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Stream: deprecated: mathematics

Topic: Profinite groups

John Baez (Jul 08 2023 at 14:10):

I'm trying to understand profinite groups in Galois theory and I'm thinking about an analogy from topology. Suppose $X$ is a compact connected manifold with a chosen basepoint $x\in X$. Are these 3 things all "the same", or are some more general?

John Baez (Jul 08 2023 at 14:10):

1) Covering spaces $p: E \to X$ where $E$ is compact.

John Baez (Jul 08 2023 at 14:10):

2) Finite sets equipped with an action of $\pi_1(X)$.

John Baez (Jul 08 2023 at 14:11):

3) Continuous actions of the profinite completion of $\pi_1(X)$ on finite sets.

John Baez (Jul 08 2023 at 14:13):

I think 1) and 2) are "the same". I.e. technically I think there's an equivalence of categories here. But less technically:

1) $\implies$ 2): every covering space with $E$ compact has finitely many sheets, so $p^{-1}(x)$ is finite and $\pi_1(X)$ acts on it.

John Baez (Jul 08 2023 at 14:20):

2) $\implies$ 1): given an action of $\pi_1(X)$ on a finite set $S$ we can use the "fundamental theorem of covering spaces" to build a covering space $p: E \to B$ such that $p^{-1}(x) = S$ with this action of $\pi_1(X)$ on it. Since $E$ is a finite-sheeted covering space of a compact manifold it is compact.

John Baez (Jul 08 2023 at 14:21):

But what I'm really wondering about is 2) $\iff$ 3).

John Baez (Jul 08 2023 at 14:24):

In general I guess if we have a group $G$, every continuous action of the profinite completion $\widehat{G}$ of $G$ on a finite set gives an action of $G$ on that finite set, but not vice versa. Right?

John Baez (Jul 08 2023 at 14:27):

We've got a homomorphism $f: G \to \widehat{G}$, and an action of $G$ on a finite set $S$ extends along $f$ to a continuous action of $\widehat{G}$ on $S$ iff the action of $G$ factors through some finite quotient of $G$. Right?

John Baez (Jul 08 2023 at 14:29):

If this is right, we get 3) $\implies$ 2), in the sense that anything of type 3) gives us something of type 2). But we only get 2) $\implies$ 3) if every action of $\pi_1(X)$ on a finite set factors through some finite quotient of $\pi_1(X)$.

David Michael Roberts (Jul 08 2023 at 14:29):

Any group acting on a finite set S factors though a finite quotient, though....

David Michael Roberts (Jul 08 2023 at 14:30):

..as $G \to Sym(S)$ has finite image.

John Baez (Jul 08 2023 at 14:31):

Okay, good that's what I thought at first... so you're saying 2) $\iff$ 3) then?

John Baez (Jul 08 2023 at 14:33):

For reason I got cold feet because this made the whole business of profinite groups looks sort of pointless.

But the situation is a bit different in Galois theory, I guess - since it's not true that every group action on a finite-dimensional vector space over a field $k$ factors through a finite quotient? (It does for a finite field $k$, though, by your argument.)

David Michael Roberts (Jul 08 2023 at 14:36):

I believe so.

You may like the paper https://arxiv.org/abs/math/0009145

John Baez (Jul 08 2023 at 14:40):

Thanks! I'm really just trying to figure out some basic stuff so I can explain it well - like, why profinite groups show up in Galois theory.

David Michael Roberts (Jul 08 2023 at 14:40):

I mean, I believe 2)<=>3)
It's too late here for me to think about the linear case.
But I note that a finite field extension is not just a finite dim vector space, but a fin dim algebra. Automorphisms are more constrained

John Baez (Jul 08 2023 at 14:41):

Right.

David Michael Roberts (Jul 08 2023 at 14:41):

You can take a minimal polynomial for the field extension, no? And then the autos of the field extension are related to the autos of the roots?

John Baez (Jul 08 2023 at 14:44):

I believe the action of the absolute Galois group $G$ of $k$ on any finite extension $K \supseteq k$ factors through a finite quotient of $G$, and it sounds like maybe you're agreeing with me.

