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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.
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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](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.170.4097&rep=rep1&type=pdf)).


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)](https://arxiv.org/pdf/0907.2710.pdf). My question is:



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> Is it true that $$ \mathrm{Th}(\xi) \in SH(X) ^{\mathrm{eff}}\Leftrightarrow r\geq 0 \hspace{.5cm} ?$$
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When is the Thom spectrum of a virtual vector bundle effective?
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 $\geq 0$.


