0
$\begingroup$

What are the chances that, for an arbitrary $p$-local harmonic spectrum $X$, if $K(n)\wedge X\simeq\ast$ for all $n$, then $X$ is contractible? This, I believe, holds for suspension spectra and finite spectra. Does anyone know of any harmonic (i.e. local to $\vee_{n\in\mathbb{N}}K(n)$) spectra for which this does not hold?

Thanks!

$\endgroup$
5
  • 5
    $\begingroup$ Isn't this pretty immediate? by assumption the map $X\rightarrow\ast$ is a $K(n)$-equivalence for all $n$. Then it is also an $E$-equivalence for the infinite wedge $E=\bigvee K(n)$. And since $X$ is $E$-local, the map is an honest equivalence. $\endgroup$ Sep 18, 2012 at 17:46
  • $\begingroup$ So in particular, it seems that any harmonic spectrum $X$ is weakly equivalent to a wedge product of its monochromatic slices? $\endgroup$ Sep 18, 2012 at 18:23
  • 3
    $\begingroup$ No, that must be false. I don't even see how you'd get a map between $X$ and the (wedge of the) $M_nX$. Here $M_nX$ = fiber $L_nX \rightarrow L_{n-1}X$ is what I'd call the monochromatic slice. $\endgroup$ Sep 18, 2012 at 18:41
  • $\begingroup$ Hrm. Good point. Too bad. I do believe that the $K(n)$ localization of $X$ is equivalent to the $K(n)$ localization of $M_nX$, but I'm not certain that map can be lifted... $\endgroup$ Sep 18, 2012 at 20:49
  • 2
    $\begingroup$ The monochromatic slices of a finite $X$ are non-connective spectra, so $X$ surely is not their wedge sum. You might find Mark Hovey's paper on Hopkins' chromatic splitting conjecture interesting: that conjecture would imply that for finite $X$ the product of the $L_{K(n)}X$ contains $X_p$ as a summand. [I think the original formulation of that conjecture is now known to be false, though.] $\endgroup$ Sep 19, 2012 at 5:53

0

Your Answer

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

Browse other questions tagged or ask your own question.