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2-Adjacency between pretzel links and the trivial link
Topology and its Applications 214 (2016) 186–191
Contents lists available at ScienceDirect
Topology and its Applications www.elsevier.com/locate/topol
2-Adjacency between pretzel links and the trivial link ✩ Zhi-Xiong Tao Department of Mathematics, School of Science, Zhejiang University of Science and Technology, Hangzhou, Zhejiang, 310023, PR China
a r t i c l e
i n f o
Article history: Received 1 August 2016 Received in revised form 25 October 2016 Accepted 25 October 2016 Available online 29 October 2016 MSC: 57M25
a b s t r a c t We study 2-adjacency between a 3-strand pretzel link with two components and the trivial link, and whether the trivial knot is 2-adjacent to a pretzel knot. By investigating mainly the coefficients of their Conway polynomials, Jones polynomials and etc., we show that any nontrivial pretzel link with two-components is not 2-adjacent to the trivial link, and vice versa. Moreover, we also show that the trivial knot is not 2-adjacent to any nontrivial pretzel knot. © 2016 Elsevier B.V. All rights reserved.
Keywords: 2-Adjacency Pretzel link Conway polynomial Jones polynomial
1. Introduction A link (knot) L is called 2-adjacent to a link (knot) W , if L admits a projection D containing two crossings c1 , c2 such that switching any 0 < s ≤ 2 of them yields a projection of W [1,10,11]. Its properties can be found in [10,11]. The 2-adjacency of classical pretzel knots has been proven that only the trefoil knot and the figure-eight knot are 2-adjacent to the trivial knot [8]. In this paper, we will study the 2-adjacent relation between a two-component pretzel link and an unlink, whether the trivial knot is 2-adjacent to a pretzel knot. We conclude the following theorem. Theorem 1.1. Any nontrivial pretzel link with two components is not 2-adjacent to the trivial link and vice versa. Theorem 1.2. The trivial knot is not 2-adjacent to any nontrivial pretzel knot. ✩
The Project Supported by Zhejiang Provincial Natural Science Foundation of China (Grant No. LY12A01025). E-mail address: [email protected].
http://dx.doi.org/10.1016/j.topol.2016.10.010 0166-8641/© 2016 Elsevier B.V. All rights reserved.
Z.-X. Tao / Topology and its Applications 214 (2016) 186–191
187
In the sequel, we assume that the reader is familiar with the definitions and the basic properties of the Conway polynomial ∇(∗), the Jones polynomial V (∗; t), and the Homfly polynomial P (∗; l, m) of links (knots). They can be found in [3–6]. Convention: We always assume that L = L1 L2 is 2-adjacent to an oriented link W = W1 W2 (here L2 and W2 are empty for the case of knots, i.e. L = L1 and W = W1 ) and the two related crossings are denoted by c1 and c2 respectively. Since L is 2-adjacent to W , so there is a 2-adjacent diagram of L denoted by D(c1 , c2 ) and D(oc1 , oc2 ) is a diagram obtained from D = D(c1 , c2 ) by opening c1 and c2 respectively. The sign of c1 (resp. c2 ) is denoted by α (resp. β) and c1 ∈ L1 . Hence, L2 is W2 . an (G) denotes the coefficient of z n in the Conway polynomial of G. Moreover, lk(G) denotes the total linking number of a link G (see p. 133 in [6]). 2. Preliminary Definition 2.1. ([5]) A pretzel link is denoted by P (−ηb; q1 , · · · , qn ) with b ≥ 0, |qi | > 1, (i = 1, · · · , n) and η = ±1. Here −ηb denotes b strands which each strand has only one crossing with sign η. The condition for P (−ηb; q1 , · · · , qn ) to be a knot is that either n ≥ 0 and all of the qi ’s and n + b are odd or just one of the qi ’s is even. We call it a pretzel knot of odd type in the former case, and a pretzel knot of even type in the latter case. Fig. 1 gives an example.
Fig. 1. P (1, −3, −4, 4) = P (−1; −3, −4, 4).
It is clear that a link (knot) is 2-adjacent to an unlink (unknot) if and only if its mirror image is and if the unlinking (unknotting) number of a pretzel link (knot) is one, then it has only three strands [7] and two of them have an even number of crossings for the pretzel link. Proposition 2.2. ([8–10]) If the notations and the conditions are as the convention, then lk(L) = lk(W ) and ∇(L) = αβz 2 ∇(D(oc1 , oc2 )) + ∇(W ).
