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Let $H^*$ and $K^*$ be two cohomology theories, and $X$ a reasonable spectrum. Here, I'm thinking that $H^*$ is singular cohomology (and for my purposes, rational cohomology will suffice), and $K$ is a Morava K-theory.

Suppose that I understand $H^*(X)$ and $K^*(X)$. Is there any way to use this (or other!) information to get a handle on $H^*(L_{K} X)$?, the $H$-cohomology of the Bousfield localisation of $X$ with respect to $K$?

I know that $L_K X$ may sometimes be defined as the totalisation of a cosimplicial spectrum $K^{\bullet +1} \wedge X$. Does the associated Eilenberg-Moore spectral sequence have good convergence properties? This seems doubtful in the case where $H$ is rational cohomology and $K$ Morava K-theory; I think that the $E_1$ term of the spectral sequence is $0$, but it's certainly the case that some $K$-localisations have torsion free homotopy.

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It certainly seems like the $E_1$ term is 0. It seems like it would be reasonable to look at the spectral sequence coming from the cosimplicial abelian group $\pi_*(H\mathbb{Z}_{(p)}\wedge K^{\wedge \bullet +1}\wedge X)$. This is an Adams spectral sequence, but is the totalization really the thing you want? –  Sean Tilson Aug 9 '13 at 14:31
This should also be pretty computable, I would think all of the formulas are worked out in that Ravenel-Wilson paper. –  Sean Tilson Aug 9 '13 at 14:32
Hey Sean, yes that's definitely the SS that I had in mind. But it's definitely 0 at $E_1$, since by Ravenel-Wilson, $H\mathbb{Z}_p \wedge K$ is null. But if we take even $X=S^0$, $L_K(S^0)$ does have some rational homotopy (in negative degrees). –  Craig Westerland Aug 9 '13 at 14:37
There are two issues here: one is that localisation $L_K(-)$ may not agree with $K$-nilpotent completion $(-)^{\hat{}}_K$ (which is the thing the cosimplicial spectrum calculates), and another is that in any case $H\mathbb{Q} \wedge -$ need not commute with totalisation. –  Oscar Randal-Williams Aug 9 '13 at 14:51
What about computing $H\mathbb{Z}_{(p)}\wedge k$ with various bockstein spectral sequences. Here $k$ is an integral lift of connective Morava $K$-theory. The issues that Oscar brings up seems pretty serious though, specifically the difference between nilpotent completion and localization. The issue of commuting with totalization seems like exactly what you are noticing with convergence. (by seems I mean is) –  Sean Tilson Aug 9 '13 at 15:12
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2 Answers

up vote 8 down vote accepted

I'm going to phrase in terms of $H$-homology instead of cohomology.

  • If $H$ is a finite spectrum, smashing with it always commutes across the limit. More generally, if you express $H$ as a (homotopy) colimit of finite spectra -- eg, by taking a cellular filtration -- then you get a directed system of spectral sequences, and the groups you're looking for are the direct limits of the abutments. However, in most circumstances you can't compute this via the direct limit of the spectral sequences and need to be sensitive to interchanging the direct and inverse limits. This might work if you can get a very tight computational handle on things.

  • If $H$ is a spectrum of finite type (connective and with finitely generated homology groups, or equivalently having a CW model with only finitely many cells in each dimension), then smashing with $H$ commutes across inverse limits of towers where there is some $r$ so that all the terms are $r$-connected. This doesn't apply in your circumstance, but would often apply when $K$ is connective.

  • Let's take an opposing case where $H = H\mathbb{Q}$ and $K = H\mathbb{F}_p$. Then the spectral sequence you're describing is the Adams spectral sequence and you certainly can't pass rationalization in through the spectral sequence. However, you can often get at the rationalization of the target because you're localizing with respect to an operation (multiplication by p) which lifts to an operation in the spectral sequence that shifts filtration (sometimes called multiplication by $h_0$). A paper of Miller's ("On relations between Adams spectral sequences, with an application to the stable homotopy of a Moore space") introduces these kinds of techniques, and they also showed up in Mahowald-Ravenel-Shick's paper on the telescope conjecture. Unfortunately integral homology can't be expressed as one of these,

Now let's get to some details about the more specific version that you asked.

