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

we see that the homotopy groups are:




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.


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.

  • $\begingroup$ thanks great +1. $\endgroup$ – wonderich Aug 13 '18 at 23:19
  • $\begingroup$ but somehow your argument cannot be directly applied to $SU(N)$ group or $SO(N)$ group, $$SU(n+1)/SU(n)=S^{2n+1}?$$ $$SO(n+1)/SO(n)=S^n?$$ So there could be similar statements one can make? $\endgroup$ – wonderich Aug 14 '18 at 1:29
  • 2
    $\begingroup$ @wonderich They show, respectively, that $\pi_k \text{Sp}(n)$ stabilizes for $k \leq 4n +1$, $\pi_k \text{SU}(n)$ stabilizes for $k \leq 2n-1$, and $\pi_k SO(n)$ for $k \leq n-2$. $\endgroup$ – user98602 Aug 14 '18 at 1:44
  • $\begingroup$ thanks - this is cool. $\endgroup$ – wonderich Aug 14 '18 at 1:46

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