What is the group of pointed homotopy classes of maps from $S^3 \times S^3$ to $S^3$? The group structure induced by the group structure on the codomain. This question is a followup to Eric's answer to another question.

Added later: The commutator map $(g,h) \mapsto ghg^{-1} h^{-1}$ descends to a map from $S^6$ to $S^3$. Which element of $\pi_6(S^3) = \mathbf{Z}/12$ is this?

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    $\begingroup$ The element of $\pi_6(S^3)$ you describe is known as the Blakers-Massey element. $\endgroup$ – Mark Grant Mar 14 '14 at 7:48

We have that $S^3 \simeq \Omega \mathbb{H}P^{\infty}$, so by adjointness we can as well consider the group of maps $[\Sigma S^3 \times S^3, \mathbb{H}P^{\infty}]$. It is well-known that $[X,\Omega Y] \simeq [\Sigma X, Y]$ are isomorphic as groups, when you define the group structure on the first one by loop composition and on the second one by the pinching map/"cogroup structure" $\Sigma X \rightarrow \Sigma X \wedge \Sigma X$.

Now it can be proved (for example in Hatcher, section 4.I), that $\Sigma (S^3 \times S^3) \simeq S^4 \vee S^4 \vee S^7$.

This suggests $$ [S^3 \times S^3, S^3] \simeq [S^4, \mathbb{H}P^{\infty}] \times [S^4, \mathbb{H}P^{\infty}] \times [S^7, \mathbb{H}P^{\infty}] \simeq \mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}/12 $$

but only as sets! This is because the splitting above is unnatural: There is a cofiber sequence $$ \Sigma (S^3 \vee S^3) \rightarrow \Sigma (S^3\times S^3) \rightarrow \Sigma (S^3\wedge S^3) $$ and this has a retraction, though not a very natural one, and apparently this cannot be retracted by a cogroup homomorphism. However, the splittability of this thing gives you already a short exact sequence $$ 0 \rightarrow \pi_6(S^3) \rightarrow [S^3\times S^3,S^3] \rightarrow \pi_3(S^3)\times \pi_3(S^3) \rightarrow 0 $$ by translating back via adjointness.

The splitting is obtained in the following way: More precisely, you can take the projections $\Sigma(S^3\times S^3) \xrightarrow{\Sigma p_k}\Sigma S^3$ for $k=1,2$ as well as the quotient map $\Sigma (S^3\times S^3) \xrightarrow{\Sigma \pi} \Sigma (S^3\wedge S^3)$ and add them together to obtain a homotopy equivalence $$\Sigma(S^3\times S^3) \rightarrow \Sigma S^3 \vee \Sigma S^3 \vee \Sigma (S^3\wedge S^3) $$ Now each of those is a nice cogroup homomorphism, but adding them together gives no cogroup homomorphism anymore since addition is not commutative here.

This gives a bijection $$ \pi_4(\mathbb{H}P^{\infty}) \times \pi_4(\mathbb{H}P^{\infty}) \times \pi_7(\mathbb{H}P^{\infty})\rightarrow [\Sigma S^3 \times S^3, \mathbb{H}P^{\infty}] $$ given explicitly by $(\alpha,\beta,\gamma) \mapsto (\alpha \circ \Sigma p_1)\bullet( \beta \circ \Sigma p_2)\bullet (\gamma \circ \Sigma \pi) $.

This in turn translates to a bijection $$ \pi_3(S^3) \times \pi_3(S^3) \times \pi_6(S^3) \rightarrow [S^3\times S^3, S^3] $$ given explicitly by $(\alpha, \beta, \gamma) \mapsto (\alpha \circ p_1)\bullet(\beta\circ p_2)\bullet(\gamma\circ \pi)$.

In particular, since precomposition induces a group homomorphism, we only need to figure out the commutators of $p_1, p_2$ and $\nu' \circ \pi$, where $\nu'$ is a generator of $\pi_5(S_3)$.

In Hatcher's book project "spectral sequences in algebraic topology", he mentions on page 67 of the part about the Serre spectral sequence the following (sadly without proof or reference): A generator of $\pi_6(S^3)$ can be constructed by considering the commutator map $S^3 \times S^3 \rightarrow S^3$ sending $(x,y)\mapsto xyx^{-1}y^{-1}$. This is constant when restricted to the $3$-skeleton, so it induces a map $S^3\wedge S^3 \rightarrow S^3$. According to him, this is a generator.

For us, this means that $[p_1,p_2] = \nu'\circ \pi$ (when we take $\nu'$ to be the map constructed above).

EDIT: I think we can obtain that $p_1,p_2$ commute with $\nu'\circ\pi$ as follows: Since there is an automorphism that exchanges $p_1$ and $p_2$ (given by flipping the factors of $S^3\times S^3$, it suffices to check this for $p_1$. Now $[p_1,\nu'\circ \pi] = [p_1,[p_1,p_2]]$ so we can write this as as the composition $$ S^3\times S^3 \xrightarrow{\Delta\times id} S^3 \times S^3 \times S^3 \xrightarrow{id\times [,]} S^3\times S^3 \xrightarrow{[,]} S^3 $$ However, since $[a,[b,c]] = 1$ whenever $a=1$ or $b=1$, the triple commutator $S^3\times S^3 \times S^3\rightarrow S^3$ factors through $(S^3\wedge S^3)\times S^3$. But since $\Delta: S^3 \rightarrow S^3\wedge S^3$ is homotopic to zero, the whole composition will be homotopic to zero now. More precisely, $[p_1,[p_1,p_2]] \simeq [1,[1,p_2]] = 1$.

This finally tells us: The extension $$ 0\rightarrow \pi_6(S^3) \rightarrow [S^3\times S^3, S^3] \rightarrow \pi_3(S^3) \times \pi_3(S^3) \rightarrow 0 $$

is central with $[p_1,p_2] = \nu'\circ \pi$.

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  • $\begingroup$ Nice answer. I gues you mean $HP^{\infty}$, not $HP^4$. $\endgroup$ – Johannes Ebert Mar 14 '14 at 7:28
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    $\begingroup$ The fact that the commutator map gives a generator of $\pi_6(S^3)$ is proved by I.M.James in "On H-spaces and their homotopy groups", Quart. Journal Math. 11 (1960) 161-179, on page 176.(qjmath.oxfordjournals.org/content/11/1/161.full.pdf) The main point is that this element (the Blakers-Massey element) is the image under the $J$-homomorphism of a generator of $\pi_3(SO(3))$. $\endgroup$ – Gustavo Granja Mar 14 '14 at 11:09

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