# Detecting the Brown-Comenetz dualizing spectrum

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 $X$?

• A silly comment, but X(1) is the sphere spectrum, so you probably don't want to include that in your list of examples of spectra that don't detect. Jun 3 '18 at 8:29
• You're right, of course. I'll fix this.
– skd
Jun 3 '18 at 16:02
• Here is a failed attempt at answering this question. Suppose $E\wedge I \neq 0$. Then the same argument as in Strickland's answer, combined with the self-duality of the generalized Moore spectra, shows that $[E, M(i_0, \cdots, i_n)] \neq 0$. If $E$ had a finite acyclic $F$, then $[E,F] = 0$. If $F$ has type $N$, then every type $N$ spectrum (in particular, $M(i_0, \cdots, i_N)$) can be obtained from $F$ via a finite number of retracts and cofiber sequences, so we would run into a contradiction. Therefore $E$ cannot have a finite acyclic.
– skd
Jun 4 '18 at 5:59
• Note that if such an $E$ with a finite acyclic existed, it would be a counterexample. Indeed, $\langle E \rangle \geq \langle X \rangle \geq \langle T(n) \rangle$ for every $n\geq N$, where $X$ has type $N$. As $\langle T(n) \rangle \geq \langle K(n) \rangle$, this implies that $\langle E \rangle \geq \langle K(n) \rangle$ for every $n\geq N$. If $E$ had a finite acyclic $F$, then $K(n)_\ast F=0$ for every $n$, i.e., $F$ would be contractible (as finite spectra are harmonic).
– skd
Jun 4 '18 at 5:59
• In fact, since $\bigvee_n \langle K(n) \rangle \geq \langle I \rangle$ (all finite spectra are harmonic), this shows that $\langle E \rangle\geq \langle I \rangle$ on finite spectra. I think that this is true on all spectra, and in fact, I believe that the question is equivalent to the following: if $E$ is a spectrum with no finite acyclics, then $\langle E\rangle \geq \langle I\rangle$. (Note that $\langle E\rangle \geq \langle I\rangle$ implies that $E$ has a finite local, namely $S/p$.) I don't know how to prove this equivalence, though.
– skd
Jun 4 '18 at 6:13

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.

• I just noticed your comments to the question - it seems you have already noticed that (1) is equivalent to (3)! Jun 4 '18 at 8:44
• Awesome, thanks for the references! This means that the question is probably really hard to solve; (3) is an infinite height analogue of the telescope conjecture. Indeed, one equivalent way to rephrase the telescope conjecture is that if the thick subcategory of $E$-acyclics consists of finite spectra of type $n+1$, then $\langle E \rangle \geq \langle K(n) \rangle$. Moreover, $\langle I \rangle \leq \bigvee_n \langle K(n) \rangle$, and I think this is the only class which is less than or equal to $\bigvee_n \langle K(n) \rangle$.
– skd
Jun 4 '18 at 12:38
• The last part of the last sentence in my comment above is not exactly correct: one also has $\langle I_\mathbf{Z} \rangle \leq \bigvee_n \langle K(n) \rangle$.
– skd
Jun 4 '18 at 23:06

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 $F(\infty)\wedge I=0$.

• Thanks! Just want to point out one other way to show that $L^f_n I = 0$. Since $L^f_n$-localization is smashing and $\langle L^f_n S \rangle \leq \langle T(0) \vee \cdots \vee T(n)\rangle$, it suffices to show that $T(n) \wedge I = 0$ for all $n$. As $\langle \mathbf{F}_p\rangle \geq \langle I\rangle$ (since the homotopy groups of $I$ are bounded above and torsion), it suffices to show that $T(n)\wedge \mathbf{F}_p = 0$. For $n=0$ this is obvious, and for $n>0$ this follows since $v_n$-self maps have positive Adams filtration.
– skd
Jun 4 '18 at 2:27
• Regarding the last paragraph: there is a trigraded spectral sequence converging to $\mathrm{Ext}^{s+n,t}(BP_\ast F(\infty), BP_\ast)$ whose $E_2$-page is $\varprojlim^n \mathrm{Ext}^{s,t}(BP_\ast F(m), BP_\ast)$. I don't know if it's any easier to show that these groups vanish.
– skd
Jun 4 '18 at 2:30