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Given two spectra $A$ and $B$, the set $[A,B]$ of homotopy classes of maps from $A$ to $B$ forms an abelian group. Can the dual abelian group $\text{Hom}([A,B],\mathbb{Q}/\mathbb{Z})$ be expressed as a group of homotopy classes of maps between spectra?

The Brown-Comenetz dual $I_{\mathbb{Q}/\mathbb{Z}}E$ of a spectrum $E$ is defined here. A key property of $I_{\mathbb{Q}/\mathbb{Z}}E$ is that its homotopy is dual of that of $E$ in the sense that $\pi_*I_{\mathbb{Q}/\mathbb{Z}}E=\text{Hom}(\pi_*E,\mathbb{Q}/\mathbb{Z})$.

Naively, I suspect that $\text{Hom}([A,B],\mathbb{Q}/\mathbb{Z})$ can be expressed in terms of $I_{\mathbb{Q}/\mathbb{Z}}A$ and $I_{\mathbb{Q}/\mathbb{Z}}B$.

What if I replace ${\mathbb{Q}/\mathbb{Z}}$ by ${\mathbb{R}/\mathbb{Z}}$ or ${\mathbb{C}/\mathbb{Z}}$?

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You seem to have some misunderstandings. The homotopy groups of $I_\mathbb{Z}E$ are not the Pontryagin duals of those of $E$ but the derived $\mathbb{Z}$-duals, up to a spectral sequence (see slide 7 in your link). When $E$ has torsion homotopy groups, the Brown-Comenetz dual $I_{\mathbb{Q}/\mathbb{Z}}E$ will have Pontryagin dual homotopy groups. In most cases, $[A,B]$ will not have any natural topology, so I don't know what you mean when you ask for it to be locally compact. –  Eric Wofsey Aug 18 '14 at 4:26
You can also define a functor $I_{\mathbb{T}}$ that represents Pontryagin duality on all spectra (by Brown representability, just like for $I_{\mathbb{Q}/\mathbb{Z}}$ and $I_\mathbb{Q}$). However, a priori these Pontryagin dual groups will just be abstract abelian groups without a natural topology, so this gives you nothing useful that you didn't already have from $I_{\mathbb{Q}/\mathbb{Z}}$ and $I_\mathbb{Q}$. Maybe you can enrich this functor to land in some more structured category such that you can recover the topology on the homotopy groups. –  Eric Wofsey Aug 18 '14 at 4:41
The category of spectra is closed symmetric monoidal, so one can ask for a hom spectrum rather than merely a hom set and then ask for a description of the Spanier-Whitehead, Anderson, or Brown-Comenetz duals of this spectrum, I guess. –  Qiaochu Yuan Aug 18 '14 at 7:44
@EricWofsey. You are correct that I had some misunderstandings. Firstly, I mixed up my duals: I am interested in the BC dual. Secondly, forget what I said about local compactness. –  Alex Turzillo Aug 18 '14 at 19:25
@QiaochuYuan, so I guess a related question is whether $I_{\mathbb{Q}/\mathbb{Z}}\text{hom}(A,B)$ can be expressed in terms of the BC duals of $A$ and $B$. –  Alex Turzillo Aug 18 '14 at 19:27

1 Answer 1

up vote 5 down vote accepted

I think this is the closest thing there is to what you are looking for. Let $G$ be any injective (i.e., divisible) abelian group. Then there is a spectrum $I_G$ representing the cohomology theory $X\mapsto \operatorname{Hom}(\pi^s_*(X),G)$. For any $A$ and $B$, we can then describe the dual $\operatorname{Hom}([A,B],G)$ as $[F(A,B),I_G]$.

In the special case that either $A$ or $B$ is a finite spectrum, however, we can do better than this. If $DA=F(A,S)$ is the Spanier-Whitehead dual of $A$, we have $$[F(A,B),I_G]=[DA\wedge B,I_G]=[DA,F(B,I_G)].$$

Since $F(B,I_G)$ is just what we call $I_GB$, we can thus describe the dual of $[A,B]$ as $[DA,I_GB]$.

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