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I'll be working in the stable world. It's an easy observation that any 2-cell complex (over the sphere) with bottom cell in dimension zero is a Thom spectrum: any such complex is the cofiber of some element $\alpha\in \pi_n S$, so it is the Thom spectrum of the map $S^{n+1} \to B\mathrm{GL}_1 S$ classifying $\alpha$. Is there an example of a finite complex (again, with bottom cell in dimension zero, which is a necessary condition) which is not a Thom spectrum? Is there such a 3-cell complex?

Edit: I think the question mark complex $Q$ gives an example of such a 3-cell complex. This is the complex constructed by lifting the stable map $\Sigma \eta:S^2\to S^1$ to $\Sigma^{−1}\mathbf{R}P^2$ by choosing a nullhomotopy of $2\eta$; it follows that if $Q$ was to be a Thom spectrum, it would be a Thom spectrum of bundle over a space with bottom cell in dimension 1 and top cell in dimension 3, along with a $\mathrm{Sq}^2$ in cohomology. But $\mathrm{Sq}^2$ (unstably) vanishes on any cohomology class in dimension 1, so this is impossible.

I guess, then, my question could be revised to asking whether there are nontrivial sufficient conditions under which a finite complex can be realized as a Thom spectrum.

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    $\begingroup$ The question mark complex might be an example, but I am not quite convinced with the argument. Just a fun-fact worth mentioning here: the upside down question mark complex i.e. Spanier-Whitehead dual of the question mark complex (up to a shift) is a Thom spectrum. This can be found in a paper of Mahowald ``Ring spectra which are Thom complexes." Also I think generalized Moore spectra, such as M(1,4), may not be a Thom spectrum. But cannot think of a quick argument! $\endgroup$ – Prasit Jan 12 at 19:21
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    $\begingroup$ Presumably you've implicitly localized at a prime, otherwise you can't get the mod p Moore complex the way you describe when p>2 since 1-p is not a unit in $\pi_0S^0$. $\endgroup$ – Dylan Wilson Jan 12 at 20:14
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    $\begingroup$ Anyway, for the actual question: can you turn Mahowald's proof that $bo$ isn't a Thom spectrum into a proof that some skeleton of it isn't? $\endgroup$ – Dylan Wilson Jan 13 at 2:08
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    $\begingroup$ I'm probably being stupid (I don't know anything about this stuff) but if $X$ is a finite complex, then $\Sigma^n X$ is is a suspension spectrum for some $n$, and in particular the Thom spectrum of a trivial bundle. Then can't you just desuspend the bundle $n$ times to exhibit $X$ as a Thom spectrum? $\endgroup$ – Tim Campion Jan 13 at 20:54
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    $\begingroup$ @TimCampion I assume they want a Thom spectrum of a rank 0 stable spherical fibration (since they mention of $BGL_1(\mathbb{S})$ and not $BGL_1(\mathbb{S})\times \mathbb{Z}$), so you're not allowed to desuspend. $\endgroup$ – Denis Nardin Jan 13 at 20:58
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The proposed argument for why $Q = S \cup_2 e^1 \cup_\eta e^3$ is not a Thom spectrum seems to use that the Thom isomorphism commutes with the Steenrod operations, which is often false. The deviation is measured by the Stiefel-Whitney classes.

If $Q$ were the Thom spectrum $B^\gamma$ of a stable spherical fibration $\gamma$ over a space $B$, then $H^*(B; Z/2) = Z/2\{1, b_1, b_3\}$ would have trivial $Sq^i$-actions (by instability) and $H^*(Q; Z/2) = Z/2\{U, Ub_1, Ub_3\}$ would be a free (right) module over $H^*(B; Z/2)$ on one generator $U \in H^0(Q; Z/2)$, the $Z/2$-orientation class. Then $Sq^i(U) = U w_i$, where $w_i$ is the $i$-th Stiefel-Whitney class of $\gamma$. Since $Sq^1(U) = U b_1$ and $Sq^2(U) = 0$ in the cohomology of $Q$, you must have $w_1 = b_1$ and $w_2 = 0$. Then $Sq^2(U b_1) = 0$ by the Cartan formula, contradicting $Sq^2(Ub_1) = Ub_3$ in $H^*(Q; Z/2)$. So $Q$ is not a Thom spectrum.

The paper https://arxiv.org/pdf/1608.08388.pdf by Basu, Sagave and Schlichtkrull gives a sufficient condition for realizing some finite $R$-modules as $R$-Thom spectra, where $R$ is even. This does not include the classical case $R = S$, but might still be of interest.

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  • $\begingroup$ Thanks! This is helpful. $\endgroup$ – skd Jan 13 at 23:19

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