1. Algebra from action groupoids

Let $G$ be a finite group acting on a finite set $X$ from the right (denoted in element as $x^{g}$). We have an algebra (of the action groupoid) over $\mathbb{C}$: the vector space

$$ H = <\delta_{x} | x \in X> \otimes <g | g \in G>$$

with the algebra structure given as $g \cdot g' = gg'$, $\delta_x \cdot \delta_{x'} = \chi_{x = x'} \delta_x$, $g \cdot \delta_{x} = \delta_{x^{(g^{-1})}} \cdot g$, and $1 = \sum_{x \in X} \delta_{x} \otimes e$, where $e$ denotes the group identity element of $G$.

2. Bi-algebra

To upgrade this algebra to a bi-algebra, it would suffice to require $X$ to be a finite group $H$ such that the action satisfies Leibniz rule:

$$ (h h')^{g} = (h^{g}) (h'^{g}).$$

Indeed, one can define the co-algebra structure by $\Delta(g) = g \otimes g$ and $\Delta(\delta_{h}) = \sum_{h' \in H} \delta_{h'} \otimes \delta_{h'^{-1}h}$. Note that this turns the representation category into a (fusion) tensor category.

3. Braided Bi-algebra

I'm interested in all ways to extend this to a braided (fusion) category. In general, this is a hard question (e.g. it was a big deal of Drinfeld's construction of interesting braided bi-algebras, leading to quantum groups.) However, there is a known extra condition that makes it possible: one requires that there exists a group homomorphism $\partial: H \to G$ such that

  1. $\partial (h^{g}) = g^{-1} (\partial h) g$ and
  2. $h^{\partial h'} = h'^{-1} h h'$

Indeed, we now have a braided structure, namely for any representation $V_{1}$ and $V_{2}$, define [1, page 4]

$$R: V_{1} \otimes V_{2} \to V_{2} \otimes V_{1}$$ $$v_{1} \otimes v_{2} \mapsto \sum_{h \in H} (\partial h)v_{2} \otimes \delta_{h} v_{1}$$

This is a miracle to me given that finding braided $R$-matrices is a hard task in general! What's even more mysterious to me is that the condition turns the pair $(G, H)$ into a crossed-module (or a strict $2$-group [2])! I wonder if there's a deeper reason why a braided structure shows up. However, having been thinking of this question for months, I have no clue.


  1. How to interpret the existence of the braided structure from the fact that the action ($G$ on $H$) and the boundary morphism $\partial$ forms a $2$-group?

  2. In general, is there any hope to fit this into some $2$-representation theory of $2$-groups?



1 Answer 1


A little background, then I'll try to answer question 1.

While attempting to encode homotopy types of connected CW-complexes $K=(K^n)_{n\in\mathbb N}$ in a purely algebraic way, J.H.C Whitehead invented crossed modules (Combinatorial homotopy. II, Bull. Am. Math. Soc. 55, 453-496 (1949). ZBL0040.38801) as a component of his `homotopy systems' wherein crossed modules describe the relationship between $\pi_2(K^2,K^1,k_0)$ and $\pi_1(K^1,k_0)$. Specifically, the map $$\partial:\pi_2(K^2,K^1,k_0)\longrightarrow\pi_1(K^1,k_0)\,,$$ which takes an embedded disk $\phi:D^2\to K$ and returns the restriction $\phi|_{\partial D^2}:S^1\to K$, gives rise to a crossed module structure with respect to the conjugation action of $\pi_1(K^1,k_0)$ on $\pi_2(K^2,K^1,k_0)$.

The theorem of Mac Lane and Whitehead (On the 3-type of a complex, Proc. Natl. Acad. Sci. USA 36, 41-48 (1950). ZBL0035.39001) is that when $\pi_n(K)=0$ for $n>2$, this crossed module structure uniquely determines the homotopy type of $K$, and moreover every crossed module can be realized by such a complex.

