18
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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 Mar 24 '18 at 19:25
  • 2
    $\begingroup$ 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. $\endgroup$ – Mike Miller Mar 24 '18 at 19:39
  • 1
    $\begingroup$ 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 Mar 24 '18 at 23:32
  • 1
    $\begingroup$ 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 Mar 27 '18 at 12:10
  • 1
    $\begingroup$ 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 Nov 9 '18 at 22:08
25
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ Maybe this is a good justification for that sequence to be on OEIS. $\endgroup$ – Mike Miller Mar 24 '18 at 19:41
  • 1
    $\begingroup$ @MikeMiller: The sequence is now on the OEIS, see A301898. $\endgroup$ – Michael Albanese May 16 '18 at 15:44
23
$\begingroup$

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)
$\endgroup$
  • 2
    $\begingroup$ 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 Mar 26 '18 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). $\endgroup$ – Mike Miller Mar 27 '18 at 1:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.