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 :



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



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> 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 ?
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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.


References:


[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



Galois extensions of ring spectra and subextensions
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)


