# Homotopy groups of $S^2$

I'd like to understand higher homotopy groups better and I guess there's no simpler way than understanding them for as simple spaces as possible; therefore $S^2$.

My question essentially has two parts, one geometrical and one computational. First part is not really about spheres per se but I think it should be easier to answer in this context

1. what do those groups really tell us about $S^2$ and where does all the complexity come from? I mean, naively, I would expect most of the groups to be trivial (as for $S^1$). This probably means that I don't yet have an intuitive grasp of higher homotopy groups; so intuition is what I am looking for in this part.

2. have those groups been computed completely (or at least, is there an algorithm to compute them)? I know the theory of higher-dimensional spheres is complicated but perhaps the case of $S^2$ (and therefore also $S^3$) might be a bit simpler.

Regarding 2., wikipedia page mentions that the problem has been reduced to combinatorial group theory of Brunnian braids. Could someone expound on this or provide additional reductions to purely combinatorial problems?

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The groups $\pi_n(S^2)$ are not known. In fact, for $n\geq 3$ they're isomorphic to $\pi_n(S^3)$; you can get this from the long exact sequence in homotopy groups associated to the Hopf fibration $S^1 \rightarrow S^3 \rightarrow S^2$. –  Aaron Mazel-Gee Jul 8 '11 at 19:51
@Aaron: I was aware of the isomorphism(s) but I don't think it's relevant to my question. Also, could you elaborate on how much is (not) known about the groups? I take your word for their not being known but is there perhaps at least an algorithm that can be used to compute them? –  Marek Jul 8 '11 at 21:02
Yes, I was just responding to the first bit of your 2nd question. These are known as <i>unstable</i> homotopy groups, about which very little is known indeed. I don't know exactly which $\pi_n(S^2)$ we do and don't know, but I'd imagine you can find the totality of what we know on e.g. Bob Bruner's website. I hadn't heard of the Brunnian braids result, but don't be fooled: problems in algebra are often impossible to be solved by algorithm (e.g. the "word problem" of determining whether two words are isomorphic in a given generators-and-relations presentation is known to be non-algorithmic). –  Aaron Mazel-Gee Jul 8 '11 at 21:32
In contrast, I think a bit more is known about <i>stable</i> homotopy groups of spheres. As it turns out, $\pi_{n+k}(S^k)$ stabilizes for $k$ large (depending on $n$); this value is denoted $\pi_n^S$ and is called the "$n^{th}$ stable homotopy group of spheres". These can be attacked directly using methods in a rich field known as "stable homotopy theory". For example, it's known that $\pi_{n+1}(S^n)\cong \mathbb{Z}/2$ for $n\geq 3$, and there are some very deep periodicity results that relate various $\pi_n^S$'s. In fact, $\pi_*^S$ can be made into a commutative ring by composition. –  Aaron Mazel-Gee Jul 8 '11 at 21:39
for some concrete intuition in a few specific cases, take a look at the hopf fibrations. also, homotopy groups of spheres are hard! it's one of the motivating problems in algebraic topology –  yoyo Jul 8 '11 at 22:08

For a pointed space $X$ let $\Omega X$ denote the loop space of loops $S^1 \to X$ respecting base-points. For non-pathological $X$ we can equip $\Omega X$ with the compact-open topology and we have the natural identification $\pi_n(\Omega X) \cong \pi_{n+1}(X)$.

Now $\pi_1$ is pretty intuitive. It tells us how complicated loops in $X$ can be up to homotopy. Equivalently, it tells us about the connected components of $\Omega X$. But it doesn't tell us how complicated those connected components themselves are. To think about that we look at $\pi_1(\Omega X) \cong \pi_2(X)$. When this is nontrivial, it means that there are loops in $\Omega X$ (or "loops between loops") that are not homotopy-equivalent.

Similarly $\pi_3(X) \cong \pi_2(\Omega X) \cong \pi_1(\Omega^2 X)$ tells us about how complicated "loops between loops between loops" are. In other words, the fact that the higher homotopy groups of $S^2$ are nontrivial tells us that the iterated loop spaces of $S^2$ are complicated.

