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 $\Delta(x)=x\otimes 1$.

  • 2
    $\begingroup$ 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. $\endgroup$ – Justin Noel May 30 '14 at 11:53
  • $\begingroup$ 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. $\endgroup$ – Jonathan Beardsley May 30 '14 at 14:07
  • $\begingroup$ 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 $MO$. $\endgroup$ – Jonathan Beardsley May 30 '14 at 14:57
  • $\begingroup$ 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. $\endgroup$ – Jonathan Beardsley Nov 14 '14 at 14:11

So this can definitely be done. It took me a while to figure out all the details, but in the end it's not so conceptually complex.

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 $M(f\circ i)\simeq \mathbb{S}/\Omega F$.

I've worked out the details in a preprint here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.