Proof. We show that \(T_i\) is incompressible and not boundary-parallel in \(S^3{\smallsetminus}L\).

First consider the inside of \(T_i\), namely \(W_i{\smallsetminus}L\). Suppose that \(T_i\) were compressible on the inside. Compressing \(T_i\) along such a disk produces a sphere in \(W_i\) enclosing the pattern component \(f_i(\ell_i)\). Hence \(f_i(\ell_i)\) would be contained in a \(3\)-ball in \(W_i\). Pulling back by \(f_i\), this would imply that \(\ell_i\) is contained in a \(3\)-ball in \(V_i\), contradicting the definition of satellite link. Thus \(T_i\) is incompressible on the inside.

Now consider the outside of \(T_i\), namely \[S^3{\smallsetminus}\operatorname{int}(W_i).\] Suppose that \(T_i\) were compressible on the outside. Compressing \(T_i\) along an outside compressing disk produces a sphere separating the components of \(L\) inside \(W_i\) from the components of \(L\) outside \(W_i\). Since both sides contain components of \(L\), this sphere is a splitting sphere for \(L\). This contradicts the assumption that \(L\) is non-split. Therefore \(T_i\) is incompressible on the outside.

It remains to rule out boundary-parallelism. Suppose that \(T_i\) were boundary-parallel in \(S^3{\smallsetminus}L\). Since the outside of \(T_i\) contains at least two components of \(L\), the torus \(T_i\) cannot be parallel through the outside to the boundary of a tubular neighborhood of a single component of \(L\). On the other hand, \(T_i\) cannot be boundary-parallel through the inside either: such a parallelism would imply that the pattern component \(f_i(\ell_i)\) is isotopic to the core of \(W_i\), contradicting the nontriviality of \(P_i\).

Therefore \(T_i\) is incompressible and not boundary-parallel in \(S^3{\smallsetminus}L\). Hence \(T_i\) is essential. ◻

Proof. Let \(L\) be a T-link. If \(L\) is a knot, then there is nothing to prove, so assume that \(L\) has at least two components.

By [3], \(L\) admits a braid-index-realizing representative which is a V-link braid. In particular, \(L\) is represented by the closure of a positive braid containing at least one positive full twist. Thus we may write this braid as \[\beta=\alpha\Delta_p^2,\] where \(\Delta_p^2\) is the positive full twist on \(p\) strands and \(\alpha\) is a positive braid.

Recall that for an oriented link \[L=K_1\cup \cdots \cup K_\mu,\] the linking number \(\operatorname{lk}(K_i,K_j)\) is defined by \[\operatorname{lk}(K_i,K_j)=\frac{1}{2}\sum_c \operatorname{sign}(c),\] where the sum is taken over all crossings between the components \(K_i\) and \(K_j\) in any diagram of \(L\). This is a link invariant.

We claim that any two distinct components of \(\widehat{\beta}\) have positive linking number. Let \(C\) and \(C'\) be distinct components of \(\widehat{\beta}\). Suppose that \(C\) is supported on \(a\ge 1\) strands and \(C'\) is supported on \(b\ge 1\) strands. Since \(\Delta_p^2\) is a full twist, each strand of \(C\) crosses each strand of \(C'\) exactly twice in the full twist, and all these crossings are positive. Hence the full twist contributes \(2ab\) positive crossings between \(C\) and \(C'\), and therefore contributes \[ab\] to \(\operatorname{lk}(C,C')\). The remaining factor \(\alpha\) is positive, so it can only add non-negative contributions to the linking number. Thus \[\operatorname{lk}(C,C')\ge ab\ge 1.\] Therefore every pair of distinct components of \(L\) has positive linking number.

