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Weak 2-groups can be classified by the data $(\pi_1,\pi_2, t, \omega)$, where $\pi_1$ is a group, $\pi_2$ an Abelian group, $t: \pi_1 \to Aut(\pi_2)$, and $\omega \in H^3(B\pi_1,\pi_2)$.

I wonder do we have a similar classification for weak 3-groups?

3-groups contain data $(\pi_1,\pi_2,\pi_3)$ (three groups). I am interested in the simple cases where $\pi_2$ is $Z_2$ or trivial and $\pi_3$ is $U(1)$ or $Z_n$.

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2 Answers 2

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Yes, but it's more complicated. First let me describe the special case where $\pi_1$ is trivial. If $G$ is such a 3-group, then $X = BG$ is a pointed connected simply connected space. It fits into a fibration

$$B^3 \pi_3 \to X \to B^2 \pi_2$$

which, as it turns out, can be delooped once into a fiber sequence

$$B^3 \pi_3 \to X \to B^2 \pi_2 \to B^4 \pi_3$$

reflecting the fact that the fibration above is classified by a map $f : B^2 \pi_2 \to B^4 \pi_3$, or equivalently by a cohomology class in $H^4(B^2 \pi_2, \pi_3)$. It's a classic result that this cohomology group is naturally isomorphic to the group of $\pi_3$-valued quadratic forms on $\pi_2$: this quadratic form corresponds to a homotopy operation $\pi_2 \to \pi_3$, represented by the Hopf fibration, which is a quadratic refinement of the Whitehead bracket $\pi_2 \times \pi_2 \to \pi_3$.

(As written, this is a classification of spaces with nontrivial $\pi_2$ and $\pi_3$. Delooping once, this is a classification of simply connected 3-groups. Delooping twice, this is a classification of grouplike braided monoidal groupoids, or said another way, braided 2-groups.)

The general case is messier. Now $X = BG$ fits into a fibration

$$\widetilde{X} \to X \to B \pi_1$$

where $\widetilde{X}$ is the universal cover of $X$, which is as above. Such fibrations are classified by actions of $\pi_1$ on $\widetilde{X}$ in a homotopy-theoretic sense, so now one has to compute the automorphisms of all possible $\widetilde{X}$, then maps from $\pi_1$ into these.

In the special case where $\pi_2$ is trivial, $X = BG$ now fits into a fibration

$$B^3 \pi_3 \to X \to B \pi_1$$

which implies that $X$ is classified by a pair consisting of an action of $\pi_1$ on $\pi_3$ and a cohomology class in $H^4(B \pi_1, \pi_3)$.

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  • $\begingroup$ Not a critics, just to understand - as far as i understand question was about group which is which is made by extensions from 3 abelian groups, or I misunderstand something ? if so where how all that looping-delooping is related ? $\endgroup$ Jun 10, 2017 at 19:15
  • $\begingroup$ @Alexander: no, the question is about 3-groups, or equivalently about pointed connected spaces $X$ with only $\pi_1(X), \pi_2(X), \pi_3(X)$ nontrivial. These can be understood as being built from "extensions" but of Eilenberg-MacLane spaces, not groups. $\endgroup$ Jun 10, 2017 at 19:24
  • $\begingroup$ Ah, sorry, I misunderstood the question. $\endgroup$ Jun 10, 2017 at 19:38
  • $\begingroup$ @Qiaochu Yuan Thank you for the answer. What is $H^4(B^2 \pi_2, \pi_3)$ when $\pi_3=U(1)$ and $\pi_2=Z_2$? $\endgroup$ Jun 10, 2017 at 22:37
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This has been worked out by Conduché in Modules croisés généralisés de longueur 2 (JPAA 34 (1984), 155-178), using simplicial group methods.

While it more or less boils down to the stuff described in the answer by Qiaochu, Conduché also describes a nice (2? 3?)-category holding the same information. Its objects are complexes$$G_2\xrightarrow{\partial}G_1\xrightarrow{\partial}G_0$$of $G_0$-groups (composite trivial, $G_0$ acting on itself by conjugation) together with the s. c. Peiffer bracket $\{,\}:G_1\times G_1\to G_2$ satisfying the elaborate but appealing identities $$ \begin{aligned} {}^{x_0}\{x_1,y_1\}&=\{{}^{x_0}x_1,{}^{x_0}y_1\}\\ \{\partial x_2,\partial y_2\}&=[x_2,y_2]\\ \partial\{x_1,y_1\}&=x_1y_1x_1^{-1}\left({}^{\partial x_1}y_1\right)^{-1}\\ \{\partial x_2,x_1\}\{x_1,\partial x_2\}&=x_2\left({}^{\partial x_1}x_2\right)^{-1}\\ \{x_1y_1,z_1\}&=\{x_1,y_1z_1y_1^{-1}\}\ \ {}^{\partial x_1}\{y_1,z_1\}\\ \{x_1,y_1z_1\}&=\{x_1,y_1\}\{x_1,z_1\}\{\partial\{x_1,z_1\}^{-1},{}^{\partial x_1}y_1\}. \end{aligned} $$ Since both $G_2$ and $G_1$ are nonabelian, it is clear that there will be lots of equivalent nonisomorphic objects, but still this description has its advantages.

Homotopy groups are, as expected, the homology groups of the complex (i. e. $\pi_1$ is the cokernel of $G_1\to G_0$, $\pi_2$ the quotient of the kernel of $G_1\to G_0$ by the image of $G_2\to G_1$, and $\pi_3$ the kernel of $G_2\to G_1$. Thus given $\pi_1$, $\pi_2$, $\pi_3$, to build a model of the above kind one also needs actions of $\pi_1$ on $\pi_2$ and $\pi_3$ as well as the bracket $\pi_2\times\pi_2\to\pi_3$. One then must build a complex as above such that its homology groups are $\pi_1$, $\pi_2$, $\pi_3$, while the actions and the bracket are induced from the structure on the complex.

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  • $\begingroup$ It appears that one also need to have two group actions $t_1: G_0 \to Aut(G_1)$ and $t_2: G_0 \to Aut(G_2)$ in the above definition. $\endgroup$ Jun 11, 2017 at 1:07
  • $\begingroup$ @Xiao-GangWen this is what I meant by $G_0$-groups. Moreover by a complex of $G_0$-groups is meant that $\partial$ is $G_0$-equivariant $\endgroup$ Jun 11, 2017 at 4:40
  • $\begingroup$ @Xiao-GangWen and these actions are denoted by ${}^{x_0}x_1$, ${}^{x_0}x_2$ in the identites $\endgroup$ Jun 11, 2017 at 13:07

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