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.



Finite complexes which are not Thom spectra
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.