(2.1)
(1) If a3 (L) = a3 (W ), then D(oc1 , oc2 ) is a two-component link. (2) If a3 (L) = a3 (W ), then either lk(L) = 0 and D(oc1 , oc2 ) is a link with two components or D(oc1 , oc2 ) is a link with four components. (3) D(oc1 , oc2 ) is a link with four components if and only if a2 (L1 ) = a2 (W1 ); D(oc1 , oc2 ) is a link with two components if and only if a2 (L1 ) = αβ + a2 (W1 ). Moreover, for the second case, 1 , oc2 )) + ∇(W1 ), ∇(L1 ) = αβz 2 ∇(D(oc
(2.2)
1 , oc2 ) is obtained by opening c1 , c2 from L1 . where D(oc (4) If L is the trivial link and is 2-adjacent to W , then D(oc1 , oc2 ) has four components. Furthermore, if 1 , oc2 )) = 0. a4 (W1 ) = 0, then lk(D(oc
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Proposition 2.3. ([6,9,10]) If the notations and conditions are as above, then (1) m2 P (D(oc1 , oc2 )) = lα+β P (L) + (lα−β + lβ−α + l−α−β )P (W ). (2) The exponent of the lowest power of m which appears in the Homfly polynomial of G (knot or link) with c(G) components is precisely 1 − c(G). Its coefficient is
c(G) 2 −lk(G)
p1−c(G) (l) = (−l )
[−(l + l
−1
c(G)−1
)]
pj0 (l).
(2.3)
j=1
Here the meanings of pj0 (l) (j = 1, · · · , c(G)) are similar to p1−c(G) (l) for every component of G. Proposition 2.4. ([9]) If the trivial knot O is 2-adjacent to a knot G, then a2 (G) = 0, V (G; 1) = −36αβlk(D(oc1 , oc2 )) and a4 (G) = −αβlk2 (D(oc1 , oc2 )). 3. The proof of Theorem 1.1 1. Suppose L to be the pretzel link with three strands and to be 2-adjacent to the trivial link (= W ). By Convention, L2 is trivial and there exist two cases: L1 is nontrivial or trivial. Case 1. L1 is nontrivial. Clearly, L1 is the connected sum of some torus knots, i.e. L1 = m j=1 T (2, 2qj + 1), so ∇(L1 ) = m ∇(T (2, 2q + 1)). Note that the unknotting number of L is one. Since the unknotting number of j 1 j=1 T (2, 2qj +1) is (|2qj +1| −1)/2 (see [3]), i.e. T (2, 2qj +1) is the trefoil knot. Thus, L1 = m j=1 T (2, 2qj +1), qj = 1 or −2 and a2 (L1 ) = m. By Proposition 2.2, we obtain m = 1 and αβ = 1, i.e. L1 is the trefoil knot. We suppose L1 to be the negative trefoil knot, 3− 1 . Hence L = P (−3, 2q, 2r) and r = 0. Since L is 2-adjacent to the trivial link and a2 (L1 ) = 1, so by Proposition 2.2, c1 , c2 ∈ L1 and D(oc1 , oc2 ) has only two components and lk(D(oc1 , oc2 )) = lk(L) = 0, i.e. q = −r and a3 (L) = αβa1 (D(oc1 , oc2 )) = αβlk(D(oc1 , oc2 )) = 0. So, in order to prove the theorem, we only need to discuss whether a3 (L) is zero or not. Noticing that the negative trefoil knot can not be changed into the unknot by switching a positive crossing, so α = β = −1.
Fig. 2. P (−3, −2r, 2r).
