Serre's finiteness theorem says if $n$ is an odd integer, then $\pi_{2n+1}(S^{n + 1})$ is the direct sum of $\mathbb{Z}$ and a finite group. By looking at the table of homotopy groups, say on Wikipedia, one empirically observes that if $n \equiv 3 \pmod 4$, then we in fact have $$ \pi_{2n+1}(S^{n+1}) \cong \mathbb{Z} \oplus \pi_{2n}(S^n). $$ This holds for all the cases up to $n = 19$. On the other hand, for $n \equiv 1 \pmod 4$ (and $n \neq 1$), the order of the finite part drops by a factor of $2$ when passing from $\pi_{2n}(S^n)$ to $\pi_{2n+1}(S^{n+1})$.

From the EHP sequence, we know that these two are the only possible scenarios. Indeed, we have a long exact sequence $$ \pi_{2n+2}(S^{2n+1}) \cong \mathbb{Z}/2\overset{P}{\to} \pi_{2n}(S^n) \overset{E}{\to} \pi_{2n+1}(S^{n+1}) \overset{H}{\to}\pi_{2n+1}(S^{2n+1}) \cong \mathbb{Z}. $$ Since $H$ kills of all torsion, one sees that the map $E$ necessarily surjects onto the finite part of $\pi_{2n+1}(S^{n+1})$. So the two cases boil down to whether or not $P$ is the zero map. What we observed was that it is zero iff $n \equiv 3 \pmod 4$, up to $n = 19$.

Since $\pi_{2n+2}(S^{2n+1}) \cong \mathbb{Z}/2\mathbb{Z}$ has only one non-zero element, which is the suspension of the Hopf map, it seems like perhaps one might be able to check what happens to this element directly. However, without a good grasp of what the map $P$ (or the preceeding $H\colon\pi_{2n+2}(S^{n+1}) \to \pi_{2n+2}(S^{2n+1})$) does, I'm unable to proceed.

Curiously, I can't seem to find any mention of this phenomenon anywhere. The closest I can find is this MO question, but this phenomenon is not really about early stabilization, since for $n = 3, 7$, the group $\pi_{2n-1}(S^n)$ is not the stable homotopy group, but something smaller. I'd imagine either this pattern no longer holds for larger $n$, or there is some straightforward proof I'm missing.

Note: Suggestions for a more descriptive title are welcome.

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    $\begingroup$ Shouldn't it be $\pi_{2n+1}(S^{n+1})$ instead of $\pi_{2n}(S^{n+1})$? $\endgroup$ Aug 7, 2017 at 21:52

1 Answer 1


$P(\eta) = [i_n,i_n] \circ \eta$, where $[i_n,i_n]: S^{2n-1} \rightarrow S^n$ is the Whitehead product of the identity map with itself. So you are asking if this composite is null.

I don't know if this is an easy problem. One sufficient condition is that $[i_n,i_n]$ be divisible by 2, but, ha, ha, this is now known to only rarely happen, thanks to the Hill-Hopkins-Ravenel theorem on the Kervaire invariant. (See [HHR, Thm 1.5].) But it does for $n=63$, so $P$ is zero in that case.

I'd search the old literature - papers of Mahowald, Barratt, James and their collaborators - for any general results, if they exist. Mahowald, in particular, has many papers with examples of families of elements on which $H$ acts nonzero.

Added 30 minutes later: [Mahowald, Some Whitehead products in $S^n$, Topology 4, 1965, Theorem 1.1.2(a)] answers your question. $P(\eta)$ is zero if $n \equiv 3 \mod 4$ and is nonzero in basically all other cases.

  • $\begingroup$ Thanks. I'm confused by the wording in the theorem 1.1.2(a). The two conditions presented are "if n ≠ 2ⁱ + 2ʲ - 3 with |i - j| ≠ 1 or n ≠ 3 mod 4" and "if n = 3 mod 4". But surely "n ≠ 3 mod 4" and "n = 3 mod 4" exhaust all possibilities. A brief skimming through the proof seems to suggest he means "and" instead of "or" in the first condition. This interpretation is also consistent with the remarks below the theorem, but it would be hard to tell for sure without actually reading through the proof. $\endgroup$ Aug 8, 2017 at 8:58
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    $\begingroup$ Welcome to the papers of Mark Mahowald. I think this got revisited also in his later Memoirs (from about 1968). $\endgroup$ Aug 8, 2017 at 13:12

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