# Why is it so hard to compute $\pi_n(S^n)$?

Of course it isn't really that hard - nowhere near as hard as $\pi_k(S^n)$ for $k>n$, for instance. The hardness that I'm referring to is based on the observation that apparently nobody knows how to do the calculation within the homotopy category of topological spaces. Approaches that I'm aware of include:

-Homology theory (the Hurewicz theorem)

-Degree theory

-The divergence theorem

and each of these reduces to a calculation within some other category (PL or Diff). My question is: is there something "wrong" with hTop that precludes a computation of $\pi_n(S^n)$ within that category?

Certainly if my assumption that such a proof does not exist is wrong, I would be very interested to know it.

I have been thinking and reading further about this question for the past couple days, and I wanted to summarize some of the main points in the answers and questions:

1. Some techniques - e.g. the Freudenthal Suspension Theorem via the James construction or the Hurewicz Theorem with singular homology - might actually lead to proofs without any approximation arguments. But so far I'm not sure we quite have it: the proof of the Freudenthal Suspension Theorem uses the fact that $J(X)$ is homotopy equivalent to the loop space of the suspension of $X$, but the only proofs I can find of this fact use a CW structure on $X$, and similarly for the proofs of the Hurewicz theorem. Can these results be proved for the sphere without PL or smooth approximation?
2. Perhaps this question is entirely wrong-headed: the techniques of PL and smooth approximation are very well adapted to homotopy theory, so why try to replace them with a language which may end up adding more complications with little additional insight? Fair enough. But the goal of this question is not to disparage or seek alternatives to existing techniques, it is to understand exactly what role they play in the theory. The statement "The identity map $S^n \to S^n$ is homotopically nontrivial and freely generates $\pi_n(S^n)$" makes no mention of CW complexes or smooth structures, yet apparently the statement is difficult or impossible to prove without that sort of language (except in the case $n=1$!) To seek an understanding of this observation is different from lamenting it.
• One option which is perhaps overkill is to use the Hopf fibration to show that pi_2(S^2) is Z and then to use Freudenthal suspension theorem (which can be proved with homotopy theoretic ideas e.g. the James construction). – Callan McGill Oct 20 '14 at 23:44
• Maybe I'm missing something, but how is homology outside of the homotopy category of topological spaces? – Denis Nardin Oct 21 '14 at 2:49
• @PaulSiegel To compute $H_n(S^n)$ I would prove the suspension isomorphism in homology (that uses only homotopy invariance and Mayer-Vietoris) and then proceed to compute $H_0(S^0)$, which is elementary. I think that this proof does not leave the homotopy category, unless you regard the subdivision isomorphism to prove excision as "PL approximation" – Denis Nardin Oct 21 '14 at 13:33
• @PaulSiegel I must admit to being slightly confused by the question. Most early calculations of $\pi_n(S^n)$ do rely heavily on hard manipulations with homotopies to reduce to continuous maps of nice forms, but the definition of hTop is in terms of such homotopies. Without some of these basic tools under our control, we have almost no way to distinguish hTop from an arbitrary category. – Tyler Lawson Oct 21 '14 at 14:47
• If you like, though, one defect in hTop is that has no categorical niceties. You can't really build new objects out of old ones or use that many elementary "building block" calculations to get larger calculations. Several decades of experience with this have resulted in a working methodology where, even if a homotopy category is our goal, it almost always comes accompanied with a "chart" (like the category of spaces, or CW-complexes, or simplicial sets, or...) where all the important constructions can be carried out; the homotopy category is just a place to examine the results. – Tyler Lawson Oct 21 '14 at 14:53

I suppose that the proof that $\pi_1(S^1) \cong \mathbb{Z}$ using covering spaces is homotopy-theoretic.

The Freudenthal Suspension Theorem (via the James construction) tells us that $\Sigma: \pi_n(S^n)\to \pi_{n+1}(S^{n+1})$ is an isomorphism for $n \geq 2$ and surjective for $n =1$.

Since $S^1$ is an H-space, the suspension map $\sigma: S^1\to \Omega\Sigma S^1$ has a retraction $r: \Omega\Sigma S^1\to S^1$. Therefore $\Sigma = \sigma_*: \pi_1(S^1)\to \pi_{2}(S^{2})$ is injective (in addition to being surjection).

Now Freudenthal completes the calculation.

Note that we don't just get an abstract isomorphism, we get that these groups are generated by $[\mathrm{id}_{S^n}]$.

EDIT: Regarding getting the James Construction homotopically: the paper

Fantham, Peter; James, Ioan(4-OX); Mather, Michael On the reduced product construction. (English summary) Canad. Math. Bull. 39 (1996), no. 4, 385–389.

derives the relevant properties using the Cube Theorems (which are about the mixing of homotopy pushouts and homotopy pullbakcs) of an earlier paper of Mather's.