The direction of L = P (−3, −2r, 2r) is as Fig. 2. Suppose sign(r) = γ. Then we have, ∇(L) − ∇(P (−3, −2γ(|r| − 1), 2r)) = −γz∇(P (−3, 2r)), · · · , ∇(P (−3, −2γ, 2r)) − ∇(P (−3, 0, 2r)) = −γz∇(P (−3, 2r)). Thus, ∇(L) = ∇(31 )∇(T (2, 2r)) − rz∇(P (−3, 2r))
Z.-X. Tao / Topology and its Applications 214 (2016) 186–191
= ∇(31 )∇(T (2, 2r)) − rz∇(T (2, 2r − 3))
189
(3.1)
= (1 + z )∇(T (2, 2r)) − rz∇(T (2, 2r − 3)). 2
According to the formula of the paper [8] (page 4): 1 3 ∇(T (2, 2x)) = sign(x)z(C|x| + C|x|+1 z 2 + z 4 · (a polynomial));
∇(T (2, 2x + 1)) = 1 +
2 z2 C|x|+1+(sign(x)−1)/2
+ z · (a polynomial). 4
(3.2) (3.3)
Here Cba indicates a combination number. If r = 1, then ∇(L) = (1 + z 2 )∇(T (2, 2)) − z = z 3 . i.e. a3 (L) = 0 or say that it is not 2-adjacent to the trivial link. If r = 2, then ∇(L) = (1 + z 2 )∇(T (2, 4)) − 2z = (1 + z 2 )(2z + z 3 ) − 2z = 3z 3 + z 5 . So a3 (L) = 0, i.e. it is not 2-adjacent to the trivial link. If r ≥ 3, from (3.2) and (3.3), 3 ∇(T (2, 2r)) = z(Cr1 + Cr+1 z 2 + · · · ) = rz + r(r2 − 1)/6z 3 + z 5 · (a polynomial), 2 z 2 + · · · = 1 + (r − 1)(r − 2)/2z 2 + z 4 · (a polynomial). ∇(T (2, 2r − 3)) = 1 + Cr−1
Thus, a3 (L) = r(r2 − 1)/6 + r − r(r − 1)(r − 2)/2 = −r(1 − 9r + 2r2 )/6. It is easy to see that it has √ only a zero point r = (9 ± 92 − 8 )/4. However, 92 − 8 is not a perfect square number. If r < 0, then 1 3 ∇(T (2, 2r)) = −z(C−r + C−r+1 z 2 + · · · ) = rz + r(r2 − 1)/6z 3 + z 5 (· · · ), 2 z 2 + · · · = 1 + (r − 1)(r − 2)/2z 2 + z 4 (· · · ). ∇(T (2, 2r − 3)) = 1 + C−r+2
It is similar to the above discussions that L is not 2-adjacent to the trivial link. Case 2. Both components of L are trivial. i.e. L = P (η, 2q, 2r). It is obvious that we just have to prove Theorem 1.1 to be true for the case r > 0. Clearly, q = −r (∵ lk(L) = 0), p−1 (L) = (−l2 )−lk(L) (−l − l−1 ) = −l − l−1 . By Proposition 2.3, p−3 (D(oc1 , oc2 )) = lα+β p−1 (L) + (lα−β + lβ−α + l−α−β )(−l − l−1 ) = −(l + l−1 )3 . So D(oc1 , oc2 ) is a four-component link, i.e. a3 (L) = 0.
Fig. 3. P (1, −4, 4).
It is similar to (3.1) that ∇(P (η, −2r, 2r)) is equal to ∇(T (2, 2r)) − rz∇(T (2, 2r + η)). i.e. a3 (L) = a3 (T (2, 2r)) − ra2 (T (2, 2r + η)) (see Fig. 3). If r = 1, solving the equation 0 = a3 (L) = 0 − a2 (2, 2 + η) = (−η − 1)/2. It is easy to see η = −1, i.e. L is an unlink. If r = 2, since a2 (T (2, 5)) = 3, a3 (T (2, 4)) = 1, a2 (T (2, 3)) = 1, so if η = −1, then a3 (L) = 1 − 6 = −5 = 0; If η = 1, then a3 (L) = 1 − 2 = −1 = 0.
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If r ≥ 3, a3 (T (2, 2r)) = r(r2 − 1)/6, a2 (T (2, 2r + η)) = (r + δ + 1)(r + δ)/2, δ = (η − 1)/2. So a3 (L) = −r(4 + 8r2 + 12rη)/24. Suppose it to be zero, then r = (−3η ± −8 + 9η 2 )/4 = (−3η ± 1)/4. It is similar to the above discussion that η = 1 or −1, i.e. r = −1 or 1. However, r = −1 contradicts the hypothesis. For r = 1, L = P (−1, −2, 2) is an unlink. 2. Suppose L to be the trivial link and to be 2-adjacent to a pretzel link with three strands (= W ). By Convention, since L2 is trivial, so is W2 , and by Proposition 2.2, lk(W ) = 0, i.e. W = P (2p + 1, −2r, 2r) and it is similar to the above discussion that p is either −1 or −2. Here if 2p + 1 > 0, we can consider its mirror image. Since whether p is −1 or −2, we have a4 (W1 ) = 0. So by Proposition 2.2, D(oc1 , oc2 ) has four components and its total linking number is zero. If p = −1, then L1 is the unknot. It is similar to the calculation of (3.1) that ∇(W ) = ∇(T (2, 2r)) − rz∇(T (2, 2r − 1)). Using the formulae (3.2) and (3.3), 3 2 ∇(W ) = [sign(r)C|r|+1 − rC[|2r−1|/2]+1 ]z 3 + z 5 · (a polynomial)
= [r(r2 − 1)/6 − r(r2 − r)/2]z 3 + z 5 · (a polynomial) So a3 (W ) = −r(2r − 1)(r − 1)/6. Since D(oc1 , oc2 ) has four components, i.e. r(2r − 1)(r − 1) = 0. So W = P (−1, −2, 2), a trivial link. If p = −2, by Proposition 2.2, the equality 1 = a2 (L1 ) = αβ + a2 (W1 ) means D(oc1 , oc2 ) to have only two components. This contradicts with the above fact. Therefore, we finish the proof of Theorem 1.1. 4. The proof of Theorem 1.2 1. Suppose L = O is the unknot and W is an even type pretzel knot P (2p + 1, 2q + 1, 2r). By the paper [8], V (W ; 1) = 3(5p + 9p2 + 4p3 + 5q + 9q 2 + 4q 3 − 12r − 24pr − 12p2 r − 24qr − 12pqr − 12q 2 r + 6r2 + 6pr2 + 6qr2 ).
