The Hopf fibration $\eta:S^3\to S^2$ represents the generator of the first stable homotopy group $\pi_1^s$.

The direct sum of the stable homotopy groups $$ \pi_*^s=\bigoplus_{k\in\mathbb{Z}} \pi_k^s $$ can be given a ring structure and $\pi_*^s$ is called the ring of stable homotopy groups. Several possibilities (listed below) of introducing such a multiplication come to my mind. My question is:

What is the correct multiplication in $\pi_*^s$? In particular, what is $\eta^2\in \pi_2^s$? Are the other possibilities listed below different and, if not, do they introduce an additional structure on $\pi_*^s$ with a name in the literature?

  1. This possibility is what I think is ''the right'' one: $$ S^4\xrightarrow{\Sigma \eta} S^3\xrightarrow{\eta} S^2 $$ represents an element of $\pi_2^s$ which may be called $\eta^2$.
  2. But how about this? Just ''smashing'' $\eta$ with itself $$ S^6\cong S^3\wedge S^3\xrightarrow{\eta\wedge \eta}S^2\wedge S^2\cong S^4 $$ represents also an element of $\pi_2^s$.
  3. In view of this other question, how about $$ S^3\xrightarrow{\Delta} S^3\wedge S^3\xrightarrow{\eta\wedge\eta} S^2\wedge S^2\cong S^4 $$ but this map is (homotopic to) the zero map, right? What is the relation to the multiplication of the other question then? Since $\pi_*^s=\bigoplus_{k\in\mathbb{Z}}[S^k,\mathbf{S}]=\bigoplus_{k\in\mathbb{Z}}[S^0,S^{-k}\wedge \mathbf{S}]=\mathbf{S}^*(*)$ where $\mathbf{S}$ is the sphere spectrum, there should be a relation to the other multiplication $\star$.
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    $\begingroup$ 1 and 2 give homotopy equivalent answers. In general, there are two ways of defining multiplication in the stable homotopy groups of spheres... both agree by an Eckmann-Hilton argument. $\endgroup$ – Dylan Wilson Apr 11 '13 at 14:19
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    $\begingroup$ Dylan is exactly right. Note that #3 can't be correct, since your first map is really $S^3 \rightarrow S^6$, and this is necessarily nullhomotopic. You might be interested in checking out (at least the introduction of) the excellent and foundational paper "Nilpotence I" by Devinatz--Hopkins--Smith; in the intro, they talk about the various definitions of multiplication in $\pi_*^{st}$, Nishida's nilpotence theorem, and some pretty neat generalizations. $\endgroup$ – Aaron Mazel-Gee Apr 11 '13 at 18:15
  • $\begingroup$ Dear @DylanWilson, thank you for the comment. I would like to accept it as an answer. Thank you for the reference, Aaron. $\endgroup$ – Ronald Bernard Apr 13 '13 at 14:59

Wikipedia seems to give the Hopf fibration as an example in the definition of the ring structure on the stable sphere homotopy ring and it seems to correspond to the composition with the suspension of itself (ie the composition of two different, composable representations of the same ring element), as you guessed in your first point. Nishida's Theorem then says that any element not in $\pi_0^s$ is nilpotent and so all prime ideals of $\pi_*^s$ are those in the zeroth grading making the ring rather complicated.

In general then, it seems that if you want to multiply two elements $[\mu]\in\pi_p^s,\:\:[\lambda]\in\pi_q^s$ with representatives $\mu\colon S^{n+p}\rightarrow S^{n},\:\:\lambda\colon S^{m+q}\rightarrow S^m$ then WLOG assuming $n\leq m+q$, a representative of the product is given by $\lambda\circ(\Sigma^{m+q-n}\mu)\colon S^{m+p+q}\rightarrow S^m$ and so $$[\mu][\lambda]=[\lambda\circ(\Sigma^{m+q-n}\mu)]\in\pi_{p+q}^s$$ which corresponds to the given grading.

I'm not sure about your other maps but they don't seem as natural as the plain composition.

  • $\begingroup$ Dear @Daniel Rust, thank you for the answer. For me, the second possibility is even more natural: In your example $[\mu\wedge \lambda]\in \pi_{p+q}^s$ is represented by $$S^{n+p}\wedge S^{m+q}\xrightarrow{\mu\wedge\lambda} S^{n+m}.$$ $\endgroup$ – Ronald Bernard Apr 11 '13 at 13:13
  • $\begingroup$ Ah I hadn't noticed that. It is a rather nice way of representing the product. $\endgroup$ – Dan Rust Apr 11 '13 at 14:06

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