## July 8, 2023

### Profinite Group Actions

#### Posted by John Baez I have three questions. I have some guesses about the answers, so don’t think I’m completely clueless. But I’m clueless enough that I’d prefer to just give the questions, not my guesses.

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 that 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 that 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?

Posted at July 8, 2023 4:19 PM UTC

TrackBack URL for this Entry:   https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/3479

For questions 1 and 2, isn’t that true for any group G, not just the fundamental groups of a manifold? And moreover, I think of this as the definition of the profinite completion of a group: as an inverse limit over all possible actions of G on finite sets, i.e., all possible homomorphisms G to S_n.

Similarly, for the last one, for any profinite group G acting on a finite set, the stabilizers of any point in the set are finite index which automatically implies that the action is continuous.

Am I missing some subtlety?

Posted by: Asvin G on July 8, 2023 7:31 PM | Permalink | Reply to this

### Profinite Group Actions

That is correct. There are finite index subgroups of profinite groups that are not open, i.e., there are profinite groups that do not equal their own profinite completion. However, by definition, the groups in Question 1 and Question 2 do equal their own profinite completion. Also, by the definition of the Krull topology on the absolute Galois group, every finite index subgroup is open.

Posted by: Jason Starr on July 9, 2023 11:33 AM | Permalink | Reply to this

### Re: Profinite Group Actions

There are uncountably many non-closed subgroups of index 2 in the absolute Galois group of the rationals. See chapter 7 of Milne’s “Field Theory” book:

https://www.jmilne.org/math/CourseNotes/ft.html

Posted by: Bruno S on July 9, 2023 6:45 PM | Permalink | Reply to this

### Re: Profinite Group Actions

Building on what Bruno S wrote, if we let $H$ be a dense index 2 subgroup of $Gal(\mathbb{Q}^{alg}/\mathbb{Q})$, then $H$ is normal and $Gal(\mathbb{Q}^{alg}/\mathbb{Q})/H$ is isomorphic to $S_2$, so $Gal(\mathbb{Q}^{alg}/\mathbb{Q})$ has a discontinuous action on a 2 element set. (The precise reference in Milne is Theorem 7.29.)

Posted by: David Speyer on July 10, 2023 2:34 AM | Permalink | Reply to this

### Re: Profinite Group Actions

Amateurs (such as myself) might be confused when someone says:

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

and someone else responds:

there are uncountably many non-closed subgroups of index 2 in the absolute Galois group of the rationals.

Since the first is talking about open subgroups and the second about closed subgroups, this reply may sound like a non sequitur.

But it’s not: a subgroup of a profinite group is open iff it is closed and has finite index!

So, if the Krull topology is a profinite topology (and it is), these non-closed subgroups of index 2 are also not open.

So I think the second commenter is actually contradicting the first.

I feel like I’m watching a chess game without quite knowing all the rules. For other people in the same position, let me quote Infinite Galois theory by Haoran Liu:

By the theorem above, we have that the Galois groups with the Krull topology are Hausdorff, compact, and totally disconnected. These are actually very familiar properties. These are the properties of the profinite groups. In fact, by Theorem 3.9, we will have that the Galois groups with the Krull topology are profinite groups.

Posted by: John Baez on July 10, 2023 12:25 PM | Permalink | Reply to this

Yes, I was wrong! I was reading the definition of the Krull topology off a webpage (a stupid thing to do).

Posted by: Jason Starr on July 10, 2023 7:43 PM | Permalink | Reply to this

### Profinite Group Actions

Here’s the kind of thing I’m worrying about: why does everyone state the “fundamental theorem of Grothendieck’s Galois theory” in this way?

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($k_s/k$).

(Emphasis mine.)

Are there discontinuous actions of the profinite group Gal($k_s/k$) on finite sets? If so, what’s an example?