(4.1)
And a2 (G) = p(p + 1)/2 + q(q + 1)/2 − r(p + q + 1); 4
4
2
2
a4 (G) = C |2p+1|+3 + C |2q+1|+3 + C |2p+1|+1 C |2q+1|+1 + 2
2
2
2
(4.2) 3 rεC|p+q+1|+1 .
(4.3)
Here ε is the sign of −p − q − 1. Proposition 2.4 tells us that the trivial knot is not 2-adjacent to a knot B with a2 (B) = 0. So by Main Theorem in [2], we will consider the following four cases (r > 0): (1) If p + q = −1, or say p = 1, q = −2, r = 1 (see Main Theorem in [2]). By (4.2), a2 (G) = 2, we know that the trivial knot is not 2-adjacent to it, since its a2 is not 0. (2) If p + q = 1, then a2 (G) = p2 − p + 1 − 2r is odd. (3) If p + q = −2, then a2 (G) = 0 ⇒ r = −(p + 1)2 < 0, it contradicts with the hypothesis. (4) If p + q = 0, then a2 (G) = 0 ⇒ p2 = r. p = ±1 means G = P (3, −1, 2) or P (−1, 3, 2). They are trivial. For p ≥ 2, by (4.3), a4 (G) = p2 (p2 − 1)/3. So p2 (p2 − 1)/3 = αβlk2 (D(oc1 , oc2 )) and αβ = 1. Putting V (G; 1) = −36lk(D(oc1 , oc2 )) in Proposition 2.4 into the above equality and using (4.1), an impossible equality is obtained: 4 = 3p2 (p2 − 1). For p ≤ −2, by (4.3), a4 (G) = p2 (p2 − 1)/3. So the result is the same as p ≥ 2.
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2. Suppose L = O is the unknot and W is an odd type pretzel knot P (2p + 1, 2q + 1, 2r + 1). So ∇(G; z) = 1 + (1 + p + q + r + pq + pr + qr)z 2 (see [8]), i.e. a2 (G) = 1 + p + q + r + pq + pr + qr and a4 (G) = 0. From the paper [8] again, we know V (G; 1) = −18(1 + 2r + r2 + 2p + 3pr + pr2 + p2 + p2 r + 2q + 3qr + qr2 + 3pq + 4pqr + p2 q + q 2 + q 2 r + pq 2 ).
(4.4)
By Proposition 2.4, a2 (G) = 0, i.e. 1 + p + q + r + pq + pr + qr = 0.
(4.5)
Since a4 (G) = 0, so V (G; 1) = 0 follows from Proposition 2.4. Using (4.5), equality (4.4) can be simplified into pqr = 0, i.e. two strands of G are 1 and −1 respectively (e.g. p = 0, from (4.5), we have (1 + q)(1 + r) = 0, i.e. r = −1 or q = −1). In other words, G is trivial. Therefore, we finish the proof. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
N. Askitas, E. Kalfagianni, On knot adjacency, Topol. Appl. 126 (2002) 63–81. D. Buck, J. Gibbons, E. Staron, Pretzel knots with unknotting number one, Commun. Anal. Geom. 21 (2) (2012) 365–408. G. Burde, H. Zieschang, Knots, de Gruyter, Berlin, New York, 1985. L.H. Kauffman, On Knots, Ann. Math. Stud., vol. 115, Princeton Univ. Press, 1987. A. Kawauchi, A Survey of Knot Theory, Birkhäuser, 1996. W.B.R. Lickorish, K.C. Millett, A polynomial invariant of oriented links, Topology 26 (1) (1987) 107–141. K. Motegi, A note on unlinking numbers of Montesinos links, Rev. Mat. Univ. Complut. Madr. 9 (1) (1996) 151–164. Zhi-xiong Tao, On 2-adjacency of classical pretzel knots, J. Knot Theory Ramif. 22 (11) (2013) 1350066. Zhi-xiong Tao, 2-Adjacency between knots, J. Knot Theory Ramif. 24 (11) (2015) 1550054. Zhi-xiong Tao, On 2-adjacency between links, Chin. Ann. Math., Ser. B 37 (5) (2016) 767–776. I. Torisu, On 2-adjacency relation of two-bridge knots and links, J. Aust. Math. Soc. 84 (1) (2008) 139–144.