Now suppose, for contradiction, that \(L\) is split. Then there is a sphere in \(S^3{\smallsetminus}L\) separating the components of \(L\) into two non-empty subsets. Choose components \(C\) and \(C'\) on opposite sides of this sphere. Since they belong to different split summands, we have \[\operatorname{lk}(C,C')=0.\] This contradicts the positive linking number established above. Hence \(L\) is non-split. ◻

Proof. It follows from [4]. ◻

Proof. Let \[L_1 = T((a+b+c, c), (a+b+2c+k(a+b+c), a+b+c)),\] which is the closure of the braid \[(\sigma_1 \cdots \sigma_{a+b+c-1})^{c} (\sigma_1 \cdots \sigma_{a+b+2c+k(a+b+c)-1})^{a+b+c}.\] Equivalently, by conjugation, \(L_1\) is the closure of the braid \[B_1 = (\sigma_1 \cdots \sigma_{a+b+2c+k(a+b+c)-1})^{a+b+c} (\sigma_1 \cdots \sigma_{a+b+c-1})^{c}.\]

We first show that the first \(a+b\) strands of \(B_1\) form parallel components in \(L_1\). To see this, track these strands through the braid. After passing once through the first factor of \(B_1\), they move to the strands indexed from \[c+k(a+b+c)+1 \quad \text{to} \quad a+b+c+k(a+b+c)\] at the bottom of \(B_1\), preserving their order from top to bottom, that is, each strand returns to the same relative position among these strands. After closing the braid, they return to the top and repeat this motion. Iterating this process \(k\) times and passing through the first factor of \(B_1\), they occupy the positions from \(c+1\) to \(a+b+c\) at the bottom of the first factor \[(\sigma_1 \cdots \sigma_{a+b+2c+k(a+b+c)-1})^{a+b+c}.\]

Next, these strands pass through the second factor \[(\sigma_1 \cdots \sigma_{a+b+c-1})^{c},\] which involves only the first \(a+b+c\) strands. After this, they occupy the first \(a+b\) positions preserving their initial order. Then they return to their original positions under the braid closure. Hence these strands trace disjoint, parallel components in \(L_1\).

Since these \(a+b\) components are parallel, they lie in a solid torus and are isotopic to its core. Removing \(a+b-2\) of these parallel components, we obtain the link \[L_2 = T((2+c, c), (2+2c+k(2+c), 2+c)),\] represented by the braid \[B_2 = (\sigma_1 \cdots \sigma_{2+2c+k(2+c)-1})^{2+c} (\sigma_1 \cdots \sigma_{2+c-1})^{c}.\]

We now analyze \(L_2\). The components corresponding to strands \(3\) through \(2+c\) of \(B_2\) are parallel. Indeed, tracking these strands through the braid shows that they move cyclically through the first factor of \(B_2\) and, after \(k+2\) passes through this factor, they occupy positions \(1\) through \(c\) preserving their initial order at the bottom of \[(\sigma_1 \cdots \sigma_{2+2c+k(2+c)-1})^{2+c}.\] Passing through the second factor \[(\sigma_1 \cdots \sigma_{2+c-1})^{c},\] they return to strands \(3\) through \(2+c\) preserving their initial order and hence to their original positions after closure. Therefore, these \(c\) components are parallel in \(L_2\).

Removing \(c-1\) of these parallel components yields the link \[L_3 = T((3, 1), (4+3k, 3)),\] which is represented by the braid \[B_3 = (\sigma_1 \cdots \sigma_{4+3k-1})^{3}(\sigma_1 \sigma_2).\]

Let \(C_1, C_2, C_3\) denote the components of \(L_3\) corresponding to the first, second, and third strands of \(B_3\), respectively. Reversing the above construction, we see that \(L_1\) is obtained from \(L_3\) by inserting \(a\), \(b\), and \(c\) parallel components in tubular neighborhoods of \(C_1\), \(C_2\), and \(C_3\), respectively. Each family of components is isotopic to the core of the corresponding solid torus. More precisely, each component of \(L_1\) is contained in a solid torus associated to a component of the companion link, and within each such torus the pattern is realized as a closed braid.

It remains to verify that the companion tori associated to the nontrivial patterns are essential in the complement of \(L_1\). Since at least one of \(a,b,c\) is greater than \(1\), at least one of the patterns is nontrivial. Moreover, the link \(L_1\) is a T-link, and hence is non-split by Proposition 2. Each companion torus separates the components of \(L_1\) lying inside the corresponding tubular neighborhood from the remaining components of \(L_1\). In the present construction, the outside of each companion torus contains at least two components of \(L_1\). Therefore, by Lemma 1, the companion tori corresponding to the nontrivial patterns are essential in the complement of \(L_1\).

