8
$\begingroup$

Given a fiber square of simplicial sets

$$\begin{array}{cc} & \hspace{-7mm} E \\ &\hspace{-7mm}\downarrow \\ \ast\longrightarrow &\hspace{-7mm} B \end{array}$$

and a homology theory $h(-)$, there is an associated Eilenberg-Moore spectral sequence converging to the homology of the homotopy fiber $F$. This is just the spectral sequence for a cosimplicial space, specifically a cobar construction $C^\bullet(E,B,\ast)$. It is claimed in this paper of Hopkins and Ravenel, in the middle of page 7, that if this Eilenberg-Moore spectral sequence converges strongly then $\Sigma_+^\infty Tot(C^\bullet(E,B,\ast))\simeq Tot(C^\bullet(\Sigma^\infty_+E,\Sigma^\infty_+ B,\mathbb{S}))$.

Why is this true? I know from an old paper of Bousfield that the EMSS for stable homotopy converges strongly in certain nice cases (e.g. if $B$ is simply connected), but does that give me an actual equivalence of objects? That only seems to tell me nice things about the algebra. For instance that there exists a finite filtration of $\pi_n^S(F)$ whose filtration quotients are the $E_{i,j}^\infty$ for $i+j=n$. That seems several steps away from making a general statement about the suspension spectrum functor commuting with totalization.

More generally, are there other functors $Top\to Spectra$ which commute in this fashion? For instance, it seems like the Thom spectrum functor (a sort of twisted suspension spectrum functor) for suitable fibrations should also commute with totalization is suspension spectrum does.

EDIT----------------------------

I've also discovered this mysterious correspondence between Tom Goodwillie and Mike Hopkins (this is how people talked about this stuff before MO I guess!) that shows that suspension commutes with Tot in certain cases.

http://www.lehigh.edu/~dmd1/tg7192

$\endgroup$
  • $\begingroup$ From looking at a few other references, it seems that this has something to do with the vanishing line you get in the EMSS when you have strong convergence... $\endgroup$ – Jonathan Beardsley Oct 6 '15 at 2:53
  • 6
    $\begingroup$ I don't think that the result holds for any homology theory $h$, only singular homology $H$. If you only ask this for a given $h$, I think that the result you want holds after $h$-localization. For an argument why this is true, take a look at Tilman Bauer's "On the Eilenberg-Moore spectral sequence for generalized cohomology theories," in particular Remark 2.3. This gives the result you want directly for $K=S$; I think it gives what you want for $K = H$ using the fact that an isomorphism in homology is a stable equivalence. $\endgroup$ – Craig Westerland Oct 6 '15 at 4:26
  • $\begingroup$ @CraigWesterland it looks for that you still have to check the lim^1 term. Is it obvious that that vanishes? $\endgroup$ – Jonathan Beardsley Oct 6 '15 at 4:41
  • 4
    $\begingroup$ A little further down the page, Tilman notes that "strong convergence" in the sense of Shipley or Bousfield is what he calls "pro-constant convergence." In Lemma 2.6, he shows this implies complete convergence, which kills the $lim^1$ term. $\endgroup$ – Craig Westerland Oct 6 '15 at 14:28
  • 2
    $\begingroup$ @CraigWesterland Oh SNAP! Thanks Craig! :-D $\endgroup$ – Jonathan Beardsley Oct 6 '15 at 14:46
8
$\begingroup$

I'm going to avoid the question and answer the edit.

Hopkins has some results on this in his Oxford thesis which were announced without proof in his Asterisque paper of 1984.

For a quite thorough analysis of a different, but related, spectral sequence, there is also the nice paper of Goerss (1993), Barratt's Desuspension Spectral Sequence and the Lie Ring Analyzer. See § 1. Barratt's spectral sequence, with a generalization by Hopkins.

See Theorem 1.1.

There is, however, a (potential?) error in Hopkins' work involving connectivity assumptions, which is addressed in Comultiplication and suspension J. Klein, R. Schwaenzl, R.M. Vogt.(1997)

Their analogous result is Theorem 1.4 of that paper.

I don't see how your claim follows from the above results, however.

$\endgroup$
  • $\begingroup$ I blame my newb-ness for posting my commentary as an "answer" instead of a "comment". $\endgroup$ – Rosona Oct 7 '15 at 17:28
  • $\begingroup$ Well, it's too long to be a comment, and with 7 upvotes I'd say it's worthy of being an answer $\endgroup$ – David White Oct 7 '15 at 22:00

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.