key: cord-0219270-9dtxdniu
authors: Lee, Sangyop; Paiva, Thiago de
title: Torus knots obtained by negatively twisting torus knots
date: 2021-08-24
journal: nan
DOI: nan
sha: 834c6eb420170fd69674998361677b1ecfa2b061
doc_id: 219270
cord_uid: 9dtxdniu

Twisted torus knots are torus knots with some full twists added along some number of adjacent strands. There are infinitely many known examples of twisted torus knots which are actually torus knots. We give eight more infinite families of such twisted torus knots with a single negative twist.

For s = 1, Lee determined twisted torus knots which are torus knots [12] and Paiva, the second author, found an infinite family of satellite twisted torus knots [17] .

In the remainder of the paper, we assume s = −1 and focus on the case that T (p, q, r, s) becomes a torus knot. Guntel found the first family of such knots by showing that the twisted torus knots T ((k + 1)q − 1, q, q − 1, −1) are the torus knots T (kq + 1, q) where q ≥ 3 and k ≥ 2 [7] . For twisted torus knots T (p, q, r, −1) whose parameters (p, q, r) are of the form (p, q, p − kq) [13] or satisfy (q <)p < r ≤ p + q [14] , Lee determined which of them are torus knots. In this paper, we find eight new infinite families of such twisted torus knots. (1) T (mn + m + 1, mn + 1, mn, −1) = T (mn + n + 1, m + 1);

(2) T (mn + m + 1, mn + 1, mn + m, −1) = T (mn + m − n, −m + 1);

(3) T (mn + m + 1, mn + 1, mn + 2, −1) = T (mn − n + 1, m − 1); (4) T (mn+m−1, mn−1, mn+m−2, −1) = T (mn+m−n−2, −m+1); (5) T (mn + m − 1, mn − 1, mn, −1) = T (mn − n − 1, m − 1); (6) T (2n + 1, n, 2n − 1, −1) = T (2n − 3, −n + 1); (7) T (3n − 1, n, n + 1, −1) = T (2n − 1, n − 1); and (8) T (3n + 1, n, 3n − 1, −1) = T (3n − 2, −2n + 1). Here, we assume that mn ≥ 2 for (4) and (5) .

For q < p ≤ 30 and r ≤ 29, we used SnapPy to verify that if T (p, q, r, −1) is a torus knot, then it is one of the knots in [ 

In this section, we prepare some braid isotopies. For this, we first simplify braid diagrams in the following way. By assigning a nonnegative integer j to a single strand, we mean j parallel strands without any twists. For positive integers k and ℓ, let (k, ℓ) denote the (k, ℓ)-torus braid, i.e., (k, ℓ) is the braid (σ 1 σ 2 · · · σ k−1 ) ℓ , where σ i is an elementary braid which is obtained from the trivial braid on k strands by letting the ith strand cross under the (i + 1)st strand (see [1, Figure 2] ). Let (k, ℓ) denote the braid (σ k−1 σ k−2 · · · σ 1 ) ℓ . Let (k, −ℓ) and (k, −ℓ) denote the mirror images of (k, ℓ) and (k, ℓ), respectively. Here, by the mirror image of a given braid β, we mean the braid obtained from β by changing all crossings. Note that (k, −ℓ) is the braid

Let ℓ k denote ℓ full twists on k strands. Then ℓ k = (k, ℓk) and ℓ k = (k, ℓk). Note that ℓ k = (k, ℓk) is the mirror image of −ℓ k . See Figure 1 . It is easy to see that if ℓ 1 and ℓ 2 are integers, then (k, ℓ 1 ) · (k, ℓ 2 ) = (k, ℓ 1 + ℓ 2 ) and (k, ℓ 1 ) · (k, ℓ 2 ) = (k, ℓ 1 + ℓ 2 ), where β 1 · β 2 is the braid obtained by stacking the braid β 1 on top of the braid β 2 . Proof. It is enough to prove that the mirror image of −1 k , which is 1 k , is isotopic to 1 k . We prove this by an induction on k. When k = 1 or 2, this is obviously true. Suppose that 1 k is isotopic to 1 k . One easily sees that 1 k+1 is isotopic to the braid on the left of Figure 2 Proof. This follows immediately from the definitions at the beginning of this section. For each x = a, b, c, d, we call the move from the left braid in Figure 4 (x) to the right a generalized destabilization and the move from the right to the left a generalized stabilization.