Posted by: John Baez on July 9, 2023 11:43 AM | Permalink | Reply to this

### Re: Profinite Group Actions

What you’ve quoted is a great way of saying the fundamental theorem of Galois theory, but I don’t think it’s fair to say “everyone state[s]” it that way: most books just talk about field extensions, and so say things at best about transitive G-sets and more often about subsets up to conjugation. In any case, I think the answer to your question is that the correct statement is not about just finite extensions.

Posted by: Theo Johnson-Freyd on July 9, 2023 12:11 PM | Permalink | Reply to this

### Re: Profinite Group Actions

I said the fundamental theorem of Grothendieck’s Galois theory, not the fundamental theorem of Galois theory.

Carboni puts it this way:

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

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

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 that 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($k_s/k$).

So, it seems the word on the street is that there’s a “fundamental theorem of Grothendieck’s Galois theory”, and this is the way to state it.

Are you folks here telling me that the word “continuous” is redundant in these statements? I still find that hard to believe. Not every finite-index subgroup of a profinite group is open (though it’s true in the topologically finitely generated case). So it seems we’d need to work a bit to show that the stabilizer of any finite separable field extension of $k$ is open in Gal($k_s/k$), if that is indeed true. I don’t know how to show that, but nor do I know a counterexample.

Posted by: John Baez on July 9, 2023 3:57 PM | Permalink | Reply to this

### Re: Profinite Group Actions

For the Galois group, it is an inverse limit of the Galois groups of finite extensions, so almost by definition, the absolute Galois group of a finite extension is open and closed (as Jason mentions).

The reason it is useful to keep track of the topology is that the Galois correspondence extends to infinite extensions: the closed subgroups of the Galois group are in bijection with extensions of the field (of possibly infinite degree).

For finite extensions, I think there is really no difference.

(I am 95% sure this is true but I didn’t check too carefully and am speaking from memory.)

Posted by: Asvin G on July 9, 2023 7:20 PM | Permalink | Reply to this

### Re: Profinite Group Actions

Asvin G writes:

For the Galois group, it is an inverse limit of the Galois groups of finite extensions, so almost by definition, the absolute Galois group of a finite extension is open and closed (as Jason mentions).

Thanks. That’s good to know and I believe it. But the heart of my question lies elsewhere. I want to know if every action of the absolute Galois group of a field $k$ on a finite set is continuous. I believe this is true iff every finite-index subgroup of this absolute Galois group is open.

But I think David Speyer is saying that not every finite-index subgroup of this absolute Galois group is open.

He says that there is a discontinuous action of the absolute Galois group of $\mathbb{Q}$ on a 2-element set. If so, the subgroup that acts trivially on the 2-element set must be an index-2 subgroup of this absolute Galois group that is not open.

Posted by: John Baez on July 10, 2023 12:06 PM | Permalink | Reply to this

### Re: Profinite Group Actions

I believe I was told once that you can use the axiom of choice to make subgroups of absolute Galois groups which are of finite index but which do not correspond to finite subextensions. So the continuity condition of Galois actions on finite sets can’t be removed.

I didn’t pay much attention to the details (maybe you come up with one of index two using complex conjugation?), since I usually prefer choice-free ways of looking at things. I only took away the moral that it’s best practice to think of Galois groups as pro-objects in the category of finite groups, rather than as mere groups which have the property of being isomorphic to projective limits of finite groups. I believe this is true for any connected topos, where the fundamental group should be thought of as a pro-discrete group.

Posted by: James Borger on July 9, 2023 11:27 PM | Permalink | Reply to this

### Re: Profinite Group Actions

James wrote:

So the continuity condition of Galois actions on finite sets can’t be removed.

Thanks. I really like it when someone comes right out and answers a question in the same language in which it was stated. I asked yes-or-no questions to make this easy.

I believe I was told once that you can use the axiom of choice to make subgroups of absolute Galois groups which are of finite index but which do not correspond to finite subextensions.

Yes! As Bruno and David pointed out above, it’s nicely explained in Proposition 7.29 of Milne’s Fields and Galois Theory. Let me summarize, for anyone reading this thread in search of answers.

