On finite type invariants of welded string links and ribbon tubes

It is an immediate consequence of Corollary 3.6 that $Cl_{R}(\mathbf{1}_{W})\cdot Cl_{R}(\mathbf{1}_{W^{\prime}})\stackrel{{\scriptstyle k+k^{\prime}}}{{\sim}}Cl_{R}(\mathbf{1}_{W}\mathbf{1}_{W^{\prime}})$. Now, since $\alpha_{d}$ is a finite type invariant of degree $d$ by Proposition 2.19, we have $\alpha_{d}(Cl_{R}(\mathbf{1}_{W})\cdot Cl_{R}(\mathbf{1}_{W^{\prime}}))=\alpha_{d}(Cl_{R}(\mathbf{1}_{W}\mathbf{1}_{W^{\prime}}))$, and the result follows from the additivity property of $\alpha_{d}$ as given in [13, Cor. 6.6]. ∎

Let us first prove the group structure. Let $F$ be a union of $w$-trees of degree $\geq l$ for $\mathbf{1}$ ($l\leq k$). Consider a union $F^{\prime}$ of $w$-trees, which consists of a parallel copy of each $w$-tree in $F$, that only differs by a twist $\bullet$ next to the head. By the Inversion move (5), we have that $(\mathbf{1},F\cup F^{\prime})$ is equivalent to $(\mathbf{1},\emptyset)$. Now, we can use Corollary 3.6 to move $F^{\prime}$ above a disk $D$ containing $F$; this introduces a union of $w$-trees $\tilde{W}$ of degree $\geq 2l$, which may intersect $D$. Next $\tilde{W}$ can in turn be moved above the disk $D$ using Corollary 3.6, and this introduces another union of $w$-trees, each of degree $\geq 3l$. Iterating this process, we eventually obtain in this way that the trivial diagram $\mathbf{1}$ is $w_{k}$-equivalent to a product $\mathbf{1}_{F}\cdot\mathbf{1}_{W}$, where $W$ is a union of $w$-trees, disjoint from $D$. We have thus built the inverse of $\mathbf{1}_{F}$ up to $w_{k}$-equivalence. Now, it remains to observe that the group $wSL(n)_{k}$ is finitely generated, since there are only finitely many $w$-trees in each finite degree. ∎

Consider the $2$-component welded string links $D$ and $D^{\prime}$ shown on the left-hand side of Figure 3.1.

On one hand, by the Reversal moves (2), the closure $Cl_{(1,2)}(D)$ is the welded long knot $K$ shown on the right-hand side of the figure. On the other hand, by the Tails exchange and Isolated arrow moves (3) and (4), the welded long knot $Cl_{(1,2)}(D^{\prime})$ is trivial. It is easily checked that $\alpha_{2}(K)=1$, thus proving that the closure invariant $\mathcal{I}_{(1,2);2}$ distinguishes $D$ and $D^{\prime}$. By Proposition 2.19, this proves that $D$ and $D^{\prime}$ are not $w_{2}$-equivalent, hence not $w_{k}$-equivalent for any $k\geq 2$. ∎

Starting with the union $U$ of four $w_{1}$-trees shown on the left-hand side below, the result follows from a sequence of Head/Tail exchange moves (8). In the figures, we indicate by an $\ast$ the place where each such move is performed. On one hand, we have the following sequence of equivalences:

$$U:\ \vbox{\hbox{\includegraphics[height=54.06006pt]{B0.pdf}}}\ \stackrel{{\scriptstyle 3}}{{\sim}}\ \vbox{\hbox{\includegraphics[height=54.06006pt]{B1.pdf}}}\ \cdot\ A_{i,j}^{-1}\stackrel{{\scriptstyle 3}}{{\sim}}\ \vbox{\hbox{\includegraphics[height=54.06006pt]{B1bis.pdf}}}\ \cdot\ A_{i,j}^{-1}\cdot C_{i,j}^{-1}\ \stackrel{{\scriptstyle 3}}{{\sim}}\ A_{i,j}^{-1}\cdot C_{i,j}^{-1}.$$

Here, the first equivalence is given by the Head/Tail exchange move (8) on the $j$th component of $\mathbf{1}$. The $w_{2}$-tree created by this move is a copy of $A_{i,j}^{-1}$, by the Twist move (11), which can be isolated from the rest of the diagram using Corollary 3.6. The second equivalence above is given by the Head/Tail exchange move (8) on the $i$th component. This introduces a $w_{2}$-tree which is a copy of $C_{i,j}^{-1}$ by the Head reversal move (2) and the Twist move (11), and which can also be isolated by Corollary 3.6. The third equivalence then follows directly from the Inversion move (5). On the other hand, starting with the same union $U$ of $w_{1}$-trees, one can perform a similar sequence of Head/Tail exchange moves, but in the opposite order. This gives the following sequence of equivalences:

$$U:\vbox{\hbox{\includegraphics[height=54.06006pt]{B0bis.pdf}}}\ \stackrel{{\scriptstyle 3}}{{\sim}}\ \vbox{\hbox{\includegraphics[height=54.06006pt]{B2.pdf}}}\ \cdot\ B_{i,j}\stackrel{{\scriptstyle 3}}{{\sim}}\ B_{i,j}\cdot D_{i,j}.$$

This shows the desired relation $A_{i,j}\cdot B_{i,j}\cdot C_{i,j}\cdot D_{i,j}\stackrel{{\scriptstyle 3}}{{\sim}}\mathbf{1}$ by Theorem 3.8. ∎

The fact that $(1)$ implies $(2)$ is given by Proposition 3.2, and since all invariants listed in (3) are degree $\leq 2$ invariants (Lemma 2.18 and Corollary 2.20), we have implication $(2)\Rightarrow(3)$. Theorem 4.3 shows that $(3)$ implies $(1)$. ∎

We start with an arbitrary $w$-tree presentation of $L$. By Corollary 3.6, we have that $L$ is $w_{3}$-equivalent to $L^{\prime}_{1}\cdot L^{\prime}_{2}$, where $L^{\prime}_{i}$ is a product of welded string links, each obtained from $\mathbf{1}$ by surgery along a single $w_{i}$-tree ($i=1,2$).

