI AND LOUIS FUNAR

Abstract. The aim of this paper is to deﬁne two link invariants satisfying cubic skein relations. In the

hierarchyofpolynomialinvariantsdeterminedbyexplicitskeinrelationstheyarethenextlevelofcomplexity

afterJones,HOMFLY,KauﬀmanandKuperberg’sG quantuminvariants. Ourmethodconsistsinthestudy2

of Markov traces on a suitable tower of quotients of cubic Hecke algebras extending Jones approach.

1. Introduction

1.1. Preliminaries. J.Conwayshowed that the Alexander polynomial of a knot, when suitably normalized,

satisﬁes the following skein relation:

! !

−1/2 1/2 ∇ −∇ = (t −t )∇

Given a knot diagram one can always change some of the crossings such that the modiﬁed diagram

represents the unknot. Therefore one can use the skein relation for a recursive computation of∇, although

this algorithm is rather time consuming, since it is exponential.

In the mid eighties V.Jones discovered another invariant verifying a diﬀerent but quite similar skein

relation, namely:

! !

−1 −1/2 1/2 t V −tV = (t −t )V ,

1/2 −1/2which was further generalized to a 2-variable invariant by replacing the factor (t −t ) with a new

variablex. The latter one was shown to specialize to both Alexander and Jonespolynomials. The Kauﬀman

polynomial is another extension of Jones polynomial which satisﬁes a skein relation, but this time in the

realm of unoriented diagrams. Speciﬁcally, the formulas:

! ! ! !!

Λ +Λ =z Λ +Λ

Λ =aΛ( )

deﬁne a regular isotopy invariant of links, which can be renormalized, by using the writhe of the oriented

diagram, in order to become a link invariant. Remark that some elementary manipulations show that Λ

veriﬁes a cubical skein relation:

1 z 1 Λ = +z Λ − +1 Λ + Λ

a a a

1991 Mathematics Subject Classiﬁcation. 16S15, 57M27, 81R15.

Key words and phrases. skein relation, cubic Hecke algebras, Markov trace.

Partially supported by a Canon grant.

12 P.BELLINGERI AND L.FUNAR

It has been recently proved ([12], and Problem 1.59 [18]) that this relation alone is not suﬃcient for a

recursive computation of Λ. Whenever the skein relations and the value of the invariant for the unknot are

suﬃcienttodetermine itsvaluesforalllinksthe systemofskeinrelationswillbe saidtobe complete. Several

results concerning the incompleteness of higher degree unoriented skein relations and their skein modules

have been obtained by J.Przytycki and his students (see e.g. [12, 28, 29]).

These invariants were generalized to quantum invariants associated to general Lie algebras and super-

algebrasand their representations. V.Turaev([33]) identiﬁed the HOMFLY and Kauﬀman polynomialswith

the invariants obtained from the series A and B ,C ,D respectively. G.Kuperberg ([19]) deﬁned then n n n

G -quantum invariant of knots by means of skein relations, by making use of trivalent graphs diagrams and2

exploited further these ideas for spiders of rank 2 Lie algebras. The skein relations satisﬁed by the quantum

invariants coming from simple Lie algebras were approached also via weight systems and the Kontsevich

integral in ([22, 23]), for the classical series, and in ([1, 2]) for the case of the Lie algebra of G .22

Notice that any link invariant coming from some R-matrix R veriﬁes a skein relation of the type:

* +n X

a j twists = 0j

j=0

which can be derived from the polynomial equation satisﬁed by the R-matrix R.

Let us mention that the skein relations are somewhat related to the representation theory of the Hopf

algebra associated to the R-matrix R. In particular, there are no other known invariants given by means

of a complete skein relations, but those from above. Moreover, one expects that the quantum invariants

associated to other Lie (super) algebras or by cabling the previous ones satisfy skein relations of degree at

least 4, as already the G invariant does.2

This makes the search for an explicit set of complete skein relations, in which at least one relation is

cubical, particularly diﬃcult and interesting. This problem was ﬁrst considered in [13] and solved in a

particular case. The aim of this paper is to complete the result of [13] by constructing a deformation of the

previously constructed quotients of the cubic Hecke algebras and of the Markov traces supported by these

(z,δ)algebras. We obtain in this way two link invariants, denoted by I and I , which are recursively(α,β)

computable and uniquely determined by two skein relations. Explicit computations show thatI detects(α,β)

the chirality of the knots with number crossing at most 10 where HOMFLY, Kauﬀman and their 2-cablings

fail. On the other hand, as HOMFLY, Kauﬀman and their 2-cablings ([24, 27]), it seems that our invariants

do not distinguish between mutant knots. We recall that the some mutant knots can be distinguished by

the 3-cablings of the HOMFLY polynomial (see [25]).

Acknowledgements. Part of this work was done during the second author’s visit to the Tokyo Institute of

Technology, whose support and hospitality are gratefully acknowledged. The authors are thankful to Chris-

tian Blanchet, Emmanuel Ferrand, Thomas Fiedler, Louis Kauﬀman, Teruaki Kitano, Soﬁa Lambropoulou,

Ivan Marin, Jean Michel, Hugh R. Morton, Luis Paris and Vlad Sergiescu for useful discussions, remarks

and suggestions.