Since $K(n)$-localizations are always $E(n)$-local, and $E(n)$-localization is smashing, $H \wedge L_{K(n)} X \simeq L_{E(n)} H \wedge L_{K(n)} X$.

Further, the map $E(n)_* H\mathbb{Z} \to E(n)_* H\mathbb{Q}$ is an isomorphism: the former is the universal ring where formal group law of $E(n)_*$ obtains a logarithm, and that ring is rational. (Otherwise the ideal $(p)$ would give a nontrivial quotient, but that quotient is $E(n)_* H\mathbb{F}_p = 0$.) Moreover, $H\mathbb{Q}$ is $E(n)$-local, and so $L_{E(n)} H\mathbb{Z} \simeq H\mathbb{Q}$.

EDIT (some elaboration): The spectrum $E(n) \wedge H\mathbb{F}_p$ has a complex orientation inherited from $E(n)$, and one from $H\mathbb{F}_p$. This gives the coefficient ring $E(n)_* H\mathbb{F}_p$ two formal group laws which must be isomorphic. The first has $p$-series satisfying $[p](x) = v_k x^{p^k}$ mod $(v_0,\cdots,v_{k-1})$; $v_0$ is $p$ by convention and $v_n$ is a unit. The second has $p$-series $[p]'(x) = 0$. A strict isomorphism $f$ implies $[p](x) = f^{-1} [p]' f(x) = 0$. This inductively implies that $v_k$ maps to zero for all $k$, and thus the unit $v_n$ is zero; the entire ring $E(n)_* H\mathbb{F}_p$ must be zero.

EDIT (correcting accidental switch to cohomology): Therefore, the integral homology of the spectrum you're asking about is actually $H\mathbb{Q}_* L_{K(n)} X$, and the universal coefficient theorem implies that the cohomology is given by applying $Ext(-,\mathbb{Z})$ to this (with a shift). This reduces the problem to the chromatic splitting conjecture that Eric described (i.e. you may want to stop working on it).

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Can you say more about the paragraph "Further the map...?" I certainly believe that the two statements $E(n)_* H\mathbb{Z} \cong E(n)_* H\mathbb{Q}$ and $E(n)_* H\mathbb{F}_p = 0$ are equivalent, but neither of them is obvious to me. Also, are you really concluding $H\mathbb{Z}^* L_{K(n)} X \cong H\mathbb{Q}^* L_{K(n)} X$? That seems incredible. –  Craig Westerland Aug 10 '13 at 17:59
Sorry. One has some elaboration now and one was a mistake. My apologies. –  Tyler Lawson Aug 11 '13 at 3:37
Hey Craig - if you don't like the formal group argument, then you can also think of it telescopically: for any finite spectrum X of type n, HF_p smash X [v_n^{-1}] vanishes because its homotopy groups vanish. Thus by Kunneth and the definition of "type n", the Morava K-theory of HF_p vanishes. So the Morava E-theory of HF_p also vanishes, as claimed. More generally we see that telescopic localization kills any spectrum which is torsion and bounded above. Of course Tyler's argument is better in that it doesn't use "nilpotence technology", but I hope the alternate perspective is helpful. –  Dustin Clausen Aug 11 '13 at 7:17
Also, it seems to me that the claim about the integral and rational cohomology of L_{K(n)}X agreeing is true: it follows from the homology statement by the universal coefficient theorem. Am I missing something? –  Dustin Clausen Aug 11 '13 at 7:27
This is wonderful -- thanks, Tyler and Dustin. –  Craig Westerland Aug 11 '13 at 21:14
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$\newcommand{\Q}{\mathbb{Q}}\newcommand{\S}{\mathbb{S}}\newcommand{\sm}{\wedge}\newcommand{\Z}{\mathbb{Z}}$Let me give not an answer but instead a long comment that says this is a hard question to which we do not know the answer in at least one "basic" case: $K = K(n)$, $H = H\Q$, $X = \S^0$. This should at least lend some context to the question.

The homotopy of the $K(n)$-local sphere is an ultra-interesting piece of homotopical data, largely because of the apparent connections to arithmetic. Not much is known about the homotopy of these spectra $L_{K(n)} \S^0$ past $n > 2$, and the story at $n = 2$ is said by experts to be sufficiently complicated as to suggest that humans will never be able to understand the raw data involved in the $n = 3$ case.

