The C-complex clasp number of links

TL;DR

This paper investigates the minimal number of clasps in C-complexes bounded by a link, relating it to linking numbers and crossing changes, thereby advancing understanding of link complexity.

Contribution

It establishes that for 2-component links, the linking number determines the minimal number of clasps, and for 3-component links, the triple linking number provides an additional bound.

Findings

Linking number determines minimal clasps for 2-component links.

Triple linking number bounds clasps in 3-component links.

C-complex clasp number relates to crossing changes to boundary links.

Abstract

In the 1980's Daryl Cooper introduced the notion of a C-complex (or clasp-complex) bounded by a link and explained how to compute signatures and polynomial invariants using a C-complex. Since then this was extended by works of Cimasoni, Florens, Mellor, Melvin, Conway, Toffoli, Friedl, and others to compute other link invariants. Informally a C-complex is a union of surfaces which are allowed to intersect each other in clasps. The purpose of the current paper is to study the minimal number of clasps amongst all C-complexes bounded by a fixed link $L$. This measure of complexity is related to the number of crossing changes needed to reduce $L$ to a boundary link. We prove that if $L$ is a 2-component link with nonzero linking number, then the linking number determines the minimal number of clasps amongst all C-complexes. In the case of 3-component links, the triple linking number provides an additional lower bound on the number of clasps in a C-complex.

Figures

Equations

$$B(L)\le C(L).$$

$$\sum _{1\le i<j\le n}|\mathrm{lk}(L_i,L_j)|\le B(L).$$

$$|\mathrm{lk}(L_1,L_2)|=|c_{+}-c_{-}|\le c_{+}+c_{-}.$$

$$e_{ij}(w_k(F))=\sum _{v=1}^m\sum _{u=1}^v\delta (i_u,i)\delta (i_v,j){ϵ}_u{ϵ}_v.$$

$${\mu }_{ijk}(L)=e_{ij}(w_k(F))+e_{jk}(w_i(F))+e_{ki}(w_j(F))$$

$$w(F_1)=x_3^{-1}{x}_2{x}_3{x}_2^{-1}.$$

$${\mu }_{123}(BR)=e_{12}(w_3(F))+e_{23}(w_1(F))+e_{31}(w_2(F))=1.$$

$${\oint }_{{\gamma }_{ij}(w)}xdy=\sum _{v=1}^m\left({\oint }_{{\gamma }_{ij}^v(w)}xdy\right).$$

$$x=\sum _{u=1}^v\delta (x_u,1){ϵ}_u.$$

$${\oint }_{{\gamma }_{ij}^v(w)}xdy=\delta (i_v,j)\sum _{u=1}^v\delta (x_u,1){ϵ}_u{ϵ}_v.$$

$$w_1(F)=x_3^{-n}{x}_2^n{x}_3^n{x}_2^{-n},w_2(F)=x_1^n{x}_1^{-n},w_2(F)=(x_1{x}_1^{-1}{)}^n$$

$$A={\oint }_{\gamma }xdy=\sum _{i=1}^v{\oint }_{{\gamma }_i^u}xdy+\sum _{i=1}^v{\oint }_{{\gamma }_i^d}xdy+\sum _{i=1}^h{\oint }_{{\gamma }_i^r}xdy+\sum _{i=1}^h{\oint }_{{\gamma }_i^{\ell }}xdy$$

$$||\gamma ||=||r||\ge 2\lceil 2\sqrt{h\cdot v}\rceil \ge 2\lceil 2\sqrt{|A|}\rceil .$$

$$2C(F)=|w_1|+|w_2|+|w_3|.$$

$$2C(F)=|w_1|+|w_2|+|w_3|\ge 2|w_3|\ge 4\lceil 2\sqrt{|{\mu }_{123}(L)|/3}\rceil$$

Full text

1. Introduction

There is a generalization of a Seifert surface to the setting of links called a C-complex or clasp-complex originally defined by Cooper [5, 6]. Informally, if $L=L_{1}\cup\dots\cup L_{n}$ is an $n$-component link then a C-complex for $L$ is a collection of Seifert surfaces, $F=F_{1}\cup\dots\cup F_{n}$ for the components of $L$ which are allowed to intersect, but only in clasps. See Figure 1 for a local picture of a clasp and Figures 3 and 4 for some examples of C-complexes. See also Definition 8.

If a C-complex, $F$, for $L$ has no clasp intersections, then $F$ is a collection of disjoint Seifert surfaces for the components of $L$. In this case $L$ is called a boundary link and $F$ is called a boundary surface. Thus, the number of clasps in a C-complex can be used the measure how far $F$ is from being a boundary surface and so how far $L$ is from being a boundary link. In this paper we shall study the minimal number of clasps amongst all C-complexes bounded by $L$. This should not be confused with the clasp number introduced by Shibuya in in [13].