1.2. The main result. The aim of this paper is to deﬁne two link invariants by means of a complete set of

skein relations. More precisely we will prove the following Theorem (see section 5):

(z,δ)Theorem 1.1. There exist two link invariants I and I which are uniquely determined by the two(α,β)

skein relations shown in (1) and (2) and their value for the unknot, which is traditionally 1. These invariants

take values in:

2 ±/2 2 ±/2Z[α, β, (α −2β) , (β +2α) ]

,

(H )(α,β)

and respectively:

±/2 ±/2Z[z ,δ ]

,

(z,δ)(P )

gPOLYNOMIAL INVARIANTS OF LINKS 3

where −1∈{0,1} is the number of components modulo 2 and:

6 5 2 4 4 4 3 3 3 2 5 2 2H := 8α −8α β +2α β +36α β−34α β +17α +8α β +32α β −(α,β)

4 6 3

−36αβ +38αβ+8β −17β +8,

and respectively

(z,δ) 23 18 16 2 14 3 9 4 7 5 6 5 7P :=z +z δ−2z δ −z δ −2z δ +2z δ +δ z +δ .

Here (Q) denotes the ideal generated by the element Q in the algebra under consideration.

2 3(1) = w + β w + wα

-2 -1 -1 -1

A w +B w +B w +C w+ +D

+E +E +F +Gw w +H w+F +G

(2)

2 2 2 2+H w +I w +L w w +M w w+L +M

43 3w w +P w = 0+N +O

The values of the polynomials A,B,C,...,P corresponding to I are given in the table below. In order(α,β)

(z,δ) 4 1/2 7 2 4to obtain those corresponding to I it suﬃces to set w = (−z /(δz)) and replace α =−(z +δ )/(z δ)

2 3and β = (δ−z )/z in the other entries of table 1.

2 2 1/2 2w =((α +2β)/(2α−β )) A =(β −α)

2 2 2 2B = (α −αβ −β) C =(α −αβ )

2 2 3 2 2 3D = (1+2αβ +α β −α ) E = (1+αβ +α β −α )

3 3 2F = (1+2αβ−β ) G =(αβ −2α−2α β)

3 2 2 4 3 2 2H =(αβ −2α−2α β +β ) I = (α −α β −2α β−3α)

3 2 2 3 2 4 2 2L =(2α β+3α −α β −αβ ) M =(β −2β−3αβ +α )

2 2 3 4 3 2 2 3 4N =(1+4αβ+3α β −α −αβ −β ) O = (1+3αβ+3α β −α −αβ )

2 5 2 3P = (3β −β −2α−3α β +4αβ )

Table 1

1.3. Properties of the invariants. The following summarize the main features of these invariants (see

section 6):

(1) theydistinguishallknotswithnumbercrossingatmost10thathavethesameHOMFLYpolynomial,

and thus they are independent from HOMFLY. However, like HOMFLY and Kauﬀman polynomials,

they seem to not distinguish among mutants knots: in particular they don’t separate the Kinoshita-

Terasaka knot from the Conway knot, which are the simplest non-equivalent mutant knots.

(2) I = I for amphicheiral knots, and I detects the chirality of all those knots with(α,β) (−β,−α) (α,β)

number crossing at most 10, whose HOMFLY, Kauﬀman polynomials as well as the 2-cabling of

HOMFLY fail to detect.

(z,δ)(3) I and I have a cubical behaviour.(α,β)4 P.BELLINGERI AND L.FUNAR

Let us explain brieﬂy what we meant by cubical behaviour.

P

jDeﬁnition 1.1. A Laurent polynomial c a is a (n,k)-polynomial (for n,k ∈ Z ) if c = 0 forj + jj∈Z

j =k(modulon).

P P

k kRemark 1.1. (1) TheHOMFLYpolynomialcanbewrittenas R (l)m andrespectivelyas S (m)l ,k kk∈Z k∈Z

where R (l) and S (m) are (2,k)-Laurent polynomials with R (l) =S (m) = 0.k k 2k+1 2k+1P P

k k(2) The Kauﬀman polynomial can be written as U (l)m (respectively as T (m)l ), wherek kk∈Z k∈Z

U (l) and T (m) are (2,k+1)-Laurent polynomials.k k

In this respect the HOMFLY and Kauﬀman polynomials have a quadratic behaviour.

(z,δ)Proposition 1.1. I and I have a cubical behaviour, i.e. for each link L there exists some l∈(α,β)

{0,1,2} so that P P

k kP (β)α M (α)βk kk∈Z k∈Z+ +I (L) = P = P ,(α,β) k kQ (β)α N (α)βk kk∈Z k∈Z+ +

where P ,Q ,M ,N are (3,k+l)-polynomials, andk k k k

X X

(z,δ) k k

I (L) = H (δ)z = G (z)δ ,k k

k∈Z k∈Z

where H ,G are (3,k)-Laurent polynomials.k k

1.4. Comments. There are three link invariants coming from Markov traces on cubic Hecke algebras,

presentlyknown. First, foreach quadraticfactorP ofthe cubic polynomialQone has aMarkovtrace whichi

factors through the usual Hecke algebra H(P ,n), yielding a re-parameterized HOMFLY invariant. Theni

(z,δ)thereistheKauﬀmanpolynomialandtheinvariantI (orI )introducedinthepresentpaper. Itwould(α,β)

be very interesting to ﬁnd whether there exists some relationship between them. The explicit computations

below show that the new invariants are independent on HOMFLY, Kauﬀman and their 2-cablings.