Using this theorem, we can assume that $H=\pi_2(K^2,K^1,k_0)$ and $G=\pi_1(K^1,k_0)$ for some connected CW-complex $K$. If you aren't familiar with CW-complexes, you can think of them as a particularly nice class of topological spaces. In this setting, I can begin to explain why the braiding is not so surprising.

  1. The second homotopy group of a space is normally abelian, but the relative group $\pi_2(K^2,K^1,k_0)$ is not abelian. The problem is that the boundaries are not the constant loop $S^1\to k_0$, but rather they are loops $S^1\to K^1$ based at $k_0$. The standard topological trick for showing commutativity $\alpha*\beta=\beta*\alpha$ involves sliding $\alpha$ and $\beta$ around each other. In other words, the trick involves conjugating $\alpha$ by the loop $\partial\beta$ to produce the class $\alpha^{\partial\beta}$; it's just that conjugating by the trivial loop never does anything. Thus in our group the best we can get is that $\alpha*\beta = \beta*\alpha^{\partial\beta}$, which is your equation 2.

Since this equation literally comes from passing one map in front of the other to swap their order, it's not surprising that it encodes some braiding information. [That a 'group with braiding' should induce a braiding on its own representations is also natural.] edit The previous statement was glib, so let me clarify how it works in this case. For these representations the $\delta_h$ are idempotent, so they act like projections on any given representation. This allows us to produce a canonical decomposition $V_1\cong\bigoplus_{h\in H} \delta_hV_1$, and we can think of a vector $\delta_hv_1$ as being concentrated in the $h$-component, or if you like, homogeneous of degree $h$.

Now that we have a way of decomposing a vector into components that are labelled by the elements of $H$, we can ask what happens when $v_2$ passes under each of these components individually. From the previous discussion, this should involve something like 'conjugating by $h$'. Since it doesn't make sense to conjugate a vector in a representation, this normally wouldn't work, but in our case $\partial h$ is a version of `conjugating by $h$' that lives in $G$... which also happens to act on our vector space. So we posit $$\delta_hv_1\otimes v_2\mapsto (\partial h)v_2\otimes \delta_hv_1\,.$$ I admit that this guess is not fully justified, but I hope it's at least plausible. Next we use the fact that $1=\sum_{h\in H}\delta_h$ to extend the map to inhomogeneous vectors to arrive at the formula you have given. \begin{align*} v_1\otimes v_2&=\left(\sum_{h\in H}\delta_hv_1\right)\otimes v_2\\ &=\sum_{h\in H}\delta_hv_1\otimes v_2\\ &\mapsto \sum_{h\in H}(\partial h)v_2\otimes \delta_hv_1\,. \end{align*}

  1. I can't quite answer this one. The representations you describe are not simply a 2-group acting on a category, at least not in any way I can make sense of. They appear to be classical representations of the algebra $H^*\rtimes G$ where $H^*=\mathsf{Fun}(H,\mathbb C)$, and where the $\rtimes$ indicates $$(\delta_m\otimes g)\cdot(\delta_n\otimes h)=(\delta_m\cdot\delta_{n^g})\otimes(gh)\,.$$
  • 1
    $\begingroup$ Thanks for your answer. (1) It's a beautiful explanation of the structure of the 2-group $\chi$ from the point of view of homotopy 2-types. But it's unclear to me how that structure $\alpha \star \beta = \beta \star \alpha^{\partial \beta}$ explains the braided structure on $Rep(\chi)$. (2) Indeed, I cannot think of a description in $(2-)$functors for the representations. In particular, if that's possible the elements in $H$ should correspond to invertible elements, which isn't the case. $\endgroup$
    – Student
    Jun 3, 2022 at 13:59
  • $\begingroup$ (1) You are right, it wasn't as clear as I had made it sound. I've edited the post to explain how to induce the braiding on the representations. $\endgroup$ Jun 4, 2022 at 3:17
  • $\begingroup$ (2) Right. If this were a genuine 2-rep, then the elements of $H$ ought to correspond to isomorphisms $\eta:F\to\text{id}_{\mathbf 1}$ in the category $\text{End}(\mathbf 1)$. $\endgroup$ Jun 4, 2022 at 3:24

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