# Geometric intuition for the condition of Galois descent

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 :)

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").
• 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... – Arrow Jul 19 '16 at 18:26
• @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? – Qiaochu Yuan Jul 19 '16 at 21:45
• Sorry, that should be $E \times G$ above. – Qiaochu Yuan Jul 19 '16 at 22:10
• 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? – ziggurism Jan 15 at 20:00