8
$\begingroup$

Let $L_E$ denote Bousfield localisation with repsect to the cohomology theory $E$. I am trying to follow through some calculations in Hovey-Strickland's paper Morava $K$-theories and localisation

Claim 7.10(e) is that

$$L_{K(n)}X = \underset{\leftarrow}{\text{holim}}_I L_{E(n)}X \wedge S/I$$

where the homotopy limit is over a tower of generalised Moore spectra (the Moore spectra $S/I$ is defined in 4.12 and the tower in 4.22)

Define $$E_*^{\vee}:=\pi_{*} (L_{K(n)} (E \wedge X))$$

the $K(n)$-local version of Morava $E$-theory (where, I believe, Morava $E$-theory here is what I might call a completed Johnson-Wilson theory, but I don't believe it really matters).

The claim (8.04) is then that we can extract a Milnor exact sequence from this:

$$0 \to \varprojlim_I {}^1 (E/I)_{\ast+1}(X) \to E_*^\vee X \to \varprojlim_I (E/I)_*X \to 0 $$

I'm not sure how to show this. I would like to think that we can get a sequence

$$0 \to \varprojlim_I {}^1 \pi_{\ast+1}(L_{E(n)}(E \wedge X) \wedge S/I)\to E_{\ast}^\vee X \to \varprojlim_I \hspace{1mm} \pi_* (L_{E(n)}(E \wedge X) \wedge S/I) \to 0 $$

and then if you drop the $E(n)$-localisation it seems to work, but I'm not really sure about this.

$\endgroup$

1 Answer 1

9
$\begingroup$

In this case, this follows because $E(n)$-localization is a smashing localization. Specifically, $$ L_{E(n)}(X) = L_{E(n)}(\mathbb{S}) \wedge X $$ for any $X$. In particular, this means that the smash product of any spectrum with an $E(n)$-local spectrum is already $E(n)$-local.

The fact that $E(n)$-localization is smashing was one of the Ravenel conjectures; specifically, conjecture 10.6 in his paper "Localization with respect to certain periodic homology theories". I don't have a copy of his orange book handy and so can't give you a reference for the theorem statement.

As a result, as the Morava $E$-theories are $E(n)$-local already, so are the smash products $E \wedge X$ in your equation.

$\endgroup$
2
  • 3
    $\begingroup$ Thanks Tyler! In its simplest form: $L_{E(n)}(E \wedge X) = L_{E(n)}(\mathbb{S}) \wedge E \wedge X = L_{E(n)} (E) \wedge X = E \wedge X$? $\endgroup$
    – Drew Heard
    Mar 30, 2012 at 2:11
  • $\begingroup$ Drew, exactly right. $\endgroup$ Mar 30, 2012 at 2:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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