- Email: [email protected]
The HOMFLYPT skein module of the lens spaces Lp,1
Topology and its Applications 175 (2014) 72–80
Contents lists available at ScienceDirect
Topology and its Applications www.elsevier.com/locate/topol
The HOMFLYPT skein module of the lens spaces Lp,1 Boštjan Gabrovšek a , Maciej Mroczkowski b,∗ a b
University of Ljubljana, Faculty of Mechanical Engineering, Aškerčeva 6, 1000 Ljubljana, Slovenia Institute of Mathematics, University of Gdansk, ul. Wita Stwosza 57, 80-952 Gdansk-Oliwa, Poland
a r t i c l e
i n f o
Article history: Received 15 April 2013 Received in revised form 11 July 2014 Accepted 11 July 2014 Available online 26 July 2014
a b s t r a c t We compute the HOMFLYPT skein module of the lens spaces Lp,1 and present a free basis of this module for each p. © 2014 Elsevier B.V. All rights reserved.
MSC: 57M27 Keywords: Knots Links Lens spaces Skein modules
1. Introduction In this paper we use diagrams of links in the solid torus with arrows, introduced in [3], to compute the HOMFLYPT skein modules of the lens spaces Lp,1 . We exhibit sets of generators of these modules and show that these sets in fact form free bases. Skein modules were introduced in 1987 by J. Przytycki [4,5]. First, we recall the definition of the HOMFLYPT skein module of an oriented 3-manifold, generalizing the HOMFLYPT polynomial of links in S 3 (see [1,7]). Let M be an oriented 3-manifold, L the set of oriented links in M up to ambient isotopy (including the empty knot) and R = Z[v ±1 , z ±1 ]. Let S be the submodule of RL generated by expressions v −1 L+ − vL− − zL0 , where L+ , L− and L0 are three identical links except inside a 3-ball where they look as in Fig. 1. We add to S the expression involving the empty knot: v −1 ∅ − v∅ − zt0 , where t0 is the trivial knot. Then the HOMFLYPT skein module of M , denoted S3 (M ), is RL/S. * Corresponding author. E-mail addresses: [email protected] (B. Gabrovšek), [email protected] (M. Mroczkowski). http://dx.doi.org/10.1016/j.topol.2014.07.003 0166-8641/© 2014 Elsevier B.V. All rights reserved.
B. Gabrovšek, M. Mroczkowski / Topology and its Applications 175 (2014) 72–80
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Fig. 1. Three links in the HOMFLYPT relation.
Fig. 2. Links t3 and t3 t−1 .
Fig. 3. Reidemeister moves.
In the next section we translate the free basis of the HOMFLYPT skein module of the solid torus (computed in [2,4,8]) into diagrams with arrows and we establish some properties of these diagrams in this skein module. In Section 3 we state our main result in Theorem 1, which is proven in Section 4. 2. The HOMFLYPT skein module of the solid torus Let T be the solid torus. In [2,4,8], it was shown that S3 (T ) is free, generated by B = {tik11 ...tikss : s ∈ N, k1 < .. < ks ∈ Z∗ , i1 ..is ∈ N} ∪ {∅}, where ∅ is the empty knot. For k > 0, tk is the oriented knot in T representing k in π1 (T ) = Z, having an ascending diagram with respect to a basepoint situated close to the exterior circle of the annulus. For k < 0, tk is t−k with reversed orientation. For example, t3 and t3 t−1 are presented in Fig. 2. Instead of using classical diagrams of links in T (i.e. projections onto the annulus with information of under and overcrossings), we will use diagrams with arrows introduced in [3]. Note, that these diagrams are very similar to the notion of gleams introduced by V. Turaev in [9]. An arrow diagram is obtained from a link L in T by cutting T along a meridional disk M , projecting the resulting vertical cylinder I × D (I = [0, 1]) onto the disk D × {0} while keeping information of over- and undercrossings for the projection of L. The intersection points in L ∩ M project onto arrows in the diagram, pointing to the part of L that is close to D × {0} in the cylinder (and away from the part of L that is close to D × {1}). Two such diagrams represent the same link in T if an only if one can get from one to the other with a sequence of five Reidemeister moves, three classical and two extra ones, presented in Fig. 3.
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Fig. 4. Interpretation of the moves Ω4 and Ω5 .
Fig. 5. Transforming a classical diagram into an arrow diagram.
Fig. 6. Arrow diagram of t3 t22 t−1 t−2 .
The interpretation of the moves Ω4 and Ω5 is illustrated in Fig. 4 (with an Ω4 move on the left of the cylinder and an Ω5 move on the right). Fig. 5 shows the translation of t3 from a classical diagram to an arrow diagram. Here, the arrow labeled with 3 stands for 3 arrows next to each other. We will use this convention with any n ∈ Z, so that n = 0 means that there are no arrows and n < 0 that there are |n| arrows opposite to the one that is pictured. Each element of B has a unique standard diagram with no crossings, non-nested ovals and such that there are no arrows in the clockwise direction (the orientation is clockwise when k < 0 in tk ). For instance t3 t22 t−1 t−2 is presented in Fig. 6. Let L be a link in T . We denote by HT (L) the expression of L in S3 (T ). Thus HT (L) is a linear expression in elements of B with coefficients in the ring R. We conclude this section with some properties of S3 (T ) that will be needed later. The equalities v −1 L+ − vL− = zL0 hold by definition in any HOMFLYPT skein module (see Fig. 1). We call such equalities HOMFLYPT relations. Denote by t¯n , n > 0, an oval with n clockwise arrows on it and clockwise orientation. For n < 0, let t¯n be t¯−n with reversed orientation. Lemma 1. In S3 (T ), one can revert clockwise arrows on an oval in the sense that for n > 0 t¯n =
i
Ai T i
or
t¯−n =
i
Ai Ti
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Fig. 7. Pushing an oval through a strand.