Contents lists available at ScienceDirect
Topology and its Applications www.elsevier.com/locate/topol
2-Adjacency between pretzel links and the trivial link ✩ Zhi-Xiong Tao Department of Mathematics, School of Science, Zhejiang University of Science and Technology, Hangzhou, Zhejiang, 310023, PR China
a r t i c l e
i n f o
Article history: Received 1 August 2016 Received in revised form 25 October 2016 Accepted 25 October 2016 Available online 29 October 2016 MSC: 57M25
a b s t r a c t We study 2-adjacency between a 3-strand pretzel link with two components and the trivial link, and whether the trivial knot is 2-adjacent to a pretzel knot. By investigating mainly the coefficients of their Conway polynomials, Jones polynomials and etc., we show that any nontrivial pretzel link with two-components is not 2-adjacent to the trivial link, and vice versa. Moreover, we also show that the trivial knot is not 2-adjacent to any nontrivial pretzel knot. © 2016 Elsevier B.V. All rights reserved.
Keywords: 2-Adjacency Pretzel link Conway polynomial Jones polynomial
1. Introduction A link (knot) L is called 2-adjacent to a link (knot) W , if L admits a projection D containing two crossings c1 , c2 such that switching any 0 < s ≤ 2 of them yields a projection of W [1,10,11]. Its properties can be found in [10,11]. The 2-adjacency of classical pretzel knots has been proven that only the trefoil knot and the figure-eight knot are 2-adjacent to the trivial knot [8]. In this paper, we will study the 2-adjacent relation between a two-component pretzel link and an unlink, whether the trivial knot is 2-adjacent to a pretzel knot. We conclude the following theorem. Theorem 1.1. Any nontrivial pretzel link with two components is not 2-adjacent to the trivial link and vice versa. Theorem 1.2. The trivial knot is not 2-adjacent to any nontrivial pretzel knot. ✩
The Project Supported by Zhejiang Provincial Natural Science Foundation of China (Grant No. LY12A01025). E-mail address: [email protected].
http://dx.doi.org/10.1016/j.topol.2016.10.010 0166-8641/© 2016 Elsevier B.V. All rights reserved.
Z.-X. Tao / Topology and its Applications 214 (2016) 186–191
187
In the sequel, we assume that the reader is familiar with the definitions and the basic properties of the Conway polynomial ∇(∗), the Jones polynomial V (∗; t), and the Homfly polynomial P (∗; l, m) of links (knots). They can be found in [3–6]. Convention: We always assume that L = L1 L2 is 2-adjacent to an oriented link W = W1 W2 (here L2 and W2 are empty for the case of knots, i.e. L = L1 and W = W1 ) and the two related crossings are denoted by c1 and c2 respectively. Since L is 2-adjacent to W , so there is a 2-adjacent diagram of L denoted by D(c1 , c2 ) and D(oc1 , oc2 ) is a diagram obtained from D = D(c1 , c2 ) by opening c1 and c2 respectively. The sign of c1 (resp. c2 ) is denoted by α (resp. β) and c1 ∈ L1 . Hence, L2 is W2 . an (G) denotes the coefficient of z n in the Conway polynomial of G. Moreover, lk(G) denotes the total linking number of a link G (see p. 133 in [6]). 2. Preliminary Definition 2.1. ([5]) A pretzel link is denoted by P (−ηb; q1 , · · · , qn ) with b ≥ 0, |qi | > 1, (i = 1, · · · , n) and η = ±1. Here −ηb denotes b strands which each strand has only one crossing with sign η. The condition for P (−ηb; q1 , · · · , qn ) to be a knot is that either n ≥ 0 and all of the qi ’s and n + b are odd or just one of the qi ’s is even. We call it a pretzel knot of odd type in the former case, and a pretzel knot of even type in the latter case. Fig. 1 gives an example.