By contrast, every connected component of $\Omega S^1$ is contractible by looking at the universal cover.

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Oh, this is a very nice point of view. I think it basically answers the first part of my question. Perhaps I should split off the second part to a separate question? –  Marek Jul 8 '11 at 19:05

In the paper [E H Brown's "Finite Computability of Postnikov Complexes" annals of Mathematics (2) 65 (1957) pp 1-20] you will find a proof that all homotopy groups of a simply-connected simplicial complex are algorithmically computable.

In particular, we know how to compute all of $\pi_n(S^2)$. The problem is that the algorithm is impossibly inefficient. Moreover, even if we actually ran the algorithm what we would get is probably not what we really want, which is to have some kind of information on the groups---and a table is not exactly that.

For example, knowing that $\pi_{n+11}(S^n)$ is a cyclic group of order 504 is a data point, but knowing that it is a cylic group of order precisely the denominator of the quotient $B_{6}/12$, where $B_6$ is the sixth Bernoulli number, and that this is in fact part of a general pattern, now that is information!

Later. You are not alone in expecting those groups to be trivial, by the way. Reading up a bit on the history of algebraic topology is helpful here: you'll see that at the time, it was quite a shock to everyone involved (that is, all those people whose names adorn theorems nowadays) that the Hopf fibration exists at all, implying that $\pi_3(S^2)$ is a non-trivial group!

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At least at $p=2$, Toda's calculations do not go nearly that far up. There are later papers that go a little further, but his Composition methods in homotopy groups of spheres'' goes up to $n=19$. It is misleading to think of $S^2$ as particularly simple. There is an old theorem (Serre?) that if $X$ is any simply connected finite CW complex that is not contractible, such as $S^2$, then for each prime $p$ there are infinitely many $n$ such that $\pi_n(X)$ has $p$-torsion.

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I would say that $S^2$ is no easy then any other group. It is not even known whether $\pi_n(S^2)$ is zero for any $n>1$. See here.

I presume you've looked at the visualisation of the hopf fibration $S^3 \to S^2$ in Hatcher's book? This should at least give some idea why $\pi_3(S^2) \ne 0$.

I believe Toda calculated the homotopy groups of $\pi_n(S^p)$ for $n \le 64$ (and some range of $p$). In general most work seems to focus on calculating the stable homotopy groups of spheres. This appears to work by calculating the p-th component at a time, and is a highly non-trivial problem.

All (most?) of the techniques involve some form of spectral sequence of varying difficulty. The 'gold-standard' is the Adams-Novikov spectral sequence, but even calculating the $E_2$ page of this spectral sequence is frighteningly difficult. For the best survey, although still a difficult read, see Chapter 1 of Ravenel's Complex Cobordism and Stable Homotopy Groups of Spheres

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I guess you mean $\pi_3(S^2)$? And yes, I am familiar with Hopf fibration and basic results that follow from it. But thanks for the other references. –  Marek Jul 8 '11 at 22:16
@Qwirk: note that $\pi_n(S^p) = 0$ for all $p > n$. (For instance, if I am not mistaken every homotopy class of maps between smooth manifolds has a smooth representative, and the image of a smooth map $S^n \rightarrow S^p$ has measure zero when $n < p$, so the map is contractible to a point.) So it's conceivable that Toda calculated these groups for all $n \leq 64$ and for all $p$. (I don't know whether he did: it's just a guess.) –  Pete L. Clark Jul 8 '11 at 23:44
@Marek - yes thanks, I have fixed the typo –  Juan S Jul 9 '11 at 4:01
@Pete: one can also appeal to simplicial approximation. –  Qiaochu Yuan Jul 9 '11 at 4:06
@Pete - indeed you are correct (at least for me the easiest way to think about it is using a cellular approximation and the usual CW structure on the spheres). I'm not sure exactly what Toda did - his papers are not online, so I didn't have an easy way to check. Would be impressive if he did do it all! –  Juan S Jul 9 '11 at 4:12