In Galois extensions of structured ring spectra, Rognes introduces the notion of a faithful $G$-Galois extension of ring spectra. Let me recall what this means:

We have a commutative ring spectrum $R$ with an action of $G$, and a $G$-equivariant morphism of commutative ring spectra $S\to R$ where $S$ has the trivial $G$-action. This is a $G$-Galois extension if:

i) The canonical map $S\to R^{hG}$ is an equivalence

ii) The canonical map $R\otimes_S R\to F(G_+,R)$ is an equivalence

iii) $R$ is a faithful $S$-module

This makes sense for stably dualizable topological groups and in fact for my question, I'm interested in topological groups that aren't finite discrete. I am, however, happy to restrict to groups that have the homotopy type of a finite CW-complex.

He proves the following result, of which I will sketch a proof at the end of the post for convenience, and to perhaps understand my question better (proposition 7.2.2.in [Rog08]):

If $K$ is an allowable subgroup of $G$, then $R^{hK}\to R$ is a faithful $K$-Galois extension.

Here, an allowable subgroup is a subgroup $K$ which is stably dualizable, such that $G_{hK}\simeq^{stably} G/K$, and such that there exists a continuous section of $G\to G/K$.

My question is the following :

Is this last condition about the existence of a continuous section necessary to get this result ?

In my sketch of the proof below I will point out precisely where we use it - I don't follow the exact same proof as Rognes, but we use the hypothesis for essentially the same thing.

An obvious remark is that when $G$ is finite discrete, this continuous section always exists, so there is no problem there (and in fact every subgroup is allowable). When $G$ splits as a product (topologically, but in particular if it splits as a product of topological groups) $K\times G/K$ then the section obviously exists, which is what allows for instance Mathew to induct on $n$ when looking at $\mathbb T^n$-Galois extensions in Torus actions on stable module categories, Picard groups and localizing subcategories.

Of course, the best answer would be a proof that doesn't use this condition, or an example where the result fails. Most of the examples I know are finite groups, so I'm not really sure what to look for.

An example where the condition fails is $S^1$ with any finite subgroup, but one would have to find $S^1$-Galois extensions.

Sketch of proof: I will use Mathew's language of descendability, which he uses in [Mat15] to give an alternative characterization of $G$-Galois extensions, cf. proposition 3.7. in that paper.

$R$ is dualizable over $S$, by 6.2.1 in [Rog08] (of course this doesn't refer to subgroups, so I won't expand on the proof of that fact), and faithful by definition. Therefore, by Thm 3.38 in The Galois group of a stable homotopy theory, $S\to R$ admits descent.

It follows that $R^{hH}\to R\otimes_S R^{hH}$ also admits descent. Let's call that new algebra $\tilde R$. We can now check $H$-Galois-ness of $R^{hH}\to R$ by tensoring with $\tilde R$ : $\tilde R\to \tilde R\otimes_{R^{hH}}R\simeq R\otimes_S R\simeq F(G_+,R)$. Now $\tilde R = R\otimes_S R^{hH}\simeq F(G/H_+,R)$ because $R$ is dualizable and $G_{hH}\simeq G/H$, so this map is $F(G/H_+,R)\to F(G_+,R)$.

If we have a section, we can identify this map with the canonical map $F(G/H_+,R)\to F(H_+,F(G/H_+,R))$ and so by proposition 3.7. in [Mat15], we get that $R^{hH}\to R$ is indeed $H$-Galois.

Of course this is just a sketch, one needs to identify the maps a bit more precisely.

I guess this reduces the question to looking at $G$-Galois extensions of the form $R\to F(G_+,R)$ i.e. the "trivial ones", in other words,

what do we need for $F(G/H_+,R)\to F(G_+,R)$ to be an $H$-Galois extension ? Are there cases with no section where this fails ?

But for instance whenever $G$ is connected, $G\to G/H$ is a principal $H$-bundle and $\pi_1(G/H)$ acts nilpotently on $H_*(G;\mathbb F_p)$, proposition 5.6.3. of [Rog08] will yield a positive answer for $R= \mathbb F_p$, so not "any" counterexample will work. For instance, these conditions are satisfied by $S^1\to S^1$ which is indeed a $\mathbb Z/p$-principal bundle.


[Rog08] : Galois extensions of structured ring spectra, arxiv:math/0502183

[Mat15]: Torus actions on stable module categories, Picard groups and localizing subcategories, arXiv:1512.01716

[Mat16]: The Galois group of a stable homotopy theory, arXiv:1404.2156

  • $\begingroup$ Note that the condition on $H$-fixed points for $F(G_+,R)$ is automatic if we assume $G_{hH}\simeq G/H$, which I want to assume, so the main part is about the non-ramification : $F(G_+,R)\otimes_{F(G/H_+,R)} F(G_+,R) \simeq F(H_+,F(G_+,R))$. Because $G\times_{G/H} G \simeq G\times H$, this is related in some sense to the Eilenberg-Moore spectral sequence, but I don't know enough about the latter $\endgroup$ Oct 27, 2020 at 9:37

1 Answer 1


It turns out that I was misapplying Rognes' 5.6.3. : when $p\neq n$, the action of $\pi_1(S^1)$ on $H_*(C_n; \mathbb F_p)$ need not be nilpotent.

In particular, the following is a counterexample : take any odd prime $p$, then the projection $S^1\to \mathbb RP^1$ is a mod $p$-equivalence, so that $\mathbb F_p^{S^1}\otimes_{\mathbb F_p^{\mathbb RP^1}} \mathbb F_p^{S^1} \simeq \mathbb F_p^{S^1}$ which is not equivalent to $\mathbb F_p^{S^1\times C_2}$, even abstractly.

This gives a counterexample where all of $G,H,G/H$ are stably dualizable, so it's better than the other counterexamples I stumbled upon, which had $G= EC_n, H =C_n$ (and so $G/H = BC_n $ which is not stably dualizable)


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