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Is the stable homotopy group of spheres a commutative ring? If not, are there easy examples?

In the Adams spectral sequence converging to the stable homotopy group of spheres, it seems that any page are with multiplication structures, are these multiplications commutative?

Thanks!

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  • $\begingroup$ Do you know anything about spectra? $\endgroup$ – fixedp Jan 27 '15 at 10:12
  • $\begingroup$ @fixedp Yes, I know a little. For $a,b \in \pi_*(S^0)$, I thought $a \wedge b$ may be different from $b \wedge a$. Do you mean $S^0$ is a commutative ring spectrum? Could you explain a little more? Thanks. $\endgroup$ – user48537 Jan 28 '15 at 1:53
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If we take the basic approach of defining the sphere spectrum as a sequence $\mathbb{S}=(S^0,S^1,...)$ with a monoidal structure given by the smash product, this is only associative and commutative up to homotopy.

From this point of view, the product $[f]\cdot [g]$ in $\pi_*^S$ is anticommutative in the sense that if we have representatives $f:S^{n+i}\to S^n$ and $g:S^{n+i+j}\to S^{n+i}$ for sufficiently large $n$, then

$$[f]\cdot [g] = (-1)^{ij}([g]\cdot [f])$$ due to the interchange of suspension coordinates.

As for your question about $\mathbb{S}$ being commutative (this is just additional info, not really relevant to your original question), you can introduce symmetric spectra (a spectrum with an action of the symmetric group on the coordinates). The symmetric sphere spectrum is commutative with a canonical isomorphism $S^n \wedge S^m \to S^{n+m}$. Here, you get a symmetric monoidal structure at the level of the maps before passing onto homotopy classes.

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