Lemma 2.5. Let β 1 , β 2 be braids on k strands. Then the closures of β 1 · β 2 and β 2 · β 1 are equivalent knots or links.

Proof. It is well known that conjugate braids yield equivalent knots or links. Also, for two group elements a and b, ab is conjugate to ba(= a −1 (ab)a). Lemma 2.6. Let j, k, ℓ be positive integers with k ≤ ℓ < j. Then the braid in the center of Figure 5 

for each x = a, b. In particular, if k = ℓ, then the central braid is isotopic to any of the lower braids. Also, if k = 1, then the five braids in Figure 5 (y) are isotopic for each y = c, d. Figure 5 (a) are isotopic as shown in Figure 6 , which illustrates the case that (j, k, ℓ) = (9, 3, 5) . Similarly for the braids in Figure 5 (b). The last two statements of the lemma follow immediately from the general case. The first/last two braids in Figure 5 Lemma 2.7. Let p, q, r be positive integers such that p ≥ r ≥ q, r + q ≥ p, and p, q are coprime. Then the following hold. • From (f) to (g): Split the family of r + q − p parallel strands into two families, one containing r − q parallel strands and the other 2q − p parallel strands. Note that r − q ≥ 0 and 2q − p ≥ 0. • From (g) to (h): Apply a generalized destabilization. • From (h) to (i): Combine the two negative full twists on r−q strands into −2 r−q . Suppose r ≤ 2q ≤ p. One can see that the closures of the braids in Figure  7 (f),(j)∼(l) are equivalent knots as follows:

• From (f) to (j): Split the family of p − r parallel strands into two families, one containing p − 2q parallel strands and the other 2q − r parallel strands. Note that p − 2q ≥ 0 and 2q − r ≥ 0. • From (j) to (k): Note that (p − 2q) + (r + q − p) = r − q. Apply a generalized destabilization. • From (k) to (l): Combine the two negative full twists on r−q strands into −2 r−q . This completes the proof. • From (f) to (g): Combine the braids 1 m+1 and (n − k) m+1 to obtain (n − k + 1) m+1 . Noting that the braid in Figure 8 (g) is β k−1 , one sees that the closures of braids β k (k = 0, 1, . . . , n) are the same knot. In particular, it is clear that β 0 is the braid in Figure 8 (h) and its closure is the torus knot T (mn + n + 1, m + 1). Lemma 2.9. Let m, n, ε be integers such that m, n are positive and ε = ±1. Let α be a braid on m − 1 strands. Let β(ε, α) and γ(ε, α) denote the braids in Figure 9(a) and (b) , respectively. Then the closures of these braids are equivalent knots or links.

Proof. For an integer k(1 ≤ k ≤ n), let β k (ε, α) denote the braid in Figure  10 (a). In particular, β n (ε, α) = β(ε, α).

One can see that the closures of the braids in Figure 8 (a)∼(g) are equivalent knots or links as follows: • From (f) to (g): Destabilize the braid in Figure 10 (f).

• From (g) to (h): Apply Lemma 2.3. One sees that the closure of the braid in Figure 10 (h) is equivalent to that of β k−1 (ε, α) by Lemma 2.5. Thus the closures of the braids β k (ε, α) are equivalent knots or links for all k = 1, . . . , n.

Consider the braid β 1 (ε, α). If ε = −1, then one easily sees that β 1 (ε, α) = γ(ε, α). Suppose ε = 1. Then β 1 (ε, α) is the braid in Figure 11 (a). One can see that the closures of the braids in Figure 11 In this section, we prove Theorem 1.1.

