Automorphisms and Bicategories - MathOverflow

Automorphisms and Bicategories

2011-10-22T23:44:57Z

Sorry if this isn't the right place for this, it hasn't gotten any answers on ME. I'm reading Lang's section on field theory and he stresses that, unlike typical "universal" constructions which are determined up to unique isomorphism, algebraic closures (and by extension, their Galois groups) are determined only up to automorphism (conjugation). It seems to me that there ought to be some interpretation of this in terms of bicategories (weak 2-categories). This intuition is supported by the fact that 2-cells are given by conjugation when we give Grp the structure of a 2-category (view groups as 1-object categories, get 2-cells via natural transformations). Is there any such interpretation?

Answer by David Roberts for Automorphisms and Bicategories

2011-10-23T01:07:14Z

Not necessarily. In the higher category-theoretic setting one asks for a 'contractible space' of choices (space might mean simplicial set or n-category) instead of uniqueness. The natural 2-category one might define may not be the 'right' one to get such a collection of choices, and so one could define a 2-category such that these things are unique in the appropriate sense, but this might just be cooked up to give that result and not of intrinsic interest. For example, one can define the 2-category of fields where the underlying 1-category is $Fields$, and there is a unique 2-arrow between any two parallel 1-arrows. This is clearly not an interesting 2-category.

(Other examples of non-unique closures are given here: http://nlab.mathforge.org/nlab/show/completion#nonunique)

Answer by Finn Lawler for Automorphisms and Bicategories

2011-10-23T01:24:20Z

I don't see bicategories coming into this in a useful way, but I think what you have is a consequence of two more general facts:

The non-uniqueness of algebraic closures is a general fact about injective hulls -- they are 'unique' up to non-unique isomorphism.
Every morphism in a groupoid yields an isomorphism of vertex groups by conjugation -- if G is a groupoid and x is an object of G, then the vertex group at x is G(x,x), and if $f \colon x \to y$ is a morphism then $g \mapsto f^{-1} g f$ is an isomorphism between G(x,x) and G(y,y).

With the objects satisfying a universal property the comparison isomorphisms between them are unique, so that the groupoid of objects satisfying the universal property is codiscrete, i.e. there is exactly one morphism between any two objects, so in particular the vertex groups of this groupoid are trivial. For an object A in a concrete category (or in a category with a chosen class of 'embeddings') there is a groupoid of injective hulls of A that is not in general codiscrete, and so it can have non-trivial vertex groups. But this groupoid, though not codiscrete, is still connected, so that each vertex group is (non-uniquely!) isomorphic to every other via conjugation by a morphism (necessarily invertible) of injective hulls.

Edit: The fundamental group of a space, as in JSE's analogy, bears much the same relationship with the fundamental groupoid of the space -- in particular, $\pi_1$s at points in the same path-component are isomorphic via conjugation in the same way.