A highly interconnected mesh of 162 paperclips.

Designed and constructed in June, 2012.

The densely packed weaving pattern is visible from this topside view.

A Borromean Box, made from 81 paperclips of 3 colors. The Pair o' Boxes sculpture has two of these intertwined.

The Borromean rings (illustrated below) is a fascinating link made of three interconnected loops. The loops cannot be separated without cutting them or passing one through another, but curiously, if any one of the loops is erased, the other two are free to separate—they're not linked together. I was first shown how to make the Borromean rings with paperclips (shown below) at the tenth Gathering for Gardner conference, and this idea seems to have been floating around the internet for some time (e.g. by Dror Bar-Natan and Leonid A. Broukhis).

The abstract Borromean rings (right), and a representation of the Borromean rings in paperclips (left).

I decided to push this idea further with my sculpture, Pair o' Boxes. This dense structure of 162 paperclips is composed of two intertwined boxes, each made from 27 Borromean ring configurations. Despite no two clips being linked, the overall structure is very interconnected.

We can make this *interconnectedness* claim more precise with the language of knot theory. Consider the link *L* in the image below as a schematic drawing of the Pair o' Boxes, where each loop is made of string that we can manipulate: we are allowed to bend, lengthen, or deform the string, but we can't cut it or pass strands through each other.

This 54-component link *L* schematically represents the Pair o' Boxes sculpture.

A nice measure of interconnectedness is called the unlinking number: if we disobey the rules and allow ourselves to pass strands through each other, what is the smallest number of such intersections needed to completely separate all loops? For our link *L*, it can be shown that 108 intersections are required! Can you figure out how to separate everything with just 108 intersections? (Hint: lift the vertical loops straight up.) Proving that 108 intersections are *necessary* is harder; one proof is sketched in the appendix below.

But there's more interconnectedness to be had! If we had two boxes that were not intertwined, the unlinking number would still be 108. How can we say that *L* is "more interconnected" than two separate boxes? Well, two boxes may be separated by a plane, so this link is called split; by contrast, the link *L* is non-split, meaning it is impossible to find a separating plane no matter how much you manipulate the strings (without cutting or intersecting them). This is a highly nontrivial fact, and the only proof I have come up with is neither simple nor short, so I'll omit it here.

When passing one strand through another, call this intersection social if the two strands come from two different loops, and antisocial if they come from the same loop.

I claim that, when untangling the Borromean rings, at least two social intersections are necessary, no matter how many antisocial intersections are used. Indeed, it is impossible to undo the Borromean rings with only antisocial intersections (in any quantity), because the link is nontrivial even under link homotopy (because its 3-fold Milnor invariant is nonzero and invariant under homotopy). Since the number of social intersections must be even (by parity of linking numbers), at least 2 are needed.

Now, each of the 54 Borromean-ring sub-links needs at least two social intersections to be unlinked, and each social intersection of *L* counts toward at most one Borromean-ring instance. So at least 108 social intersections are necessary.