David Michael Roberts (Jul 08 2023 at 14:44):

In the covering space version you have the nice ones that are principal bundles, and every finite-sheeted covering space is an associated bundle to one of these.

David Michael Roberts (Jul 08 2023 at 14:45):

So do you get an analogous construction with Galois extensions and arbitrary finite extensions?

David Michael Roberts (Jul 08 2023 at 14:46):

Or is it more like having a transitive pi1 action on the fibre of the covering space, hence being a quotient of a principal bundle? I think it's this one. This is the same as being a connected compact covering space

David Michael Roberts (Jul 08 2023 at 14:47):

Then an arbitrary compact covering space is a finite disjoint union of these.

John Baez (Jul 08 2023 at 14:48):

Right.

John Baez (Jul 08 2023 at 14:48):

Ultimately I'm trying to understand the "fundamental theorem of Grothendieck Galois theory", namely

Theorem - the category of commutative separable algebras over $k$ (= finite products of finite separable field extensions of $k$) is equivalent to the opposite of the category of continuous actions of the absolute Galois group (as a profinite group) on finite sets.

And now I'm trying to understand why you can't say it this "simpler" way:

Fake Theorem - the category of commutative separable algebras over $k$ (= finite products of finite separable field extensions of $k$) is equivalent to the opposite of the category of actions of the absolute Galois group on finite sets.

John Baez (Jul 08 2023 at 14:50):

If this sounds too technical you might prefer

Theorem - the category of finite separable field extensions of $k$ is equivalent to the opposite of the category of continuous transitive actions of the absolute Galois group (as a profinite group) on finite sets.

This is analogous to thinking about connected covering spaces rather than general ones.

David Michael Roberts (Jul 08 2023 at 14:50):

So general finite extensions are like a (blah) of extensions contained in Galois extensions. The amalgamation? Can't remember the word.

Just trying to give an argument why the field autos form a finite group

David Michael Roberts (Jul 08 2023 at 14:52):

Actions of the Galois group on finite sets are all continuous, no?

David Michael Roberts (Jul 08 2023 at 14:55):

Isn't every finite-index subgroup open?

David Michael Roberts (Jul 08 2023 at 14:58):

Oh, it seems not automatically https://people.math.wisc.edu/~nboston/FLTJensen.pdf

David Michael Roberts (Jul 08 2023 at 15:01):

Gotta sleep, might come back to this.

John Baez (Jul 08 2023 at 15:03):

Isn't every finite-index subgroup [of a profinite group] open?

Right, this is a somewhat different thing to wonder about, but closely related: if this were true all actions of a profinite group on a finite set would be continuous, so saying "continuous" would be redundant. But it's not true.

(That article shows it's true for "topologically finitely generated groups", but the Galois group of $\mathbb{Q}$ is not finitely generated, and I don't see why it would be topologically finitely generated either.)

John Baez (Jul 08 2023 at 15:11):

But in case anyone is getting lost, which would be very easy to do, I guess this is what I need to understand now:

Question 1: given a topological space $X$, does every action of $\pi_1(X)$ on a finite set extend to a continuous action of the profinite completion of $\pi_1(X)$ on a finite set? If not, what's a counterexample?

Question 2: given a compact manifold $X$, does every action of $\pi_1(X)$ on a finite set extend to a continuous action of the profinite completion of $\pi_1(X)$ on a finite set? If not, what's a counterexample?

Question 3: given a field $k$, is every action of the absolute Galois group of $k$ (which is a profinite group) on a finite set continuous? If not, what's a counterexample?

John Baez (Jul 08 2023 at 15:12):

I have opinions about these, but I'll just lay them out here.

Morgan Rogers (he/him) (Jul 09 2023 at 08:09):

The answer to question 1 is yes for abelian groups. For more general groups it may not be true, but for me this is because the completion is taken in the wrong category: the profinite completion replaces a group with the inverse limit of its finite quotients, but to get continuity of finite actions in general we need to take a limit indexed by the finite actions (these two things coincide in the abelian case). The result of the latter is a group with a profinite topology, and it's annoying that these don't coincide with profinite groups.