Definition 1*.*

For a link $L$ we define the clasp number of $L$, $C(L)$, to be the minimum number of clasps amongst all C-complexes bounded by $L$.

For a 2-component link $L=L_{1}\cup L_{2}$ and any C-complex $F=F_{1}\cup F_{2}$ bounded by $L$, the linking number, $\operatorname{lk}(L_{1},L_{2})$, can be computed as the number of positive clasps in $F$ minus the number of negative. It follows that $C(L)\geq|\operatorname{lk}(L_{1},L_{2})|$. Our first main result is that for most $2$-component links, $C(L)=|\operatorname{lk}(L_{1},L_{2})|$.

Theorem 2**.**

Let $L=L_{1}\cup L_{2}$ be a 2-component link. If $\operatorname{lk}(L_{1},L_{2})\neq 0$ then $C(L)=|\operatorname{lk}(L_{1},L_{2})|$. If $\operatorname{lk}(L_{1},L_{2})=0$ then $C(L)\in\{0,2\}$.

We mentioned that the number of clasps in a C-complex for $L$ measures how far that link is from being a boundary link. We take a moment and make that explicit. Any link can be reduced to a boundary link by a finite sequence of crossing changes. Indeed, that boundary link can be taken to be the unlink. Let $B(L)$ be the minimum number of crossing changes needed to reduce $L$ to a boundary link. If $F$ is a C-complex for $L$ admitting $C(L)$ total clasps, then by changing a crossing at each clasp as in Figure 2 one reduces $F$ to a boundary surface and so $L$ to a boundary link. Therefore

[TABLE]

On the other hand, changing a crossing of $L$ changes only one linking number of $L$ and that by at most $1$. As any boundary link has vanishing pairwise linking numbers, we conclude that if $L=L_{1}\cup\dots\cup L_{n}$ is an $n$-component link then

[TABLE]

According to Theorem 2 if $L=L_{1}\cup L_{2}$ has only two components and $\operatorname{lk}(L_{1},L_{2})\neq 0$ then $C(L)=|\operatorname{lk}(L_{1},L_{2})|$. Thus, the discussion of the preceding paragraph yields the following corollary

Corollary 3**.**

Let $L=L_{1}\cup L_{2}$ be a 2-component link. If $\operatorname{lk}(L_{1},L_{2})\neq 0$ then there exists a sequence of $|\operatorname{lk}(L_{1},L_{2})|$ crossing changes reducing $L$ to a boundary link. If $\operatorname{lk}(L_{1},L_{2})=0$ then either $L$ is a boundary link or there exists a sequence of $2$ crossing changes reducing $L$ to a boundary link.

Example 4*.*

In order to demonstrate that Theorem 2 and Corollary 3 are surprising, consider the link of Figure 3. The depicted C-complex has three clasps. Since $\operatorname{lk}(L_{1},L_{2})=1$, there exists a C-complex bounded by $L$ with a single clasp and perhaps more surprisingly there exists a single crossing change reducing $L$ to a boundary link.

According to Theorem 2, the linking number determines the clasp number of 2-component links. This behavior does not extend to links of more than 2 components. In [11], Milnor introduced a family of higher order linking invariants. The first of these, the triple linking number, $\mu_{ijk}$, is well defined when the pairwise linking numbers vanish and measures the linking of the $i$’th, $j$’th, and $k$’th components. According to Mellor-Melvin [10], $\mu_{ijk}(L)$ can be computed in terms of the clasps of a C-complex bounded by $L$. Thus, it comes as no surprise that $\mu_{123}(L)$ can be used to deduce a bound on $C(L)$. We explicitly compute this bound.

Theorem 5**.**

Let $L=L_{1}\cup L_{2}\cup L_{3}$ be a 3-component link with vanishing pairwise linking numbers. Then $C(L)\geq 2\left\lceil 2\sqrt{|\mu_{123}(L)|/3}\right\rceil$. Here $\left\lceil-\right\rceil$ is the ceiling function.

In order to illustrate the power of this theorem we compute the clasp number of some examples. The Boromean rings, denoted $BR$, has $\mu_{123}(BR)=1$ and so by Theorem 5, $C(BR)\geq 4$. Figure 4 (a) depicts a C-complex bounded by $BR$ with four clasps. Thus, $C(BR)=4$. For any $n\in\mathbb{N}$ the generalized Boromean rings $BR^{n}$ of Figure 4 (b) bounds a C-complex with $4n$ clasps and has $\mu_{123}(BR^{n})=n^{2}$. We make this computation in Proposition 10. As a consequence we get the following corollary, producing links with vanishing pairwise linking numbers and arbitrarily large clasp number.

Corollary 6**.**

For any $n\in\mathbb{N}$, consider generalized Boromean rings $BR^{n}$ of Figure 4 (b). The pairwise linking numbers of $BR^{n}$ vanish and yet $2\left\lceil 2n/\sqrt{3}\right\rceil\leq C(L)\leq 4n$.

