For many years, my mother has baked a challah nearly every Friday for Shabbat. Occasionally, however, she asks me to do some portion of the challah making, possibly including the braiding. For reasons we’ll see later, I don’t like her braiding algorithm. This post includes several algorithms, written in a way that someone who knows what a braid group is can understand. (I never had much success following those series of diagrams I sometimes see; I always wished someone would write out the braiding process in terms of generators of the braid group, so that’s what I’m going to do here after I give the preliminary definitions.)

Wikipedia’s page on braid groups has lots of interesting things, so I’ll only write a few essential points here, leaving the reader to explore Wikipedia at eir leisure. I’m finding it a bit tricky to give a good informal definition of braids, so I’ll just assume that my reader knows roughly what a braid is and skip to the formal definition.

The braid group on $n$ strands is the group $B_n=\langle a_1,\ldots,a_{n-1}\mid a_ia_{i+1}a_i=a_{i+1}a_ia_{i+1} \text{ for } 1\le i\le n-2, \ a_ia_j=a_ja_i \text{ for } |i-j|>1\rangle$. In terms of actually playing with braid strands, $a_i$ means interchanging strand $i$ with strand $i+1$ by putting strand $i$ over strand $i+1$. It is pretty simple to see that these generators do indeed induce all possible braids (although I haven’t yet said what a braid is), and that the relations ought to hold. Now, of course, a braid is an element of the braid group.

The braid groups become rather complicated quite quickly. While $B_0=B_1=0$ and $B_2=\mathbb{Z}$, already $B_3$ is nonabelian, and it’s isomorphic to the fundamental group of the complement of a trefoil knot in $\mathbb{R}^3$.

Note also that there is a natural homomorphism $\pi:B_n\to S_n$ that tells us where the strand that started in the $i^\text{th}$ position ends up.

Okay, now it’s time for some challah braiding algorithms. My mother’s usual challah has four strands on the bottom and three on the top. The algorithm for the top braid is pretty natural: $(a_1a_2^{-1})^n$, where $n$ is decided by the length of the dough ropes.

I’m more concerned about the element of $B_4$ used for the bottom braid. She uses $(a_1a_2^{-1}a_3^{-1}a_2)^n$.  If $n=1$, we have $\pi(a_1a_2^{-1}a_3^{-1}a_2)=(142)(3)$ (in cycle notation). This is already bad news to me: one step of the algorithm produces a single fixed point! I think one step of the algorithm ought to give an $n$-cycle (here a 4-cycle) or else a pure braid (i.e. a braid in the kernel of $\pi$). But it gets worse: the strand that starts in position 3 has no undercrossings. So when we’re done, it sits on top of every other strand.

It turns out not to be so bad because the three-strand braid sits on top of the four-strand braid, so the central portion of the four-strand braid is not visible in the finished bread. But aesthetically (and mathematically), this feels like a serious flaw to me.

Fortunately, I found an alternate algorithm for four-strand braiding that lacks these flaws: $(a_2a_1a_3^{-1})^n$. If $n=1$, $\pi(a_2a_1a_3^{-1})=(1243)$, which is nice. Furthermore, every strand has both overcrossings and undercrossings. So this is my new preferred braid.

Sometimes, however, it is preferable to braid with six strands. There was an article in the newspaper that explained how to do it, but I was unable to follow it. Fortunately, I found a YouTube video that shows someone doing it (possibly the same way; I can’t tell). I was able to transcribe this method in terms of generators of the braid group. However, I’m not quite sure where it is supposed to end, so my braid may be slightly different from the one shown in the video. The braid is the video is $(e^{-1}d^{-1}c^{-1}b^{-1}a^{-1}(bcdeabd^{-1}c^{-1}b^{-1}a^{-1}e^{-1}d^{-1})^n$, except that it might stop somewhere in the middle of the $(\cdot)^n$. I don’t really want to calculate $\pi$ of this braid (computations like this have never been that easy for me), but I would guess that it is a 6-cycle if it stops at an appropriate moment. (Also, it’s not as complicated as the formula would make it seem, since there’s a lot of stuff like moving the strand on the right all the way over to the left, and it takes a lot of generators to express that, even though it’s not complicated when you’re actually braiding dough.)