For modular lattices there is a canonical example of the "smallest" non-modular lattice, $\mathbf N_5$. Is there a similar example for 3-categories which are not equivalent to any strict 3-category? I have read so many times that there do exist such 3-categories but I have never seen an explicit example of one.
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The homotopy $3$-groupoid of $S^2$ is not equivalent to a strict $3$-category. While it may or may not be the strictly smallest weak 3-category with this property, this seems to be the standard example.
This was, as far as I know, first proven by Clemens Berger. For a reference, see Carlos T. Simpson's book "Homotopy theory of higher categories", Corollary I.4.3.3. and section I.4.4. in this draft.
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Simply connected 3-groupoids $X$ have two homotopy groups $\pi_2(X), \pi_3(X)$, and are classified by this pair of homotopy groups together with a $k$-invariant, namely a cohomology class in $H^4(B^2 \pi_2, \pi_3)$. It turns out that $X$ is strictifiable iff the $k$-invariant vanishes iff $X$ is just the product $B^2 \pi_2 \times B^3 \pi_3$.
The case of the fundamental $3$-groupoid of $S^2$ corresponds to $\pi_2 = \pi_3 = \mathbb{Z}$ with $k$-invariant the element of
$$H^4(B^2 \pi_2, \pi_3) \cong H^4(\mathbb{CP}^{\infty}, \mathbb{Z}) \cong \mathbb{Z}$$
which as a cohomology operation $H^2(-, \mathbb{Z}) \to H^4(-, \mathbb{Z})$ is just the cup square. A "smaller" example is to take $\pi_2 = \pi_3 = \mathbb{Z}_2$ with $k$-invariant the cup square $H^2(-, \mathbb{Z}_2) \to H^4(-, \mathbb{Z}_2)$.
answered Nov 13, 2015 at 21:33
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2$\begingroup$ Is there a way to understand this without any knowledge of homological algebra? Maybe a finite presentation? Just a guess but, your "smaller" example has 1 object $x$, 1 arrow $id_x = a $, 2 2-arrows $id_a$ and $b$ (with $a^2 = id_b$), and... 8 3-arrows? (2 in each of $hom(id_a, id_a)$, $hom(b,b)$, $hom(b,id_a)$, $hom(id_a,b)$ )? Also, what is a "cup square"? I can't find any reference for this. $\endgroup$ Nov 17, 2015 at 2:47
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$\begingroup$ @Justaskin_: "cup square" refers to the squaring operation $x \mapsto x^2$ in a cohomology ring (whose multiplication is called the "cup product"). You're not going to get very far studying 3-groupoids without knowing any homological algebra... but in any case, you can think about simply connected 3-groupoids as "braided 2-groups": explicitly, they can be modeled as braided monoidal groupoids in which every object is invertible. Strictness corresponds to replacing "braided" with "symmetric" (or something like that; maybe it's even worse than this). $\endgroup$ Nov 17, 2015 at 3:07
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$\begingroup$ The point here is that in this case the interesting data of the 3-groupoid is not in the morphisms but in the coherence data being required on various compositions, which in the braided 2-group model corresponds to the braiding. For strict 3-groupoids you replace these coherences with identities and things are much less interesting. $\endgroup$ Nov 17, 2015 at 3:07