Therefore, \[T((a+b+c, c), (a+b+2c+k(a+b+c), a+b+c))\] is a braided satellite link with companion \[T((3, 1), (4+3k, 3)). \qedhere\] ◻

Proof. It follows from [4]. ◻

Proof. Let \[L = T((3,1),(4+3k,3)),\] which is the closure of the braid \[B = (\sigma_1 \cdots \sigma_{4+3k-1})^{3}(\sigma_1 \sigma_2).\] The link \(L\) has three components by Lemma 2.

Let \(L_3\) denote the component corresponding to the third strand of \(B\). We describe how this strand moves through the braid. Starting at the third strand at the top, it passes through the braid \((\sigma_1 \cdots \sigma_{4+3k-1})^{3}\) and appears at the last strand (i.e., strand \(4+3k\)) at the bottom. After closing the braid, it returns to the same position at the top and repeats this process. Iterating this motion, after \(k\) repetitions the strand reaches position \(4\) at the top, and then moves down to position \(1\) at the bottom of \((\sigma_1 \cdots \sigma_{4+3k-1})^{3}\). It then passes through the factor \((\sigma_1 \sigma_2)\), where it becomes the overstrand and ends at the third strand at the bottom of \(B\). Finally, under the braid closure, it returns to its original position.

From this description, we see that the component \(L_3\) is supported on a subset of \(k+2\) strands. In particular, it contains the last overstrand of \((\sigma_1 \cdots \sigma_{4+3k-1})^{3}\) and the overstrand of \((\sigma_1 \sigma_2)\). Therefore, \(L_3\) lies inside an embedded trivial solid torus \(V\) whose boundary is a trivial torus \(T\) intersecting the disc bounded by the braid axis once (see Figure 1).

Now consider the link formed by the core of \(V\) together with the other two components of \(L\). This link is represented by the braid \[B_1 = (\sigma_1 \cdots \sigma_{3+2k-1})^{2}(\sigma_1 \sigma_2),\] which is precisely the T-link \[T((3,1),(3+2k,2)).\] The core of \(V\) corresponds to the third strand of \(B_1\).

It follows that \(L\) is obtained from \(T((3,1),(3+2k,2))\) by inserting braided patterns inside tubular neighborhoods of its components. More precisely, each component of \(L\) lies inside a solid torus corresponding to a component of the companion link, and the pattern is given by closed braids in these solid tori.

It remains only to check essentiality of the companion tori associated to nontrivial patterns. Since \(L\) is a T-link, it is non-split by Proposition 2. The construction above shows that each such torus separates a nontrivial braided pattern from the remaining components of \(L\), with at least two components on the outside. Therefore Lemma 1 applies, and these tori are essential in the complement of \(L\).

Therefore, \[T((3,1),(4+3k,3))\] is a braided satellite link with companion \[T((3,1),(3+2k,2)). \qedhere\] ◻

Proof. By Proposition 3, the link \[L=T((a+b+c,c),(a+b+2c+k(a+b+c),a+b+c))\] is a braided satellite link with companion \[C_1=T((3,1),(4+3k,3)).\] That is, \(L\) is obtained by inserting braided patterns inside tubular neighborhoods of the components of \(C_1\).

On the other hand, by Proposition 4, the link \(C_1\) is itself a braided satellite link with companion \[C_0=T((3,1),(3+2k,2)).\] Thus \(L\) is an iterated braided satellite link: first \(C_1\) is obtained from the companion \(C_0\) by a braided satellite construction, and then \(L\) is obtained from \(C_1\) by another braided satellite construction.

Equivalently, composing the two braided satellite structures gives a braided satellite structure on \(L\) whose companion is \(C_0\). The essentiality of the resulting companion tori follows from the essentiality established in Propositions 3 and 4. Therefore \[T((a+b+c,c),(a+b+2c+k(a+b+c),a+b+c))\] is a braided satellite link with companion \[T((3,1),(3+2k,2)). \qedhere\] ◻

Proof. The equivalence between \(L\) and \(V\) follows from [3], obtained by applying Proposition 2.1 and the isotopy described in Figure 5 of [3]. In particular, this shows that \(L\) admits a positive braid representative with \(a+b+c\) strands and \((k+1)\) full twists.

