# Finite complexes which are not Thom spectra

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.

• 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! – Prasit Jan 12 at 19:21
• 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 $\pi_0S^0$. – Dylan Wilson Jan 12 at 20:14
• 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? – Dylan Wilson Jan 13 at 2:08
• 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? – Tim Campion Jan 13 at 20:54
• @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. – Denis Nardin Jan 13 at 20:58

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.