Further, one expects that our invariants belong to a family of genuine two-parameter invariants, as

expressed in the following:

Conjecture1.1. There exists a Markov trace on H(Q,n) taking values in an algebraic extension ofZ[α, β],

which lifts the Markov trace underlying I .(α,β)

In other words the non-determinacy H in I can be removed. Notice that the polynomials H(α,β) (α,β)

and P deﬁne irreducible planar algebraic curves which are not rational. In particular, one cannot express

explicitly the invariants as one variable polynomials.

1.5. Cubic Hecke algebras. The form of the ﬁrst skein relation (1) explains the appearance of cubic

quotients of braid group algebras C[B ]. Recall that the braid group B on n strands is given by then n

presentation:

B =hb ,...,b |b b =b b ,|i−j|> 1 and b b b =b b b , i<n−1i.n 1 n−1 i j j i i i+1 i i+1 i i+1

Furthermore we deﬁne the cubic Hecke algebraby analogywith the usual (i.e. quadratic) Hecke algebra (see

[9]), as follows:

H(Q,n)=C[B ]/(Q(b ); j =1,...,n−1),n j

3 2where Q(b ) = b −αb −βb −1, α, β∈ C is a cubic polynomial, which will be ﬁxed through out thisj jj j

paper.

Our purposeis to constructMarkovtracesonthe towerofcubic Hecke algebrassince these will eventually

lead to link invariants. This method was pioneered by V.Jones ([16]) and A.Ocneanu, who applied it to the

case of usual Hecke algebras and obtained the celebrated HOMFLY polynomial. Later on several authors

(see [14, 21, 15, 26]) employed more sophisticated algebraic and combinatorial tools in searching for Markov

traceson other Iwahori-Heckealgebras,for instance those of typeB, which are leadingto invariantsfor links

in a solid torus.

The cubic Hecke algebras are particular cases of the generic cyclotomic Hecke algebras introduced by

M.Brou´e and G.Malle (see [6]) and studied in [7, 8] in connection with braid group representations. Recall

the following results concerning the structure of the cyclotomic Hecke algebras with Q(0)= 0 (according to

[10, 6, 7, 8] and [11], p.148-149):

66POLYNOMIAL INVARIANTS OF LINKS 5

(1) dim H(Q,3)= 24, and H(Q,3) is isomorphic to the group algebra of the binary tetrahedral groupC

h2,3,3i of order 24, or equivalently, the linear group SL(2,Z ).3

(2) dim H(Q,4)= 648, andH(Q,4) is the group algebraofG in the Shepard-Todd classiﬁcation (seeC 25

[32]).

(3) H(Q,5) is the cyclotomic Hecke algebra of group G , whose order is 155520. It is conjectured that32

this algebra is free of ﬁnite dimension which would imply (by using the Tits deformation theorem)

that it is isomorphic to the group algebra of G .32

(4) dim H(Q,n) =∞ for n≥ 6.C

Thus a direct deﬁnition of the trace on H(Q,n) for n≥6 is highly a nontrivial matter, because it would

involvein particular, the explicit solutionofthe conjugacyproblemin these algebras. In orderto circumvent

thesediﬃcultiesoneintroducesatowerofsmallerquotientsK (α,β)byaddingonemorerelationtoH(Q,3),n

as follows:

2b b b +R = 0,2 2 01

where

2 2 2 2 2 2 2 2 2 2 2 2R = A b b b +B b b b +B b b b +C b b b +D b b b +E b b b +E b b b +0 1 1 2 1 1 1 2 2 11 2 1 2 1 1 2 1 1 2 1 1

2 2 2 2 2 2 2 2F b b +F b b +G b b +G b b +H b b +H b b +I b b b +L b b +2 2 1 1 1 2 1 2 12 1 1 2 1 1 2 2

2 2L b b +M b +M b +N b +O b +P,1 2 1 21 2

and A,B,...,P are the polynomials from table 1.

Remark 1.2. ThemainfeatureofthesequotientsisthefactthatthealgebrasK (α,β) areﬁnitedimensional,n

for all values ofn. Moreover, these algebras do not collapse for largen, thus yielding an interesting tower of

algebras.

Remark 1.3. Let us explain the heuristics behind that choice for the additional relation. For generic Q the

⊕33algebra H(Q,3) is semi-simple and decomposes as C ⊕M ⊕M , where M is the algebra of m×m3 m2

2matrices. The quadratic Hecke algebra H (3) =C[B ]/(b +(1−q)b −q) arises as a quotient of H(Q,3),q 2 ii

⊕2by killing the factorC⊕M ⊕M . It is known that Jones and HOMFLY polynomials can be derived from32

the unique Markov trace on the tower H (n). In a similar way, the rank 3 Birman-Wenzl algebra ([5]) -q

which supports an unique Markov trace inducing the Kauﬀman polynomial - is the quotient of H(Q,3) by

2the factorC⊕M . In our case we introduced the extra relation above which kills precisely the central factor2

3C of H(Q,3).

