Thoughts on Braids

Define $G(n)$ to be the free groupoid on the following graph:

We have two objects, $\nn$ and $\ss$, and $n+1$ morphisms running between them, $m_0, \ldots, m_n$, as well as their inverses $m^{-1}_0, \ldots, m^{-1}_n$.

Let's define a sequence of functors $(\sigma_i : G(n) \to G(n) \st i \in \{1, \ldots, n-1\})$ like so: \[ \begin{array}{rcl} \sigma_i(\nn) &=& \nn\\ \sigma_i(\ss) &=& \ss\\ \sigma_i(m_j) &=& j \qquad (i \ne j)\\ \sigma_i(m_i) &=& m_{i-1}m_i^{-1} m_{i+1}\\ \end{array} \] This maps one of the morphisms in the category into a zig-zag involving the preceding and succeeding morphisms:

We can define functors $\sigma^{-1}_i$ very similarly: \[ \begin{array}{rcl} \sigma^{-1}_i(\nn) &=& \nn\\ \sigma^{-1}_i(\ss) &=& \ss\\ \sigma^{-1}_i(m_j) &=& j \qquad (i \ne j)\\ \sigma^{-1}_i(m_i) &=& m_{i+1}m_i^{-1} m_{i-1}\\ \end{array} \]

Claim

I claim that these functors behave just like the generators of the Artin braid group on $n$ strands. The functor/generator $\sigma_i$ realizes the crossing of the $i^{th}$ and the $(i+1)^{th}$ strands in one direction, and $\sigma_i^{-1}$ the other direction.

We have the following properties:

$\sigma_i \sigma^{-1}_i = \mathsf{id} = \sigma^{-1}_i \sigma_i$.
Let's see why, for example, $\sigma_i \sigma^{-1}_i m_i = m_i$. \[ \begin{array}{rcl} \sigma_i \sigma^{-1}_i m_i &=& \sigma_i (m_{i+1}m_i^{-1} m_{i-1}) \\ &=& (\sigma_i m_{i+1})(\sigma_i (m_i^{-1}))(\sigma_i m_{i-1}) \\ &=& m_{i+1}(\sigma_i (m_i^{-1})) m_{i-1} \\ &=& m_{i+1}(\sigma_i m_i)^{-1} m_{i-1} \\ &=& m_{i+1}(m_{i-1}m_i^{-1} m_{i+1})^{-1} m_{i-1} \\ &=& m_{i+1}m_{i+1}^{-1} m_i m_{i-1}^{-1} m_{i-1} \\ &=& (m_{i+1}m_{i+1}^{-1}) m_i (m_{i-1}^{-1} m_{i-1}) \\ &=& m_i \\ \end{array} \]
$\sigma_i \sigma_j = \sigma_j \sigma_i$ when $|i-j| > 1$.
$\sigma_i \sigma_j \sigma_i = \sigma_j \sigma_i \sigma_j$ when $|i-j| = 1$. ("the third reidemeister move")
Let's check for example $\sigma_1\sigma_2 \sigma_1 = \sigma_2 \sigma_1 \sigma_2$.
For brevity, we write $\mathsf{0123\cdots}$ for $m_0 m_1 m_2 m_3\cdots$ and $\mathsf{\bar 0\bar1 \bar2 \bar3\cdots}$ for $m_0^{-1}m_1^{-1}m_2^{-1}m_3^{-1}\cdots$. \[\begin{array}{rcl} \sigma_1\sigma_2 \sigma_1 (\mathsf{1}) &=&\sigma_1\sigma_2 (\mathsf{0 \bar 12}) \\ &=&\sigma_1 (\mathsf{0 1(1\bar2 3)}) \\ &=&\sigma_1 (\mathsf{0 \bar2 3}) \\ &=&\mathsf{0 \bar2 3} \\ &=&\mathsf{0\bar 1 (1 \bar 2 3)} \\ &=&\sigma_2 (\mathsf{0\bar 1 2}) \\ &=&\sigma_2\sigma_1 (\mathsf{1}) \\ &=&\sigma_2\sigma_1 \sigma_2 (\mathsf{1}) \end{array} \qquad\qquad \begin{array}{rcl} \sigma_1\sigma_2 \sigma_1 (\mathsf{2}) &=&\sigma_1\sigma_2 (\mathsf{2}) \\ &=&\sigma_1 (\mathsf{1\bar2 3}) \\ &=& (\mathsf{(0\bar 1 2) \bar2 3}) \\ &=&\mathsf{0 \bar1 3} \\ &=&\sigma_2 \mathsf{0 \bar1 3} \\ &=&\sigma_2 (\mathsf{(0\bar 1 2) \bar2 3}) \\ &=&\sigma_2\sigma_1 (\mathsf{1\bar 2 3}) \\ &=&\sigma_2\sigma_1 \sigma_2 (\mathsf{2}) \end{array} \]

Characterizing the Braid Subset

The above properties let us map the braid group into the automorphisms of the groupoid $G(n)$, but of course there are lots of other functors $G(n) \to G(n)$ that are not generated by the $\sigma_i, \sigma^{-1}_i$. For one thing, we could swap $\nn$ and $\ss$. But let's fix an $n$ and try to note some properties that are possessed by all functors $F$ that are composites of some sequence of $\sigma_i, \sigma^{-1}_i$. Let's do the easy ones first:

$F \nn = \nn$
$F \ss = \ss$
$F m_0 = m_0$
$F m_n = m_n$

Now let's think about some ways that $G(1)$ can map into $G(n)$. The groupoid $G(1)$ is just the circle. Define $\mathcal E_i : G(1) \to G(n)$ to be the functor that takes \[ \begin{array}{rcl} \mathcal E_i(\nn) &=& \nn\\ \mathcal E_i(\ss) &=& \ss\\ \mathcal E_i(m_0) &=& m_i\\ \mathcal E_i(m_1) &=& m_{i+1}\\ \end{array} \]

A property that we observe holds for any $F$ that arises as a composite of some sequence $\sigma_i, \sigma^{-1}_i$ is:

There exists a permuation $\pi : \{0,\ldots,n-1\} \to \{0,\ldots,n-1\}$ such that for every $i \in \{0,\ldots,n-1\}$ there is a natural isomorphism $F \circ \mathcal E_i \cong \mathcal E_{\pi i}$

Finally, we guess that these properties are enough to characterize an embedding of the braid group:

Conjecture: The subgroup of automorphisms of $G(n)$ that consist of the $F$ that satisfy properties (1)-(5) above is isomorphic to the Artin braid group $B_n$. If (5) is replaced by the stronger property

For every $i \in \{0,\ldots,n-1\}$ there is a natural isomorphism $F \circ \mathcal E_i \cong \mathcal E_i$

then we obtain instead the pure braid group $P_n$.