Not Only Lines Can be Interlaced

We return to the definition of an interlacing of lines. We used this term to denote a finite set of pairwise skew lines in three-dimensional space. That is, among all possible sets of lines, we look at sets in general position which form an everywhere dense open subset of the space of all sets of lines.

The same can be done with other types of configurations. For example, we can consider finite sets of points in three-dimensional space. We say that such a set is nonsingular if for $k\le 4$ there is no set of $k$ points lying in a $(k-2)$-dimensional subspace (i.e., a four-tuple does not lie in a plane, a triple does not lie on a line, and all points are distinct). By an isotopy of such a set we mean a motion in the course of which these conditions are not violated. We say that a nonsingular set of points is amphicheiral if it is isotopic to its mirror image.

We shall not treat the problem of classifying nonsingular sets of points, but rather turn our attention to the amphicheiral problem.

Theorem. A nonsingular set of $q$ points in three-dimensional space is nonamphicheiral if $q\equiv 6\mod8$ or $q\equiv3\mod4$ and $q\ge7$.

Proof. Given a nonsingular set of points, we define $s$ to be the sum of the linking numbers of all triples of pairwise skew lines determined by pairs of points in our set. If our set has $q$ points, then the number of such triples is $\frac16\binom{q}{2}\binom{q-2}{2}\binom{q-4}{2}$. If $q\equiv 6$ or $7\mod8$, then this number is odd, and so $s$ is also odd, since it is a sum of an odd number of terms each of which is $\pm1$. Clearly, $s$ is preserved under isotopies of the set of points, and it is multiplied by $-1$ under mirror reflection. Hence, $s=0$ for an amphicheiral set. We conclude that if $q\equiv 6$ or $7\mod8$, a nonsingular set of $q$ points cannot be amphicheiral. To treat the case $q\equiv 3\mod8$, $q\ge11$, we introduce another numerical invariant of a nonsingular set of points. We first note that, given any two points $A$ and $B$ of our configuration, one can determine two opposite cyclic orderings of the remaining $q-2$ points, namely, the order in which a plane rotating around the axis $AB$ passes through them. If a triple of lines consists of the line $AB$ along with two lines joining four successive points in this ordering (more precisely, one line joins the first point to the second and the other one joins the third point to the fourth), then we say that the triple is cyclic. Our numerical invariant of a nonsingular set of points will then be the sum of the linking numbers of all cyclic triples of lines with distinguished first line. If $q\ge7$, then there are $(q-2)\binom{q}{2}$ terms in this sum, and so the sum is odd if $q\equiv3\mod4$, $q\ge7$. On the other hand, the sum is clearly equal to zero if we have an amphicheiral set. $\qedsymbol$

It is natural to ask questions about amphicheirality for nonsingular sets of points which are analogous to the four questions discussed above in connection with amphicheiral interlacings of lines. We do not know complete answers to those questions.

In the same spirit one can consider a mixed situation: configurations of both lines and points. There are various ways of defining a nonsingular configuration of this type, but the most natural definition is to require that the lines in the configuration be pairwise skew, the points not lie on the lines, and no two points lie in a common plane with one of the lines. Even less is known about the classification and amphicheirality of mixed configurations.

When investigating problems related to geometrical objects in Euclidean space, it is often useful to extend the space to a projective space. Projective space has even been called the ``great simplifier''. Passing to a projective space normally enables us to find a simpler projective classification problem inside our original classification problem, and this projective problem is usually interesting in its own right. The case of interlacings of lines is, however, an exception to this rule. When one goes from $\mathbb{R}^3$ to the projective space $\mathbb{R}P^3$, an interlacing of lines corresponds to a set of disjoint projective lines, and in this way one obtains all possible configurations of disjoint lines in $\mathbb{R}P^3$ in which no line is contained in the plane at infinity. Isotopy of interlacings is equivalent to the existence of an isotopy between the corresponding configurations of lines in $\mathbb{R}P^3$ in the course of which the lines remain disjoint. ¹Here one need not concern oneself with the plane at infinity. Thus, the problem of classifying interlacings up to isotopy is actually equivalent to the corresponding problem for configurations of lines in projective space. ²We do not get a simpler problem. But in the case of the problem of classifying nonsingular sets of points in three-dimensional space, passing from $\mathbb{R}^3$ to $\mathbb{R}P^3$ leads to a splitting up of the problem; however, we shall not discuss this here.

Instead we consider the following counter-part of the question on existing of amphicheiral interlacings of a given number of lines: For which pairs of non-negative numbers $p$, $q$ there exist an amphicheiral nonsingular configuration of $p$ lines and $q$ points in $\mathbb{R}P^3$? We proved above the following partial results:

• If $q=0$, a necessary and sufficient condition for this is $p\not\equiv3\bmod4$.
• If $p=0$, it is necessary that $q\not\equiv6\bmod8$.
• If $p=0$ and $q\ge7$, it is necessary that $q\not\equiv3\bmod4$.

The following complete answer was found by Podkorytov [10], after the previous version [2] of this paper was written:

Podkorytov Theorem. (See [10].) An amphicheiral nonsingular set of $q$ points and $p$ lines in three-dimensional real projective space exists if and only if either

$$q\le 3$$       and     $$p\equiv0$$ or $$1\bmod4,$$

or

$$q\equiv0$$    or $$1\bmod4$$       and     $$p\equiv0\bmod2.$$

$\qedsymbol$

Even in the case of interlacings of lines, passing to $\mathbb{R}P^3$ is not completely pointless. In $\mathbb{R}P^3$ we can see more clearly the topological reasons why interlacings are nonisotopic. As we showed at the very beginning of the article, any interlacing can be deformed into a set of parallel lines, and so there exists a homeomorphisms of $\mathbb{R}^3$ under which any interlacing is taken to any other given interlacing with the same number of lines. In $\mathbb{R}P^3$ this is no longer the case. The linking number introduced above for oriented skew lines can be interpreted in terms of the usual linking number in algebraic topology, applied to the corresponding lines in $\mathbb{R}P^3$ (except that we must double the topological invariant, which takes the values $\pm1/2$, since for us the values $\pm1$ are more convenient). Moreover, in all cases we know of, the nonisotopy of two interlacings of lines is proved using topological invariants of the corresponding sets of projective lines in $\mathbb{R}P^3$, although there probably exist nonisotopic interlacings of lines for which the corresponding sets of projective lines can be taken to one another by means of a homeomorphism of the ambient space.

Perhaps we should show greater caution and make our definitions in accordance with the accepted topological terminology, i.e., call interlacings of lines isotopic if the corresponding sets of projective lines can be taken into one another by a homeomorphism of $\mathbb{R}P^3$ which is isotopic to the identity (recall that an isotopy of the homeomorphism $h\colon X\to Y$ is a family of homeomorphisms $h_t\colon X\to Y$ with $t\in[0,1]$, $h_0=h$, such that the map $X\times[0,1]\to Y\colon(x,t)\mapsto h_t(x)$ is continuous). Then what we earlier called isotopies would be called rigid isotopies. Our cavalier attitude about this is permissible only because at the present level of knowledge we do not have examples of interlacings which show that these two types of isotopies actually lead to different equivalence relations. In some related situations, however, we do know such examples. We now discuss one such case.