Fig. 8. Reducing the number of arrows on the left of s.
Fig. 9. Reducing the number of arrows on the kink.
for some Ai , Ai ∈ R. Any Ti has the form tik11 ...tikss , where k1 > 0, ..., ks > 0 and i1 k1 + ... + is ks = n. Similarly, any Ti has the form tik11 ...tikss , where k1 < 0, ..., ks < 0 and i1 k1 + ... + is ks = −n. Proof. The proof is by induction. For n = 1, t¯1 is transformed into t1 with a sequence of Ω1 , Ω5 and Ω1 moves. Similarly, t¯−1 is transformed into t−1 . Suppose that the statement of the lemma is true for k < n. Let c be an oval with n clockwise arrows on it. We perform an Ω1 move on c, then push one arrow on the kink thus created with Ω5 (this arrow is counterclockwise on the kink). Using HOMFLYPT relation and Ω5 we push another arrow on the kink getting also, by smoothing the crossing, two ovals with 1 and n − 1 arrows on which induction applies. Using HOMFLYPT and Ω5 again, we push another arrow on the kink or split the oval into two ovals with n − 2 and 2 arrows on which induction applies. Continuing in this way, all arrows are pushed on the kink, becoming counterclockwise. 2 Lemma 2. Let D be a diagram of a link L with an oval c containing n arrows, n ∈ Z, and a strand s adjacent to c, that may contain arrows outside the drawn region (Fig. 7). The oval c can be pushed over s in the sense that either
holds in S3 (T ) for some Ai ∈ R, depending on the orientation of the components c and s. Proof. We consider the case where c is oriented counter-clockwise and s is oriented “upwards”. For n = 0 the proof is trivial. For n > 0 we first push an arc of c over s by Ω2 and successively apply the moves and HOMFLYPT relations in Fig. 8 until there are no arrows left on the left side of s. We resolve the last term by successively applying the moves and HOMFLYPT relations in Fig. 9. It follows that the link can be expressed as
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Fig. 10. A 1-gon and a 2-gon.
Fig. 11. The Ω(p) move.
A similar calculation shows that the above formula holds for any n < 0. The formula also holds if we reverse both orientations of c and s. If c is oriented clockwise and s is oriented “upwards” or if c is oriented counter-clockwise and s is oriented “downwards”, then one gets similarly for any n ∈ Z:
2 Lemma 3 (1- and 2-gon lemma). Let D be a diagram of a link in T and suppose that either a strand of D forms a 1-gon or two strands of D form a 2-gon (Fig. 10). Then, D can be expressed in S3 (T ) with diagrams which have less crossings than D and no more arrows than D. Proof. Let d be the disk bounded by the 1-gon or the 2-gon. We may assume that there are no 1-gons or 2-gons inside d (in the sense that the discs they bound would be included in d). Otherwise, we consider a most nested 1-gon or 2-gon instead (with no 1-gons or 2 gons inside). • Case 1: Suppose d bounds a 1-gon. By assumption, there may be only ovals inside d, which we push outside of d using Lemma 2. We continue by pushing the arrows lying on the strand of the 1-gon outside of d using HOMFLYPT relations and Ω5 . Finally, we remove the crossing of the 1-gon with Ω1 . • Case 2: Suppose d bounds a 2-gon. If there are ovals inside d, we push them outside of d using Lemma 2. We continue by pushing the arrows lying on the strands of the 2-gon outside of d using HOMFLYPT relations and Ω5 . If the interior of d is not empty, there must be a braid going from one strand of the 2-gon to the other. In this case we push the arrows on the braid outside of d and continue by pushing the crossings of the braid outside of d using HOMFLYPT relations and Ω3 . We are left with a trivial braid whose strands can be pushed outside of d through the double points of the 2-gon using HOMFLYPT relations and Ω3 . Once there are no strands inside of d, we can perform an Ω2 move (using possibly a HOMFLYPT relation before) and remove the two crossings of the 2-gon. 2
3. Construction of H and main theorem A diagram of a link in Lp,1 is the same as a diagram of a link in T . The difference comes from one extra Reidemeister move between such diagrams, the Ω(p) move shown in Fig. 11. This move corresponds to sliding an arc of the link over the meridional disk of the solid torus T that is glued to T in order to get Lp,1 .
