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Hey everybody,

I think this question might be just a simple oversight on my part, but this has been bugging me a few days.

I am reading Hatcher's Spectral Sequences book, and trying to understand his example where he computes $\pi_*^s$ for $p=2$ (page 21-23), and I'm a bit confused about a certain step. He claims that the element corresponding to $h_3^2$ must have order 2 in $\pi_{14}^s$, because of "the commutativity property of the composition product, since $h_3$ has odd degree". Now, I see why $h_3^2$ can have order at most 4, because $h_3^2h_0^2=0 \in E_2$, but why must it have order 2 exactly? What does the odd degree have to do with it? If I am not mistaken, the Yoneda product on $Ext_A(Z/2,Z/2)$ induces the composition product on $\pi_*^s$, which, mod 2, is commutative, but the Yoneda product has $h_3h_0=h_0h_3$ in the $E_2$ page, so I can't from that derive the induced composition product is 0. Do I need to use a fact about $\pi_s^*$ that doesn't come from this spectral sequence?

Thanks for the help everybody! -Joseph Victor

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The point is that the element $h_3$ in odd degree lifts to an element (typically called $\sigma$) in $\pi_7^s$ whose square must be 2-torsion. This is true for all elements in odd degree because the stable homotopy groups of spheres are graded-commutative. It does not necessarily have to be exactly 2-torsion a priori - it could be 1-torsion (i.e. trivial) - but it must be annihilated by 2.

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  • $\begingroup$ Thank you, but I guess thats a fact about $\pi^s_*$ I didn't know. (I am by no means an expert). Do you know of a good reference for these facts? $\endgroup$ – Joseph Victor Jul 7 '12 at 7:16
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    $\begingroup$ You can find a proof in Hatcher's Algebraic Topology book - Proposition 4.56 on pp. 385 $\endgroup$ – Drew Heard Jul 7 '12 at 8:58

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