Detecting the Brown-Comenetz dualizing spectrum

Question

The Brown-Comenetz dualizing spectrum $I_{\mathbf{Q/Z}}$ is not detected by very many spectra: it is $BP, \mathbf{Z}, \mathbf{F}_2, X(n)$ for $n\geq 2$, and even $I_{\mathbf{Q/Z}}$-acyclic. However, if $X$ is any nontrivial finite spectrum, then $X\wedge I_{\mathbf{Q/Z}}$ is not contractible. This motivates a natural question: let $E$ be any spectrum such that $E\wedge I_{\mathbf{Q/Z}}$ is not contractible. Then, is it true that $\langle E\rangle \geq \langle X\rangle$ for some nontrivial finite spectrum $X$?

Answer 1

Your question appears to be equivalent to the 'dichotomy conjecture' of Hovey, which I believe is still open.

First, note that any finite spectrum has a type, and all finite spectrum of type $n$ have the same Bousfield class, usually denoted $F(n)$. In Hovey and Strickland's memoir (Appendix B) they conjecture that if $E \wedge I \ne 0$, then $\langle E \rangle \ge \langle F(n) \rangle$ for some $n$. This is precisely your question.

In his paper on the chromatic splitting conjecture, Hovey conjectured that every spectrum has either a finite acyclic or a finite local (the dichotomy conjecture). With Palmieri (The structure of the Bousfield lattice) he proved that the following conjectures are equivalent:

(1) If $E \wedge I \ne 0$, then $\langle E \rangle \ge \langle F(n) \rangle$ for some $n$.

(2) The dichotomy conjecture.

(3) If $E$ has no finite acyclics, then $\langle E \rangle \ge \langle I \rangle$.

The analog of the dichotomy conjecture is known to hold in other categories with a good theory of support satisfying the tensor product property - see 'The Bousfield lattice of a triangulated category and stratification' by Iyengar and Krause.

Answer 2

I don't think that the answer is known. However, here are some comments. I will work everywhere with $p$-local spectra, for some fixed prime $p$, and write $I$ for the $p$-local Brown-Comenetz spectrum, so that $[W,I]=\text{Hom}(\pi_0(W),\mathbb{Q}/\mathbb{Z}_{(p)})$.

I first claim that $I\wedge W=0$ iff $[W,S_p]_*=0$ (where $S_p$ is the $p$-adic completion of the $0$-sphere spectrum). Indeed, we have $I\wedge W=0$ iff $\pi_*(I\wedge W)=0$ iff $\text{Hom}(\pi_*(I\wedge W),\mathbb{Q}/\mathbb{Z}_{(p)})=0$ iff $[W\wedge I,I]_*=0$ iff $[W,F(I,I)]_*=0$. Moreover, the evident map $S\to F(I,I)$ extends uniquely to give a map $S_p\to F(I,I)$, and one can check that this is an equivalence. The claim follows from this.

In most of the cases that you have quoted where $I\wedge W=0$, the proof actually uses the Adams Spectral Sequence to show that $[W,S_p]_*=0$, and then deduces $I\wedge W=0$ by the argument given above.

Next, you have mentioned that $I\wedge X(2)=0$. Here $X(2)$ is a ring spectrum, and it follows that $I\wedge M=0$ whenever $M$ is an $X(2)$-module. By construction, there is a ring map $X(2)\to MU$, so any $MU$-module is an $X(2)$-module, and so annihilates $I$. In particular we have $E\wedge I=0$ whenever $E$ is one of the standard complex-orientable spectra, such as $BP$, $E(n)$, $K(n)$, $P(n)$, $B(n)$ and so on. Also, as $L_nS$ has the same Bousfield class as $E(n)$, we see that $L_nS\wedge I=0$.

Next, recall that $L_n^fS$ denotes the finite localisation of $S$ of chromatic height $n$. I claim that $L_n^fS\wedge I$ is also zero. If $n=0$ then $L_n^fS=H\mathbb{Q}$, and this is complex-orientable, so we have already seen that $L_n^fS\wedge I=0$. Suppose instead that $n>0$. One can reduce in a standard way to the claim that $(v^{-1}X)\wedge I=0$ whenever $X$ is a finite spectrum of height $n$, and $v$ is a $v_n$-self map of $x$. Equivalently, we must show that $[\Sigma^m v^{-1}X,S_p]=0$ for all $m$. If $|v|=d>0$ then $v^{-1}X$ is the homotopy colimit of the spectra $\Sigma^{-id}X$, and for fixed $m$ we have $[\Sigma^{m-id}X,S_p]=0$ for large $i$ by a connectivity argument. The claim follows from this by the Milnor exact sequence.

I'll now explain one example where I am not sure whether $I\wedge W=0$. Take $F(0)=S^0$, then define $F(n+1)$ recursively to be the cofibre of some $v_n$-self map of $F(n)$, inducing multiplication by $v_n^{i_n}$ in $BP$-theory say. There are then evident maps $F(n)\to F(n+1)$, so we can let $F(\infty)$ denote the homotopy colimit. There is an Adams-Novikov spectral sequence $$ \text{Ext}^{**}(BP_*F(\infty),BP_*) \Longrightarrow [F(\infty),S_p]_*. $$ I suspect that the $E_2$ page is already zero, so $[F(\infty),S_p]_*=0$, so $I\wedge F(\infty)=0$. However, I have not proved this.

UPDATE: Tobias Barthel suggested to me the following proof that $I\wedge F(\infty)=0$ (and that $I\wedge W=0$ for yet another large collection of spectra $W$). Put $K(\mathbb{N})=\bigvee_{n\in\mathbb{N}}K(n)$, and recall that $K(\mathbb{N})$-local spectra are called harmonic, whereas $K(\mathbb{N})$-acyclic spectra are called dissonant. It is a theorem of Hopkins and Ravenel that all suspension spectra are harmonic. We can construct $S/p^n$ as the desuspension of a Moore space, so it is harmonic. We can construct $S_p$ as the homotopy inverse limit of the spectra $S/p^n$, so it is also harmonic. This means that whenever $W$ is dissonant we have $[W,S_p]_*=0$ and so $W\wedge I=0$. In particular, it is not hard to see that $K(\mathbb{N})_*(F(\infty))=0$ and so $F(\infty)\wedge I=0$.