For question 2, the simplest example of a space with a non-abelian fundamental group is the wedge product of two circles, whose fundamental group is $\Z \ast \Z$, but I have a hunch that this may be too nice to produce a counterexample. We can try though! I know the universal cover of that space is a 4-regular tree (I encountered it in a geometric group theory course) so we can figure it out with a little work.

Morgan Rogers (he/him) (Jul 09 2023 at 08:13):

For Question 3 I similarly suspect the answer is no, but you'll need a number theorist to provide a counterexample :sweat_smile:

John Baez (Jul 09 2023 at 10:48):

Thanks!

1) Is every finite-index normal subgroup of a profinite group $G$ open?

2) To get a finite-index subgroup $H \subset G$ that's not open, does it suffice to find a finite-index subgroup $H$ such that the normal subgroup

$\bigcap_{g \in G} gHg^{-1}$

does not have finite index?

(I'm proceeding completely by instinct here; I haven't checked this yet.)

David Michael Roberts (Jul 09 2023 at 11:01):

The thing that's missing from this conversation is the definition of the profinite topology on the absolute Galois group.

It's very much not the profinite completion of a discrete group.

I could imagine the open subgroups are precisely those that stabilise some finite field extension!

Morgan Rogers (he/him) (Jul 09 2023 at 11:30):

Ah if that's the case then there will be no conflict, great!

John Baez (Jul 09 2023 at 15:21):

David Michael Roberts said:

The thing that's missing from this conversation is the definition of the profinite topology on the absolute Galois group.

It's very much not the profinite completion of a discrete group.

I could imagine the open subgroups are precisely those that stabilise some finite field extension!

Aaron Starr said something to this effect on the n-Cafe:

Also, by the definition of the Krull topology on the absolute Galois group, every finite index subgroup is open.

John Baez (Jul 09 2023 at 15:23):

I don't understand that remark yet. But if it's true, does this mean the word "continuous" is redundant in all these quotes?
Carboni says this:

Recalling the fundamental theorem of Grothendieck Galois Theory that the dual category of the category of commutative separable algebras over a field $k$ is equivalent to the topos of continuous representations in finite sets of the profinite fundamental group of $k$, the natural question arises of giving an abstract proof of such a theorem.

Lastaria says this in his paper "Separable algebras in Grothendieck Galois theory":

Theorem 2.1. (Grothendieck). There is an equivalence of categories

$(SepAlg_K)^{\rm op} \simeq Set^G_{\rm{fin}}$

between the dual of the category of separable $K$-algebras over a field $K$ and the category of continuous actions on finite sets of the profinite Galois group $G = Gal(K_s/ K)$ of a separable closure $K_s$ of $K$.

And a questioner on MathOverflow said this:

I've been learning about Grothendieck's Galois theory, and I just haven't been able to understand the fundamental theorem properly. Let's phrase the fundamental theorem in the case of fields:

Let $k$ be a field, and $k_s$ its separable closure. There is an anti-equivalence between the category of finite separable $k$-algebras and the category of finite sets equipped with a continuous Gal($k_s/k$)-action. Under this equivalence, separable field extensions of $k$ correspond to sets with a transitive action, and Galois extensions of $k$ correspond to finite quotients of Gal($k_s/k$).

Morgan Rogers (he/him) (Jul 09 2023 at 15:32):

While lacking absolute confidence, my answer is yes, continuity is redundant in these contexts once one has restricted to actions on finite sets. The reason why continuity is mentioned is that the category of continuous actions on arbitrary sets is generated by these finite actions: any continuous action of the Galois group is such that each finite subset of elements generates a finite subset. But the topology is defined precisely to make all of the finite actions continuous, so it's redundant!

David Michael Roberts (Jul 10 2023 at 01:38):

Well, it seems we have an interesting situation: https://math.stackexchange.com/questions/1256911/why-is-gal-omega-k-a-topological-group-under-the-krull-topology?rq=1 the filter of open subgroups of Gal(k) in the Krull topology is generated by the subgroups fixing finite Galois extensions of k. If every finite extension is contains in a finite Galois extension, then this is saying that open subgroups are those that merely fix a finite extension of k, not necessarily Galois. This feels close to what I was saying.