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Let Bp = {tik11 ...tikss : s ∈ N, k1 < .. < ks ∈ Z∗ , − p2 < k1 < .. < ks ≤ p2 , i1 ..is ∈ N} ∪ {∅}. Let D be the set of arrow diagrams of links in Lp,1 . We now construct a function H : D → RBp . It will be shown, that H induces an isomorphism from S3 (Lp,1 ) to RBp . H is defined on standard diagrams of elements of B, such as the one in Fig. 6, and extended linearly to all diagrams of links in T by the formula H(D) = H(HT (D)). Here, HT (D) is to be understood as a linear expression in standard diagrams of elements of B, rather than as a linear expression in elements of B. Let g be a standard diagram of tik11 ...tikss ∈ B. Let m be the maximum of the |ki |-s. If m < p2 or m = p2 and no ki equals −m, we define H(g) = g. Otherwise, let c be a component t−m of g or, if there are no such components, let c be a component tm of g. Let g be obtained from g by performing an Ω(p) move on the component c. We say, that an arrow is bad if its orientation is opposite to the orientation of the component on which it lies. Notice, that there are less arrows in g than in g or, if this number is unchanged, there are less bad arrows in g than in g (this occurs only when m = p2 ). We define recursively H(g) = H(HT (g )). From Lemmas 1 and 2 it follows that no diagram of HT (g ) has more arrows than g (HT (g ) is computed by pushing ovals out of the oval c transformed by Ω(p) , then possibly changing clockwise arrows into anticlockwise arrows). Following the proof of these lemmas, it is also clear that no additional bad (or good) arrows will appear in the computation of HT (g ). Thus, all terms in HT (g ) have less arrows than g or the same number of arrows but less bad arrows, and the recursive process of defining H must end. Theorem 1. H induces an isomorphism from S3 (Lp,1 ) to RBp . Thus, S3 (Lp,1 ) is free with basis Bp . Proof. HT is invariant under all Reidemeister moves except Ω(p) , so H is also invariant under these moves. Also, HT satisfies the HOMFLYPT relation, so H satisfies it as well. H is invariant under some Ω(p) moves, namely the ones between pairs of diagrams g and g used in the definition of H above. Thus, H will be an isomorphism from S3 (Lp,1 ) to RBp provided it is invariant under all Ω(p) moves. This fact will be proven in the next section (see Propositions 1 and 2). 2 4. Invariance of H under Ω(p) moves Let D be a diagram consisting of an element of B to which is added an oval e in such a way that there are no nested ovals in D. The oval e has b ∈ Z arrows and it may have any orientation. Let D be obtained from D by an Ω(p) move performed on oval e. We call such a move from D to D standard (see Fig. 12). Proposition 1. H is invariant under all Ω(p) moves if it is invariant under all standard Ω(p) moves. Proof. Let D be a diagram on which we mark by s the small part on one of its arcs where Ω(p) can be performed on; s is chosen in such a manner that no arrows lie on it. Let D be obtained from D by performing Ω(p) on s. It is enough to prove that by freezing s we can express D as a linear expression of elements of B with an added oval e on which s lies. If D has at least one crossing, then it must contain a 1-gon or a 2-gon. If s lies on it, then it is easy to check, that there must be another 1-gon or 2-gon on which s does not lie (s also does not lie in the disc this 1-gon or 2-gon bounds, otherwise one could not perform an Ω(p) move on it). One may therefore apply Lemma 3 to this 1-gon or 2-gon not containing s. In this way we will eventually remove all crossings in D. Using Lemma 2 we push all ovals out of the oval on which s lies. Using Lemma 1 we revert the ovals with clockwise arrows on them, except for the oval on which s lies. All these operations (removal of crossings, pushing and reverting of ovals) can be performed simultaneously on D and D . In this way the Ω(p) move from D to D can be expressed with several standard Ω(p) moves. 2
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Fig. 12. Standard Ω(p) move.
Fig. 13. Diagrams E and E .
n We denote by Ω(p) a standard Ω(p) move from D to D (such as in Fig. 12), where one of the diagrams has n arrows and the other one has at most n arrows. Here, we are counting all the arrows as positive, even clockwise arrows (which may only lie on the component on which the move is performed).
Proposition 2. H is invariant under all standard Ω(p) moves. Proof. We will prove for any n: n (In ): if Ω(p) is a standard move from D to D , then H(D) = H(D ). This will prove the statement of the proposition. The proof is by induction on n. If n = 0 there are no Ω(p) moves, so (I0 ) is true. Suppose now that n > 0 and (Ik ) holds for all k < n. n Let Ω(p) be a standard move from D to D . Let e be the component of D on which the move is performed. We will use an inner induction on l, the number of arrows on components of D except for e. Notice, that if l = 0 there are no arrows on components outside e, so these components are mapped by HT to elements in −1 the ring R, namely v z−v , and H(D) = H(D ) by definition. There are 3 cases to consider: • Case 1: D has n arrows and D has less than n arrows. If e is a component of D with the maximal number, say max, of arrows and, moreover, bad arrows if there is any other component of D with max bad arrows on it, then H(D ) = H(D) by definition. Otherwise, there is a component d of D such that H(D) = H(D ) by definition, where D is obtained from D by performing an Ω(p) move on d. In this case, either d has more arrows than e, or it has the same number of arrows as e, but the arrows on d are bad and the ones on e are good. Let a and b be the numbers of arrows respectively on d and e. Let E be the diagram obtained from D by applying Ω2 and Ω(p) moves on d (see left of Fig. 13). Similarly, let E be the diagram obtained from D by applying Ω2 and Ω(p) moves on e (see right of Fig. 13). Now, E and E are diagrams of the same link in the torus T : one gets E from E by applying 2p times an Ω5 move on the right crossing of E , thus pushing p arrows next to the a arrows and p arrows away from the b arrows. Thus H(E ) = H(E ). To show that H(D ) = H(E ), we first transform D and E with Ω2 and Ω5 moves into Dt and Et as shown in Fig. 14.
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Fig. 14. Showing that H(D ) = H(E ) (when D has less arrows than D).