By Lemma 4, the link \(L\) is a braided satellite link with companion \[C = T((3,1),(3+2k,2)).\] By Lemma 3, the link \(C\) has three components, and hence \(\partial N(C)\) consists of three tori.

From the construction of \(C\) in the proofs of Propositions 3 and 4, two of these tori lie around collections of strands corresponding to the first \(a+b\) strands of the braid representing \(L\). In particular, these tori are disjoint from a circle encircling these strands, and intersect the corresponding spanning disk in two simple closed curves.

The isotopies used to transform \(L\) into \(V\) preserve this circle and its spanning disk. Therefore, after applying these isotopies, the corresponding two tori remain disjoint from the braid axis of \(V\).

The third torus is carried to a torus that remains disjoint from the braid axis and intersects the disk bounded by the axis in a single simple closed curve, corresponding to a circle encircling the first \(a+b\) strands of \(V\), as illustrated in Figure 2.

It follows that all companion tori can be isotoped to be disjoint from the braid axis of \(V\), as claimed. ◻

Proof. By [3], the link \(L\) is equivalent to the V-link \[V((2,\overline{2k}),(3,3)).\] Applying the isotopy given by a \(180^\circ\) rotation of the projection plane about a horizontal axis, as illustrated in [3], transforms this diagram into \[V((2,2k),(3,3)).\] This proves the equivalence.

Since \(V((2,2k),(3,3))\) is represented by a positive braid with a single full twist, the result follows. ◻

Proof. It follows from Lemmas 4 and 6. ◻

Proof. The link \(L\) is represented as the closure of the braid \[(\sigma_1)^{2k}(\sigma_1\sigma_2)^3.\]

Now observe that \((\sigma_1\sigma_2)^3=\Delta_3^2\) is the full twist on three strands, and its closure is the torus link \(T(3,3)\), which has three components, each pair having linking number equal to \(1\). Since \(\sigma_1^{2k}\) is a pure braid, the closure of \[(\sigma_1)^{2k}(\sigma_1\sigma_2)^3\] also has three components. Moreover, each factor \(\sigma_1^2\) contributes exactly \(1\) to the linking number between the first and second components, and does not affect the linking numbers involving the third component. Hence the pairwise linking numbers of \(L\) are \[\operatorname{lk}(K_1,K_2)=k+1,\qquad \operatorname{lk}(K_1,K_3)=1,\qquad \operatorname{lk}(K_2,K_3)=1.\] In particular, \(L\) has a component whose linking number with each of the other two components is equal to \(1\).

Suppose now that \(L\) can be represented as the closure of a positive braid with more than one full twist. Then there exists a positive braid \[\beta=\alpha(\Delta_n^{2})^{r},\qquad r\ge 2,\] with \(\alpha\) positive, such that \(\widehat{\beta}\cong L\).

Let \(C\) and \(C'\) be two distinct components of \(\widehat{\beta}\). Suppose that \(C\) is obtained from \(a\ge 1\) strands and \(C'\) from \(b\ge 1\) strands of the braid. Since \((\Delta_n^{2})^{r}\) is a pure braid, during the \(r\) full twists each strand of \(C\) crosses each strand of \(C'\) exactly \(2r\) times, all positively. Therefore the contribution of \((\Delta_n^{2})^{r}\) to the crossing number between \(C\) and \(C'\) is \[2r\,ab,\] so its contribution to the linking number is \[\frac{1}{2}(2r\,ab)=rab.\] Because \(r\ge 2\) and \(a,b\ge 1\), we get \[\operatorname{lk}(C,C')\ge rab\ge 2.\] The remaining factor \(\alpha\) is also positive, so it can only increase the linking number. Thus every pair of distinct components of the closure of \(\beta\) must satisfy \[\operatorname{lk}(C,C')\ge 2.\]

But this is impossible for \(L\), since we already computed that \(L\) has pairs of components with linking number equal to \(1\). This contradiction shows that \(L\) cannot be represented by any positive braid with more than one full twist. ◻