In [10], Mellor-Melvin provides a means of computing $\mu_{123}(L)$ in terms of any collection of Seifert surfaces for the components of $L$. We shall use this result in the special case of a C-complex. While a more complete description appears in Section 3, we recall it informally now. Start with a link $L=L_{1}\cup L_{2}\cup L_{3}$ bounding a C-complex $F=F_{1}\cup F_{2}\cup F_{3}$, follow a component $L_{k}$ of $L$, and record a word $w_{k}(F)$ in $x_{1}^{\pm 1},x_{2}^{\pm 1},x_{3}^{\pm 1}$ capturing the order and sign of the clasps $L_{k}$ encounters. Set $e_{ij}(w_{k}(F))\in\mathbb{Z}$ to be the signed count of the number of $x_{i}$’s appearing in $w_{k}$ before an $x_{j}$. The triple linking number is given by $\mu_{123}(L)=e_{12}(w_{3}(F))+e_{23}(w_{1}(F))+e_{31}(w_{2}(F))$.

A technical result we use in our proof of Theorem 5 is a new geometric strategy to compute $e_{ij}(w)$. For any word $w$ in letters $x_{1}^{\pm 1},x_{2}^{\pm 1},x_{3}^{\pm 1}$ and any $i,j\in\{1,2,3\}$ construct a piecewise linear curve $\gamma_{ij}(w)$ in $\mathbb{R}^{2}$ as follows. Start at the origin $(0,0)$. Each time you see an $x_{i}$ (respectively $x_{i}^{-1}$, $x_{j}$, $x_{j}^{-1}$) in $w$ travel right (respectively left, up, down) a length of 1. The following reveals that $e_{ij}(L)$ is the area enclosed by this curve.

Theorem 7**.**

Let $w=\displaystyle\prod_{n=1}^{m}x_{i_{n}}^{\epsilon_{n}}$ be a word in letters $x_{1}^{\pm 1},x_{2}^{\pm 1},x_{3}^{\pm 1}$. For any $i\neq j\in\{1,2,3\}$, $e_{ij}(w)=\displaystyle\oint_{\gamma_{ij}(w)}x\,dy.$ Additionally, if $\gamma_{ij}(w)$ is a simple closed curve with counterclockwise orientation, then $e_{ij}(w)$ is the area enclosed by $\gamma_{ij}(w)$.

1.1. Questions

Theorem 2 states that any 2-component link with nonzero linking number has a C-complex admitting precisely $|\operatorname{lk}(L_{1},L_{2})|$ clasps. However, our proof makes no attempt to minimize the first Betti number of the C-complex, which is the measure of complexity most directly accessible using the tools like Alexander polynomial or signature [1, 2]. We pose the following question.

Question 1*.*

Suppose that $L=L_{1}\cup L_{2}$ is a 2-component link with nonzero linking number. Amongst all C-complexes $F$ bounded by $L$ admitting precisely $|\operatorname{lk}(L_{1},L_{2})|$ clasps, what is the minimal value for $\beta_{1}(F)$? Is it possible to simultaneously minimize the number of clasps in $F$ as well as $\beta_{1}(F)$?

Theorem 2 almost completely determines $C(L)$ for 2-component links. Theorem 5 concludes that $C(L)\geq 2\left\lceil 2\sqrt{|\mu_{123}(L)|/3}\right\rceil$ for three component links with vanishing linking numbers. One might ask if equality holds.

Question 2*.*

Let $L=L_{1}\cup L_{2}\cup L_{3}$ be a 3-component link with vanishing pairwise linking numbers and $\mu_{123}(L)\neq 0$. Does it follows that $C(L)=2\left\lceil 2\sqrt{|\mu_{123}(L)|/3}\right\rceil$?

More specifically, for any $n\in\mathbb{N}$, consider the generalized Boromean rings $BR^{n}$ of Figure 4 (b). Corollary 6 concludes that $2\left\lceil 2n/\sqrt{3}\right\rceil\leq C(BR^{n})\leq 4n$. When $n=2$ this gives $6\leq C(BR^{2})\leq 8$.

Question 3*.*

What is $C(BR^{n})$?

One might ask about the clasp number of links of more than three components.

Question 4*.*

Let $n\geq 3$ and let $L=L_{1}\cup\dots\cup L_{n}$ be an $n$-component link with vanishing pairwise linking numbers and $\mu_{ijk}(L)\neq 0$ for some $i,j,k$. Is there a formula for $C(L)$ in terms of the set of all triple linking numbers of $L$?

In the case of links of more than 2 components with nonvanishing pairwise linking numbers, the triple linking numbers are not well defined. Instead by [7] there is a total triple linking number recording all of the individual triple linking numbers taking values in some quotient $\mathcal{M}$.

