8
$\begingroup$

By nullification with respect to $K(A,n)$, I mean the Bousfield localization $L_{A,n}$ where a spectrum $E$ is $L_{A,n}$-local if and only if $\tilde E^\ast(K(A,n)) = 0$.

Question: Is there a good description of which spectra are $L_{A,n}$-local? Is there a good description of the localization functor $L_{A,n}$?

Notes:

  • I'd be happy to understand the case $A = \mathbb Z/p$ when $E$ is also $p$-local.

  • For example, $K(n)$-local spectra are $L_{\mathbb Z/p,m}$-local for $m \geq n+1$.

  • If $E$ is $L_{A,n}$-local, then $E$ is also $L_{A,n+1}$-local. This follows from the Zabrodsky lemma. As a consequence, $HA$ is $L_{A,n}$-acyclic for all $n$.

  • It's a fact (due to Bousfield I think) that $K(\mathbb Z/p, n)$ is always either $E$-local or $E$-acyclic (as a space -- note that $E$-acyclicicity the same whether we consider $K(\mathbb Z/p,n)$ as a space or suspension spectruem, but $E$-locality may be different I think).

  • And I believe any nontrivial homological localization of $p$-local spectra [EDIT: assuming it kills some suspension spectrum -- e.g. not the harmonic localization!] must kill some $K(\mathbb Z/p,n)$. So I'm asking about the "minimally nontrivial" localizations of $p$-local spectra. This follows from the theorem of Bousfield that if a space $X$ is $E$-acyclic, then $K(\pi_n(X), n)$ is $E$-acyclic, and then you can show that if some $K(A,n)$ is $E$-acyclic, so is $K(\mathbb Z/p, n)$ for some $p$.

  • One form of the Sullivan conjecture (proved by Miller says that in the unstable case, every finite-dimensional complex is $L_{A,n}^{unst}$-local for every finite $A$ and $n \geq 1$.

  • A related theorem of Lee says that in the stable case, every finite spectrum is $L_{A,n}$-local for every finite $A$ and $n \geq 2$. Unlike the unstable case, I don't think this extends to finite-dimensional spectra -- e.g. I think it already fails for any infinite wedge of equidimensional spheres. Certainly it fails even for the finite case at $n=1$ by the Segal conjecture (proved by Carlsson).

  • Nick Kuhn points out in his answer below that in the unstable case, Neisendorfer showed the following:

    Let $X$ be the $p$-completion of a simply-connected finite space with $\pi_2(X)$ finite. Then for all $m \in \mathbb N$, we have $L_{\mathbb Z/p, 1}^{unst} \tau_{\geq m} X = X$.

    I believe this extends to say that if $X$ is the $p$-completion of an $n$-connected finite space with $\pi_{n+1}(X)$ finite, then for all $m$ we have $L_{\mathbb Z/p, n}^{unst} \tau_{\geq m} X = X$.

  • The stable analog of Neisendorfer's theorem then says that if $L$ is a localization of $p$-complete spectra which kills $H\mathbb Z/p$, then for any spectrum $E$, $\tau_\geq n E \to E$ is an $L$-equivalence for all $n \in \mathbb N$.

$\endgroup$
8
  • $\begingroup$ What about MU-localization? Isn't this an example of a nontrivial localization which kills no $K(\mathbb{Z}/p,n)$? $\endgroup$ Apr 24, 2019 at 6:32
  • $\begingroup$ @SaalHardali I thought MU-localization was the same as $H\mathbb Z$-localization, which is to say the identity... $\endgroup$
    – Tim Campion
    Apr 24, 2019 at 12:16
  • $\begingroup$ Maybe I'm forgetting some hypotheses for this to be true... $\endgroup$
    – Tim Campion
    Apr 24, 2019 at 12:22
  • 4
    $\begingroup$ $H\mathbb{Z}$-localization is not the identity. (For example it kills K-theory). $\endgroup$ Apr 24, 2019 at 13:53
  • 1
    $\begingroup$ I would like to add that I gave $MU$ instead of the simpler $H\mathbb{Z}$ because its also not true that localization w.r.t. all the morava $K$-theories (i.e. the harmonic localization) is the same as $MU$-localization which indicates in some sense that there are baziilions of non-trivial loclization of this sort. $\endgroup$ Apr 24, 2019 at 14:06

1 Answer 1

3
$\begingroup$

This question is more fun at the space level: see [Neisendorfer, Joseph A. Localization and connected covers of finite complexes. The Čech centennial (Boston, MA, 1993), 385–390, Contemp. Math., 181, Amer. Math. Soc., Providence, RI, 1995.]

$\endgroup$
1
  • 13
    $\begingroup$ @NickolasKuhn: Your answer is somewhat cryptic. For the benefit of those who don't have the time to check the reference, a tiny bit of explanation of what happens at the space level would be very welcome. $\endgroup$ Apr 23, 2019 at 22:00

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.