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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?


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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. – Christian Nassau Sep 18 '12 at 17:46
So in particular, it seems that any harmonic spectrum $X$ is weakly equivalent to a wedge product of its monochromatic slices? – Jon Beardsley Sep 18 '12 at 18:23
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. – Christian Nassau Sep 18 '12 at 18:41
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... – Jon Beardsley Sep 18 '12 at 20:49
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.] – Christian Nassau Sep 19 '12 at 5:53

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