# Thom Spectra and Hopf-Galois Extensions of Ring Spectra

So I've been fiddling with this for a long time, so apologies to anyone that's already heard me talk about this ad nauseum. I haven't been able to get anywhere with it, and it seemed that as such, it might be worth posting here, since if it's true, I think it'd be interesting.

Suppose I've got an inclusion of (at least) $\mathbb{E}_2$-groups $j:H\hookrightarrow G$ such that I've got a fibration $H\overset{j}\to G\overset{q}\to G/H$, and I've got a Thom spectrum $Mf$ induced by a continuous map $f:G\to BGL_1(\mathbb{S})$. From Fridolin Roth's thesis, (and really ultimately from ABGHR) we know that under nice circumstances (e.g. if $Mf$ is $H\mathbb{Z}$-oriented) then $Mf$ is a Hopf-Galois extension of the sphere spectrum with associated Hopf-algebra the spherical group ring $\mathbb{S}[G_+]$.

In particular, we have that the spectral sequence which computes the cofixed points of the coaction (in this case just the Thom diagonal) $Mf\to Mf\wedge \mathbb{S}[G_+]$, which is just the BKSS of the cosimplicial spectrum computing that limit, converges to the homotopy of the sphere spectrum (it's just a sort of Adams spectral sequence).

Now, the map $G\to G/H$ would seem to induce a coaction $$Mf\to Mf\wedge \mathbb{S}[G_+]\to Mf\wedge\mathbb{S}[(G/H)_+].$$ And thus one might ask what the cofixed points of this new coaction are. There should be at least a sort of "unit" map $\mathbb{S}\to Mf^{hco(G/H)_+}$, and in an attempt to determine a sort of "intermediate" Hopf-Galois extension, one could hope to identify this object in between $\mathbb{S}$ and $Mf$.

In the case that the morphism $G\to BGL_1(\mathbb{S})$ is the trivial one, or in other words $Mf=\mathbb{S}[G_+]$, one can identify this cofixed point object as $\mathbb{S}[H_+]$ by noticing that the cosimplicial object defining the cofixed points is precisely the diagram defining the homotopy limit of the diagram $\ast\leftarrow \mathbb{S}[G_+]\to \mathbb{S}[(G/H)_+]$, or, in other words, the suspension spectrum of the homotopy fiber of the quotient map $G\to G/H$. This, and other vague considerations, have led me to believe that perhaps there is a general way of realizing the Thom spectrum $M(f\circ j)$ as the cofixed points of $Mf$ under the induced coaction of $\mathbb{S}[(G/H)_+]$. I am concerned that there is some elementary consideration that would cause this to be true that I am just not seeing.

I have attempted to get at this thing by thinking about the bundles of spectra defining $Mf$ and $M(f\circ j)$, and using some kind of six-functor yoga, as well as just straight up computation, but haven't really gotten anywhere. Any thoughts would be dearly appreciated.

One last remark: note that by cofixed points I do not mean homotopy orbits. In algebra, for a group $X$ which is coacted upon by a Hopf-algebra $C$, $\Delta:X\to X\otimes C$, the cofixed points of the coaction are the elements $x\in X$ such that $\Delta(x)=x\otimes 1$.

• I would like to point out that $Mf$ being $\mathbb{Z}$-orientable is not a very severe restriction and follows from $G$ being simply connected. The condition is equivalent to $f$ lifting to the simply connected cover $BSL_1 S$ of $BGL_1 S$. Assuming your Hopf-Galois condition is essentially the convergence of the associated Adams spectral sequence, when $Mf$ is not $\mathbb{Z}$-orientable then you obtain a Hopf-Galois extension of the 2-adic sphere spectrum. Also in the algebraic context of coactions, what you are calling cofixed points are traditionally called primitives. – Justin Noel May 30 '14 at 11:53
• Ah thanks @JustinNoel I had seen that word (primitives) used in some places. Perhaps it will be less confusing if I start using that rather than cofixed points. – Jonathan Beardsley May 30 '14 at 14:07
• I should also mention that that convergence I mention definitely holds for a LOT of interesting Thom spectra: $MU$, $MSO$, $MSU$, $X(n)$, Baker and Richter's $M\xi$. And the alternate situation (being an extension of the 2-adic sphere spectrum) holds for $MO$. – Jonathan Beardsley May 30 '14 at 14:57
• I should perhaps also add that in certain nice cases I have been able to work this out (the write-ups are on my website). But that relies on the collapse of a certain Kunneth spectral sequence which seems unlikely in general. It would be nice to have more general conditions. – Jonathan Beardsley Nov 14 '14 at 14:11

The basic idea is that if you've got a fibration $F\overset{i}\to E\overset{p}\to B$ of connected $\mathbb{E}_n$-spaces and a map $f:E\to BGL_1(\mathbb{S})$ then you want to take the left Kan extension along $p$ as in the following (very poorly typeset) diagram:
$$\array{F\to & \!\!\!\!\!\!E\to &\!\!\!\!\!BGL_1(\mathbb{S})\to Spectra\\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\downarrow & \nearrow\\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!B&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\longrightarrow}$$
This left Kan extension realizes the colimit of the morphism $B\to Spectra$ as $M(f\circ i)/\Omega B$, which can be thought of as $M(f\circ i)/(\Omega E/\Omega F)$ or even as $(\mathbb{S}/\Omega F)/(\Omega E/\Omega F)$. But we can also (following the answer to this question) recognize this as the Kan extension along $E\to \ast$, giving $M(f\circ i)/\Omega B\simeq Mf$. Moreover, this can be shown to produce $Mf$ as a Thom spectrum over $M(f\circ i)$, which immediately gives you the $B$-coaction on $Mf$ and the torsor condition of Hopf-Galois extensions $Mf\otimes_{M(f\circ i)}Mf\simeq Mf\otimes \mathbb{S}[B]$.
To get that $M(f\circ i)$ can be recovered as the primitives of the $B$-coaction, notice that if everything is $H\mathbb{Z}$-oriented then in homology the descent, or primitives, spectral sequence computing the cotensor product $Mf\Box_B\mathbb{S}$ is equivalent to the Eilenberg-Moore spectral sequence computing $E\Box_B\ast\simeq E\times_B \ast\simeq F$. In other words (and this is very rough), in homology, both spectral sequences converge to the homology of $F$, which (if we assume $H\mathbb{Z}$-orientation) is exactly the homology of $M(f\circ i)\simeq \mathbb{S}/\Omega F$.