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Remark: My question is valid in the classic setting of the stable homotopy category of spectra of CW-complexes. An answer on that setting will also be valid.

Denote as $SH(X)$ Voevodsky's stable homotopy category over a scheme $X$. Denote $SH(X)^{\mathrm{eff}}$ its effective variant, that is to say, the smalles triangulated subcategory of $SH(X)$ which is closed under direct sums and contains suspension spectra of spaces but not their $\mathbb{P}^1$- desuspensions (cf section 2).

Let $V\to X$ be a vector bundle. Denote as $\mathrm{Th}(V)$ its Thom space, and denote as well its infinite supension, which belongs to $SH(X)$. More concretely, it belongs to $SH(X)^{\mathrm{eff}}$.

Let $\xi$ be a virtual vector bundle over $X$ of rank $r\in \mathbb{Z}$ and denote $\mathrm{Th} (\xi)$ its associated Thom spectrum (cf. section 4.1 of this paper). My question is:

Is it true that $$ \mathrm{Th}(\xi) \in SH(X) ^{\mathrm{eff}}\Leftrightarrow r\geq 0 \hspace{.5cm} ?$$

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    $\begingroup$ Neither implication holds For example, $S^0$ is the Thom spectrum of the virtual bundle $-1$ on $S^1$ so $\Rightarrow$ does not hold. On the other hand, $H\mathbb{F}_2$ is a Thom spectrum on $\Omega^2S^3$ for a virtual bundle of rank 0, so $\Leftarrow$ does not hold. $\endgroup$
    – Dylan Wilson
    Commented Jan 8, 2018 at 18:37
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    $\begingroup$ actually, maybe I'm confused about your definition of 'effective' in the case of classical stable homotopy theory. How can you be triangulated, contain suspension spectra, and not contain their desuspensions? $\endgroup$
    – Dylan Wilson
    Commented Jan 8, 2018 at 18:39
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    $\begingroup$ In motivic homotopy theory there are two possible definitions for $SH^{eff}(X)$: the smallest stable subcategory containing suspension spectra and closed under colimits (this is the classical one, defined by Voevodsky) and simply the smallest subcategory containing suspension spectra and closed under colimits (sometimes called the subcategory of "very effective" spectra). It seems to me that you are mixing the two. $\endgroup$
    – Denis Nardin
    Commented Jan 8, 2018 at 18:56
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    $\begingroup$ Well, if you use the first definition in the classical case you just get everything... $\endgroup$
    – Dylan Wilson
    Commented Jan 8, 2018 at 18:58
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    $\begingroup$ @DylanWilson Yes, the "you" in my comment was towards the OP, sorry for not making it clear. Actually I now realized that when s/he says "desuspensions" he means "desuspensions by $\mathbb{P}^1$, so at least his description is consistent with the "classical" case of the slice filtration. It still doesn't help with what he means for "usual" spectra. $\endgroup$
    – Denis Nardin
    Commented Jan 8, 2018 at 18:59

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Yes.

A bit more generally, if $\xi$ is a perfect complex of rank $\geq 0$, then $Th(\xi)$ is effective (even very effective): the question is Nisnevich-local on $X$ and $\xi$ is locally a complex of trivial vector bundles.

Conversely, suppose $Th(\xi)$ is effective. Since pullback preserves effective spectra, we can assume $X$ the spectrum of a field, so $Th(\xi)=(\mathbb P^1)^{\wedge r}$ and $r$ must be $\geq 0$.

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  • $\begingroup$ What is the Thom spectrum of the trivial rank -1 bundle on P^1? $\endgroup$
    – Dylan Wilson
    Commented Jan 9, 2018 at 16:16
  • $\begingroup$ Ah, maybe I was ignoring a basepoint thing... the trivial bundle introduces a disjoint basepoint, which then gets desuspended and you land outside the effective stuff? (Same for the classical case). Whoops! $\endgroup$
    – Dylan Wilson
    Commented Jan 9, 2018 at 16:30
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    $\begingroup$ @DylanWilson Yes, but also $X$ is the base in the question, so this is a bit different. These arguments should apply as well to sheaves of spectra over a base space, with effective replaced by connective. I'm not sure about the implication $\Rightarrow$ for Thom spectra over smooth $X$-schemes though. $\endgroup$
    – Marc Hoyois
    Commented Jan 9, 2018 at 16:33

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