Saying that every finite-index subgroup is open then implies that every finite-index subgroup is the stabiliser of some finite extension, no? Can we see this algebraically?

David Michael Roberts (Jul 10 2023 at 01:40):

Let $G$ be a profinite group where every finite-index subgroup is open and $S$ a finite set. Then a homomorphism (not necc. cts for the discrete topology on the codomain) $\alpha\colon G\to Sym(S)$ has kernel of finite index, which means that $\alpha$ must be continuous.

David Michael Roberts (Jul 10 2023 at 01:47):

Oh, I'm being silly. A Galois extension is precisely a field extension corresponding to a subgroup of Gal(k). And of course the statement of Galois theory is that the lattice of intermediate subgroups of a Galois group Gal(K|k) is precisely the lattice of intermediate fields between k and K.

So now I understand Jason's comment on the n-Lab.

John Baez (Jul 10 2023 at 09:14):

You mean on the n-Category Cafe?

John Baez (Jul 10 2023 at 09:16):

Morgan Rogers (he/him) said:

While lacking absolute confidence, my answer is yes, continuity is redundant in these contexts once one has restricted to actions on finite sets.

Meanwhile David Speyer says Gal($\mathbb{Q}^{\mathrm{alg}}/\mathbb{Q}$) has a discontinuous action on a 2 element set.

John Baez (Jul 10 2023 at 09:17):

So I'm pretty confused... also because I was talking about Gal($\mathbb{Q}^{\mathrm{sep}}/\mathbb{Q}$), not Gal($\mathbb{Q}^{\mathrm{alg}}/\mathbb{Q}$).

Reid Barton (Jul 10 2023 at 09:25):

In characteristic 0, all field extensions are separable so it makes no difference.

David Michael Roberts (Jul 10 2023 at 09:35):

@John Baez yes, silly me

John Baez (Jul 10 2023 at 09:57):

@Reid Barton yes, silly me

John Baez (Jul 10 2023 at 12:16):

Okay, I'm now pretty sure that the absolute Galois group of $\mathbb{Q}$ has a discontinuous action on a 2-element set! Following Milne I give the argument here.

John Baez (Jul 10 2023 at 12:18):

One interesting thing, pointed out by James Borger and also Milne, is that to get this discontinuous action we use the axiom of choice!

Morgan Rogers (he/him) (Jul 10 2023 at 15:07):

James Borger has also referred to Galois groups as pro-finite groups, but this time as formal pro-objects rather than limits in spaces! I'm sure there is a standard, reassuring result guaranteeing that these different notions of profinite can be safely conflated, but I don't know where to find it...

Mike Shulman (Jul 10 2023 at 15:13):

I'm pretty sure that the category of pro-objects in finite sets is equivalent to the category of topological spaces that are cofiltered limits of finite discrete spaces. I was confused because I thought I heard someone earlier in this thread contradict that, but the nLab agrees with me at [[profinite space]]. It's not true for more general [[pro-sets]], but I think it is for finite ones.

Morgan Rogers (he/him) (Jul 10 2023 at 15:33):

Thanks Mike! The problem I pointed out is that the finite coset actions of a group need not themselves be groups, so there is no guarantee a priori that the topology making these continuous will make $G$ a profinite group. Or so I thought, but there is a characterization of profinite groups as compact totally disconnected groups which does the trick (it's pointed out in a quote that John shared on the same n-cat-café page).

Mike Shulman (Jul 10 2023 at 16:50):

Ah, so the question is about bridging the gap between pro-objects in finite groups and group objects in profinite sets?

John Baez (Jul 10 2023 at 17:09):

Borger said something interesting about these issues.

John Baez (Jul 10 2023 at 17:12):

By the way, I don't think the real meat of Milne's example requires anything about Galois theory; it amounts to taking an infinite product of copies of $\mathbb{Z}/2$, giving it the sort of obvious profinite topology, and finding an index-2 subgroup that is not open. Since we can think of this infinite product as a vector space over $\mathbb{F}_2$, and an index-2 subgroup is just a codimension-1 vector subspace, this amounts to a puzzle in "profinite vector spaces".