6
$\begingroup$

Continuing in my attempts to understand bits and pieces of Borceux and Janelidze's Galois Theories, I've just realized that I don't have any geometric intuition for the most convenient characterization of Galois descent for covering morphisms.

The general setting is a complete category of the form $\mathsf{Fam}(\mathsf A)$, or alternatively, a complete category with a functorial choice of connected components for each object. The geometric intuition (I hope) comes from looking at locally connected spaces and I want it to understand Galois field extensions geometrically.

The idea is that an arrow $p:E\rightarrow B$ is a morphism of Galois descent if and only if the following two conditions hold (see Galois Theories between Prop 6.6.6 and 6.6.7).

  • $p$ is an effective descent morhpsim
  • $p$ is trivialized by itself, i.e $p^\ast p$ is a trivial covering morphism (trivial fiber bundle with discrete fiber).

Of course the first condition makes sense, but the second one? What's the idea there? Why would I want $p$ to trivialize itself?

For fields all arrows are effective descent morphisms so being of Galois descent amounts to asking for the square below to be a pullback, where the top left corner is a finite coproduct of copies of $\operatorname{Spec} E$: $$\require{AMScd} \begin{CD} \operatorname{Spec} E^{\amalg{n}} @>>> \operatorname{Spec}E \\ @VVV @VV{p}V\\ \operatorname{Spec}E @>>{p}> \operatorname{Spec} B \end{CD}$$ i.e as usual an $E$-algebra isomorphism $E\otimes_BE\cong E^n$.

Added. I dug out this paper by Schauenburg on Hopf-Galois and Bi-Galois extensions, and in the long paragraph above Lemma 2.4.2 he says a lot of seemingly nice things which are way over my head. The words 'principal bundle' appear often enough to warrant some hope. In particular, it's written:

This is the algebro-geometric version of a principal fiber bundle with structure group $G$, or a $G$-torsor.

Now I do not know almost any algebraic geometry beyond the very basics, and I am looking for topological/geometric intuition of the original condition I asked about, so if this addition is irrelevant to the question, just ignore it :)

$\endgroup$

1 Answer 1

6
$\begingroup$

Let $G \to E \to B$ be a principal bundle. It's classified by a map $f : B \to BG$ in the sense that $E$ is the homotopy fiber of this map. This means that $E$ has a certain universal property: namely, it is the universal map to $B$ such that the composition with $f$ is equipped with a nullhomotopy. This says precisely that $E$ is the universal space over $B$ equipped with a trivialization of the pullback of the principal bundle to $E$: in other words, $E \to B$ is the universal thing that trivializes itself.

The relationship to Galois extensions is that $K \to L$ is a finite Galois extension with Galois group $G$ iff $\text{Spec } L \to \text{Spec } K$ is a principal $G$-bundle in a suitable sense ("$G$-torsor").

$\endgroup$
5
  • $\begingroup$ This looks great but has some words I don't understand. If you have the time/desire to expand your answer to spell things out some more (like the relation to Galois extensions) or maybe include an example of what happens geometrically for spectra of fields, that would be incredible! $\endgroup$
    – Arrow
    Jul 19, 2016 at 16:30
  • $\begingroup$ Also I'm not following how the universal property yields a trivialization :( If $B$ is connected I take it the by trivialization you mean $E\times_BE$ is a coproduct of $E$'s... $\endgroup$
    – Arrow
    Jul 19, 2016 at 18:26
  • $\begingroup$ @Arrow: no, I mean that $E \times_B E$ is isomorphic, as a principal bundle over $E$, to the trivial bundle $E \times BG$. I've added an edit, but it's hard to say more without knowing what your background is. Do you know what a principal bundle is? $\endgroup$ Jul 19, 2016 at 21:45
  • $\begingroup$ Sorry, that should be $E \times G$ above. $\endgroup$ Jul 19, 2016 at 22:10
  • $\begingroup$ Is it correct to say that $E \times_B E \to E$ is a trivial bundle because it admits a canonical global section $E\to E \times_B E$ namely the diagonal map? $\endgroup$
    – ziggurism
    Jan 15, 2020 at 20:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.