Fig. 1. P (1, −3, −4, 4) = P (−1; −3, −4, 4).
It is clear that a link (knot) is 2-adjacent to an unlink (unknot) if and only if its mirror image is and if the unlinking (unknotting) number of a pretzel link (knot) is one, then it has only three strands [7] and two of them have an even number of crossings for the pretzel link. Proposition 2.2. ([8–10]) If the notations and the conditions are as the convention, then lk(L) = lk(W ) and ∇(L) = αβz 2 ∇(D(oc1 , oc2 )) + ∇(W ).
(2.1)
(1) If a3 (L) = a3 (W ), then D(oc1 , oc2 ) is a two-component link. (2) If a3 (L) = a3 (W ), then either lk(L) = 0 and D(oc1 , oc2 ) is a link with two components or D(oc1 , oc2 ) is a link with four components. (3) D(oc1 , oc2 ) is a link with four components if and only if a2 (L1 ) = a2 (W1 ); D(oc1 , oc2 ) is a link with two components if and only if a2 (L1 ) = αβ + a2 (W1 ). Moreover, for the second case, 1 , oc2 )) + ∇(W1 ), ∇(L1 ) = αβz 2 ∇(D(oc
(2.2)
1 , oc2 ) is obtained by opening c1 , c2 from L1 . where D(oc (4) If L is the trivial link and is 2-adjacent to W , then D(oc1 , oc2 ) has four components. Furthermore, if 1 , oc2 )) = 0. a4 (W1 ) = 0, then lk(D(oc
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Proposition 2.3. ([6,9,10]) If the notations and conditions are as above, then (1) m2 P (D(oc1 , oc2 )) = lα+β P (L) + (lα−β + lβ−α + l−α−β )P (W ). (2) The exponent of the lowest power of m which appears in the Homfly polynomial of G (knot or link) with c(G) components is precisely 1 − c(G). Its coefficient is
c(G) 2 −lk(G)
p1−c(G) (l) = (−l )
[−(l + l
−1
c(G)−1
)]
pj0 (l).
(2.3)
j=1
Here the meanings of pj0 (l) (j = 1, · · · , c(G)) are similar to p1−c(G) (l) for every component of G. Proposition 2.4. ([9]) If the trivial knot O is 2-adjacent to a knot G, then a2 (G) = 0, V (G; 1) = −36αβlk(D(oc1 , oc2 )) and a4 (G) = −αβlk2 (D(oc1 , oc2 )). 3. The proof of Theorem 1.1 1. Suppose L to be the pretzel link with three strands and to be 2-adjacent to the trivial link (= W ). By Convention, L2 is trivial and there exist two cases: L1 is nontrivial or trivial. Case 1. L1 is nontrivial. Clearly, L1 is the connected sum of some torus knots, i.e. L1 = m j=1 T (2, 2qj + 1), so ∇(L1 ) = m ∇(T (2, 2q + 1)). Note that the unknotting number of L is one. Since the unknotting number of j 1 j=1 T (2, 2qj +1) is (|2qj +1| −1)/2 (see [3]), i.e. T (2, 2qj +1) is the trefoil knot. Thus, L1 = m j=1 T (2, 2qj +1), qj = 1 or −2 and a2 (L1 ) = m. By Proposition 2.2, we obtain m = 1 and αβ = 1, i.e. L1 is the trefoil knot. We suppose L1 to be the negative trefoil knot, 3− 1 . Hence L = P (−3, 2q, 2r) and r = 0. Since L is 2-adjacent to the trivial link and a2 (L1 ) = 1, so by Proposition 2.2, c1 , c2 ∈ L1 and D(oc1 , oc2 ) has only two components and lk(D(oc1 , oc2 )) = lk(L) = 0, i.e. q = −r and a3 (L) = αβa1 (D(oc1 , oc2 )) = αβlk(D(oc1 , oc2 )) = 0. So, in order to prove the theorem, we only need to discuss whether a3 (L) is zero or not. Noticing that the negative trefoil knot can not be changed into the unknot by switching a positive crossing, so α = β = −1.
Fig. 2. P (−3, −2r, 2r).