Question 5*.*

Let $L=L_{1}\cup\dots\cup L_{n}$ be an $n$-component link with either a nonvanishing pairwise linking number or a nonvanishing triple linking number. Is there a formula for $C(L)$ in terms of the linking numbers and the total triple linking number?

2. C-complexes and the proof of Theorem 2

Throughout this paper all knots will be smoothly embedded curves in $S^{3}$, and all surfaces will be smoothly embedded in $S^{3}$, compact, connected, and oriented. A smoothly embedded compact oriented surface with boundary equal to a knot $K$ will be called a Seifert surface for $K$. We begin by recalling the formal definition of a C-complex.

Definition 8*.*

[2, Section 2.1] Given a link, $L=L_{1}\cup\dots\cup L_{n}$, a C-complex for $L$ is a collection of Seifert surfaces, $F=F_{1}\cup\dots\cup F_{n}$, for the components of $L$ which may intersect transversely with the following constraints:

1. (1)

For each $i,j\in\{1,\dots,n\}$, $F_{i}\cap F_{j}$ is a union of simple arcs running from a point in $L_{i}=\partial F_{i}$ to a point in $L_{j}=\partial F_{j}$. These arcs are called clasps. See Figure 1. 2. (2)

For any three distinct $i,j,k$, $F_{i}\cap F_{j}\cap F_{k}=\emptyset$.

For this section we restrict to the case that the number of components is $n=2$. A clasp between $F_{1}$ and $F_{2}$ has endpoints given by a point in $F_{1}\cap L_{2}$ and a point in $L_{1}\cap F_{2}$. We call a clasp positive (or negative, respectively) if these points of intersection are positive (or negative, respectively). See Figure 1 for a local picture. If $F_{2}$ is any Seifert surface for $L_{2}$, then the linking number $\operatorname{lk}(L_{1},L_{2})$ is given by counting with sign how many times $L_{1}$ passes through $F_{2}$. See for example [12, 5D]. If $F_{1}\cup F_{2}$ is a C-complex for $L_{1}\cup L_{2}$ then this is precisely the same as the signed count of the clasps shared by $F_{1}$ and $F_{2}$. We are now ready to prove Theorem 2.

Theorem 2.

Let $L=L_{1}\cup L_{2}$ be a 2-component link. If $\operatorname{lk}(L_{1},L_{2})=0$ then $C(L)\in\{0,2\}$. If $\operatorname{lk}(L_{1},L_{2})\neq 0$ then $C(L)=|\operatorname{lk}(L_{1},L_{2})|$

Proof of Theorem 2.

Let $L=L_{1}\cup L_{2}$ be a 2-component link and $F=F_{1}\cup F_{2}$ be any C-complex bounded by $L$. Let $c_{+}$ be the number of positive clasps in $F$ and $c_{-}$ be the number of negative clasps. By the triangle inequality,

[TABLE]

So that $F$ has at least $|\operatorname{lk}(L_{1},L_{2})|$ many clasps. As $F$ is an arbitrary C-complex bounded by $L$, $C(L)\geq|\operatorname{lk}(L_{1},L_{2})|$. Thus, we need only to show that $C(L)\leq|\operatorname{lk}(L_{1},L_{2})|$. Since $C(L)$ is the minimum number of clasps amongst all C-complexes bounded by $L$, it suffices to exhibit a C-complex with precisely $|\operatorname{lk}(L_{1},L_{2})|$ clasps or 2 clasps in the case that $\operatorname{lk}(L_{1},L_{2})=0$. Without loss of generality we shall assume that $\operatorname{lk}(L_{1},L_{2})\geq 0$.

We begin by producing a pair of Seifert surfaces $F_{1}$ and $F_{2}$ for $L_{1}$ and $L_{2}$ which will have no negative clasps in their intersection but which may have some non-clasp intersections. Let $F_{1}$ be any Seifert surface for $L_{1}$. Suppose $F_{1}$ is transverse to $L_{2}$ and $F_{1}\cap L_{2}$ contains $n_{+}$ positive points of intersection and $n_{-}$ points of negative intersection. If both of $n_{+}$ and $n_{-}$ are nonzero then as you follow $L_{2}$ you will at some point encounter a positive point of intersection with $F_{1}$ followed by a negative, as in Figure 5 (a). By adding a tube to $F_{1}$ as in Figure 5 (b) we see a new Seifert surface bounded by $L_{1}$ which intersects $L_{2}$ in two fewer points. Iterating, we see a Seifert surface for $L_{1}$, which we persist in calling $F_{1}$, bounded by $L_{1}$ which either intersects $L_{2}$ in only positive points or only negative points of intersection. Thus, $n_{+}=0$ or $n_{-}=0$. Since $n_{+}-n_{-}=\operatorname{lk}(L_{1},L_{2})\geq 0$ by assumption we see that $n_{-}=0$. By the same process, we find a Seifert surface $F_{2}$ which intersects $L_{1}$ in only positive points of intersection.

