The Braid Category

Braid Category

The braid category $(\mathsf{Braid},+,0,\tau )$ is a braided monoidal category whose objects are natural numbers $n$ (equivalently, collections of $n$ dots), and whose morphisms are automorphisms represented by elements of the braid groups $B_n$. The monoidal product is given by addition on objects and juxtaposition on morphisms. The braiding $\tau :n+m\to m+n$ is represented by the braid

Remark

One can see that $\beta$ indeed satisfies the braiding axioms by the braid relation. A graphical proof is provided in \cite[Theorem XIII.2.1]{kasselQuantumGroups1995}.

e.g. For example, the following is a morphism $3\to 3$:

This means in $\mathsf{Braid}$, there are no morphisms other than automorphisms from the braid group. Thus this is a discrete category, and by Artin’s presentation \cite[Sec. 5.4]{prasolovKnotsLinksBraids1997} of the braid group, $\mathsf{Braid}$ is therefore generated by the following morphisms

Theorem

The braid category $\mathsf{Braid}$ is the free (strict) braided monoidal category generated by one object.

Proof For any strict braided monoidal category $(\text{C},\otimes ,1,\tau )$, let $X\in {\text{C}}_0$ be an object. To define a strict braided monoidal functor $F:\mathsf{Braid}\to \text{C}$, there is no choice but to set $F(n):=\underset{n\text{ factors}}{\underbrace{X\otimes X\otimes \cdots \otimes X}}.$
Moreover, the elementary braid generator ${\sigma }_j\in B_n$, which crosses the $j$-th and $(j+1)$-st strands, is obtained by vertically stacking $(j-1)$ straight strands, a single crossing, and $(n-j-1)$ additional straight strands. Therefore, it must be sent to $F({\sigma }_j)={\text{id}}_X^{\otimes (j-1)}\otimes {\tau }_{X,X}\otimes {\text{id}}_X^{\otimes (n-j-1)}.$
It remains to verify that this assignment respects the braid relations: ${\sigma }_i{\sigma }_j={\sigma }_j{\sigma }_i\text{for }|i-j|>1,{\sigma }_j{\sigma }_{j+1}{\sigma }_j={\sigma }_{j+1}{\sigma }_j{\sigma }_{j+1}.$The first relation holds because braidings acting on disjoint tensor factors commute. To illustrate this, consider the object $4\in {\mathsf{Braid}}_0$ and the braid generators ${\sigma }_1,{\sigma }_3:4\to 4$. Then

F(\sigma_1\sigma_3) &= (\tau_{X,X} \otimes \id_X \otimes \id_X) \circ (\id_X \otimes \id_X \otimes \tau_{X,X}) \\ &= (\tau_{X,X} \circ (\id_X \otimes \id_X)) \otimes ((\id_X \otimes \id_X) \circ \tau_{X,X}) \\ &= (\tau_{X,X} \circ \id_{X\otimes X}) \otimes (\id_{X\otimes X} \circ \tau_{X,X}) \\ &= \tau_{X,X} \otimes \tau_{X,X}, \end{aligned} $$and similarly, $$ \begin{aligned} F(\sigma_3\sigma_1) &= (\id_X \otimes \id_X \otimes \tau_{X,X}) \circ (\tau_{X,X} \otimes \id_X \otimes \id_X) \\ &= ((\id_X \otimes \id_X) \circ \tau_{X,X}) \otimes (\tau_{X,X} \circ (\id_X \otimes \id_X)) \\ &= \tau_{X,X} \otimes \tau_{X,X}. \end{aligned} $$ For the second relation, note that by the [[Braided Monoidal Categories#^d7b677|corollary]], $\tau_{X,X}$ is a Yang--Baxter operator. Hence $F(\sigma_j)$ satisfies the second braid relation by the [[Yang-Baxter Equation#^7c4536|proposition]].