$\pi_3(Sp(N))=\mathbb{Z}$, $\pi_4(Sp(N))=\mathbb{Z}_2$, $\pi_5(Sp(N))=\mathbb{Z}_2$?

Question

From the computation of some lower dimension $N$ of $Sp(N)$ group,

we see that the homotopy groups are:

$\pi_3(Sp(N))=\mathbb{Z}$,

$\pi_4(Sp(N))=\mathbb{Z}_2$,

$\pi_5(Sp(N))=\mathbb{Z}_2$,

at least for $N=1,2,3,4,5.$

Question: Are these results generally true for any $N$? If so, are there some simple explanations and intuitions behind?

• One attempt may be using the Bott periodicity theorem.

• What else topological properties of $Sp(N)$ to gain us some better intuitions/explanations?

p.s. My notations may be different from Stable homotopy groups of $Sp(2n)$, the above $Sp(N)$ means $Sp(2n)$ there.

Answer 1

Remember the fibration $\text{Sp}(n) \to \text{Sp}(n+1) \to S^{4n+3}$, where the fiber is the stabilizer of the transitive action of $\text{Sp}(n+1)$ on the unit sphere of $\Bbb H^{n+1}$.

Running the long exact sequence in homotopy groups, along with the fact that $\pi_k S^{4n+3} = 0$ for all $k < 7$ and $n \geq 1$, shows that the map $\text{Sp}(n) \to \text{Sp}(n+1)$ induces an isomorphism on $\pi_k$ for all $n \geq 1$ and $k \leq 5$. Thus you know the first 5 homotopy groups of all symplectic groups if you know them for $\text{Sp}(1)$.

Now $\text{Sp}(1) = S^3$, the group of unit quaternions. By reading off a table of the first 5 homotopy groups of $S^3$ we see that your list of the first 5 homotopy groups of $\text{Sp}(n)$ is correct.