Let $X$ be a spectrum. Is there a canonical construction/functor that would associate to this spectrum, an inverse spectrum $X'$, in the sense that $$\pi_*(X)\cong \pi_{-*}(X')?$$

To be more precise, such a spectrum $X'$ can always be constructed by attaching cells to produce the right homotopy groups, but is there a more conceptual way of creating it?


1 Answer 1


Not exactly what you asked for, but there is the Brown-Comenetz dual of $X$, $I_{\mathbb{Q}/\mathbb{Z}}X$, which has the property that its homotopy groups are Pontryagin dual to the homotopy groups of $X$ (in the negative degree):

$\pi_{-\ast}(I_{\mathbb{Q}/\mathbb{Z}}X) \simeq Hom(\pi_\ast(X),\mathbb{Q}/\mathbb{Z})$

So, if the homotopy groups of $X$ are all finite, then the Pontragin dual groups will be (non-canonically) isomorphic and this will have the required property.

See here for more details

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    $\begingroup$ There's a question that always bothered me. Maybe you know the answer. How canonical is the assignment $X\mapsto I_{\mathbb Q/\mathbb Z}X$? More precisely, is the assignment $X\mapsto I_{\mathbb Q/\mathbb Z}X$ a functor from the $\infty$-groupoid of spectra to itself? It's obviously a functor at the level of the homotopy category. But what about at the $\infty$-groupoid level? $\endgroup$ Dec 23, 2018 at 21:42
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    $\begingroup$ @AndréHenriques I've probably misunderstood your question, but does this answer it? If $I_\mathbf{Q/Z} := I_\mathbf{Q/Z}(S^0)$ denotes the Brown-Comenetz dualizing spectrum, then $I_\mathbf{Q/Z}(X) = F(X, I_\mathbf{Q/Z})$, and this is functorial in $X$. $\endgroup$
    – skd
    Dec 24, 2018 at 5:47
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    $\begingroup$ @skd. Yes, of course... that does answers my question. But I now realise that I haden't asked the right question. Here's the right question: what is $I_{\mathbb Q/\mathbb Z}$? More precisely, is there a way to define it up to contractible space of choices? $\endgroup$ Dec 24, 2018 at 16:18

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