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?
$\begingroup$
$\endgroup$
3
-
2$\begingroup$ If this helps, you should think of the group of automorphisms of a fixed algebraic closure K as analogous to the fundamental group of a space X based at a point *. The choice of algebraic closure is the choice of basepoint. And the "Galois group" of a non-algebraically closed field is something like "the fundamental group of X" defined without reference to a basepoint -- an object that is really defined only up to conjugation. $\endgroup$– JSECommented Oct 23, 2011 at 1:12
-
1$\begingroup$ @JSE: Continuing the analogy, is the concept of "Galois groupoid" then useful? $\endgroup$– Yuri SulymaCommented Oct 23, 2011 at 2:32
-
$\begingroup$ I don't have a concrete example, but I have seen papers using the groupoid language. In particular, morphisms in the Galois groupoid of $X$ are described by isomorphisms of fiber functors on the category of étale sheaves of finite sets on $X$. $\endgroup$– S. Carnahan ♦Commented Oct 23, 2011 at 4:59
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
4
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.
answered Oct 23, 2011 at 1:24
-
1$\begingroup$ RE injective hull: Ah, that was the nLab page I vaguely remembered and couldn't find. Thanks! There is also either a comment at the n-category cafe or an answer here on MO by Minhyong Kim about how the Galois group is much more analogous to the fundamental group in that there is a secret 'choice of basepoint' for an interpretation of that phrase that makes sense for field extensions. Sorry I can't find it in a hurry. $\endgroup$– David Roberts ♦Commented Oct 23, 2011 at 10:25
-
3$\begingroup$ Do you mean this one: mathoverflow.net/questions/546/… ? $\endgroup$ Commented Oct 23, 2011 at 19:01
-
$\begingroup$ Thanks, Finn. If I'd had a little more time before bed last night I would have tracked it down myself. $\endgroup$– David Roberts ♦Commented Oct 24, 2011 at 3:27
-
1$\begingroup$ I should have said: thanks for mentioning it, it's a fantastic piece. $\endgroup$ Commented Oct 24, 2011 at 10:32
$\begingroup$
$\endgroup$
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)
answered Oct 23, 2011 at 1:07