The direction of L = P (−3, −2r, 2r) is as Fig. 2. Suppose sign(r) = γ. Then we have, ∇(L) − ∇(P (−3, −2γ(|r| − 1), 2r)) = −γz∇(P (−3, 2r)), · · · , ∇(P (−3, −2γ, 2r)) − ∇(P (−3, 0, 2r)) = −γz∇(P (−3, 2r)). Thus, ∇(L) = ∇(31 )∇(T (2, 2r)) − rz∇(P (−3, 2r))
Z.-X. Tao / Topology and its Applications 214 (2016) 186–191
= ∇(31 )∇(T (2, 2r)) − rz∇(T (2, 2r − 3))
189
(3.1)
= (1 + z )∇(T (2, 2r)) − rz∇(T (2, 2r − 3)). 2
According to the formula of the paper [8] (page 4): 1 3 ∇(T (2, 2x)) = sign(x)z(C|x| + C|x|+1 z 2 + z 4 · (a polynomial));
∇(T (2, 2x + 1)) = 1 +
2 z2 C|x|+1+(sign(x)−1)/2
+ z · (a polynomial). 4
(3.2) (3.3)
Here Cba indicates a combination number. If r = 1, then ∇(L) = (1 + z 2 )∇(T (2, 2)) − z = z 3 . i.e. a3 (L) = 0 or say that it is not 2-adjacent to the trivial link. If r = 2, then ∇(L) = (1 + z 2 )∇(T (2, 4)) − 2z = (1 + z 2 )(2z + z 3 ) − 2z = 3z 3 + z 5 . So a3 (L) = 0, i.e. it is not 2-adjacent to the trivial link. If r ≥ 3, from (3.2) and (3.3), 3 ∇(T (2, 2r)) = z(Cr1 + Cr+1 z 2 + · · · ) = rz + r(r2 − 1)/6z 3 + z 5 · (a polynomial), 2 z 2 + · · · = 1 + (r − 1)(r − 2)/2z 2 + z 4 · (a polynomial). ∇(T (2, 2r − 3)) = 1 + Cr−1
Thus, a3 (L) = r(r2 − 1)/6 + r − r(r − 1)(r − 2)/2 = −r(1 − 9r + 2r2 )/6. It is easy to see that it has √ only a zero point r = (9 ± 92 − 8 )/4. However, 92 − 8 is not a perfect square number. If r < 0, then 1 3 ∇(T (2, 2r)) = −z(C−r + C−r+1 z 2 + · · · ) = rz + r(r2 − 1)/6z 3 + z 5 (· · · ), 2 z 2 + · · · = 1 + (r − 1)(r − 2)/2z 2 + z 4 (· · · ). ∇(T (2, 2r − 3)) = 1 + C−r+2
It is similar to the above discussions that L is not 2-adjacent to the trivial link. Case 2. Both components of L are trivial. i.e. L = P (η, 2q, 2r). It is obvious that we just have to prove Theorem 1.1 to be true for the case r > 0. Clearly, q = −r (∵ lk(L) = 0), p−1 (L) = (−l2 )−lk(L) (−l − l−1 ) = −l − l−1 . By Proposition 2.3, p−3 (D(oc1 , oc2 )) = lα+β p−1 (L) + (lα−β + lβ−α + l−α−β )(−l − l−1 ) = −(l + l−1 )3 . So D(oc1 , oc2 ) is a four-component link, i.e. a3 (L) = 0.
Fig. 3. P (1, −4, 4).
It is similar to (3.1) that ∇(P (η, −2r, 2r)) is equal to ∇(T (2, 2r)) − rz∇(T (2, 2r + η)). i.e. a3 (L) = a3 (T (2, 2r)) − ra2 (T (2, 2r + η)) (see Fig. 3). If r = 1, solving the equation 0 = a3 (L) = 0 − a2 (2, 2 + η) = (−η − 1)/2. It is easy to see η = −1, i.e. L is an unlink. If r = 2, since a2 (T (2, 5)) = 3, a3 (T (2, 4)) = 1, a2 (T (2, 3)) = 1, so if η = −1, then a3 (L) = 1 − 6 = −5 = 0; If η = 1, then a3 (L) = 1 − 2 = −1 = 0.
