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Thanks to the fibrations

\begin{align*} SO(n) \to SO(n+1) &\to S^n\\ SU(n) \to SU(n+1) &\to S^{2n+1}\\ Sp(n) \to Sp(n+1) &\to S^{4n+3} \end{align*}

we know that

\begin{align*} \pi_i(SO(n)) \cong \pi_i(SO(n+1)) \cong \pi_i(SO), \quad i &\leq n-2\\ \pi_i(SU(n)) \cong \pi_i(SU(n+1)) \cong \pi_i(SU), \quad i &\leq 2n - 1 = (2n+1) - 2\\ \pi_i(Sp(n)) \cong \pi_i(Sp(n+1)) \cong \pi_i(Sp), \quad i &\leq 4n+1 = (4n + 3) - 2. \end{align*}

These values of $i$ are known as the stable range. So the first unstable groups are $\pi_{n-1}(SO(n))$, $\pi_{2n}(SU(n))$, and $\pi_{4n+2}(Sp(n))$ respectively.

I was able to find $\pi_{n-1}(SO(n))$ for $1 \leq n \leq 16$ by combining the tables on the nLab page for the orthogonal group and appendix A, section 6, part VII of the Encyclopedic Dictionary of Mathematics. The groups are

$$0, \mathbb{Z}, 0, \mathbb{Z}\oplus\mathbb{Z}, \mathbb{Z}_2, \mathbb{Z}, 0, \mathbb{Z}\oplus\mathbb{Z}, \mathbb{Z}_2\oplus\mathbb{Z}_2, \mathbb{Z}\oplus\mathbb{Z}_2, \mathbb{Z}_2, \mathbb{Z}\oplus\mathbb{Z}, \mathbb{Z}_2, \mathbb{Z}, \mathbb{Z}_2, \mathbb{Z}\oplus\mathbb{Z}.$$

There doesn't seem to be any pattern here, so I guess that there is no general result for $\pi_{n-1}(SO(n))$. (Feel free to correct me if I'm wrong.) I just noticed that every second term contains a copy of $\mathbb{Z}$, while every fourth term contains two copies.

The case of $SU(n)$ is completely different: in The space of loops on a Lie group, Bott proved, among other things, that $\pi_{2n}(SU(n)) \cong \mathbb{Z}_{n!}$, see Theorem 5.

Again consulting the Encyclopedic Dictionary of Mathematics, I was able to find $\pi_{4n+2}(Sp(n))$ for $n = 1, 2, 3$. The groups are $\mathbb{Z}_{12}$, $\mathbb{Z}_{120}$, and $\mathbb{Z}_{10080}$. This seems to suggest that this case is more similar to $SU(n)$ than $SO(n)$, so one might hope there is a Bott-type result.

Is there an analogue of Bott's result for $Sp(n)$? That is, is there some increasing function $f : \mathbb{N} \to \mathbb{N}$ such that $\pi_{4n+2}(Sp(n)) \cong \mathbb{Z}_{f(n)}$?

OEIS has no sequences beginning $12, 120, 10080$, so I have no guess what $f(n)$ could be. It is interesting to note that $12 \mid 120$ and $120 \mid 10080$ which is another similarity with the $SU(n)$ case.

Of course, three groups is not much to go on, so this may be a completely misguided guess. Some questions that would be nice to answer before seriously hoping for such a result are:

  • Is $\pi_{4n+2}(Sp(n))$ always cyclic?
  • Is $\pi_{4n+2}(Sp(n))$ always finite?
  • Is $|\pi_{4n+2}(Sp(n))|$ increasing in $n$?

Any information regarding these three questions would also be interesting to know.

Falling short of answering any of these questions, have any more of these groups (namely $\pi_{18}(Sp(4)), \pi_{22}(Sp(5)), \dots$) been computed?

Update: I added the sequence $|\pi_{4n+2}(Sp(n))|$ to the OEIS: A301898.

Also, the answer to the question I asked was also in the Encyclopedic Dictionary of Mathematics on page 1746.

