Question

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

Answer 1

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.