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If r ≥ 3, a3 (T (2, 2r)) = r(r2 − 1)/6, a2 (T (2, 2r + η)) = (r + δ + 1)(r + δ)/2, δ = (η − 1)/2. So a3 (L) = −r(4 + 8r2 + 12rη)/24. Suppose it to be zero, then r = (−3η ± −8 + 9η 2 )/4 = (−3η ± 1)/4. It is similar to the above discussion that η = 1 or −1, i.e. r = −1 or 1. However, r = −1 contradicts the hypothesis. For r = 1, L = P (−1, −2, 2) is an unlink. 2. Suppose L to be the trivial link and to be 2-adjacent to a pretzel link with three strands (= W ). By Convention, since L2 is trivial, so is W2 , and by Proposition 2.2, lk(W ) = 0, i.e. W = P (2p + 1, −2r, 2r) and it is similar to the above discussion that p is either −1 or −2. Here if 2p + 1 > 0, we can consider its mirror image. Since whether p is −1 or −2, we have a4 (W1 ) = 0. So by Proposition 2.2, D(oc1 , oc2 ) has four components and its total linking number is zero. If p = −1, then L1 is the unknot. It is similar to the calculation of (3.1) that ∇(W ) = ∇(T (2, 2r)) − rz∇(T (2, 2r − 1)). Using the formulae (3.2) and (3.3), 3 2 ∇(W ) = [sign(r)C|r|+1 − rC[|2r−1|/2]+1 ]z 3 + z 5 · (a polynomial)
= [r(r2 − 1)/6 − r(r2 − r)/2]z 3 + z 5 · (a polynomial) So a3 (W ) = −r(2r − 1)(r − 1)/6. Since D(oc1 , oc2 ) has four components, i.e. r(2r − 1)(r − 1) = 0. So W = P (−1, −2, 2), a trivial link. If p = −2, by Proposition 2.2, the equality 1 = a2 (L1 ) = αβ + a2 (W1 ) means D(oc1 , oc2 ) to have only two components. This contradicts with the above fact. Therefore, we finish the proof of Theorem 1.1. 4. The proof of Theorem 1.2 1. Suppose L = O is the unknot and W is an even type pretzel knot P (2p + 1, 2q + 1, 2r). By the paper [8], V (W ; 1) = 3(5p + 9p2 + 4p3 + 5q + 9q 2 + 4q 3 − 12r − 24pr − 12p2 r − 24qr − 12pqr − 12q 2 r + 6r2 + 6pr2 + 6qr2 ).
(4.1)
And a2 (G) = p(p + 1)/2 + q(q + 1)/2 − r(p + q + 1); 4
4
2
2
a4 (G) = C |2p+1|+3 + C |2q+1|+3 + C |2p+1|+1 C |2q+1|+1 + 2
2
2
2
(4.2) 3 rεC|p+q+1|+1 .
(4.3)
Here ε is the sign of −p − q − 1. Proposition 2.4 tells us that the trivial knot is not 2-adjacent to a knot B with a2 (B) = 0. So by Main Theorem in [2], we will consider the following four cases (r > 0): (1) If p + q = −1, or say p = 1, q = −2, r = 1 (see Main Theorem in [2]). By (4.2), a2 (G) = 2, we know that the trivial knot is not 2-adjacent to it, since its a2 is not 0. (2) If p + q = 1, then a2 (G) = p2 − p + 1 − 2r is odd. (3) If p + q = −2, then a2 (G) = 0 ⇒ r = −(p + 1)2 < 0, it contradicts with the hypothesis. (4) If p + q = 0, then a2 (G) = 0 ⇒ p2 = r. p = ±1 means G = P (3, −1, 2) or P (−1, 3, 2). They are trivial. For p ≥ 2, by (4.3), a4 (G) = p2 (p2 − 1)/3. So p2 (p2 − 1)/3 = αβlk2 (D(oc1 , oc2 )) and αβ = 1. Putting V (G; 1) = −36lk(D(oc1 , oc2 )) in Proposition 2.4 into the above equality and using (4.1), an impossible equality is obtained: 4 = 3p2 (p2 − 1). For p ≤ −2, by (4.3), a4 (G) = p2 (p2 − 1)/3. So the result is the same as p ≥ 2.
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2. Suppose L = O is the unknot and W is an odd type pretzel knot P (2p + 1, 2q + 1, 2r + 1). So ∇(G; z) = 1 + (1 + p + q + r + pq + pr + qr)z 2 (see [8]), i.e. a2 (G) = 1 + p + q + r + pq + pr + qr and a4 (G) = 0. From the paper [8] again, we know V (G; 1) = −18(1 + 2r + r2 + 2p + 3pr + pr2 + p2 + p2 r + 2q + 3qr + qr2 + 3pq + 4pqr + p2 q + q 2 + q 2 r + pq 2 ).
(4.4)
By Proposition 2.4, a2 (G) = 0, i.e. 1 + p + q + r + pq + pr + qr = 0.
(4.5)
Since a4 (G) = 0, so V (G; 1) = 0 follows from Proposition 2.4. Using (4.5), equality (4.4) can be simplified into pqr = 0, i.e. two strands of G are 1 and −1 respectively (e.g. p = 0, from (4.5), we have (1 + q)(1 + r) = 0, i.e. r = −1 or q = −1). In other words, G is trivial. Therefore, we finish the proof. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
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