As D and D have less arrows than D, one has a ≥ b > p2 . On Dt the number of arrows on component p p−1 p e is p−1 2 + | 2 + 1 − b| = 2 + b − 1 − 2 which equals b − 1 if p is odd and b − 2 if p is even. On component d there are a arrows. Thus the number of arrows on Dt is n − 1 or n − 2. Similarly, on Et there are less than n arrows (b − 1 or b − 2 on e and a or a − 1 on d). We claim that H(Et ) = H(Dt ). Indeed, using the same methods as in Lemma 2, one can push out the a − p+1 2 arrows on d, push out any ovals out of e and, using Lemma 1, revert all ovals with clockwise arrows except for d, all this being done without increasing the number of arrows. Thus, Dt and Et are simultaneously expressed with diagrams with no crossings, Dt = i Ai Di and Et = i Ai Ei , where Ai ∈ R and Ei is obtained from Di with a standard Ω(p) move. As the number of arrows in all Di and Ei is less than n, we have H(Di ) = H(Ei ) by induction on n. Thus H(E ) = H(Et ) = H(Dt ) = H(D ). A similar argument shows that H(E ) = H(D ). Indeed, by exchanging the roles of a and b, E is obtained from E (except for the crossings, but both can be switched by going around a component with an arrow and applying Ω5 twice). Combining preceding equalities one gets H(D ) = H(D ) = H(D). • Case 2: D has less than n arrows and D has n arrows. Recall that l is the number of arrows in D outside e. If l > 0, there are ovals with arrows other than e. We push these ovals through e at the same time in D and D (i.e. in D we push them into e and in D we push them out of e). From Lemma 2, in the resulting linear expressions for D and D two situations can occur: all ovals are pushed through with their arrows, and we can apply Case 1 by reverting the roles of D and D ; some arrows are transfered onto e, which decreases l and we apply the induction on l. In the second case it may happen that the total number of arrows is decreased and the global induction on n can be applied. • Case 3: both D and D have n arrows (this occurs when p is even and the number of arrows on e is b = p2 ). Two subcases are possible: – There is a component d in D with a arrows, a > b = p2 , such that H(D) = H(D ) by definition, where D is obtained from D by performing an Ω(p) move on d. Using the same notations as in Case 1, one gets again H(E ) = H(E ) (see Fig. 13). Showing that H(D ) = H(E ) is done exactly as in Case 1: in both cases in order to apply induction on n we only use the fact that a > p2 (b can be equal to p2 ). Using several Ω5 moves D and E can be transformed simultaneously into diagrams Dt and Et shown in Fig. 15. Notice that the number of arrows in Dt is n (because a > p/2 by assumption) and the number of arrows in Et is less than n. As in Case 1, Dt and Et are simultaneously expressed with diagrams with no crossings, Dt = i Ai Di and Et = i Ai Ei , where Ai ∈ R and Ei is obtained from
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Fig. 15. Showing that H(D ) = H(E ) (when D and D have n arrows).
Di with a standard Ω(p) move. As the number of arrows in all Ei is less than n, we have H(Di ) = H(Ei ) by Case 1, if there are n arrows in Di , or by induction on n otherwise. Thus H(Dt ) = H(Et ) and H(D ) = H(Dt ) = H(Et ) = H(E ). From preceding equalities it follows that H(D ) = H(D ) = H(D). – There are no components in D with more than p2 arrows. If the number of bad arrows is decreased from D to D , then H(D) = H(D ) by definition. Otherwise, we use again induction on l. If l > 0, one pushes all the ovals into e in D (and out of e in D ), as it was done in Case 2, obtaining again two situations: all ovals with their arrows are pushed and H(D ) = H(D) by definition; some arrows are transfered onto e, so that either induction on l is applied, or one gets Case 1 or Case 2. 2 The problem of computing S3 (Lp,1 ) was studied by S. Lambropoulou and J. Przytycki (see [L-P] on p. 13 of Chapter IX of the book [6]) but their paper is not yet available. The result they announce agrees with our result. Finally, the methods used in our paper should also work for the computation of the Kauffman 2-variable skein module of the lens spaces Lp,1 . Acknowledgements The authors were supported by the Polish–Slovenian grant BI-PL-2010-2011-1 and the Slovenian Research Agency grant P1-0292-0101. References [1] P. Freyd, D. Yetter, J. Hoste, W.B.R. Lickorish, K. Millett, A. Ocneanu, A new polynomial invariant of knots and links, Bull. Am. Math. Soc. 12 (1985) 239–246. [2] J. Hoste, M. Kidwell, Dichromatic link invariants, Trans. Am. Math. Soc. 321 (1) (1990) 197–229. [3] M. Mroczkowski, M. Dabkowski, KBSM of the product of a disk with two holes and S 1 , Topol. Appl. 156 (2009) 1831–1849. [4] J.H. Przytycki, Skein modules of 3-manifolds, Bull. Acad. Pol. Math. 39 (1–2) (1991) 91–100, e-print: arXiv:math/0611797. [5] J.H. Przytycki, Fundamentals of Kauffman bracket skein modules, Kobe J. Math. 16 (1) (1999) 45–66, e-print: arXiv:math/ 9809113. [6] J.H. Przytycki, KNOTS: From Combinatorics of Knot Diagrams to the Combinatorial Topology Based on Knots, Cambridge University Press, 2015, 650 pp., to appear, Chapter IX, e-print: http://arxiv.org/abs/math.GT/0602264. [7] J.H. Przytycki, P. Traczyk, Invariants of links of the Conway type, Kobe J. Math. 4 (1988) 115–139. [8] V.G. Turaev, The Conway and Kauffman modules of a solid torus, J. Sov. Math. 52 (1990) 2799–2805. [9] V.G. Turaev, Shadow links and face models of statistical mechanics, J. Differ. Geom. 36 (1) (1992) 35–74.