Proof. By Lemma 7, for positive integers \(a,b,c,k\), with at least one of \(a,b,c\) greater than \(1\), the link \[L = T((a+b+c, c), (a+b+2c+k(a+b+c), a+b+c))\] is a braided satellite link with companion \[C = V((2,2k),(3,3)).\]

By Lemma 5, the link \(L\) admits a positive braid representative \(\beta\) on \(a+b+c\) strands containing \((k+1)\) full twists. Since \(k\) is arbitrary, this produces positive braid representations of \(L\) with arbitrarily many full twists. Furthermore, by Lemma 5, the companion tori \(\partial N(C)\) are disjoint from the braid axis of \(\beta\).

On the other hand, by Proposition 5, the companion link \[C = V((2,2k),(3,3))\] does not admit any positive braid representation with more than one full twist.

Thus, for infinitely many choices of \(k\), we obtain satellite links \(L\) admitting positive braid representations with arbitrarily many full twists, while their companion links \(C\) admit no positive braid representation with more than one full twist. ◻

Proof. Let \(A\) be the braid axis of the given positive braid representative \(\beta\) of \(L\), and let \[T_1,\dots,T_n \subset \partial N(C)\] be the family of companion tori associated to \(C\), where \(n \ge 1\).

By [2], after isotopy we may assume that each \(T_i\) is disjoint from the braid axis \(A\).

By the definition of a satellite link, for each \(T_i\), all other tori in \(T_1,\dots,T_n\) lie entirely in one of the two components of \(S^3 {\smallsetminus}T_i\).

Among the tori \(T_1,\dots,T_n\), let \[T'_1,\dots,T'_j\] be those which separate \(S^3\) into two solid tori. For each \(T'_i\), write \[S^3 {\smallsetminus}T'_i = V^i_1 \cup V^i_2,\] where \(V^i_2\) is the solid torus containing the braid axis \(A\). We now distinguish two cases.

Case 1. For every \(T'_i\), all the remaining companion tori lie in \(V^i_2\).

In this case every companion torus is nested on the same side of the braid axis. Therefore the argument of [2] applies directly: taking the cores of the tori \[T_1,\dots,T_n\] produces a link isotopic to the companion \(C\), and this link is represented by a positive braid with \(k\) full twists and braid axis \(A\). This gives conclusion (1).

Case 2. There exists one \(T'_i\) such that all the remaining companion tori lie in \(V^i_1\).

Remove from \(L\) all link components contained in \(V^i_2\), and remove the torus \(T'_i\) from the family \[T_1,\dots,T_n.\] Denote by \(L'\) the resulting satellite link, and let \(C'\) be the link given by the cores of the remaining tori. By construction, \(L'\) is still represented by a positive braid with \(k\) full twists.

Moreover, the choice of \(T'_i\) implies that for every companion torus of \(L'\), all the other companion tori lie on the side containing the braid axis. Hence Case 1 applies to \(L'\). It follows that \(C'\) is represented by a positive braid, say \(\beta''\), with \(k\) full twists.

Now view \(C'\) inside the solid torus \(V^i_1\). Since \(T'_i=\partial V^i_1\), and \(C'\) is braided in \(V^i_1\), the torus \(T'_i\) is boundary parallel to the boundary of a tubular neighbourhood of the braid axis of the braid representative \(\beta''\) of \(C'\). Equivalently, the original companion \(C\) is obtained by adjoining to \(C'\) the core of the solid torus bounded by \(T'_i\) that contains the braid axis \(A\), that is, the braid axis of \(\beta''\).

Thus \(C\) is the link formed by \(\beta''\) together with its braid axis. Hence \(C\) admits a positive braid representative of the form \[B_0(\sigma_1 \cdots \sigma_{a-2})^{(k-1)(a-1)}(\sigma_1 \cdots \sigma_{a-1})^{a},\] where \[B_0(\sigma_1 \cdots \sigma_{a-2})^{(k-1)(a-1)} \in B_a\] and \[\mathrm{span}\!\left(B_0(\sigma_1 \cdots \sigma_{a-2})^{(k-1)(a-1)}\right)\subset [1,a-1].\] This is exactly conclusion (2).

Therefore one of the two alternatives must hold. ◻