Thanks to Freudenthal we know that $\pi_{n+k}(S^n)$ is independent of $n$ as soon as $n \ge k+2$. However, I was looking at the table on Wikipedia of some of the homotopy groups of spheres and noticed that $\pi_2^S$ (the second stable stem) and $\pi_6^S$ are achieved earlier than required by the suspension theorem. So my question is:

What is known about this phenomenon of early stabilization in the homotopy groups of spheres? Does it occur finitely many times or infinitely many times? Also, is there any sort of intuitive reason why stabilization might occur early?

Also, I spotted a few times where stabilization almost appeared, but there was a pesky copy of $\mathbb{Z}$ that appeared and then disappeared right before the stable range. I'm talking about the cases of $\pi_{11+n}(S^n)$, $\pi_{15+n}(S^n)$, and $\pi_{19+n}(S^n)$. I assume this has something to do with Hopf elements showing up somewhere?

I know pretty much nothing about this except the basic homotopy theory, a few very small calculations, and the Pontrjagin construction relating all of this to framed manifolds- so any references or illuminating insights would be helpful!

  • $\begingroup$ Another close call is $\pi_{9+n}(S^n)$. $\endgroup$ – Dylan Wilson Dec 26 '10 at 22:26

Nice question! Funny enough, the answer is that you've already found all of the examples of early stabilization (excepting, of course, the fact that you didn't mention $\pi_n S^n$ stabilizing early). This is true even if you ignore odd torsion.

The "pesky copy of ℤ" is indeed related to the Hopf invariant. The fact that these classes exist on the edge of the stable range goes back to Serre's work where he shows exactly which homotopy groups of spheres contain a free summand.

More generally, though, the way one can see what's going on at the edge of the stable range is to use the EHP sequence - see, for example, http://web.math.rochester.edu/people/faculty/doug/mypapers/ehp.pdf. For any integer $n > 0$, there is (2-locally) a homotopy fiber sequence $$ S^n \to \Omega S^{n+1} \to \Omega S^{2n+1} $$ and this gives rise to a long exact sequence of homotopy groups (after 2-localization) $$ \cdots \to \pi_{2n} S^n \mathop\to^E \pi_{2n+1} S^{n+1} \mathop\to^H \pi_{2n+1} S^{2n+1} \mathop\to^P \pi_{2n-1} S^n \mathop\to^E \pi_{2n} S^{n+1} \to 0. $$ Here the maps labelled "E" are for suspension, the maps labelled "H" are for the Hopf invariant, and the maps labelled "P" are for something related to a Whitehead product. We know the middle term, so we can rewrite this: $$ \cdots \to \pi_{2n} S^n \mathop\to^E \pi_{2n+1} S^{n+1} \mathop\to^H \mathbb{Z} \mathop\to^P \pi_{2n-1} S^n \mathop\to^E \pi_{2n} S^{n+1} \to 0. $$ The last map on the right is the "edge" of the stable range and so you have early stabilization if and only if this map is an isomorphism, or equivalently if the map "P" is zero. The map "P" is zero if and only if the map "H" is surjective.

However, "H" in this case really is the classical Hopf invariant: if you have an element $f:S^{2n+1} \to S^{n+1}$ viewed as an attaching map, the element $Hf$ detects which element of $H^{2n+2}$ is the square of the generator of $H^{n+1}$ in the space $Cf$ obtained by using $f$ to attach a cell. J.F. Adams proved that the only time $Hf$ can take the value 1 is if $n+1$ is 1, 2, 4, or 8. So the only time you could have early stability is when the stable stem $n-1$ is -1, 0, 2, or 6. (EDIT: Had an indexing error at the end. Sorry.)

  • $\begingroup$ Wonderful! I'm glad this had a satisfying answer :) Thank you! $\endgroup$ – Dylan Wilson Dec 27 '10 at 6:32
  • $\begingroup$ Also- I guess we get a "close call" whenever P is an injection, i.e. when H is zero. Do we know when the Hopf map is zero? $\endgroup$ – Dylan Wilson Dec 27 '10 at 6:44
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    $\begingroup$ Sure. With a little more work, you can show that: the exact sequence at the edge that I listed exists integrally rather than just 2-locally; the map P is an injection precisely when n is even and it hits a direct summand except in the cases n=2,4,8; and the image of P is of order exactly 2 when n is odd and not equal to 1, 3, or 7. $\endgroup$ – Tyler Lawson Dec 27 '10 at 14:14

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