The geometric interpretation of these relations is now obvious: the ﬁrst skein relation (1) is the cubical

relation corresponding to taking the quotient H(Q,n) while the main skein relation (2) deﬁnes the smaller

quotient algebras K (α,β).n

Our main theorem is a consequence of the more technical result below (see sections 2, 3 and 4).

Theorem 1.2. There are precisely four values of (z, z¯) (formal expressions in α and β) for which there

exists a Markov traceT on K (α,β) with parameters (z, z¯) i.e. verifying the following conditions:n

(1)T(xy) =T(yx),

(2)T(xb ) =zT(x),n−1

−1(3)T(xb ) =z¯T(x).n−1

The ﬁrst pair (z, z¯) is

2 2z = (2α−β )/(αβ +4), z¯=−(α +2β)/(αβ +4),

and the corresponding trace takes values as follows:

−1Z[α, β, (αβ +4) ]

T :K (α,β)→ .α,β n

(H )(α,β)

The other three solutions are not rational functions on α, β, but nevertheless one can express α, β and z¯ as

2rational functions of z,δ, where δ =z (βz+1). Speciﬁcally, we have a Markov trace:

±1 ±1Z[z ,δ ](z,δ)

T :K (α, β)→ ,∗ (z,δ)(P )6 P.BELLINGERI AND L.FUNAR

where

2 3 7 2 4 4β =(δ−z )/z , α =−(z +δ )/(z δ) and z¯=−z /δ.

Remark 1.4. Forparticularvaluesof(α,β)∈Conemightﬁndthattheindeterminacyidealfortherespective

Markov traces is smaller than the specialization of the ideal above. A speciﬁc example is the Z/6Z-valued

z,δinvariant, corresponding to the values α = β = 0 in [13], which is a specialization of the invariant I for

3 2z =−1, and δ = z . We can reﬁne the general Markov trace in order to restrict to a Z/3Z-valued trace

(see section 6), but this reﬁnement does not survive the deformation process.

There is a natural way to convert a Markov traceT into a link invariant, by setting:

n−1 e(x) 2 21 z¯

I(x) = T(x),

zz¯ z

where x∈B is a braid representative of the link L and e(x) is the exponent sum of x.n

(z,δ)Therefore we derive two invariants I and I from the previous Markov traces, which satisfy the(α,β)

claimed skein relations.

1.6. Outline of the proof. We will prove by recurrence on n that a Markov trace on K (α,β) extendsn

to a Markov trace on K (α,β). Since there is a nice system of generators for K (α,β) constructedn+1 n+1

inductively starting from a generatorssystem forK (α,β), such an extension, if it exists, it must be unique.n

This is a consequence of the special form of the skein relation (2). However, the most diﬃcult step is to

provethat the canonicalcombinatorialextension fromK (α,β) toK (α,β) is indeed a well-deﬁned linearn n+1

functional which moreover satisﬁes the condition of trace commutativity.

The method of proofis greatlyinspired by [3]. One deﬁnes a graphwhose verticesarelinear combinations

on the elements of the Abelian semi-group associated to the free group inn−1 letters (in ﬁrst instance) and

whose edges correspond to pairs of elements which diﬀer by exactly one relation, from the set of relations

which present the algebras K (α,β).n

One gives an orientation on part of the edges of this graph and look for the existence of minimal elements

in each connected component of the graph. If there is an unique minimal element in each component then

one is able to derive a basis for K (α,β). In order to achieve the uniqueness one adds suﬃciently many∞

relations, which are formal consequences of the basic ones.

The usual procedure to obtain the existence of minimal elements is to consider the lexicographic order

on the free semi-group on n−1 letters and to use the relations as replacements of some word by a linear

combination of smaller ones, in such a way that the initial word is inductively simpliﬁed until one reaches a

normal form.

In our situation the simpliﬁcation procedure is encoded in the oriented paths of the graph. Speciﬁcally,

these are given by the following monomial substitutions:

3 2(3) (C0)(j) : ab b→αab b+βab b+ab,jj j

(4) (C1)(j) : ab b b b→ab b b b,j+1 j j+1 j j+1 j

2(5) (C2)(j) : ab b b b→aS b,j+1 j+1 jj

2 2(6) (C12)(j) : ab b b b→aC b,j+1 jj j+1

2 2(7) (C21)(j) : ab b b b→aD b,j+1 jj+1 j

2 2 2 2 2 2where E = αb +βb +1, S = b b b −R (j) , C = b b b +α(b b b −b b b )+j+1 j+1 j j+1 j+1 0 j j j+1 j+1 j jj+1 j j j+1 j j+1

2 2 2 2 2 2 2 2 2 2β(b b −b b ) and D = b b b +α(b b b −b b b )+β(b b −b b ), j∈{0,...,n−2}.j+1 j j j j j j+1 jj j+1 j+1 j j+1 j j+1 j+1 j j+1

Here R (j) is the result of translating the indices of all letters in R by j−1 units.0 0

Several edges of our graph will remain unoriented. They correspond to the following monomial substitu-

tions:

(8) (P ) : ab b b→ab b b, whenever |i−j|>1.ij i j j iPOLYNOMIAL INVARIANTS OF LINKS 7

The transformations (3-8) will be called reduction or simpliﬁcation transformations.