There is no reason to expect that $F_{1}\cup F_{2}$ is a C-complex. After a small isotopy of $F_{1}$ and $F_{2}$ we may assume that they intersect transversely. Therefore $F_{1}\cap F_{2}$ consists of a collection of:

• •

Arcs with one endpoint in $L_{1}=\partial F_{1}$ and the other in $L_{2}=\partial F_{2}$. (Clasps.)

• •

Arcs with both endpoints in $L_{1}=\partial F_{1}$ or both endpoints in $L_{2}=\partial F_{2}$. (Ribbons.)

• •

Simple closed curves interior to $F_{1}$ and interior to $F_{2}$. (Loops.)

See Figure 6. Since $F_{1}$ has no negative points of intersection with $L_{2}$, there can be no negative clasps in $F_{1}\cap F_{2}$. The endpoints of a ribbon intersection are intersection points between $F_{1}$ and $L_{2}$ (or $F_{2}$ and $L_{1}$) with opposite signs. Since we have already arranged that there are no negative points of intersection, there can be no ribbon intersections in $F_{1}\cap F_{2}$. Thus, $F_{1}\cap F_{2}$ consists only of loops and positive clasps. It remains to further modify $F_{1}$ and $F_{2}$ to eliminate all loops.

Assume that $\operatorname{lk}(L_{1},L_{2})\neq 0$ so that there there is at least one clasp in $F_{1}\cap F_{2}$. Let $c$ be one such clasp. Consider any loop intersection $\ell\subseteq F_{1}\cap F_{2}$. Without loss of generality we may assume that there exists an arc $\alpha$ in $F_{2}$ running from a point in $c$ to a point in $\ell$. Moreover, we may assume that $\alpha$ connects two points pushed off from $F_{1}$ in the same normal direction. Figure 7 reveals how one may add a tube to $F_{1}$ following $\alpha$ to combine $c$ and $\ell$ into a single simple arc. The resulting arc has one endpoint in $L_{1}$ and the other in $L_{2}$. In other words, it is a clasp. Thus, we have reduced the number of loop intersections by 1 and preserved the number of clasp intersections. Iterating, we eliminate all loop intersections and produce a C-complex for $L=L_{1}\cup L_{2}$ with number of clasps equal to $\operatorname{lk}(L_{1},L_{2})$, as claimed.

In the case that the linking number is zero, $F_{1}\cap F_{2}$ contains no clasps. If $F_{1}\cap F_{2}$ also has no loops, then $F_{1}\cup F_{2}$ is a C-complex with no clasps and $C(L)=0$. Otherwise, modify $F_{1}\cup F_{2}$ as in Figure 8 to add a positive and a negative clasp. Now we use the move of Figure 7 just as in the previous paragraph to remove all loop intersections and produce a C-complex with precisely 2 clasps, so $0\leq C(L)\leq 2$. In order to see that $C(L)$ cannot be $1$, notice that since $c_{+}-c_{-}=\operatorname{lk}(L_{1},L_{2})=0$, it must be that $c_{+}=c_{-}$. In particular, $F$ has an even number of clasps. This completes the proof.

∎

3. Triple linking numbers via clasps and polyominos

In this section we recall an invariant of links called the triple linking number and provide a formula in terms of the area of a polyomino. A polyomino is a region of $\mathbb{R}^{2}$ consisting of a union of closed unit squares with vertices at points in $\mathbb{Z}^{2}$.

In [10] Mellor-Melvin produces a formula for the triple linking number for any union of Seifert surfaces for the components of $L$. We shall recall it in the special case of a C-complex. Let $L=L_{1}\cup\dots\cup L_{n}$ be an $n$-component link and $F=F_{1}\cup\dots\cup F_{n}$ be a C-complex bounded by $L$. We associate to each $k=1,\dots,n$ a word $w_{k}(F)$ called a clasp word as follows. Pick a basepoint $p_{k}$ on $L_{k}$ and follow $L_{k}$ in the positive direction starting at $p_{k}$. Record an $x_{j}$ whenever $L_{k}$ crosses through $F_{j}$ at a positive clasp and $x_{j}^{-1}$ when $L_{k}$ crosses $F_{j}$ at a negative clasp. Let $e_{ij}(w_{k}(F))$ be given by counting with sign how often in $w_{k}(F)$ $x_{i}$ appears before $x_{j}$. More formally, if $w_{k}(F)=\displaystyle\prod_{v=1}^{m}x_{i_{v}}^{\epsilon_{v}}$ then

[TABLE]

Here $\delta(a,b)=\begin{cases}1&\text{if }a=b\\ 0&\text{otherwise}\end{cases}$ indicates the Kronecker $\delta$. We encourage the reader to take a moment and use this definition to compute $e_{12}(x_{1}x_{2}x_{1}^{-1}x_{2}^{-1})=1$. The triple linking number is given by

[TABLE]

When $L$ is a link with vanishing pairwise linking numbers, $\mu_{ijk}(L)$ is independent of the choice of $F$ and of the choice of basepoints.