(1) Consider the twisted torus knot T (mn + m + 1, mn + 1, mn, −1). It is the closure of the braid in Figure 12 (a). The torus braid (mn+m+1, mn+1) splits into two torus braids (mn + m + 1, 1) and (mn + m + 1, mn) as shown in Figure 12 (b). The lower left isotopy in Figure 5 (a) with (j, k) = (mn + m + 1, mn) yields the braid in Figure 12 (c). The braids 1 mn and −1 mn are merged into a trivial braid on mn strands, so we get the braid in Figure  12 (d), which is β n in Lemma 2.8. Hence T (mn + m + 1, mn + 1, mn, −1) = T (mn + n + 1, m + 1) by Lemma 2.8.

(2) Let p = mn+m+1, q = mn+1, r = mn+m. Then p ≥ r ≥ q, r+q ≥ p and 2q > p. By Lemma 2.7(1) T (mn+m+1, mn+1, mn+m, −1) is obtained by closing the braid in Figure 7 Figure 7 (i) is the mirror image of the braid β(1, 2 m−1 ) in Lemma 2.9 after a π-rotation. It is easy to see that the braid γ(1, 2 m−1 ) in the lemma is closed to be the torus knot T (m − 1, mn + m − n). Thus T (mn + m + 1, mn + 1, mn + m, −1) is the mirror image of T (m − 1, mn + m − n) by Lemma 2.9, i.e., T (mn + m + 1, mn + 1, mn + m, −1) = T (mn + m − n, −m + 1).

(3) Let p = mn+m+1, q = mn+1, r = mn+2. Then p ≥ r ≥ q, r +q ≥ p and 2q > p. By Lemma 2.7(1) T (mn+m+1, mn+1, mn+2, −1) is obtained by closing the braid in Figure 7 (i). Note that

Hence one can see that the braid in Figure 7 (i) is the braid β(1, 1 m−1 ) in Lemma 2.9. It is easy to see that the braid γ(1, 1 m−1 ) in the lemma is closed to be the torus knot T (m − 1, mn − n + 1). Thus T (mn + m + 1, mn + 1, mn+2, −1) = T (m−1, mn−n+1) = T (mn−n+1, m−1) by Lemma 2.9.

(4) Let p = mn + m − 1, q = mn − 1, r = mn + m − 2, where mn ≥ 2. Then p ≥ r ≥ q, r + q ≥ p and 2q > p except when (m, n) = (1, 2) or n = 1: in the former, both of T (p, q, r, −1) = T (2, 1, 1, −1) and T (mn + m − n − 2, −m + 1) = T (−1, 0) are the unknot, and in the latter, T (p, q, r, Theorem 1] . By Lemma 2.7(1) T (mn+m+1, mn+ 1, mn + 2, −1) is obtained by closing the braid in Figure 7 (i). Note that

Hence by using Lemma 2.2, one can see that the braid in Figure 7 (i) is the mirror image of the braid β(−1, 2 m−1 ) in Lemma 2.9 after a π-rotation. It is easy to see that the braid γ(−1, 2 m−1 ) in the lemma is closed to be the torus knot T (m−1, mn+m−n−2). Thus T (mn+m−1, mn−1, mn+m−2, −1) = T (mn + m − n − 2, −m + 1) by Lemma 2.9.

(5) Let p = mn+m−1, q = mn−1, r = mn, where mn ≥ 2. Then p ≥ r ≥ q, r + q ≥ p and 2q > p except when (m, n) = (1, 2) or n = 1: in the former, both of T (p, q, r, −1) = T (2, 1, 2, −1) and T (mn − n − 1, m − 1) = T (−1, 0) are the unknot, and in the latter, T (p, q, r,

Fibonacci numbers with f 1 = f 2 = 1). By Lemma 2.7(1) T (mn + m − 1, mn − 1, mn, −1) is obtained by closing the braid in Figure 7 (i). Note that

Hence one can see that the braid in Figure 7 (i) is the braid β(−1, 1 m−1 ) in Lemma 2.9. It is easy to see that the braid γ(−1, 1 m−1 ) in the lemma is closed to be the torus knot T (m − 1, mn − n − 1). Thus T (mn + m − 1, mn − 1, mn, −1) = T (mn − n − 1, m − 1) by Lemma 2.9.