Remark that we introduced someextra relations, namely (5) and (6) which arenot among the relations of

thegivenpresentationofK (α,β), butwhichareneverthelesssatisﬁedinK (α,β). Thisnewrelationsmaken n

the reduction process ambiguous. The reason for introducing them is to insure the existence of descending

paths towards some minimal elements even in the case when the graph might contain closed oriented loops.

The next step consists of checking the existence and uniqueness of minimal elements in this semi-oriented

graph by means of so-called Pentagon Lemma (see section 2). One notices that one cannot always ﬁnd a

unique minimal element, by using directed paths issued from a ﬁxed vertex. Furthermore we shall enlarge

our graph to a tower of graphs modeling not one particular algebra K (α,β) for ﬁxed n, but the set ofn

∞linear functionals deﬁned on the whole tower∪ K (α,β) and satisfying certain compatibility conditionsnn=2

which relate the values taken on K (α,β) to those on K (α,β). The main feature of the tower is thatn n+1

now one can simplify further the minimal elements by recurrence on the level n, until we ﬁnd elements

in K (α,β). Here the Colored Pentagon Lemma (see section 2) can be applied and the uniqueness of the0

minimalelementsinthetowerofgraphisreducedtoﬁnitelymanyalgebraicconditions. Wewillﬁndactually

thatthemainobstructionslieinK (α,β), asitmightbeinferredfromthestudyofquadraticHeckealgebras.4

∞From a diﬀerent perspective we actually proved that a certain linear functional on the tower∪ K (α,β)nn=2

is well-deﬁned.

Eventually one has to verify whether the linear functional obtained above satisﬁes the commutativity

conditions for being a Markov trace. One proves that there is only one obstruction to the commutativity,

which lies also in K (α,β).4

Summarizing, there are two types of obstructions to the existence of a Markov traces:

• CPC obstructions, coming from the Colored Pentagon Condition, and

• commutativity obstructions.

Thesealgebraicobstructionsarepolynomialswithintegercoeﬃcientsinthevariablesαandβ, andhavebeen

computed by using a computer code, by using formal calculus. The output of these computations is a set of

explicit polynomials, which belong to the principal ideal generated by H , showing that the functional(α,β)

deﬁned above is indeed a Markov trace, when restricting its values to the quotient by this principal ideal.

2. Markov traces on K (α,β)n

2.1. The cubic Hecke H(Q,3) algebra revisited. The generalized Hecke algebras H(P,3) could be

considered for polynomialsP ofhigher degree, by using the same deﬁnition as in the cubic case. One notices

however that dim H(P,3) =∞ as soon as the degree of P is at least 6.C

The structure of the algebras H(P,n) is well-known in the classical case (see [9]) when P is quadratic.

They are ﬁnite dimensional semi-simple algebras of dimension n!, isomorphic (for generic P) to the group

algebra of the permutation group on n elements. There is no general theory for higher degree polynomials

P, due to their considerable complexity.

In the particular case of cubic Q and n = 3 it was shown in [13] the following:

Proposition 2.1. For all cubic polynomials Q with Q(0) = 0 we have dim H(Q,3) = 24. A convenientC

base of the vector space H(Q,3) is

2 2 2 2 2e = 1, e = b , e = b , e = b , e = b , e = b b , e = b b , e = b b , e = b b , e = b b , e =1 2 1 3 4 2 5 6 1 2 7 2 1 8 2 9 2 10 1 111 2 1 1 2

2 2 2 2 2 2 2 2 2 2 2b b , e = b b , e = b b , e = b b b , e = b b b , e = b b b , e = b b b , e = b b b , e =1 12 13 14 1 2 1 15 2 1 16 1 2 17 1 18 2 192 1 2 2 1 1 1 2 1 1 1

2 2 2 2 2 2 2 2 2 2 2 2b b b , e = b b b , e = b b b , e = b b b , e = b b b b = b b b b , e = b b b b = b b b b b =1 20 1 1 21 22 2 2 23 2 2 1 1 2 2 24 2 2 1 2 2 11 2 2 1 2 1 1 1 1 1 1 1

2 2b b b b .2 21 1

⊕33Proposition2.2. H(Q,3) is a semi-simple algebra which decomposes generically asC ⊕M ⊕M , where32

3M is the algebra ofn×n matrices. The morphism intoH(Q,3)→C is obtained via the abelianization map.n

2Each one ofthe three projectionsH(Q,3)→M factors throughthe projectionH(Q,3)→H(P ,3) =C⊕M2 i 2

onto the quadratic Hecke algebra H(P ) deﬁned by one divisor P of Q.i i

Proof. The proof is a direct computation, making use of the following identities ([13]):

