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 $L_{A,n}$?


  • 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 $n \in \mathbb N$.

  • $\begingroup$ What about MU-localization? Isn't this an example of a nontrivial localization which kills no $K(\mathbb{Z}/p,n)$? $\endgroup$ Commented 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
    Commented Apr 24, 2019 at 12:16
  • $\begingroup$ Maybe I'm forgetting some hypotheses for this to be true... $\endgroup$
    – Tim Campion
    Commented Apr 24, 2019 at 12:22
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    $\begingroup$ $H\mathbb{Z}$-localization is not the identity. (For example it kills K-theory). $\endgroup$ Commented Apr 24, 2019 at 13:53
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    $\begingroup$ 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. $\endgroup$ Commented Apr 24, 2019 at 14:06

1 Answer 1


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.]

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    $\begingroup$ @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. $\endgroup$ Commented Apr 23, 2019 at 22:00

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