By the involutivity of twists, the Reversal moves (2) and the Isolated arrow move (3), we can freely assume that each $w_{1}$-tree in $L^{\prime}_{1}$ is a copy of either $Z_{i,j}$ or $Z_{i,j}^{-1}$. By Remark 3.9, we thus have

for some coefficients $e_{i,j}\in\mathbb{Z}$. Observe that the welded linking numbers are additive under stacking, and recall that they are $w_{2}$-equivalence invariants by Lemma 2.18 and Proposition 3.2. Hence, applying $\mu(kl)$ (for some $k\neq l$) to this equivalence gives

An elementary computation gives that $\mu_{Z_{i,j}}(kl)=\delta_{ik}\delta_{jl}$, hence $e_{i,j}=\mu_{L}(ij)$ for any pair $(i,j)$ of distinct integers.

We now focus on $L^{\prime}_{2}$. Consider a $w_{2}$-tree $T$ for $\mathbf{1}$, such that $\mathbf{1}_{T}$ is a factor of $L^{\prime}_{2}$. Let us first show that, up to $w_{3}$-equivalence, $T$ can be assumed to be a copy of $E_{i}^{\pm 1}$, $A_{i,j}^{\pm 1}$, $B_{i,j}^{\pm 1}$, $C_{i,j}^{\pm 1}$ or $G_{i,j,k}^{\pm 1}$. Suppose first that all three endpoints of $T$ are on the same component, say component $i$. Then by the Fork move (10), $\mathbf{1}_{T}$ is nontrivial only if the head of $T$ is located between both tails of $T$, and the Reversal moves (2), AS move (9) and Twist relation (11) ensure that $T$ is necessarily a copy of either $E_{i}$ or $E_{i}^{-1}$. In the case where $T$ is attached to exactly two components of $\mathbf{1}$, say $i$ and $j$, then the same combinatorial arguments give that $T$ can be freely assumed to be a copy of $A_{i,j}^{\pm 1}$, $B_{i,j}^{\pm 1}$, $C_{i,j}^{\pm 1}$ or $D_{i,j}^{\pm 1}$. Hence by Proposition 4.1, we can further assume that $T$ is either $A_{i,j}^{\pm 1}$, $B_{i,j}^{\pm 1}$ or $C_{i,j}^{\pm 1}$ with $i<j$. Finally, in the case where the three endpoints of $T$ lie on pairwise distinct components $i,j,k$, then the same considerations show that $T$ is a copy of $G_{i,j,k}^{\pm 1}$ for pairwise distinct indices $i,j,k$ such that $j<k$. Moreover by Corollary 3.6, any two factors obtained from $\mathbf{1}$ by surgery along a $w_{2}$-tree, commute up to $w_{3}$-equivalence. Summarizing, we have proved that

for some $g_{i,j,k}$, $a_{i,j}$, $b_{i,j}$, $c_{i,j}$ and $e_{i}$ in $\mathbb{Z}$. In what follows, for convenience we shall call basic factor any factor appearing in the product (4.1).
Consider the invariant $\mathcal{I}_{(i);2}$, which is the normalized Alexander coefficient $\alpha_{2}$ of the $i$th component. By [13, Lem. 6.4], we have that $\mathcal{I}_{(i);2}(E_{i}^{\pm 1})=\pm 1$ and $\mathcal{I}_{(i);2}$ vanishes on any other basic factor. Recall that $\mathcal{I}_{(i);2}$ is an invariant of $w_{3}$-equivalence (Theorem 3.2). By the additivity property of Proposition 3.7, evaluating $\mathcal{I}_{(i);2}$ on $(\ref{eqnw3nbr})$ thus gives us that $e_{i}=\mathcal{I}_{(i);2}(L)$.
Next we evaluate the closure invariants $\mathcal{I}_{(i,\overline{j})}$, $\mathcal{I}_{(\overline{i},j)}$ and $\mathcal{I}_{(i,j)}$ ($i<j$) on the following basic factors:

Moreover, these three closure invariants vanish on $L_{1}$, and on all basic factors $G_{i,j,k}$. By Theorem 3.2 and Proposition 3.7, evaluating these invariants on $(\ref{eqnw3nbr})$ gives:

Consequently, we have

1. $a_{i,j}=\mathcal{I}_{(i,\overline{j});2}(L)+\mathcal{I}_{(\overline{i},j);2}(L)-\mathcal{I}_{(j,i);2}(L)-\mathcal{I}_{(i);2}(L)-\mathcal{I}_{(j);2}(L)$;

2. $b_{i,j}=\mathcal{I}_{(\overline{i},j);2}(L)-\mathcal{I}_{(i);2}(L)-\mathcal{I}_{(j);2}(L)$;

3. $c_{i,j}=\mathcal{I}_{(\overline{i},j);2}(L)-\mathcal{I}_{(j,i);2}(L)$.

Finally, the closure invariant $\mathcal{I}_{(\overline{j},i,k);2}$ takes the following values on basic factors:

where $\star$ stands for any other basic factor. Hence by Theorem 3.2 and Proposition 3.7, we have:

Here however, unlike in the preceding computation, $\mathcal{I}_{(\overline{j},i,k);2}$ does not vanish on $L_{1}$. Indeed, the closure $Cl_{(\overline{j},i,k)}$ of the following diagram