2 2b b b b =b b b b ,j+1 j+1 j j j+1 j+1j j

2 2 2 2 2 2 2 2b b b =b b b +α(b b b −b b b )+β(b b −b b ),j+1 j j+1 j+1 j j j+1 jj+1 j j+1 j j j+1 j j+1

68 P.BELLINGERI AND L.FUNAR

2 2 2 2 2 2 2 2b b b =b b b +α(b b b −b b b )+β(b b −b b ).j+1 j j+1 j+1 j j j+1 jj j+1 j j+1 j j+1 j j+1

2.2. ThealgebrasK (α,β). ThequotientP(∞)ofH(Q,∞)ishomogeneousifanyidentityF(b ,b ,...,b ) =n i i+1 j

0, which holds in P(∞) remains valid under the translation of indices i.e. also F(b ,b ,...,b ) = 0,i+k i+k+1 j+k

fork∈Z,k≥1−i. If one seeks for Markovtraces on towersof quotients ofC[B ] it is convenient to restrictn

ourselves to the study of homogeneous quotients.

We deﬁne K (α,β) as the homogeneous quotient H(Q,n)/I , where I is the two-sided ideal generatedn n n

by:

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2b b b +(β −α)b b b +(α−αβ −β)b b b +(α−αβ −β)b b b +(α−αβ )b b b +j j j−1 j−1 jj−1 j−1 j j−1 j j−1 j−1 j j−1 j−1

2 2 3 2 2 2 3 2 2 2 3 2(1+2αβ +α β −α )b b b +(1+αβ +α β− α )b b b +(1+αβ +α β −α )b b b +j−1 j−1 j−1 j j j−1j j−1 j−1

3 2 2 3 2 2 3 2 2 3 2 2(1 + 2αβ− β )b b + (1 + 2αβ− β )b b + (αβ − 2α− 2α β)b b + (αβ − 2α− 2α β)b b +j jj j−1 j−1 j j−1 j−1

3 2 2 2 3 2 2 2 4 3 2 2 3 2(αβ −2α−2α β+β )b b +(αβ −2α−2α β+β )b b +(α−α β−2α β−3α)b b b +(2α β+3α−j−1 j−1 j−1 j j−1j j

2 3 2 3 2 2 3 2 4 2 2 2 4 2 2 2α β −αβ )b b +(2α β+3α−α β −αβ )b b +(β −2β−3αβ +α )b +(β −2β−3αβ +α )b +j j−1 j−1 j j−1 j

2 2 3 4 3 2 2 3 4 2 5 2 3(1+4αβ+3α β −α −αβ −β )b +(1+3αβ+3α β −α −αβ )b +3β −β −2α−3α β+4αβ ,j−1 j

for j∈{1,...,n−1}.

3 ⊕3∼Proposition 2.3. Under the identiﬁcation H(Q,3) =C ⊕M ⊕M , the quotient K (α,β) corresponds3 32

⊕3to M ⊕M .32

Proof. In fact it suﬃces to show that the ideal I is a vector space of dimension 3. Let R be the span of3

R ,R ,R , where0 1 2

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2R := b b b + (β − α)b b b + (α − αβ − β)b b b + (α − αβ − β)b b b +(α − αβ )b b b +0 2 2 1 21 1 2 1 2 1 1 2 1 1 1

2 2 3 2 2 2 3 2 2 2 3 2(1 + 2αβ +α β −α )b b b + (1 +αβ +α β −α )b b b + (1 +αβ +α β −α )b b b +(1 + 2αβ−1 1 1 2 2 12 1 1

3 2 2 3 2 2 3 2 2 3 2 2 3 2 2 2β )b b +(1+2αβ−β )b b + (αβ −2α−2α β)b b +(αβ −2α−2α β)b b +(αβ −2α−2α β+β )b b2 2 12 1 1 2 1 1 2

3 2 2 2 4 3 2 2 3 2 2 3 2 3 2+(αβ −2α−2α β+β )b b +(α −α β −2α β−3α)b b b +(2α β+3α −α β −αβ )b b +(2α β+3α −1 1 2 1 2 12

2 3 2 4 2 2 2 4 2 2 2 2 2 3 4 3α β −αβ )b b + (β −2β−3αβ +α )b +(β −2β−3αβ +α )b +(1+4αβ+3α β −α −αβ −β )b1 2 11 2

2 2 3 4 2 5 2 3+(1+3αβ+3α β −α −αβ )b +3β −β −2α−3α β+4αβ ,2

2 2 2 2 2 2 2 2 2 2 2 2 2R :=b R =b b b b −βb b b +(1+αβ)b b b +(1+αβ)b b b +(1+αβ)b b b (−α β−2α)b b b +1 1 0 1 2 2 1 2 1 11 1 2 1 2 1 1 2 1 1 1 2

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2(−α β−2α)b b b +(−α β−2α)b b b +(β −α)b b +(β −α)b b +(α −αβ )b b +(α −αβ )b b +(α −1 2 2 1 2 21 1 2 1 1 2 1 1

2 2 2 2 2 3 2 2 2 3 2 2 3αβ −β)b b +(α−αβ −β)b b +(α β+β+3α )b b b +(1+αβ+α β −α )b b +(1+αβ+α β −α )b b +1 1 1 2 1 2 1 1 22 2