The answer to Question 3 in my post is no, and here is a counterexample.

Take the subfield $E \subset \mathbb{C}$ generated by $\sqrt{-1}$ and the square roots of all the rational primes. This is Galois over $\mathbb{Q}$, and its Galois group over $\mathbb{Q}$, say

$G = Gal(E/\mathbb{Q}),$

is a countable product of copies of $\mathbb{Z}/2$. He forms the subgroup $H \subset G$ consisting of lists of elements of $\mathbb{Z}/2$ where all but finitely many are zero. This subgroup $H$ is dense in $G$ in the profinite topology.

We can think of $G$ and $H$ as vector spaces over $\mathbb{F}_2$. Thus $G/H$ is an infinite-dimensional vector space over $\mathbb{F}_2$, and using the Axiom of Choice we can pick a basis for it. This lets us find a group

$H \subset G' \subset G$

such that $G'$ has index 2 in $G$. If $G'$ were open in $G$ it would be closed (by the miracle of profinite groups), but this would contradict the fact that $H$ is dense in $G$.

So, we’ve found an index-2 subgroup $G'$ of the Galois group $G = Gal(E/\mathbb{Q})$ that is not open. So, by letting $G$ act on $G/G'$, we get a discontinuous action of $Gal(E/\mathbb{Q})$ on a 2-element set!

This is not exactly what I was asking for: I was wanting to find a discontinuous action of the absolute Galois group $Gal(\mathbb{Q}^{alg}/\mathbb{Q})$ on a finite set. But I think our discontinuous action of $Gal(E/\mathbb{Q})$ on a finite set gives a discontinuous action of $Gal(\mathbb{Q}^{alg}/\mathbb{Q})$ on the same finite set. (If I’m wrong someone please alert me.)

So yeah: this is all about the axiom of choice. I hadn’t expected that!

Posted by: John Baez on July 10, 2023 1:12 PM | Permalink | Reply to this

### Re: Profinite Group Actions

Here is an example of an absolute Galois group with a nonopen index 2 subgroup that is easy enough to understand. A result of of Douady shows that the absolute Galois group $G$ of $\mathbf C(x)$ is a free profinite group of rank $2^{\aleph_0}$. What this means is $G$ has a subset $X$ of size $2^{\aleph_0}$ such that, for any map $f\colon X\to H$ with $H$ a profinite group such that every nbhd of $1$ in $H$ contains all but finitely many elements of $f(X)$, there is unique extension of $f$ to a cts homomorphism $G\to H$.

In particular, we see that there is a bijection between open subgroups of index $2$ and non-empty finite subsets of $X$ (since a surjective homomorphism to $\mathbf Z/2\mathbf Z$ corresponds to choosing the nonempty finite subset of $X$ mapping to the nontrivial element), and so there are $2^{\aleph_0}$ such open subgroups.

In particular, we can map $G$ onto $(\mathbf Z/2\mathbf Z)^{2^\aleph_0}$ (viewed as a profinite group being a direct product of finite groups) by bijecting $X$ with the vectors with a $1$ in one coordinate and $0$ in the remaining coordinates. (A basic nbhd of the identity $0$ is the kernel of a projection to finitely many of the $\mathbf Z/2\mathbf Z$.) Clearly this map is surjective as the image of $X$ generates a dense subgroup. But $(\mathbf Z/2\mathbf Z)^{2^{\aleph_0}}$ is a vector space over $\mathbf Z/2\mathbf Z$ of dimension strictly larger then $2^{\aleph_0}$, and so it has more than $2^{\aleph_0}$ homomorphisms onto $\mathbf Z/2\mathbf Z$ and so some are discontinuous. Therefore, $G$ also has discontinuous homomorphisms onto $\mathbf Z/2\mathbf Z$.

Posted by: Benjamin Steinberg on July 27, 2023 3:30 AM | Permalink | Reply to this

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