Cohomotopy group of $[\Sigma^7 Q_2, S^7]$

Question

Let $Q_2$ be the symplectic quasi-projective space of rank2. How to calculate 3-local of cohomotopy group $[\Sigma^7 Q_2, S^7]$. Thank you.

Answer 1

Your space $Q=Q_2$ is the cofibre of some map $\alpha\colon S^6\to S^3$, so $\alpha$ lies in the group $\pi_6(S^3)$, which is isomorphic to $\mathbb{Z}/12$ (see https://en.wikipedia.org/wiki/Homotopy_groups_of_spheres and the references cited there). We therefore have a cofibration sequence $$ S^{13} \xrightarrow{\Sigma^7\alpha} S^{10} \to \Sigma^7Q \to S^{14} \xrightarrow{\Sigma^8\alpha} S^{11}, $$ giving rise to an exact sequence $$ \pi_{11}S^7 \xrightarrow{(\Sigma^7\alpha)^*} \pi_{14}S^7 \xrightarrow{} [\Sigma^7Q,S^7] \to \pi_{10}S^7 \xrightarrow{(\Sigma^8\alpha)^*} \pi_{13}S^7 $$ Inserting the calculations of the above homotopy groups, we get an exact sequence $$ 0 \xrightarrow{} (\mathbb{Z}/120)\sigma' \xrightarrow{} [\Sigma^7Q,S^7] \to (\mathbb{Z}/24)\nu \xrightarrow{(\Sigma^8\alpha)^*} (\mathbb{Z}/2)\nu^2 $$ At the prime $2$ we see from Lemma 5.4 of Toda's "Composition methods in homotopy groups of spheres" that $\pi_6(S^3)$ is generated by $\nu'$ with $\Sigma^2(\nu')=2\Sigma\nu$. It follows that $\Sigma^8\alpha$ is divisible by $2$, which forces the above map $(\Sigma^8\alpha)^*$ to be zero. Thus, the first three terms above form a short exact sequence. This means that the $2$-primary part of $[\Sigma^7Q,S^7]$ is $\mathbb{Z}/8\oplus\mathbb{Z}/8$ or $\mathbb{Z}/16\oplus\mathbb{Z}/4$ or $\mathbb{Z}/32\oplus\mathbb{Z}/2$ or $\mathbb{Z}/64$, and the $3$-primary part is $\mathbb{Z}/3\oplus\mathbb{Z}/3$ or $\mathbb{Z}/9$, and the $5$-primary part is $\mathbb{Z}/5$. It may be possible to use Steenrod operations or Adams operations to resolve the extension problem, but I have not attempted that.