3 2 3 2 3 2 2 3 2 4 2 2(1+2αβ−β )b +(1+2αβ−β )b +(αβ −2α−2α β+β )b +(αβ −2α−2α β)b +β −2β−3αβ +α ,1 21 2

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2R :=b R =b b b b +b b b −αb b b −αb b b −αb b b +α b b b +(α +β)b b b +(α +β)b b b +2 1 1 2 2 1 2 1 1 1 2 2 11 1 1 2 1 2 1 1 2 1 1 1 2 1 1

2 2 2 2 2 2 2 2 3(−β)b b +(−β)b b +(1+αβ)b b + (1+αβ)b b +(1+αβ)b b + (1+αβ)b b + (−α β−αβ+1)b b b +2 2 1 1 1 2 12 1 1 2 1 1 2 2

2 2 2 2 2 2 2 2 2 2 3(−α β−2α)b b +(−α β−2α)b b +(β −α)b +(β −α)b +(−αβ +α−β)b +(−αβ +α )b +1+2αβ−β .2 1 1 2 1 21 2

∼Lemma 2.1. There is an isomorphism of vector spaces R =I .3

Proof. Remark ﬁrst that the following identities hold true in H(Q,3):

b R =R b =R , b R =R b =R , b R =R b =R +βR +αR .1 0 0 1 1 1 1 1 1 2 1 2 2 1 0 1 2

Then by direct computation we obtain that:

b R =R b =R , b R =R b =R , b R =R b =R +βR +αR .2 0 0 2 1 2 1 1 2 2 2 2 2 2 0 1 2

From these relations we derive that xR y∈R for all x,y∈H(Q,3), and hence I ⊂R. The other inclusion0 3

is immediate.

The proposition is then a consequence of the previous lemma.

2.3. Uniqueness of the Markov trace on the tower K (α,β). From now on we will work with then

group ringZ[α,β][B ] instead ofC[B ].∞ ∞

Deﬁnition2.1. Letz,z¯∈Z(α,β)berationalfunctionsinthevariablesαandβ andRaZ[α,β,z,z¯]module.

∞The linear functionalT :∪ K (α,β)→R is said to be an admissible functional (with parameters z andnn=1

z¯) on K (α, β) if the following conditions are fulﬁlled:∞

T(xb y) =zT(xy) for all x,y∈K (α,β),n nPOLYNOMIAL INVARIANTS OF LINKS 9

−1T(xb y) =z¯T(xy) for all x,y∈K (α,β).nn

An admissible functionalT is a Markov trace if it satisﬁes the following trace condition:

T(ab) =T(ba) for any a,b∈K (α,β).n

Remark 2.1. The tower of quadratic Hecke algebras admits an unique Markov trace ([16]). Similarly the

tower of Birman-Wenzl algebras ([5]) admits an unique Markov trace.

k kDeﬁnition 2.2. The admissible functionalT is multiplicative ifT(xb ) =T(x)T(b ) holds for all x∈n n

H(Q,n) and k∈Z.

Remark 2.2. The Markov trace on the quadratic Hecke algebras is multiplicative, and henceT(xy) =

T(x)T(y), for anyx∈H(Q,n) andy∈h1,b ,b ,...,b i. Howeverone cannot expect that this propertyn n+1 n+k

holds true for Markov traces on arbitrary higher degree Hecke algebras.

Proposition2.4. Admissible functionals on the tower of cubic Hecke algebras are multiplicative. In partic-

ular

2T(ab b) =tT(ab) a,b∈B ,nn

holds true, where t =αz+β+z¯.

2 −1Proof. One uses the identityb =αb +β+b for provingthe multiplicativity fork =2, and then continuenn n

by recurrence for all k.

One can state now the unique extension property of Markov traces.

Proposition 2.5. For ﬁxed (z,t) there exists at most one Markov trace on K (α,β) with parameters (z,t).n

Proof. Deﬁne recursively the modules L as follows:n

L =H(Q,2),2

ji kL =Chb b b| where i,j,k∈{0,1,2}i,3 1 12

εL =Chab b| where a,b are elements of the basis ofL , and ε∈{1,2}i⊕ L .n+1 n nn

Lemma 2.2. The natural projection π :L →K (α,β) is surjective.n n

2 2 2 2Proof. For n = 2 it is clear. For n = 3 we know that b b b , b b b b , b b b b ∈ π(L ), from the exact2 2 1 2 2 2 2 31 1 1 1

form of the relations R ,R ,R generating the ideal I . We shall use a recurrence on n, and assume that0 1 2 3

the claim holds true for n.

Consider now w∈ K (α,β) represented by a word in the b ’s having only positive exponents. Wen+1 i

assume that the degree of the word in the variable b is minimal among all linear combinations of wordsn

(with positive exponents) representing w.

(1) If this degree is less or equal to 1 then there is nothing to prove.

2(2) If the degree is 2 then either w = ub v, u,v∈ K (α,β) so using the induction hypothesis we arenn

εdone, or else w =ub zb v, where u,z,v∈K (α,β). Therefore z =xb y where x,y∈K (α,β)n n n n−1n−1

by the induction hypothesis and ε∈{0,1,2}.