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    $\begingroup$ If you tensor the $\pi_{n-1} (SO(n))$ with $\mathbb Q$, it is $4$-periodic, and there is no torsion for $4n$, at least in your data. So there may be some degree of pattern here. $\endgroup$
    – Will Sawin
    Commented Mar 24, 2018 at 19:25
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    $\begingroup$ Follow the boundary map in the edge case. We have $\pi_{4n+3}\text{Sp}(n+1) \to \Bbb Z \to \pi_{4n+2} \text{Sp}(n) \to \pi_{4n+2} \text{Sp}(n+1)$. The other terms are in the computable range: this reduces to $\Bbb Z \to \Bbb Z \to G \to 0$. In particular $G$ is cyclic and computing its order is equivalent to computing the image of that boundary map. $\endgroup$
    – mme
    Commented Mar 24, 2018 at 19:39
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    $\begingroup$ Please feel free to extend the nLab table, Michael. I think it's important to have such data freely available, rather than locked away in the EDM, or scattered across multiple papers. $\endgroup$
    – David Roberts ♦
    Commented Mar 24, 2018 at 23:32
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    $\begingroup$ The metastable groups $\pi_{4n+i}Sp(n)$, $i=2,\dots, 8$ are stated, with references, in Mimura's article "Homotopy Theory of Lie Groups", which appears in the Handbook of Algebraic Topology. I'm also aware that Kaoru Morisugi looked at the 2-primary homotopy of $Sp(n)$ above the metastable range, giving (pieces of) information up to $\pi_{4n+15}Sp(n)$ and $\pi_{8n+4}Sp(n)$. $\endgroup$
    – Tyrone
    Commented Mar 27, 2018 at 12:10
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    $\begingroup$ I just wanted to add that $\pi_{2k}(G)\otimes \mathbb{Q} = 0$ for any $k> 0$ and any Lie group $G$. To see this, one can reduce to connected and compact groups using the fact that a non-compact connected Lie group is topologically a product of $\mathbb{R}^n$ with a compact Lie group. Every compact Lie group is know to have the rational homotopy type of a product of spheres odd dimension (and we know exactly what spheres appear for each simple group), so we know how to compute the rational homotopy groups of any Lie group. $\endgroup$
    – Jason DeVito - on hiatus
    Commented Nov 9, 2018 at 22:08

2 Answers 2

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The answer appears to be in the paper Homotopy groups of symplectic groups by Mimura and Toda. They claim the calculation was already in a paper of Harris, but that was stated in terms of a symmetric space and it's not immediately obvious to me how to translate into information about the groups.

They state that the group is $\mathbb Z_{(2n+1)!}$ if $n$ is even and $\mathbb Z_{(2n+1)! \cdot 2}$ if $n$ is odd, which agrees with your data.

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    $\begingroup$ Maybe this is a good justification for that sequence to be on OEIS. $\endgroup$
    – mme
    Commented Mar 24, 2018 at 19:41
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    $\begingroup$ @MikeMiller: The sequence is now on the OEIS, see A301898. $\endgroup$
    – Michael Albanese
    Commented May 16, 2018 at 15:44
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The first unstable homotopy groups of $SO(n)$ are actually 8-periodic (except for some junk at the beginning). Some more unstable homotopy groups of $SO(n)$ can be found in:

The 8-periodicity for the orthogonal group comes about as follows: the relevant piece of the stabilization sequence is $$ \pi_n S^n\to \pi_{n-1}SO(n)\to \pi_{n-1}SO(n+1)\to 0. $$ The unstable homotopy groups $\pi_{n-1}SO(n)$ are then direct sums of the stable stuff from $\pi_{n-1}SO(\infty)$ plus a cyclic quotient of $\pi_n S^n\cong \mathbb{Z}$. The 8-periodicity effectively comes from the stable summand (check the list of homotopy groups of the infinite orthogonal group). The cyclic quotient of $\pi_n S^n$ is only 2-periodic, alternating between $\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$; I think this basically comes from the corresponding Euler class of the sphere alternating between 2 and 0.

The description of the unstable homotopy of the symplectic groups given in Will Sawin's answer can also be found in

  • B. Harris. Some calculations of homotopy groups of symmetric spaces. Trans. Amer. Math. Soc. 106 (1963), 174-184. (link to journal website)
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    $\begingroup$ Do you have any intuition for why the first unstable homotopy groups of $SO(n)$ act so differently to the first unstable homotopy groups of $SU(n)$ and $Sp(n)$? I find this surprising because in the stable case, $SO$ and $Sp$ are related, and $SU$ is the isolated one. $\endgroup$
    – Michael Albanese
    Commented Mar 26, 2018 at 15:56
  • $\begingroup$ @Michael If you try to follow with my comment in the O(n) case, the exact sequence’s left term is the unstable homotopy group one dimension up instead of Z (which came from the stable range). This happens because the n in the sphere’s dimension grows at the same rate as the n in O(n). $\endgroup$
    – mme
    Commented Mar 27, 2018 at 1:20

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