(6) Let p = 2n + 1, q = n, r = 2n − 1. Then p ≥ r ≥ q, r + q ≥ p and p ≥ 2q ≥ r except when n = 1: in this case, both of T (p, q, r, −1) = T (3, 1, 1, −1) and T (2n − 3, −n + 1) = T (−1, 0) are the unknot. By Lemma 2.7(2) T (2n + 1, n, 2n − 1, −1) is obtained by closing the braid in Figure  7 (l). This braid is the first braid in Figure 13 . We obtain the second braid in the figure after a second Redemeister move and then the third braid after a destabilization. By using [13, Lemma 2.4], Lemma 2.5, and the fact that a full twist commutes with any braid, one easily sees that the closure of the third braid is the torus knot T (n − 1, −2n + 3). Thus T (2n + 1, n, 2n − 1, −1) = T (2n − 3, −n + 1). Figure 13 . Braid isotopy (7) Consider the twisted torus knot T (3n − 1, n, n + 1, −1). If n = 1, then T (3n−1, n, n+1, −1) = T (2, 1, 2, −1) and T (2n−1, n−1) = T (1, 0) are both the unknot. Hence we may assume n ≥ 2. The knot T (3n − 1, n, n + 1, −1) (8) Consider the twisted torus knot T (3n + 1, n, 3n − 1, −1). If n = 1, then T (3n + 1, n, 3n − 1, −1) = T (4, 1, 2, −1) and T (3n − 2, −2n + 1) = T (1, −1) are both the unknot. Hence we may assume n ≥ 2. The knot T (3n + 1, n, 3n − 1, −1) is the closure of the braid in Figure 15 (a). One can Figure 15 . The closures of these braids are the same knot. see that the closures of the braids in Figure 14 (a)∼(k) are equivalent knots as follows:

• From (a) to (b): Apply the upper left isotopy in Figure 5 (a) with letting (j, k, ℓ) = (3n + 1, n, 3n − 1). • From (b) to (c): Combine the braids (3n − 1, n) and −1 3n−1 , and split the family of n parallel strands into two families, one containing 2 parallel strands and the other n − 2 parallel strands. • From (c) to (d): Apply a generalized destabilization.

The braids (n, 2) and 1 2 are the mirror images of the braids (n, −2) and −1 2 , respectively. Thus one can see that the mirror images of the braids in (d) and (e) are isotopic by using the lower right isotopy in Figure 5(b)

Split the torus braid (2n − 1, −(3n − 1)) into two torus braids

Split the torus braid (n, 2) into two of torus braids (n, 1), and apply the lower left isotopy in Figure 5(b) with letting (j, k) = (2n − 1

Apply the lower left isotopy in Figure 5(b) with letting (j, k) = (2n − 1, n − 1), and split the torus braid (2n − 1, −2n) into two torus braids

By using Lemma 2.5, delete the upper (n, 1) and attach it to the bottom. Also, by applying the right isotopy in Figure 5(d) with letting (j, ℓ) = (2n − 1, n), divide the torus braid

−1) in the braid in Figure 15(l) can be canceled

Braids: A survey

Bridge spectra of twisted torus knots

The simplest hyperbolic knots

Volume bounds for generalized twisted torus links

The next simplest hyperbolic knots

Hyperbolic knots with small Seifert-fibered Dehn surgeries

Knots with distinct primitive/primitive and primitive/Seifert representatives

Twisted torus knots T (p, q; kq, s) are cable knots

Twisted torus knots that are unknotted

Torus knots obtained by twisting torus knots

Satellite knots obtained by twisting torus knots: hyperbolicity of twisted torus knots

Positively twisted torus knots which are torus knots

Twisted torus knots T (p, q, p − kq, −1) which are torus knots

Twisted torus knots T (mn+m+1, mn+1, mn+m+2, −1) and T (n+1, n, 2n− 1, −1) are torus knots

Heegaard splittings of twisted torus knots

Knots and Links

Unexpected essential surfaces among exteriors of twisted torus knots

Satellites and Lorenz knots

On the knot Floer homology of twisted torus knots

is the closure of the braid in Figure 14 (a). One can see that the closures of One easily sees that the closure of the braid in Figure 14 (k) is the torus knot T (n − 1, 2n − 1). Thus T (3n − 1, n, n + 1, −1) = T (2n − 1, n − 1).