Example 9*.*

For the sake of clarity, we provide an example computing the triple linking number of the Borromean Rings $BR=BR_{1}\cup BR_{2}\cup BR_{3}$ using the C-complex $F$ of Figure 4.

• •

Following $BR_{1}$ starting at the arrow we encounter in order a negative clasp with $F_{3}$, a positive clasp with $F_{2}$, a positive clasp with $F_{3}$ and a negative clasp with $F_{2}$. Therefore,

[TABLE]

Similarly, $w(F_{2})=x_{1}^{-1}x_{1}$ and $w(F_{3})=x_{1}^{-1}x_{1}$.

• •

Count with sign how many times you see $x_{2}$ before $x_{3}$ in $w(L_{1})$ to get $e_{23}(w_{1}(F))=+1$. Similarly, $e_{12}(w(F_{3}))=e_{31}(w(F_{2}))=0$.

• •

The triple linking number is given by summing,

[TABLE]

Our next goal is the statement and proof of Theorem 7, which computes $e_{ij}(w_{k}(F))$ in terms of some curve $\gamma_{ij}(w_{k}(F))$ in the plane. We begin by explaining the construction of $\gamma_{ij}(w_{k}(F))$. Let $w$ be any word in the letters $x_{1}^{\pm 1},\dots,x_{n}^{\pm 1}$. We give a procedure which associates to $w$ a curve in the plane. Start at the point $(0,0)\in\mathbb{R}^{2}$. Each time you encounter $x_{i}$ in $w$ travel right a length of $1$. When $x_{i}^{-1}$ is encountered travel left. When $x_{j}$ or $x_{j}^{-1}$ is encountered travel up or down respectively. Call the resulting curve $\gamma_{ij}(w_{k})$. For instance, when $w=x_{i}x_{j}x_{i}x_{j}x_{i}^{-2}x_{j}^{-2}$, $\gamma_{ij}(w)$ appears in Figure 9. The assiduous reader will now compute $e_{ij}(w)=3$ using equation (1) which suggestively agrees with the area of the region enclosed by $\gamma_{ij}(w)$.

We are now ready to prove Theorem 7.

Theorem 7.

Let $w=\displaystyle\prod_{v=1}^{m}x_{i_{v}}^{\epsilon_{v}}$ be a word in letters $x_{1}^{\pm 1},\dots,x_{n}^{\pm 1}$. For any $i\neq j\in\{1,\dots,n\}$, ${e_{ij}(w)=\displaystyle\oint_{\gamma_{ij}(w)}x\,dy.}$ Additionally, if $\gamma_{ij}(w)$ is a simple closed curve with counterclockwise orientation, then $e_{ij}(w)$ is the area enclosed by $\gamma_{ij}(w)$.

Proof of Theorem 7.

Let $w=\displaystyle\prod_{v=1}^{m}x_{i_{v}}^{\epsilon_{v}}$ be a word in the letters $x_{1}^{\pm},\dots,x_{n}^{\pm 1}$. Let $|w|=m$ be the length of $w$. Then $\gamma_{ij}(w)$ consists of a concatenation of $|w|$ many curves, $\gamma_{ij}^{1}(w),\dots,\gamma_{ij}^{m}(w)$ where $\gamma_{ij}^{v}(w)$ is constant if $i_{v}\notin\{i,j\}$ and is a length 1 line segment traveling in a cardinal direction otherwise. Therefore, the integral in question breaks up as

[TABLE]

If $i_{v}\neq j$ then $\gamma_{ij}^{v}(w)$ is either constant or parametrizes a horizontal line segment. In either case $dy=0$ so that $\displaystyle\oint_{\gamma_{ij}^{v}(w)}x\,dy=0$. If $i_{v}=j$ then $\gamma_{ij}^{v}(w)$ is a vertical line segment parametrized by $\gamma_{ij}^{v}(t)=(x,t\cdot\epsilon_{v}+c)$ with $x$ and $c$ constants and $t$ running from [math] to $1$. In particular $dy=\epsilon_{v}\,dt$. The fixed $x$-coordinate over which this vertical line sits is the signed count of $u<v$ with $i_{u}=i$:

[TABLE]

Thus, in the case that $i_{v}=j$, we have $\displaystyle\oint_{\gamma_{ij}^{v}(w)}x\,dy=\displaystyle\int_{0}^{1}x\cdot\epsilon_{v}\,dt=x\cdot\epsilon_{v}=\displaystyle\sum_{u=1}^{v}\delta(x_{u},1)\epsilon_{u}\epsilon_{v}$. Combining the cases $i_{v}=j$ and $i_{v}\neq j$, we see for for all $v$,

[TABLE]

Summing over all values of $v$, $\displaystyle\oint_{\gamma_{ij}(w)}x\,dy=\displaystyle\sum_{v=1}^{n}\delta(i_{v},j)\displaystyle\sum_{u=1}^{v}\delta(x_{u},i)\epsilon_{u}\epsilon_{v}$. An application of the distributive law reduces this to the definition of $e_{ij}(w)$ appearing in equation (1). This completes the proof of the first claim.

