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Let $X$ be a finite spectrum. Say that $X$ has characteristic two if multiplication by two on $X$ is nullhomotopic.

Does there exist a noncontractible finite spectrum of characteristic two?

(This is mentioned as an open problem in Barratt's 1959 paper "Spaces of finite characteristic.")

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    $\begingroup$ It feels like nilpotence or the thick subcategory theorem could be useful here... $\endgroup$
    – Dylan Wilson
    Commented Nov 3, 2013 at 6:34
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    $\begingroup$ @Dylan: I do not find it obvious that this class of spectra is closed under cofibers. $\endgroup$
    – Lennart Meier
    Commented Nov 4, 2013 at 16:02
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    $\begingroup$ Damn... if only we knew Freyd's generating hypothesis were true. Then closure under cofibers would be 'obvious.' ;) $\endgroup$
    – Dylan Wilson
    Commented Nov 5, 2013 at 15:00

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This doesn't seem very hard. Am I missing something?

Let $X$ be a non-trivial finite spectrum of characteristic $2$. Then let $R=\mathrm{Hom}(X,X)$, the function spectrum of maps $X$ to $X$. This $R$ is an associative $S$-algebra, with $0=2$ in $\pi_0R$; furthermore, $R$ is finite.

If $X\neq0$, then $H_*(X;F)\neq0$, where $F=\mathbb{F}_2$, and hence $H_*(R;F)\neq0$. Note that in this case the unit map $S\to R$ has non-trivial image in homology (which is the unit of the ring $H_*(R;F)$), and thus $H^0(R;F)\to F$ is surjective.

Consider the forgetful functor $\mathrm{Ass}_S\to \mathrm{Unital}_S$ from associative $S$-algebras to unital spectra. This has a (homotopical) left adjoint $T_*$: given a unital spectrum $M=(M,f\colon S\to M)$, let $T_*(M)$ be the homotopy pushout in $\mathrm{Ass}_S$ of $$ T(M) \xleftarrow{T(f)} T(S)\to S, $$ where $T$ is the free associative algebra functor.

You can also build $T_*(M) = \mathrm{colim}_n T_*^n(M)$ iteratively a la the James construction: to get $T_*^n(M)$, glue the $n$-fold smash product $M^{\wedge n}$ to $T_*^{n-1}(M)$ along a suitable map $F^n(M)\to M^{\wedge n}$. [I.e., consider the $n$-cubical diagram obtained from smashing $S\to M$ with itself $n$-times; $F^n(M)\to M^{\wedge n}$ is the map from the hocolim of the deleted $n$-cube to the object at the terminal position of the $n$-cube.]

Let $M=S^0\cup_2 e^1\to R$ be the map of unital spectra which exists by our hypothesis on $R$. The map extends to an associative algebra map $$ T_*(M) \to R. $$ In homology, $H_*(T_*(M);F) \approx F[x]$ with $|x|=1$, since $M^{\wedge n}/F_n(M) \approx S^n$.

In mod 2 cohomology, the generator $u\in H^0(T_*(M);F_2)$ must satisfy $Sq^1(u)\neq0$, and an easy argument shows that therefore $Sq^n(u)\neq0$ for all $n$. But $H^0(R;F) \to H^0(T_*(M);F) \xrightarrow{\sim} H^0(S;F)=F$ is surjective, so there exists $v\in H^0(R;F)$ such that $Sq^n(v)\neq0$ for all $n$, contradicting finiteness.

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    $\begingroup$ Huh. That's... much easier than I expected a proof to be. $\endgroup$
    – Tyler Lawson
    Commented Nov 14, 2013 at 5:31
  • $\begingroup$ At odd primes, how many relations to you have to impose in an associative algebra beyond $p=0$ to detect $\xi_1$? $\endgroup$
    – Tyler Lawson
    Commented Nov 14, 2013 at 5:36
  • $\begingroup$ @TylerLawson It does seem easy. I guess the same argument shows that if $R$ is an associative ring of "characteristic $\alpha$", then you have $\xi_1$. $\endgroup$
    – Charles Rezk
    Commented Nov 14, 2013 at 15:19
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    $\begingroup$ I asked @TylerLawson on chat to explain the easy argument referred to in the last paragraph, and here's what he said (edited for length): "the cohomology of T_*(M) inherits a coalgebra structure from the ring structure, so there's a map H^*(T_*M) -> H^*(T_*M) ⊗ H^*(T_*M)... however, the Cartan formula tells you that Sq^n(u) maps to Sq^n(u⊗u) = sum Sq^p(u)⊗Sq^q(u) in particular, by induction we know Sq^{n-1}(u)⊗Sq^1(u) is nonzero thus Sq^n(u) must have been nonzero in the first place." $\endgroup$
    – Saul Glasman
    Commented Nov 14, 2013 at 15:49
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    $\begingroup$ Is associativity necessary? It seems to me that this argument shows that any ring spectrum (up to homotopy, and not necessarily associative) with $2 = 0$ cannot be a finite spectrum (which is a more general version of the statement that the mod $2$ Moore spectrum fails to be a ring spectrum). $\endgroup$
    – Akhil Mathew
    Commented Nov 14, 2013 at 17:10

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