2(a) If ε =0 then w can be reduced to uzb v.n

(b) If ε = 1 then w = ub xb yb v = uxb b b yv hence the degree of w can be lowered byn n−1 n n−1 n n−1

one, which contradicts our minimality assumption.

2(c) If ε =2 then w =uxb b b yv. One derives that:n nn−1

2 i j kb b b ∈Chb b b , i,j,k∈{0,1,2}i,n nn−1 n−1 n n−1

0 2 0hence we reduced the problem to the case when w is a word of type ub v .n

(3) If the degreeofw is at least3 wewill contradictthe minimality assumption. In fact, in this situation

0 a bw will contain either a sub-word w = b ub , with u∈ K (α,β) and a+b≥ 3, or else a sub-wordnn n

00w =b ub vb , with u,v∈K (α,β).n n n n

ε(a) In the ﬁrst case using the induction we can write u=xb y, x,y∈K (α,β).n−2n−1

0 a+b a+b−1 a+b−2 a+b−3(i) Furthermore, ifε = 0 thenw =b xy =αb xy+βb xy+b xy, and hencen n n n

the degree of w can be lowered by one.10 P.BELLINGERI AND L.FUNAR

0 a−1 b−1 a−1 b−1(ii) If ε = 1 then w = b xb b b yb = b xb b b yb , and again its degreen n−1 n n−1 n n−1n n n n

can be reduced by one unit.

(iii) If ε =2 then either a or b is equal 2. Assume that a = 2. We can therefore write:

0 2 2 b−1 2 2 b−1 2 2 b−1w = xb b b yb =xb b b yb +α(b b b −b b b )yb +n n−1 n n n−1 n−1n n−1 n n n−1 n n−1 n n

2 2 b−1β(b b −b b )yb ,n n−1n−1 n n

contradicting again the minimality of the degree of w.

ε δ(b) In the second case we can write also u=xb y, v =rb s with x,y,r,s∈K (α,β).n−1n−1 n−1

00(i) Ifεorδ equals1then,aftersomeobviouscommutationthewordw containsthesub-word

b b b which can be replaced by b b b and hence diminishing its degree.n n−1 n n−1 n n−1

00 2 2 2(ii) Ifε =δ = 2thenw =xb b b yrb b s. Weusethehomogeneitytoreplaceb b bn n n n nn−1 n−1 n−1

i j k 00by a sum of elements of typeb b b . Each term of the expressionofw which comesn−1 n n−1

from a factor having j < 2 has the degree less than it had before. The remaining terms

i 2 k 2 2are xb b b yrb b s, so they contains a sub-word b ub whose degree we alreadyn nn−1 n n−1 n−1 n

know that it can be reduced as above. This proves our claim.

εEventuallyrecallthattheMarkovtracesT onH(Q,∞)aremultiplicativeandhencetheysatisfy:T(xb y) =n

εT(b )T(yx). Therefore there is a unique extension ofT from K (α,β) to K (α,β). This ends the proofn n+1n

of our proposition.

Proposition 2.6. The admissible functionals on the tower of algebras K (α,β) satisfy the identities:n

T(xuv) =T(u)T(xv) for x,v∈H(Q,m) and u∈h1,b ,b ,...,b i.m m+1 m+k

Proof. For k = 0 this is the multiplicativity of the trace. We will reason by recurrence on k, and assume the

claimholdstruefork. Bylemma2.2onecanreduceinK (α,β)utoa(non-necessarilyunique)normalm+k+1

εform u=u b u , where u ∈h1,b ,b ,...,b i, j∈{1,2}, with ε∈{0,1,2}. The multiplicativity of1 2 j m m+1 m+km+k

the admissible functionals show that

εT(xuv) =T(b )T(xu u v).1 2m+k

By recurrence, the following holds true:

T(xu u v) =T(u u )T(xv),1 2 1 2

and again the multiplicativity implies that:

εT(u) =T(b )T(u u ),1 2m+k

which ends the proof.

3. CPC Obstructions

3.1. The pentagonal condition. The following lemma is also a consequence of the previous proposition:

Lemma 3.1. There is a surjection of (K (α,β),K (α,β))-bimodulesn n

K (α,β)⊕K (α,β)⊗ K (α,β)⊕K (α,β)⊗ K (α,β)−→K (α,β)n n n n n n+1K (α,β) K (α,β)n−1 n−1

given by:

2x⊕y⊗z⊕u⊗v→x+yb z+ub v.n n

Remark 3.1. In particular, admissible functionals on the tower K (α,β) are unique up to the choice ofn

T(1)∈R.

Wewantnowtousethetransformations(3-7)inordertosimplifythepositivewordsfromK (α,β),insuchn

awaythatthedegreeofb isassmallaspossible. AccordingtothepreviouslemmaeverywordinK (α,β)n−1 n

εican be written as a linear combination of words of type x b y , with ε ∈ 0,1,2 and x ,y ∈ K (α,β).i i i i i n−1n−1

Unfortunately we are forced to use also the transformations from (8), P : b b ↔ b b , for |i−j| > 1,ij i j j i

which have to be used in both directions.