The second claim follows from a standard application of Green’s theorem. ∎

As an illustration of the efficacy of Theorem 7 we use it to make some computations.

Proposition 10**.**

For any $n\in\mathbb{N}$, the generalized Boromean rings $BR^{n}$ of Figure 4 has triple linking number $n^{2}$.

Proof.

Using the C-complex of Figure 4 (b) we get clasp words

[TABLE]

The curve $\gamma_{23}(w_{1}(F))$ traces a counterclockwise $n\times n$ square. The curve, $\gamma_{31}(w_{2}(F))$ lies in the vertical line $x=0$ so that $e_{31}(w_{2}(F))=0$. Finally, $\gamma_{23}(w_{1}(F))$ lies in the horizontal line $y=0$ so that $e_{12}(w_{3}(F))=0$. Therefore, $\mu_{123}(BR^{n})=n^{2}$. ∎

4. The proof of Theorem 5

We now turn our attention to a lower bound on the number of clasps in a C-complex in terms of the triple linking number. Notice that the curve $\gamma_{ij}(w(L_{k}))$ of Section 3 has length equal to the number of clasps in $F_{k}\cap F_{i}$ plus the number of clasps in $F_{k}\cap F_{j}$. By Theorem 7, $\displaystyle\oint_{\gamma_{ij}(w)}x\,dy=e_{ij}(w(L_{k}))$. Thus, we will begin the proof of Theorem 5 by studying how $\displaystyle\oint_{\gamma}x\,dy.$ provides a lower bound on the length of $\gamma$.

For the lemma below, a polyomino curve is a closed curve in $\mathbb{R}^{2}$ given by a concatenation of straight lines of length 1 between points in $\mathbb{Z}^{2}$. The length of a curve, $\gamma$, is denoted by $||\gamma||$.

Lemma 11**.**

Let $\gamma$ be a polyomino curve in $\mathbb{R}^{2}$. Let $A=\displaystyle\oint_{\gamma}x\,dy$. Then $||\gamma||\geq 2\left\lceil 2\sqrt{|A|}\right\rceil$.

Proof.

Let $\gamma$ be a polyomino curve in $\mathbb{R}^{2}$ and let $A=\displaystyle\oint_{\gamma}x\,dy$. If $\gamma$ is a simple closed curve then a standard application of Green’s theorem shows that $|A|=\underset{A}{\displaystyle\iint}1\,dxdy$ is the area of the region $R$ enclosed by $\gamma$. In [8], Harary-Harborth shows that the minimum perimeter amongst all polyominos with a fixed area $|A|$ is given by $\left(2\left\lceil 2\sqrt{|A|}\right\rceil\right)$. Thus, $||\gamma||$, which is the perimeter of $A$, is at least $2\left\lceil 2\sqrt{|A|}\right\rceil$, as the lemma claims.

It remains to deal with the case that $\gamma$ is not simple. Recall that by assumption, $\gamma$ consists of a concatenation of vertical and horizontal line segments of length 1. Denote the rightward pointing horizontal line segments as $\gamma^{r}_{1}(t),\dots,\gamma^{r}_{h}(t)$, the leftward pointing as $\gamma^{\ell}_{1}(t),\dots,\gamma^{\ell}_{h}(t)$, the upward as $\gamma^{u}_{1}(t),\dots,\gamma^{u}_{v}(t)$ and the downward as $\gamma^{d}_{1}(t),\dots,\gamma^{d}_{v}(t)$. As $\gamma$ is a closed curve, the number of rightward and leftward pointing segments must be equal to each other as must the number of upward and downward pointing segments.