That hasn't stopped us from making qualitative guesses at how parts of the data will play out, and the Chromatic Splitting Conjecture (CSC) is one such guess. Without going into specifics, the CSC makes predictions about the degrees in which the rational homotopy of the $K(n)$-local sphere is supported. This is something you could quickly determine if you knew about $H\Q^*(L_{K(n)} \S^0)$, and certainly $H\Q^*(\S^0)$ and $K(n)^*(\S^0)$ are things we should be able to feed as inputs into the answer this question hopes for. So, that the CSC is in fact a C(onjecture) should be chilling.

Having said that, let me also say why the facts I outline in the comments on the question are directly against the spirit of the CSC, making the situation even more bleak. Firstly, there are two nice models for chromatic localization which are good to know about: $$L_{K(n)} X = L_{K(n)}^{fin} L_n X = \lim{}_I \{L_n(M^0(p^{I_0}, \ldots, v_{n-1}^{I_{n-1}}) \sm X)\},$$ $$L_n X = \operatorname{colim}{}_I \{L_n M^0(p^{I_0}, \ldots, v_{n-1}^{I_{n-1}}) \sm X\}.$$ Furthermore, the localization functor $L_n$ is smashing, so $L_n X = X \sm L_n \S^0$. These identities let you prove lots of interesting things about $K(n)$-localization, so they're useful things to have at the top of your head.

They also indicate a process by which the rational homotopy of $L_{K(n)} S^0$ comes about. Avoiding the prime $p = 2$, which has e.g. $\Z/4$ in $\pi_* M(2)$, the homotopy of $M(p^n)$ has the property that it is all $p^n$-torsion. The way that rational homotopy of $L_{K(n)} \S^0$ thus arises is by finding torsion-free components of $\pi_* L_{K(n)} \S^0$ itself, and the essential way that a system of $p^n$-torsion groups can assemble into a non-torsion group as $n$ varies is the $p$-adic system $$\Z_p = \lim \{\cdots \to \Z/p^2 \to \Z/p\}.$$

Oscar points out that taking $E$-homology (or: smashing) doesn't commute with inverse limits. However, there's a spectral sequence (attributed at least to Sadofsky, possibly further back to Hopkins) in the case that the spectra $X_*$ in the inverse system are all $E$-local, which has the rough form $$\lim{}^r_{E_* E\text{-comodules}}(E_* X_*) \Rightarrow E_* \lim X_*.$$ There are some assumptions on $E$ to get this spectral sequence to go, including at least something about the flatness of $E_* E$, and perhaps more as well that I've forgotten. Whatever these are, I promise they're satisfied for $E = H\Q$, and so one might attempt (as I did in the comments) to mix these two ideas: the Moore spectrum system and this spectral sequence.

The trouble is that the Moore spectra are not $H\Q$-local, and this is somehow the essential point of the CSC. Let's suppose that maybe I can fix this locality problem later, and for now I should just rationalize the system so that I can compute the inverse limit spectral sequence and see what comes out. Let's first call out an analogy to classical, algebraic rationalization and consider that $p$-adic system above. The objects $\Z/p^n$ in the system are not rational, and if I try to rationalize them and study the resulting system, what I'll find remains is $$0 = \lim \{ \cdots \to 0 \to 0 \}.$$ The same thing will happen when studying the homotopy of the Moore spectra: if they're all $p^n$-torsion, then they're also all rationally acyclic, and so I learn no information whatsoever.

In short, the CSC is some statement about accumulation behavior in the homotopy of $L_n M^0(p^n, v_1^\infty, \ldots, v_{n-1}^\infty)$, and I have destroyed exactly that information by rationalizing.

So: these are the general tools we have, and this is an explanation of why they are not as helpful as you'd like. Maybe there are $X$ other than $\S^0$ for which there is a better answer, but I wouldn't know about that.

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A strictly better discussion of these exact things around the CSC can be found in section 14 of the soon-to-be-released transcripts of the MIT E-theory seminar notes: math.berkeley.edu/~ericp/latex/mit-ethy.pdf . –  Eric Peterson Aug 10 '13 at 7:07
Cool! Thanks, Eric. –  Craig Westerland Aug 11 '13 at 21:15
Sure. I'm sure I didn't tell you much you didn't already know, but I thought it would be appreciated by onlookers. –  Eric Peterson Aug 11 '13 at 21:54
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