Let $S/2$ be the mod 2 Moore spectrum (i.e. the cofiber of $2: S \to S$). Then multiplication by 2 acts nontrivially on this spectrum: the homotopy groups of $S/2$ are all $\mathbb{Z}/4$-modules by a formal argument, but not $\mathbb{Z}/2$-modules. For instance, $\pi_2(S/2) = \mathbb{Z}/4$. The long exact sequence in homotopy groups enables one to determine the homotopy groups of $S/2$ (in the range that the stable homotopy groups of spheres are known) up to extension, but the extension problems are generally nontrivial: $\pi_2(S/2)$ is a case in point (as the aforementioned exact sequence $0 \to \mathbb{Z}/2 \to \pi_2(S/2) \to \mathbb{Z}/2 \to 0$ shows).

Is there a general technique for resolving these kinds of extension problems? I can get it in this case using some computation in the cobar complex to get the $t-s = 1$ line of the Adams spectral sequence for $S/2$ and see that the Bockstein acts nontrivially. I'm curious about a more efficient method of doing this. (Another example I had in mind was $bo \wedge \mathbb{CP}^2$: this is $bu$ by the "theorem of Reg Wood" and this is visible in the ASS, but directly computing $KO$-groups of complex projective spaces yields nontrivial extension problems, I think.)

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    $\begingroup$ An alternative technique: Consider $(S/2)/2$, the cofiber of multiplication by 2 on $S/2$ (and is equivalent to $S/2 \wedge S/2$). You get two different answers for what $\pi_2$ of this should be depending on how your extension problem was solved, and you have a nontrivial $Sq^2$ in its cohomology which should tell you that one of those answers is wrong. $\endgroup$ – Tyler Lawson Aug 23 '13 at 3:47

This is a comment, not an answer, I suppose. Just a reference to Adams and Walker "On complex Stiefel manifolds''. This follow up to Adams' "Vector fields on spheres'' directly computes the $KO$-groups of complex projective spaces (see Theorem 2.2) by the methods of VFS, which computed the complex $K$-theory of complex and real projective spaces and the real $K$-theory of real projective spaces. There are no extension problems in sight in the calculation of $KO(\mathbf{C}P^n)$.

Aside from the obvious, multiplication by $h_0$ in the ASS, method of detection of multiplication by $2$, there aren't a whole lot of systematic methods for detecting multiplication by $2$, let alone less simple extensions. Massey products/Toda brackets can help.

  • $\begingroup$ Thanks. I'll take a look at Adams's paper -- I had only seen VFS. $\endgroup$ – Akhil Mathew Jun 22 '12 at 20:57
  • $\begingroup$ Unless I'm mistaken, there is an extension problem in calculating $KO_∗(\mathbb{CP}^2)$ (specifically, there's an extension of $\mathbb{Z}$ and $\mathbb{Z}/2$ for $KO_6(\mathbb{CP}^2)$ in the AHSS). $\endgroup$ – Akhil Mathew Jun 22 '12 at 21:03
  • $\begingroup$ (It seems the paper does $KO^*$ rather than $KO_*$.) $\endgroup$ – Akhil Mathew Jun 22 '12 at 21:06
  • $\begingroup$ I should have realized you meant KO homology, but you just wrote KO, and so did I. $\endgroup$ – Peter May Jun 23 '12 at 1:05
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    $\begingroup$ I should have realized you meant KO homology, but you just wrote KO, and so did I. $\endgroup$ – Peter May Jun 23 '12 at 1:06

Often these extension problem are solved using Toda brackets (as Peter May already mentioned). I will first give the general statement, not only for spectra but also for module spectra. The statement may sound a bit complicated, but is quite useful.

Let $R$ be a strictly associative ring spectrum. Let $x\in \pi_n R$ be an element in the coefficients and denote by $Cx$ the cone of $\Sigma^n R \xrightarrow{x} R$. Then we have a long exact sequence \begin{eqnarray*}\cdots\to\pi_* \Sigma^n R\to \pi_* R \to \pi_* Cx \to \pi_{*-1} \Sigma^n R\to \pi_{*-1}R\to\cdots\end{eqnarray*} which splits into short exact sequences of the form \begin{eqnarray*}0\to \pi_* R/x\pi_*R \xrightarrow{\alpha} \pi_* Cx \xrightarrow{\beta} \{\pi_{*-n-1}R\}_x\to 0\end{eqnarray*} where $\{\pi_{*-n}R\}_x$ denotes all elements which are annihilated by $x$.

Let $y\in \pi_m R$ and $z\in \pi_k R$ be elements in the coefficients of $R$ with $xy=0$ and $yz=0$. Let $\widetilde{y} \in \pi_*Cx$ be an element with $\beta(\widetilde{y}) = y$. Let $w\in \pi_*R$ be an element such that the projection of $w$ is mapped to $\widetilde{y}z$ under $\beta$. Then $w\in \langle x,y,z\rangle$.

Back to your example: We take $R = \mathbb{S}$ and $x = 2\in\pi_0 \mathbb{S}$. We have now a short exact sequence $$ 0 \to \pi_2\mathbb{S} \to \pi_2 \mathbb{S}/2 \to \pi_1\mathbb{S} \to 0 $$ where the outer groups are isomorphic to $\mathbb{Z}/2$ and are generated by $\eta^2$ and $\eta$, respectively. Let now $y = \eta$ and $z =2$. Lift $\eta$ to an element $\widetilde{\eta} \in \pi_2 \mathbb{S}/2$. Then the statement above tells us that $2\cdot \widetilde{\eta}$ is in the image of $\langle 2, \eta, 2 \rangle = \eta^2$. In particular, $2\cdot \widetilde{\eta}$ is non-zero. This solves the extension problem.

This technique can be applied to many cases, including $KO\wedge C\eta$. I did some calculations in the category of $TMF$-modules with this.

The problem remains how to compute Toda brackets. In general, this might be difficult, but often methods to compute $\pi_*R$ give also methods to compute the Toda brackets in $\pi_* R$. For example, Massey products in the $E^2$-term of the Adams spectral sequence converge to Toda brackets (if there are no "crossing differentials"). The Massey product $\langle 2,\eta, 2\rangle$ can, for example, be computed via cobar representatives.

Regarding references: The statement about Toda brackets and cofiber sequences follows more or less directly from the definition of Toda brackets in the framework of triangulated categories. See for example, Section 4.6 of my Thesis -- although I was a little bit lazy there with signs. Regarding the convergence of Massey products to Toda brackets one finds a statement in Kochman's Bordism, Stable Homotopy, and Adams Spectral Sequences. Note that he uses another definition of Toda bracket, which agrees with the one in triangulated categories in the case of triple brackets (and also else if one ignores indetermenancy) as proven in some paper by Kochman.

  • $\begingroup$ Interesting! Thanks for indicating the connection with tmf. $\endgroup$ – Akhil Mathew Aug 23 '13 at 22:20
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    $\begingroup$ Toda brackets. This is a favorite question of mine for students: Show there is no complex with cohomology exactly {x, Sq^1 x , Sq^2Sq^1x, Sq^1Sq^2Sq^1x}. This is instant from the Adem relation for Sq^2 Sq^2. Then this implies 0 \notin <2,eta,2>. Then the attempt to attach the top cell must fail and this implies \pi_2 of the Moore space is Z/4. Now I can't use this question again :-) $\endgroup$ – Robert Bruner Mar 21 '15 at 12:06

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