Even when computing $\pi_k(S^n), k < n$ all of the hard work, as far as I can tell, comes from showing that continuous maps behave reasonably up to homotopy; there is no difficulty once you show, whatever way you like (simplicial approximation, smooth approximation, etc.), that you can ignore space-filling curves. Until you know that it's necessary to consider the possibility that continuous maps behave in totally ungeometric ways that make them unsuitable for modeling homotopy theory.

Topological spaces and continuous maps are both absurdly general objects, and in fact too general to model homotopy theory: instead of working with the homotopy category of topological spaces we should of course be working with the homotopy category of CW complexes, and of course there are other ways to describe the homotopy category that don't involve topological spaces at all. In some of these descriptions (starting from simplicial sets, I would guess) it may be quite easy to compute $\pi_k(S^n), k \le n$.

Morally this computation should be easy starting from a heuristic picture of $\infty$-groupoids: $S^n$ is the free $\infty$-groupoid on an $n$-morphism, and in particular has no interesting $k$-morphisms for $k < n$. For $k = n$, again heuristically, all you can do starting from an $n$-morphism is compose it with itself a lot; the $\mathbb{Z}$ appearing here is the free group on one generator. Possibly this is the sort of reasoning that homotopy type theory is supposed to make precise.

Edit: That heuristic reasoning above may not seem too convincing because it doesn't seem to say anything about the higher homotopy groups, so let me spell out what it suggests about $\pi_3(S^2)$.

To compute this it suffices to ask what the free $3$-groupoid on a $2$-morphism is. A $3$-category with one object and one $1$-morphism is precisely a braided monoidal category, so the question now is what the free grouplike braided monoidal groupoid on an object $X$ is like. Well, it has a dual $X^{\ast}$ (which must in fact be its inverse) and there are unit and counit maps $1 \to X \otimes X^{\ast}, X^{\ast} \otimes X \to 1$. Finally there is a braiding $X \otimes X^{\ast} \to X^{\ast} \otimes X$. These three maps can be composed, and we get a map $1 \to 1$ which in no way is required to be the identity; in fact it can be identified with the generator of $\pi_3(S^2)$.

Incidentally, the corresponding argument for $\pi_2(S^1)$ fails because we don't have a braiding; the corresponding question is what the free grouplike monoidal groupoid on an object is like. The key is to be extremely careful what the unit and counit look like in a monoidal category which is not assumed to be symmetric, and once we are, the argument correctly suggests that $\pi_2(S^1)$ is trivial.

Edit #2: And the Freudenthal suspension theorem appears here in the fact that for $\pi_{n+1}(S^n), n \ge 3$ the question stabilizes to looking at the free grouplike symmetric monoidal groupoid on an object.

• My question (which I may not be able to pose very well) is: what demands does your heuristic picture place on the underlying category of spaces? Where would the heuristic run aground if nobody had ever invented CW complexes or simplicial sets and we were forced to contend with space filling curves? – Paul Siegel Oct 21 '14 at 9:05
• @Paul: the specific heuristic argument I wrote down above requires more or less that there be a nerve / geometric realization functor $X \mapsto |X|$ from $n$-groupoids to spaces and that this functor have the property that $\pi_k(|X|, x)$ is the automorphism group of the identity automorphism of the identity... of $x$ in $X$. The simplest version of this demand is that we should be able to create Eilenberg-MacLane spaces, although that isn't enough to run the heuristic argument. – Qiaochu Yuan Oct 21 '14 at 17:53
• In other words I'm taking for granted that the homotopy hypothesis provides the correct model-independent description of homotopy theory and that everything else is the quest for usable models, which may or may not involve topological spaces. – Qiaochu Yuan Oct 21 '14 at 17:56

This example and many other illustrate that geometric arguments cannot always be completely replaced by algebraic ones, much like the fundamental theorem of algebra does not seem to have a simple purely algebraic proof. (I'm out on a limb with this statement.)

It looks to me that a large part of the fundamental functors of algebraic topology have a geometric origin; think homotopy, (co)homology, cobordism, $K$-theory. I cannot imagine formal arguments, devoid of geometric intuition leading to such concepts. Obviously geometric arguments alone cannot get you very far; think homotopy, (co)homology, cobordism theory, $K$-theory without long exact or spectral sequences.

Being a mathematical "mutt" myself, I always favor impure arguments. They give me the comforting feeling of not being isolated. Also, they broaden my sources of inspiration.

One can show that $\Omega^n S^n$ is the group completion of the configuration space of distinct unordered particles in $R^n$. Now $\pi_0$ of this configuration space is the natural numbers. This shows that $\pi_0(\Omega^n S^n)$ is the Grothendieck group of the natural number. Thus, $Z=\pi_0(\Omega^n S^n)=\pi_n(S^n)$.

• The arguments I've seen for this would fail the "internal to the homotopy category" criterion. – Ryan Budney Oct 21 '14 at 4:31