Contents lists available at ScienceDirect
Topology and its Applications www.elsevier.com/locate/topol
The HOMFLYPT skein module of the lens spaces Lp,1 Boštjan Gabrovšek a , Maciej Mroczkowski b,∗ a b
University of Ljubljana, Faculty of Mechanical Engineering, Aškerčeva 6, 1000 Ljubljana, Slovenia Institute of Mathematics, University of Gdansk, ul. Wita Stwosza 57, 80-952 Gdansk-Oliwa, Poland
a r t i c l e
i n f o
Article history: Received 15 April 2013 Received in revised form 11 July 2014 Accepted 11 July 2014 Available online 26 July 2014
a b s t r a c t We compute the HOMFLYPT skein module of the lens spaces Lp,1 and present a free basis of this module for each p. © 2014 Elsevier B.V. All rights reserved.
MSC: 57M27 Keywords: Knots Links Lens spaces Skein modules
1. Introduction In this paper we use diagrams of links in the solid torus with arrows, introduced in [3], to compute the HOMFLYPT skein modules of the lens spaces Lp,1 . We exhibit sets of generators of these modules and show that these sets in fact form free bases. Skein modules were introduced in 1987 by J. Przytycki [4,5]. First, we recall the definition of the HOMFLYPT skein module of an oriented 3-manifold, generalizing the HOMFLYPT polynomial of links in S 3 (see [1,7]). Let M be an oriented 3-manifold, L the set of oriented links in M up to ambient isotopy (including the empty knot) and R = Z[v ±1 , z ±1 ]. Let S be the submodule of RL generated by expressions v −1 L+ − vL− − zL0 , where L+ , L− and L0 are three identical links except inside a 3-ball where they look as in Fig. 1. We add to S the expression involving the empty knot: v −1 ∅ − v∅ − zt0 , where t0 is the trivial knot. Then the HOMFLYPT skein module of M , denoted S3 (M ), is RL/S. * Corresponding author. E-mail addresses: [email protected] (B. Gabrovšek), [email protected] (M. Mroczkowski). http://dx.doi.org/10.1016/j.topol.2014.07.003 0166-8641/© 2014 Elsevier B.V. All rights reserved.
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Fig. 1. Three links in the HOMFLYPT relation.
Fig. 2. Links t3 and t3 t−1 .
Fig. 3. Reidemeister moves.
In the next section we translate the free basis of the HOMFLYPT skein module of the solid torus (computed in [2,4,8]) into diagrams with arrows and we establish some properties of these diagrams in this skein module. In Section 3 we state our main result in Theorem 1, which is proven in Section 4. 2. The HOMFLYPT skein module of the solid torus Let T be the solid torus. In [2,4,8], it was shown that S3 (T ) is free, generated by B = {tik11 ...tikss : s ∈ N, k1 < .. < ks ∈ Z∗ , i1 ..is ∈ N} ∪ {∅}, where ∅ is the empty knot. For k > 0, tk is the oriented knot in T representing k in π1 (T ) = Z, having an ascending diagram with respect to a basepoint situated close to the exterior circle of the annulus. For k < 0, tk is t−k with reversed orientation. For example, t3 and t3 t−1 are presented in Fig. 2. Instead of using classical diagrams of links in T (i.e. projections onto the annulus with information of under and overcrossings), we will use diagrams with arrows introduced in [3]. Note, that these diagrams are very similar to the notion of gleams introduced by V. Turaev in [9]. An arrow diagram is obtained from a link L in T by cutting T along a meridional disk M , projecting the resulting vertical cylinder I × D (I = [0, 1]) onto the disk D × {0} while keeping information of over- and undercrossings for the projection of L. The intersection points in L ∩ M project onto arrows in the diagram, pointing to the part of L that is close to D × {0} in the cylinder (and away from the part of L that is close to D × {1}). Two such diagrams represent the same link in T if an only if one can get from one to the other with a sequence of five Reidemeister moves, three classical and two extra ones, presented in Fig. 3.
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Fig. 4. Interpretation of the moves Ω4 and Ω5 .
Fig. 5. Transforming a classical diagram into an arrow diagram.
Fig. 6. Arrow diagram of t3 t22 t−1 t−2 .
The interpretation of the moves Ω4 and Ω5 is illustrated in Fig. 4 (with an Ω4 move on the left of the cylinder and an Ω5 move on the right). Fig. 5 shows the translation of t3 from a classical diagram to an arrow diagram. Here, the arrow labeled with 3 stands for 3 arrows next to each other. We will use this convention with any n ∈ Z, so that n = 0 means that there are no arrows and n < 0 that there are |n| arrows opposite to the one that is pictured. Each element of B has a unique standard diagram with no crossings, non-nested ovals and such that there are no arrows in the clockwise direction (the orientation is clockwise when k < 0 in tk ). For instance t3 t22 t−1 t−2 is presented in Fig. 6. Let L be a link in T . We denote by HT (L) the expression of L in S3 (T ). Thus HT (L) is a linear expression in elements of B with coefficients in the ring R. We conclude this section with some properties of S3 (T ) that will be needed later. The equalities v −1 L+ − vL− = zL0 hold by definition in any HOMFLYPT skein module (see Fig. 1). We call such equalities HOMFLYPT relations. Denote by t¯n , n > 0, an oval with n clockwise arrows on it and clockwise orientation. For n < 0, let t¯n be t¯−n with reversed orientation. Lemma 1. In S3 (T ), one can revert clockwise arrows on an oval in the sense that for n > 0 t¯n =
i
Ai T i
or
t¯−n =
i
Ai Ti
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Fig. 7. Pushing an oval through a strand.
