Isotopic Lines of an Interlacing

The next step--the classification of interlacings of five lines--requires a more careful study of the inner structure of interlacings. The reader has undoubtedly noticed the striking difference between amphicheiral and nonamphicheiral interlacings--compare the sets of lines in Figure 23. The left and the center interlacings both have the feature that any line of the interlacing can be taken to any other line by means of an isotopy. This is not the case for a amphicheiral interlacing (the right one in Figure 23). We shall say that two lines of an interlacing are isotopic if there exists an isotopy of the interlacing which takes one of the lines to the other one. To be sure, strictly speaking this is not an isotopy, because at the last moment the two lines come together. Instead of changing the meaning of the word ``isotopy'', we are better off leaving the meaning unchanged and adopting the following definition of isotopic lines of an interlacing: there is an isotopy of the entire interlacing which makes the two lines approach one another so that one can separate them from the other lines of the interlacing by a hyperboloid (in which case there is nothing to stop us from bringing the two lines together).

Isotopic lines have the same location relative to the other lines in the interlacing. Hence, if $a$ and $b$ are isotopic lines and $c$ and $d$ are two other lines of the same interlacing, then the triples $a,c,d$ and $b,c,d$ have the same linking number. Using this necessary condition for lines to be isotopic, we can easily show that in the interlacing on the right in Figure 23 the line $l$ is not isotopic to $m$. In fact, the triple $l,n,k$ has linking number $+1$, while the triple $m,n,k$ has linking number $-1$.

It is clear that, given any two isotopic lines in an interlacing, an isotopy can be found which interchanges them and causes all of the other lines to end up in the same place as before. Hence, isotopy of lines in an interfacing is an equivalence relation, and the set of all lines in an interfacing is partitioned into isotopy equivalence classes. The left and center interlacings in Figure 23 each has only one equivalence class, while the right interlacing has two: the lines $k$ and $l$ are in one class, and $m$ and $n$ are in another.

If we choose one line from each equivalence class in an interlacing, then the isotopy type of the resulting interlacing does not depend on our choice of our choice of line in each equivalence class. This interlacing is called the derived interlacing.

It is useful to pass to the derived interlacing if it contains fewer lines than the original interlacing. In order to recover the original interlacing from the derived one, one needs a relatively small amount of additional information, namely, how many lines were in each class and how they were linked to one another. In fact, by means of an isotopy one can reduce the original interlacing to a state in which the lines of each equivalence class are generatrices of the same family on a one-sheeted hyperboloid, and the hyperboloids containing the lines of the different equivalence classes do not intersect. We leave it as an exercise to construct an isotopy that does this.

The derived interlacing determines the relative location of the hyperboloids. To recover the original interlacing it remains only to specify one of the two families of generatrices on each hyperboloid. Here one does not have to do this at all if the class has only one line or if there is only one class in all and it has two lines. Otherwise the choice of a family of generatrices can be specified by means of a numerical invariant $\varepsilon =\pm1$ for each isotopy class of lines in the interlacing; this is defined to be the linking number of the triple of lines $a,b,x$, whee $a$ and $b$ are lines in the equivalence class and $x$ is any line distinct from $a$ and $b$. We shall prove that this invariant depends only on the class of lines isotopic to $a$ and $b$. The proof will use certain formulas in which we will use the following notation: the linking number of lines $a,b,c$ will be denoted by $lk(a,b,c)$.

Lemma. For any lines $a,b,c,d$ one has

$$lk(a,b,c)lk(a,b,d)lk(a,c,d)lk(b,c,d)=1.$$

This identity follows immediately from the definition of the linking number of a triple of lines as the product of the linking numbers of the three pairs of lines in the triple furnished with certain orientations. If we give orientations to the lines $a,b,c,d$ and then compute the left side of the above equality, we obtain the product of the squares of the linking numbers of all possible pairs of lines $a,b,c,d.$$\qedsymbol$

We now prove that $lk(a,b,x)$ does not depend on $x$ when $a$ and $b$ are isotopic lines of the interlacing. Let $y$ be any line of the interlacing which is distinct from $a,b,x$. By the lemma we have

$$lk(a,b,x)=lk(a,b,y)lk(a,x,y)lk(b,x,y).$$

Since the lines $a$ and $b$ are isotopic, we have $lk(a,x,y)=lk(b,x,y)$, and hence $lk(a,b,x)=lk(a,b,y)$.

It remains to show that $lk(a,b,x)$ does not depend on the choice of representatives $a$ and $b$ of the isotopy class of lines. In fact, if $c$ is a line which is isotopic to $a$ and distinct from $b$, then, as already proved, we have

$$lk(a,b,x)=lk(a,b,c)=lk(a,c,b)=lk(a,c,x).\qed$$

A class of isotropic lines of an interlacing whose invariant is $\varepsilon$ $(=\pm1)$ will be called an $\varepsilon$-class.

Some interlacings can be brought to the form of an interlacing of one line by successively taking the derived interlacing. Such an interlacing is said to be completely decomposable. A completely decomposable interlacing can be characterized up to isotopy by the invariants associated with each transition from an interlacing to its derived interlacing. We shall introduce some notation for this characterization. An interlacing of $p$ generatrices of a hyperboloid which form an $\varepsilon$-class of isotopic lines will be denoted by $\langle \varepsilon p\rangle$.

We now consider $p$ hyperboloids which encompass disjoint regions and which have the lines of the interlacing $\langle \varepsilon p\rangle$ as their axes. An interlacing made up of $p$ subinterlacings $A_1,\dotsc,A_p$, each of which is in the region bounded by the corresponding hyperboloid, will be denoted by $\langle +A_1,\dotsc,A_p\rangle$ if $\varepsilon =+1$ and $\langle -A_1,\dotsc,A_p\rangle$ if $\varepsilon =-1$. In situations where the signs do not matter to us, we shall omit them from the notation. For example, the interlacings in Figure 23 are characterized by the symbols $\langle +4\rangle$, $\langle -4\rangle$, and $\langle \langle +2\rangle ,\langle -2\rangle \rangle$. The interlacings in Figure 16 are given by the symbols $\langle +4\rangle ,\langle +5\rangle ,\langle +6\rangle$. The amphicheiral interlacing of an even number $p$ of lines that was constructed above is given by $\langle \langle +p/2\rangle ,\langle -p/2\rangle \rangle$. In particular, the interlacing in Figure 17 is $\langle \langle +2\rangle ,\langle -2\rangle \rangle$.

Not every interlacing is completely decomposable. For example, the derived interlacing for the interlacing in Figure 19 coincides with the original interlacing, and it cannot be placed on a hyperboloid (otherwise it would not be an amphicheiral interlacing). This is the simplest example of an interlacing which is not completely decomposable.