Up to a translation and a reparametrization preserving $||\gamma||$ and $\displaystyle\oint_{\gamma}x\,dy$, we may assume that $\gamma$ is parametrized by some $(x(t),y(t))$ such that the minimum value of $x(t)$ is $x(0)=0$. It follows for all $t$ that $0\leq x(t)\leq h$, where $h$ is the number of rightward pointing length 1 line segments in $\gamma$. Breaking the integral up as a sum,

[TABLE]

Since $\gamma^{\ell}_{i}$ and $\gamma^{r}_{i}$ are horizontal line segments, they each have $dy=0$ so that $\displaystyle\oint_{\gamma^{r}_{i}}x\,dy=\displaystyle\oint_{\gamma^{\ell}_{i}}x\,dy=0.$ Since $\gamma^{u}_{i}$ is an upward pointing length 1 line segment, we may parametrize $\gamma^{u}_{i}$ as $(x,t+c)$ where $x$ and $c$ are constant and $t$ runs from [math] to $1$. Therefore, $dy=dt$ and $0\leq x\leq h$. Thus, $\displaystyle\oint_{\gamma_{i}^{u}}x\,dy=\displaystyle\int_{0}^{1}xdt=x$ and in particular $0\leq\displaystyle\oint_{\gamma_{i}^{u}}x\,dy\leq h.$ Similarly, $-h\leq\displaystyle\oint_{\gamma_{i}^{d}}x\,dy\leq 0$. Therefore, $0\leq\displaystyle\sum_{i=1}^{v}\displaystyle\oint_{\gamma_{i}^{u}}y\,dx\leq h\cdot v$ and $-h\cdot v\leq\displaystyle\sum_{i=1}^{v}\displaystyle\oint_{\gamma_{i}^{d}}y\,dx\leq 0$. Applying these bounds to the rightmost expression in (2) we see that $-h\cdot v\leq A\leq h\cdot v$, so that $|A|\leq h\cdot v$.

Let $R$ be an $h\times v$ rectangle and let $r$ be the curve traversing its boundary counterclockwise. As $r$ is made up of the same number of length 1 line segments as $\gamma$, $||\gamma||=||r||$. Since $R$ is a polyomino of area $h\cdot v$, [8] applies and $||r||\geq 2\left\lceil 2\sqrt{h\cdot v}\right\rceil$. Summarizing,

[TABLE]

This completes the proof.

∎

If $w=\displaystyle\prod_{v=1}^{m}x_{i_{v}}^{\epsilon_{v}}$ is a word in $x_{1}^{\pm 1},\dots,x_{n}^{\pm 1}$ for which the signed count of $x_{i}$’s and $x_{j}$’s are both zero then $||\gamma_{ij}(w)||$ is the same as the length of the word $w$ after deleting all letters other than $x_{i}^{\pm 1}$ and $x_{j}^{\pm 1}$, while $e_{ij}(w)=\displaystyle\oint_{\gamma_{ij}(w)}y\,dx$ by Theorem 7. Thus, Lemma 11 has the following corollary.

Corollary 12**.**

Let $w=\displaystyle\prod_{n=1}^{m}x_{i_{n}}^{\epsilon_{n}}$ be a word in $x_{1}^{\pm 1},\dots x_{n}^{\pm 1}$. Fix some $i\neq j\in\{1,\dots,n\}$ and assume the signed count of $x_{i}$’s and $x_{j}$’s are both zero. If $e_{ij}(w)=A$ then $|w|\geq 2\left\lceil 2\sqrt{|A|}\right\rceil$

We are now ready to prove Theorem 5 giving a lower bound on $C(L)$ in terms of $\mu_{ijk}(L)$.

Theorem 5.

Let $L=L_{1}\cup L_{2}\cup L_{3}$ be a 3-component link with vanishing pairwise linking numbers. Then $C(L)\geq 2\left\lceil 2\sqrt{|\mu_{123}(L)|/3}\right\rceil$.

Proof.

Let $L$ be a 3-component link with vanishing pairwise linking numbers and $F$ be a C-complex bounded by $L$. Let $C(F)$ be the number of clasps between the components of $F$. Let $w_{1}=w_{1}(F)$, $w_{2}=w_{2}(F)$ and $w_{3}=w_{3}(F)$ be the resulting clasp words. Each clasps corresponds to a letter in two of these words, and so

[TABLE]

Let $e_{1}=e_{23}(w_{1})$, $e_{2}=e_{31}(w_{2})$, and $e_{3}=e_{12}(w_{3})$. Then $\mu_{123}(L)=e_{1}+e_{2}+e_{3}$. Assume without loss of generality that $|e_{1}|\leq|e_{2}|\leq|e_{3}|$. Then it must be that $|e_{3}|\geq\frac{|\mu_{123}(L)|}{3}$. Corollary 12 concludes that $|w_{3}|\geq 2\left\lceil 2\sqrt{|e_{3}|}\right\rceil\geq 2\left\lceil 2\sqrt{|\mu_{123}(L)|/3}\right\rceil$.

Now, each letter of $w_{3}$ corresponds to either a clasp in $F_{3}\cap F_{1}$ or a clasp in $F_{3}\cap F_{2}$. Each of these clasps produces a letter in $w_{1}$ or in $w_{2}$. As a consequence $|w_{3}|\leq|w_{1}|+|w_{2}|$. Putting this together,

[TABLE]

division by 2 completes the proof.

∎