Fig. 8. Reducing the number of arrows on the left of s.
Fig. 9. Reducing the number of arrows on the kink.
for some Ai , Ai ∈ R. Any Ti has the form tik11 ...tikss , where k1 > 0, ..., ks > 0 and i1 k1 + ... + is ks = n. Similarly, any Ti has the form tik11 ...tikss , where k1 < 0, ..., ks < 0 and i1 k1 + ... + is ks = −n. Proof. The proof is by induction. For n = 1, t¯1 is transformed into t1 with a sequence of Ω1 , Ω5 and Ω1 moves. Similarly, t¯−1 is transformed into t−1 . Suppose that the statement of the lemma is true for k < n. Let c be an oval with n clockwise arrows on it. We perform an Ω1 move on c, then push one arrow on the kink thus created with Ω5 (this arrow is counterclockwise on the kink). Using HOMFLYPT relation and Ω5 we push another arrow on the kink getting also, by smoothing the crossing, two ovals with 1 and n − 1 arrows on which induction applies. Using HOMFLYPT and Ω5 again, we push another arrow on the kink or split the oval into two ovals with n − 2 and 2 arrows on which induction applies. Continuing in this way, all arrows are pushed on the kink, becoming counterclockwise. 2 Lemma 2. Let D be a diagram of a link L with an oval c containing n arrows, n ∈ Z, and a strand s adjacent to c, that may contain arrows outside the drawn region (Fig. 7). The oval c can be pushed over s in the sense that either
holds in S3 (T ) for some Ai ∈ R, depending on the orientation of the components c and s. Proof. We consider the case where c is oriented counter-clockwise and s is oriented “upwards”. For n = 0 the proof is trivial. For n > 0 we first push an arc of c over s by Ω2 and successively apply the moves and HOMFLYPT relations in Fig. 8 until there are no arrows left on the left side of s. We resolve the last term by successively applying the moves and HOMFLYPT relations in Fig. 9. It follows that the link can be expressed as
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Fig. 10. A 1-gon and a 2-gon.
Fig. 11. The Ω(p) move.
A similar calculation shows that the above formula holds for any n < 0. The formula also holds if we reverse both orientations of c and s. If c is oriented clockwise and s is oriented “upwards” or if c is oriented counter-clockwise and s is oriented “downwards”, then one gets similarly for any n ∈ Z:
2 Lemma 3 (1- and 2-gon lemma). Let D be a diagram of a link in T and suppose that either a strand of D forms a 1-gon or two strands of D form a 2-gon (Fig. 10). Then, D can be expressed in S3 (T ) with diagrams which have less crossings than D and no more arrows than D. Proof. Let d be the disk bounded by the 1-gon or the 2-gon. We may assume that there are no 1-gons or 2-gons inside d (in the sense that the discs they bound would be included in d). Otherwise, we consider a most nested 1-gon or 2-gon instead (with no 1-gons or 2 gons inside). • Case 1: Suppose d bounds a 1-gon. By assumption, there may be only ovals inside d, which we push outside of d using Lemma 2. We continue by pushing the arrows lying on the strand of the 1-gon outside of d using HOMFLYPT relations and Ω5 . Finally, we remove the crossing of the 1-gon with Ω1 . • Case 2: Suppose d bounds a 2-gon. If there are ovals inside d, we push them outside of d using Lemma 2. We continue by pushing the arrows lying on the strands of the 2-gon outside of d using HOMFLYPT relations and Ω5 . If the interior of d is not empty, there must be a braid going from one strand of the 2-gon to the other. In this case we push the arrows on the braid outside of d and continue by pushing the crossings of the braid outside of d using HOMFLYPT relations and Ω3 . We are left with a trivial braid whose strands can be pushed outside of d through the double points of the 2-gon using HOMFLYPT relations and Ω3 . Once there are no strands inside of d, we can perform an Ω2 move (using possibly a HOMFLYPT relation before) and remove the two crossings of the 2-gon. 2
3. Construction of H and main theorem A diagram of a link in Lp,1 is the same as a diagram of a link in T . The difference comes from one extra Reidemeister move between such diagrams, the Ω(p) move shown in Fig. 11. This move corresponds to sliding an arc of the link over the meridional disk of the solid torus T that is glued to T in order to get Lp,1 .
