Absence of Maps Between p-local and q-local spectra

2012-01-31T21:16:55Z

Suppose $X$ and $Y$ are spectra (or homotopy classes thereof) such that $X$ is p-local and $Y$ is q-local, for primes $p\neq q$. Is it indeed true then, and if so how would one show that $[X,Y]_\ast=0$?

Thanks!

Answer by Charles Rezk for Absence of Maps Between p-local and q-local spectra

2012-01-31T22:25:44Z

The rational Eilenberg-Mac Lane spectrum $H\mathbb{Q}$ is $p$-local for every prime $p$, but certainly $[H\mathbb{Q}, H\mathbb{Q}]_*\neq 0$.

If you replace "$p$-local" and "$q$-local" with "$p$-complete" and "$q$-complete", then your conclusion does hold.

Answer by Tom Goodwillie for Absence of Maps Between p-local and q-local spectra

2012-02-01T01:43:28Z

Responding to your comment on Charles's answer:

[I wish I could delete my comment to Charles's answer. My comment seems to have exploded and taken your comment down with it.]

Both localization of a spectrum at a prime and completion of spectra at a prime are examples of Bousfield localization. The one is (a little circularly) localization with respect to $p$-localized homotopy theory, while the other is localization with respect to mod $p$ homotopy theory.

Localizing a spectrum at a prime $p$ (or at a set of primes) is the kind of Bousfield localization that is most like localization in algebra. In this case $\pi_n$ of the localization of $X$ is the algebraic localization of $\pi_nX$, i.e. the result of inverting all the primes other than $p$. The local spectra are the spectra $X$ such that for every $n$ the abelian group $\pi_nX$ is one on which every prime other than $p$ acts invertibly, and the acyclic objects are the ones which become trivial upon localizing, i.e. the spectra $X$ such that for every $n$ the group $\pi_nX$ is a torsion group without $p$-torsion. Localization of $X$ is the same as smash product of $X$ with the localization of the sphere spectrum. If you want to describe the acyclic objects as the $E_*$-acyclic objects for some $E$, you can of course let $E$ be the localization of the sphere. Note that we are not starting with an $E$ and using the general machine of Bousfield to make the localization functor, although of course we could.

Completing at a prime is again not too far from being a purely algebraic matter. Here we can choose $E$ to be the mod $p$ Moore spectrum, i.e. the homotopy cofiber of the map $p:S\to S$ from the sphere spectrum to itself. The acyclic objects are the spectra for which each homotopy group has $p$ acting invertibly. The localization $LX$ (which is called the $p$-completion) can be described as the holim, over natural numbers $k$, of the smash product of $X$ with the mod $p^k$ Moore spectrum. If each homotopy group of $X$ is finitely generated, then $\pi_nLX$ can be described as the tensor product of $\pi_nX$ with the group $\mathbb Z_p$ of $p$-adic integers, and also in this case $LX$ is the smash product $X\wedge LS$. But neither of these statements is true for general $X$: for example, the $p$-completion of $H\mathbb Q$ is trivial while $\mathbb Q\otimes \mathbb Z_p$ is the field of $p$-adic numbers and $X\wedge LS=H(\mathbb Q\otimes \mathbb Z_p)$. And (therefore) $LH(\mathbb Q/\mathbb Z)=\Omega H\mathbb Z_p$.

Note that a spectrum which is local at a prime $q$ different from $p$ is trivial mod $p$. In general maps from $E$-acyclic objects to $E$-local objects are trivial, so we see that maps from a $q$-local spectrum to a $p$-complete spectrum are trivial. In particular, as Charles says, maps from a $q$-complete spectrum to a $p$-complete spectrum are trivial.

Absence of Maps Between p-local and q-local spectra - MathOverflow

Absence of Maps Between p-local and q-local spectra

Answer by Charles Rezk for Absence of Maps Between p-local and q-local spectra

Answer by Tom Goodwillie for Absence of Maps Between p-local and q-local spectra