I got this very nice answer to one of my previous questions to which I have the following follow up question:

What I understand, or let's say, think I understand so far is: one main problem of interest in algebraic topology is to determine $\pi_n (S^m)$ for all $n$ and all $m$ $\geq 0$. We know some cases, for example we know that for $m > n$, $\pi_n (S^m) = \{0\}$ and for $n = m$, $\pi_n(S^m) = \mathbb Z$. We also know, due to the Freudenthal suspension theorem, that $\pi_{n + k}(S^n) \cong \pi_{n + k + 1}(S^{n + 1})$ for $n > k + 1$. We also know some other combinations of $n$ and $m$ but we have not managed to determine all of them.

Quite recently, in 2009, the following theorem, known as the Kervaire Invariant One problem was proved:

The Kervaire invariant $\kappa : \Omega_{4k + 2}^{fr} \to \mathbf{Z}_2$ is trivial unless $4k + 2 = 2,6,14,30,62$.

Here $\Omega_{4k + 2}^{fr}$ denotes the group of cobordism classes of framed $4k + 2$ manifolds.

Question: How does that let me compute $\pi_n(S^m)$? When I write compute, I really mean "gather information". It would be good enough to show that there are no surjective maps $S^n \to S^m$ then we'd have $\pi_n(S^m) = \{0\}$. But I don't see how to use the Kervaire invariant to gather any information about $\pi_n(S^m)$. Thanks for your help.

  • 1
    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ Sep 8, 2012 at 20:11
  • $\begingroup$ The Kervaire problem isn't completely solved: it might still be nontrivial in dimension 126! $\endgroup$ Sep 8, 2012 at 23:13
  • $\begingroup$ Knowing the solution of the Kervaire invariant problem is a crucial step for knowing the group of smooth structures on a sphere (en.wikipedia.org/wiki/…). The last chapters of Kosinski's differential manifolds treat this problem. Once the solution for the Kervaire invariant problem is given, the problem of determining the group of smooth structures reduces to a calculation of homology groups of spheres, which is a problem in its own. $\endgroup$
    – user17786
    Sep 9, 2012 at 2:35
  • $\begingroup$ @AaronMazel-Gee I know : ) $\endgroup$ Sep 9, 2012 at 7:01

2 Answers 2


Perhaps it's worth mentioning how exactly homotopy theorists think about the Kervaire invariant (at least I think - I'm not really one, but this is my impression!). First of all, as Qiacohu notes, there is an isomoprhism between the stable homotopy groups $\pi_k^s(S^0)$ and the group of cobordism classes of framed $k$-manifolds. This is why the Kervarire invaraint has something to do with stable homotopy groups of spheres.

So what we want is a way to compute $\pi_k^s(S^0)$. One way to do this is through the Adams spectral sequence $$\text{Ext}^{p,q}_\mathcal{A}(\mathbb{Z}/p,\mathbb{Z}/p) \Rightarrow \pi_{p-q}^s(S^0) \otimes \mathbb{Z}_p$$

The $E_2$-term of this is well known in certain ranges. For example there are classes $h_j \in \text{Ext}^{1,2^j}_\mathcal{A}(\mathbb{Z}/p,\mathbb{Z}/p)$ and in turn classes $h_j^2 \in \text{Ext}_\mathcal{A}^{2,2^{j+1}}(\mathbb{Z}/p,\mathbb{Z}/p)$ (If you want to see some Ext charts see Chrisitian Nassau's homepage). Browder proved that the existence of Kervarire invariant one classes is equivalent to the statement that the $h_j^2$ survive the Adams spectral sequence (i.e. give rise to homotopy classes). So for example it is pretty easy to see that $h_1^2$ and $h_2^2$ survive; there are simply no classes that can possibly 'kill' these elements. But as $j$ increases the structure of $\text{Ext}_\mathcal{A}^{p,q}(\mathbb{Z}/p,\mathbb{Z}/p)$ rapidly become unweildly. The remarkable result of Hopkins-Hill-Ravenel is that for $j \ge 7$ the class $h_j^2$ is not a permanent cycle in the Adams spectral sequence. The kind of cool thing is that it is not even known what the differential is that kills these classes.

Here is a link to a big $E_2$ page: http://www.nullhomotopie.de/charts/bigpng.png. The vertical axes is $p$ and the horizontal is $p-q$. The classes $h_j$ are the classes with $p=1$.

  • $\begingroup$ Thanks for adding this; this clarifies my claim that the Kervaire invariant is a higher analog of the Hopf invariant. $\endgroup$ Sep 9, 2012 at 9:47
  • $\begingroup$ True! I hadn't thought about it that way before $\endgroup$
    – Juan S
    Sep 9, 2012 at 10:45

I haven't thought so much about the Kervaire invariant problem, but I don't think the point is to actually compute any particular homotopy group. (However, definitely check out the Snaith article that Juan S has linked if you're still curious; if there are direct applications of a resolution of the Kervaire problem to computing the homotopy groups of spheres, they'll certainly be in there. IIRC from skimming its introduction a few days ago, he makes the case that the Kervaire invariant is something like "the next Hopf invariant"$^1$.) Rather, the ring $\pi_*^S$ is one of the fundamental objects of stable homotopy theory for a number of reasons, and so it's exciting any time we find anything that relates to it or helps us explain some aspect of its structure. Some of the reasons it's fundamental are themselves deep and take a while to explain, but the first thing to say is that the "stable sphere" is the simplest nontrivial "stable space"$^2$, since it's just the stabilization of the two-point space $S^0$.

$^1$: Here are some notes I wrote for a talk I gave on the Steenrod algebra. In them, I describe the Hopf invariant (which detects the nontriviality of the Hopf map $S^3 \rightarrow S^2$, and can be defined on any element of $\pi_{2n-1}(S^n)$), and use the Steenrod operations to easily prove that it can only be 1 in dimensions of the form $2^k$. (This might be seen as analogous to Browder's result that the Kervaire invariant can only be nontrivial in dimensions of the form $2^k-2$.) As I go on to discuss, this actually is intimately related to the existence of a division algebra structure on $\mathbb{R}^n$, or equivalently the existence of mutually-linearly-independent families of vector fields on spheres. Eventually, Adams proved that the only Euclidean spaces which admit division algebra structures are $\mathbb{R}^1\cong \mathbb{R}$, $\mathbb{R}^2 \cong \mathbb{C}$, $\mathbb{R}^4 \cong \mathbb{H}$, and $\mathbb{R}^8 \cong \mathbb{O}$ (the real numbers, complex numbers, quaternions, and octonions). This solved the Hopf invariant one problem entirely.

$^2$: For a gentle introduction to the basic objects of stable homotopy theory -- "stable spaces", so to speak (they're actually called spectra) -- you can read a brief note I've written here, which as a bonus ends with one of the more concrete and accessible reasons why $\pi_*^S$ is fundamental (beyond what I've already said, which from a category-theoretic point of view is kind of bogus, at least as it stands now).


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