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Let Bp = {tik11 ...tikss : s ∈ N, k1 < .. < ks ∈ Z∗ , − p2 < k1 < .. < ks ≤ p2 , i1 ..is ∈ N} ∪ {∅}. Let D be the set of arrow diagrams of links in Lp,1 . We now construct a function H : D → RBp . It will be shown, that H induces an isomorphism from S3 (Lp,1 ) to RBp . H is defined on standard diagrams of elements of B, such as the one in Fig. 6, and extended linearly to all diagrams of links in T by the formula H(D) = H(HT (D)). Here, HT (D) is to be understood as a linear expression in standard diagrams of elements of B, rather than as a linear expression in elements of B. Let g be a standard diagram of tik11 ...tikss ∈ B. Let m be the maximum of the |ki |-s. If m < p2 or m = p2 and no ki equals −m, we define H(g) = g. Otherwise, let c be a component t−m of g or, if there are no such components, let c be a component tm of g. Let g be obtained from g by performing an Ω(p) move on the component c. We say, that an arrow is bad if its orientation is opposite to the orientation of the component on which it lies. Notice, that there are less arrows in g than in g or, if this number is unchanged, there are less bad arrows in g than in g (this occurs only when m = p2 ). We define recursively H(g) = H(HT (g )). From Lemmas 1 and 2 it follows that no diagram of HT (g ) has more arrows than g (HT (g ) is computed by pushing ovals out of the oval c transformed by Ω(p) , then possibly changing clockwise arrows into anticlockwise arrows). Following the proof of these lemmas, it is also clear that no additional bad (or good) arrows will appear in the computation of HT (g ). Thus, all terms in HT (g ) have less arrows than g or the same number of arrows but less bad arrows, and the recursive process of defining H must end. Theorem 1. H induces an isomorphism from S3 (Lp,1 ) to RBp . Thus, S3 (Lp,1 ) is free with basis Bp . Proof. HT is invariant under all Reidemeister moves except Ω(p) , so H is also invariant under these moves. Also, HT satisfies the HOMFLYPT relation, so H satisfies it as well. H is invariant under some Ω(p) moves, namely the ones between pairs of diagrams g and g used in the definition of H above. Thus, H will be an isomorphism from S3 (Lp,1 ) to RBp provided it is invariant under all Ω(p) moves. This fact will be proven in the next section (see Propositions 1 and 2). 2 4. Invariance of H under Ω(p) moves Let D be a diagram consisting of an element of B to which is added an oval e in such a way that there are no nested ovals in D. The oval e has b ∈ Z arrows and it may have any orientation. Let D be obtained from D by an Ω(p) move performed on oval e. We call such a move from D to D standard (see Fig. 12). Proposition 1. H is invariant under all Ω(p) moves if it is invariant under all standard Ω(p) moves. Proof. Let D be a diagram on which we mark by s the small part on one of its arcs where Ω(p) can be performed on; s is chosen in such a manner that no arrows lie on it. Let D be obtained from D by performing Ω(p) on s. It is enough to prove that by freezing s we can express D as a linear expression of elements of B with an added oval e on which s lies. If D has at least one crossing, then it must contain a 1-gon or a 2-gon. If s lies on it, then it is easy to check, that there must be another 1-gon or 2-gon on which s does not lie (s also does not lie in the disc this 1-gon or 2-gon bounds, otherwise one could not perform an Ω(p) move on it). One may therefore apply Lemma 3 to this 1-gon or 2-gon not containing s. In this way we will eventually remove all crossings in D. Using Lemma 2 we push all ovals out of the oval on which s lies. Using Lemma 1 we revert the ovals with clockwise arrows on them, except for the oval on which s lies. All these operations (removal of crossings, pushing and reverting of ovals) can be performed simultaneously on D and D . In this way the Ω(p) move from D to D can be expressed with several standard Ω(p) moves. 2
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Fig. 12. Standard Ω(p) move.
Fig. 13. Diagrams E and E .
n We denote by Ω(p) a standard Ω(p) move from D to D (such as in Fig. 12), where one of the diagrams has n arrows and the other one has at most n arrows. Here, we are counting all the arrows as positive, even clockwise arrows (which may only lie on the component on which the move is performed).
Proposition 2. H is invariant under all standard Ω(p) moves. Proof. We will prove for any n: n (In ): if Ω(p) is a standard move from D to D , then H(D) = H(D ). This will prove the statement of the proposition. The proof is by induction on n. If n = 0 there are no Ω(p) moves, so (I0 ) is true. Suppose now that n > 0 and (Ik ) holds for all k < n. n Let Ω(p) be a standard move from D to D . Let e be the component of D on which the move is performed. We will use an inner induction on l, the number of arrows on components of D except for e. Notice, that if l = 0 there are no arrows on components outside e, so these components are mapped by HT to elements in −1 the ring R, namely v z−v , and H(D) = H(D ) by definition. There are 3 cases to consider: • Case 1: D has n arrows and D has less than n arrows. If e is a component of D with the maximal number, say max, of arrows and, moreover, bad arrows if there is any other component of D with max bad arrows on it, then H(D ) = H(D) by definition. Otherwise, there is a component d of D such that H(D) = H(D ) by definition, where D is obtained from D by performing an Ω(p) move on d. In this case, either d has more arrows than e, or it has the same number of arrows as e, but the arrows on d are bad and the ones on e are good. Let a and b be the numbers of arrows respectively on d and e. Let E be the diagram obtained from D by applying Ω2 and Ω(p) moves on d (see left of Fig. 13). Similarly, let E be the diagram obtained from D by applying Ω2 and Ω(p) moves on e (see right of Fig. 13). Now, E and E are diagrams of the same link in the torus T : one gets E from E by applying 2p times an Ω5 move on the right crossing of E , thus pushing p arrows next to the a arrows and p arrows away from the b arrows. Thus H(E ) = H(E ). To show that H(D ) = H(E ), we first transform D and E with Ω2 and Ω5 moves into Dt and Et as shown in Fig. 14.
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Fig. 14. Showing that H(D ) = H(E ) (when D has less arrows than D).
