2-groups are to crossed modules as 2-categories are to...? - MathOverflow

2-groups are to crossed modules as 2-categories are to...?

2010-08-15T02:36:53Z

Given a 2-group $\mathcal{G}$, you can construct a crossed module $(G,H,t,\alpha)$ and vice versa.

Is there something similar you can say for strict 2-categories?

In a personal attempt to understand strict 2-categories, I ended up constructing a speculative conceptual tool (whose validity remains to be seen) that I call the boundary of a 2-morphism. I've written up some raw notes here:

Boundary of a 2-Morphism and the Interchange Law

The basic idea is that given morphisms $f,g:x\to y$ and a 2-morphism $\alpha:f\Rightarrow g$, we define its boundary as an endomorphism

$$\partial\alpha:y\to y$$

satisfying

$$\partial\alpha\circ f = g.$$

When the source of the 2-morphism is an identity morphism, then we have

$$\partial\alpha = t(\alpha),$$

which seems to relate things well to cross modules when all morphisms are invertible.

I'm curious if there is anything like a crossed module, but where we're not dealing with groups and morphisms are not invertible. What I'm trying to cook up seems like it might be related to such a thing if it exists.

Any thoughts and/or any comments on my notes would be greatly appreciated.

PS: Apologies in advance if my writing is not very clear. I'm not a mathematician, but am trying to teach myself some basic higher category theory.

Answer by Theo Johnson-Freyd for 2-groups are to crossed modules as 2-categories are to...?

2010-08-16T04:56:33Z

This is not an answer, but rather a "no go" observation. I claim that you should not expect 2-categories in general to have "crossed-module" like descriptions, or at least not any such description that's any easier to think about than "2-category". Part of what makes 2-groups easy is that they have lots and lots of symmetry. Ignoring the 2-morphisms (and 2-composition), the 1-morphisms form a group, so by group translation you can relate the structure between any two 1-morphisms to the structure between some 1-morphism and the identity. And that structure is group- or torsor-like, since if you ignore the 0-morphism and the 1-composision, the 1-morphisms are a groupoid.

I expect that you can construct something for a 2-category with (1) only one 0-morphism and (2) all 1-morphisms invertible. I.e. this is a 2-group but relaxing the invertibility condition on arbitrary 2-morphisms. Then I would expect that this should correspond to a "crossed module of groups" where the second "group" $H$ need only be a monoid, although I haven't thought about the details.

Answer by Chris Schommer-Pries for 2-groups are to crossed modules as 2-categories are to...?

2010-08-16T13:50:11Z

I think Theo's "no go" is exactly right. Here is an example which might make things easier to understand: Let X be any category. I am going to construct an interesting 2-category with one object which is like a 2-group, but without the invertibility. So there is a single 0-morphism p. The morphisms from p to itself form a category which is a disjoint union of X and two points: $$ 0 \sqcup X \sqcup \infty $$ This is the disjoint union of categories so X and these other points don't interact. That completely describes the vertical composition. Now I need to tell you the horizontal composition. The element 0 is the (strict) identity for the horizontal composition. The point $\infty$ has the property that $z \cdot \infty = \infty = \infty \cdot z$ for any z. Finally the horizontal composite of any two things in X results in $\infty$.

Equivalently we can describe this as a monoidal structure on $ 0 \sqcup X \sqcup \infty $. It is actually strictly commutative too.

The reason this an important example is that we have embedded the category X fully-faithfully into this monoidal category. So any sort of algebraic description of monoidal categories or 2-categories or even strict 2-categories must be at least as complicated as the theory of all categories. This is in severe contrast with the situation for 2-groups for the reasons that Theo pointed out.

This example is also related to Reid Barton's answer to my question: Hom alg for comm. monoids. See also the related questions: A peculiar model strcture on simplicial sets? and simplicial commutative monoids group completion. The example I just described also works to give a simplicial commutative monoid where now X is any simplicial set. However when you apply the "Dold-Kan correspondence" you always get the zero chain complex. This shows that the Dold-Kan correspondence fails to be an equivalence for commutative monoids. It also says that in order to describe higher categories in terms of something like a chain complex (e.g. something like a crossed module) you absolutely need some invertablity.