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 $p=2$, 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_*^2$, \pi_{14}^s$, 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

2 apparently it doesn't like {*,*} in math mode...

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 $p=2$, 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 $\pi_*^2$, 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 $h_3^2h_0^2=0\in E_2^{h_3^2h_0^2=0 \in E_2$, }$, but why must it have order 2 exactly? What does the odd degree have to do with it? If I am not mistaken, the Yoneda product on$Ext_A^{,}(Z/2,Z/2)$Ext_A(Z/2,Z/2)$ 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

1

# How do you know when something must die in the Adams Spectral Sequence for $\pi_*^s$

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 $p=2$, 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 $\pi_*^2$, 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