As D and D have less arrows than D, one has a ≥ b > p2 . On Dt the number of arrows on component p p−1 p e is p−1 2 + | 2 + 1 − b| = 2 + b − 1 − 2 which equals b − 1 if p is odd and b − 2 if p is even. On component d there are a arrows. Thus the number of arrows on Dt is n − 1 or n − 2. Similarly, on Et there are less than n arrows (b − 1 or b − 2 on e and a or a − 1 on d). We claim that H(Et ) = H(Dt ). Indeed, using the same methods as in Lemma 2, one can push out the a − p+1 2 arrows on d, push out any ovals out of e and, using Lemma 1, revert all ovals with clockwise arrows except for d, all this being done without increasing the number of arrows. Thus, Dt and Et are simultaneously expressed with diagrams with no crossings, Dt = i Ai Di and Et = i Ai Ei , where Ai ∈ R and Ei is obtained from Di with a standard Ω(p) move. As the number of arrows in all Di and Ei is less than n, we have H(Di ) = H(Ei ) by induction on n. Thus H(E ) = H(Et ) = H(Dt ) = H(D ). A similar argument shows that H(E ) = H(D ). Indeed, by exchanging the roles of a and b, E is obtained from E (except for the crossings, but both can be switched by going around a component with an arrow and applying Ω5 twice). Combining preceding equalities one gets H(D ) = H(D ) = H(D). • Case 2: D has less than n arrows and D has n arrows. Recall that l is the number of arrows in D outside e. If l > 0, there are ovals with arrows other than e. We push these ovals through e at the same time in D and D (i.e. in D we push them into e and in D we push them out of e). From Lemma 2, in the resulting linear expressions for D and D two situations can occur: all ovals are pushed through with their arrows, and we can apply Case 1 by reverting the roles of D and D ; some arrows are transfered onto e, which decreases l and we apply the induction on l. In the second case it may happen that the total number of arrows is decreased and the global induction on n can be applied. • Case 3: both D and D have n arrows (this occurs when p is even and the number of arrows on e is b = p2 ). Two subcases are possible: – There is a component d in D with a arrows, a > b = p2 , such that H(D) = H(D ) by definition, where D is obtained from D by performing an Ω(p) move on d. Using the same notations as in Case 1, one gets again H(E ) = H(E ) (see Fig. 13). Showing that H(D ) = H(E ) is done exactly as in Case 1: in both cases in order to apply induction on n we only use the fact that a > p2 (b can be equal to p2 ). Using several Ω5 moves D and E can be transformed simultaneously into diagrams Dt and Et shown in Fig. 15. Notice that the number of arrows in Dt is n (because a > p/2 by assumption) and the number of arrows in Et is less than n. As in Case 1, Dt and Et are simultaneously expressed with diagrams with no crossings, Dt = i Ai Di and Et = i Ai Ei , where Ai ∈ R and Ei is obtained from
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Fig. 15. Showing that H(D ) = H(E ) (when D and D have n arrows).
Di with a standard Ω(p) move. As the number of arrows in all Ei is less than n, we have H(Di ) = H(Ei ) by Case 1, if there are n arrows in Di , or by induction on n otherwise. Thus H(Dt ) = H(Et ) and H(D ) = H(Dt ) = H(Et ) = H(E ). From preceding equalities it follows that H(D ) = H(D ) = H(D). – There are no components in D with more than p2 arrows. If the number of bad arrows is decreased from D to D , then H(D) = H(D ) by definition. Otherwise, we use again induction on l. If l > 0, one pushes all the ovals into e in D (and out of e in D ), as it was done in Case 2, obtaining again two situations: all ovals with their arrows are pushed and H(D ) = H(D) by definition; some arrows are transfered onto e, so that either induction on l is applied, or one gets Case 1 or Case 2. 2 The problem of computing S3 (Lp,1 ) was studied by S. Lambropoulou and J. Przytycki (see [L-P] on p. 13 of Chapter IX of the book [6]) but their paper is not yet available. The result they announce agrees with our result. Finally, the methods used in our paper should also work for the computation of the Kauffman 2-variable skein module of the lens spaces Lp,1 . Acknowledgements The authors were supported by the Polish–Slovenian grant BI-PL-2010-2011-1 and the Slovenian Research Agency grant P1-0292-0101. References [1] P. Freyd, D. Yetter, J. Hoste, W.B.R. Lickorish, K. Millett, A. Ocneanu, A new polynomial invariant of knots and links, Bull. Am. Math. Soc. 12 (1985) 239–246. [2] J. Hoste, M. Kidwell, Dichromatic link invariants, Trans. Am. Math. Soc. 321 (1) (1990) 197–229. [3] M. Mroczkowski, M. Dabkowski, KBSM of the product of a disk with two holes and S 1 , Topol. Appl. 156 (2009) 1831–1849. [4] J.H. Przytycki, Skein modules of 3-manifolds, Bull. Acad. Pol. Math. 39 (1–2) (1991) 91–100, e-print: arXiv:math/0611797. [5] J.H. Przytycki, Fundamentals of Kauffman bracket skein modules, Kobe J. Math. 16 (1) (1999) 45–66, e-print: arXiv:math/ 9809113. [6] J.H. Przytycki, KNOTS: From Combinatorics of Knot Diagrams to the Combinatorial Topology Based on Knots, Cambridge University Press, 2015, 650 pp., to appear, Chapter IX, e-print: http://arxiv.org/abs/math.GT/0602264. [7] J.H. Przytycki, P. Traczyk, Invariants of links of the Conway type, Kobe J. Math. 4 (1988) 115–139. [8] V.G. Turaev, The Conway and Kauffman modules of a solid torus, J. Sov. Math. 52 (1990) 2799–2805. [9] V.G. Turaev, Shadow links and face models of statistical mechanics, J. Differ. Geom. 36 (1) (1992) 35–74.