# Picture Frame

While I was working with a friend mathematician, I showed him a couple ways `physicists use math in strange ways.

One of the things that I showed him was the usage of commutators in quantum mechanics. They look like this:

$$[A, B] = AB - BA$$

That is zero if $A$ and $B$ commute, that is $AB = BA$. If they do not do that, like general square matrices or operators, then the commutator gives you the way they do not commute.

## The Problem

He told me that they actually come from group theory and that there is a really neat problem that can be solved with them:

You want to put a framed picture onto the wall with a cord to the picture frame. The cord is a single one, and both ends are attached to the frame. How do you need to put the cord around $n$ nails, so that the picture can be taken down by taking out

anyof the nails?

For that, he told me, that I should write the commutator like that:

$$[A,B] = ABA^{-1}B^{-1}$$

In the previous revision of this article, prior to 2013-08-18, I used the commutators $AB - BA$ with the implicit $-BA = A^{-1} B^{-1}$. Writing the commutator like he said is cleaner, since there is only one operation on this group of windings, namely performing them after each other.

I have seen a website where they give you the solution, but I would like to show the derivation he gave me. He told me, that the solution for $n = 2$ is simply $[A, B]$. Later on, I tried to understand what that actually means. This is what I came up with.

Start with two nails, A and B. There is a basic loop around each nail, $A$ and $B$ respectively:

If you do $A$ and then $B$ directly, and wiggle on the cord a little, you will get the following, which I will call `AB`:

If you go through one of the basic loops in reverse, it will count as its inverse, say $A^{-1}$.

## The Solution

The solution is supposed to be:

$$[A, B] = ABA^{-1}B^{-1}$$

That means doing $A$, then $B$, then $A^{-1}$ ($A$ backwards) and then $B^{-1}$ ($B$ backwards). After some wiggling, I get this:

Look closely. If you pull out nail A, the loop that is tight around A will become free, go around B and will finally release the picture frame. If you release B, the loop that goes below B will become free. The loop above B will just fall down. That will also release the frame. So this is the solution to that problem.

## Higher Orders

The really interesting part about the group theory approach is that it scales to higher orders fairly easily. So what about three nails? The definition for the commutator of higher orders is the following:

$$[A_1,A_2,...,A_n] = \left[[A_1,...,A_{n-1}],A_n \right]$$

With the basic loops $A$, $B$ and $C$, the solution to the three nail problem should be the following:

$$\begin{aligned} \begin{aligned} \big[ [A, B], C \big] &= [ A B A^{-1} B^{-1}, C ] \ &= A B A^{-1} B^{-1} C (A B A^{-1} B^{-1})^{-1} C^{-1} \ &= A B A^{-1} B^{-1} C (B A B^{-1} A^{-1}) C^{-1} \ &= A B A^{-1} B^{-1} C B A B^{-1} A^{-1} C^{-1} \end{aligned} \end{aligned}$$

I tried to create a picture with this. It is not really easy to see what is going on there.

As far as I checked, the picture falls down if one removes either A, B or C, so the answer seems to be correct. In principle, it should be possible to do this with four and more nails, but I think that it will be almost impossible to present this in a clear way.