The Whitehead product and $\pi_{\leq 3} S^2$

Question

Why does "the non-vanishing of the Whitehead bracket" imply that the fundamental 3-groupoid $\pi_{\leq 3} S^2$ of the two-shere cannot be strictified (as claimed here)?

Answer 1

I have looked at the draft version of Simpson's book, and one can put the argument as that strict 3-groupoids cannot model Whitehead products. Another argument for this is that strict 3-groupoids are equivalent to crossed complexes (over groupoids) of length 3. This is the case $n=3$ of a result proved in this paper. This structure has no quadratic information. Thus strict globular groupoids model only a restricted range of homotopy types.

However cat$^3$-groups, which can be seen as strict $3$-fold groupoids in which one structure is that of a group, do model pointed homotopy 3-types, as shown by Loday in 1982, and an account of how these and related structures model Whitehead products $\pi_2 \times \pi_2 \to \pi_3$ is given in this essay, Theorem 2.4. Such structures are in some cases calculable because of a related van Kampen type theorem, published in 1987, referred to in the previous op. cit. This theorem also led, through considering some particular pushouts, to a nonabelian tensor product of groups (which act on each other and on themselves by conjugation) which has proved of interest especially to group theorists, so that a current bibliography has now 169 items dating from 1952. In particular there can be deduced many calculations of the first non trivial Whitehead product for $SK(G,1)$ for groups $G$ (see [119] by G. Ellis in that bibliography).

It seems to me important that such strict higher van Kampen theorems have been proved only for certain structured spaces, namely filtered spaces, and $n$-cubes of pointed spaces. Of course, homotopy groups themselves are defined only for spaces with a base point, a structure which does not contain much information on the space.

Grothendieck was of course unaware of such uses of strict $n$-fold groupoids when writing Pursuing Stacks.