# “Why the heck are the homotopy groups of the sphere so damn complicated?”

This is a quote from a dear friend asking the rest of us on Facebook. I gave him some half-baked response, but the truth is I don't really know enough about this to give him a good response.

So why ARE they so complicated? The topologists here want to give a few responses so I can give him some feedback to his desperate query?

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I've been told that it was only conjectured that they are complicated. It wasn't until the "Doomsday Conjecture" was proved (except one case) in the past year that we knew for a fact that it was hopeless for us to ever get a grip on them. I have absolutely no idea what the Doomsday Conjecture says or how it relates, this is just what I've heard. – Matt Apr 27 '10 at 6:26
I think mathematical objects should be complicated until proven simple. – Qiaochu Yuan Apr 27 '10 at 6:35
I think the question needs extra explanation. calculations show that the homotopy groups are complicated, well, but this is not the answer, right? perhaps you are looking for an empirical argument that spheres have nontrivial higher homotopy groups? – Martin Brandenburg Apr 27 '10 at 8:08
@hilbertthm90: As Charles posted, the original Doomsday conjecture was proven false by Mahowald in ~1971. The homotopy groups of spheres have been computable through a range since Serre's work on spectral sequences and have resisted most attempts to find simple systematic patterns ever since. – Tyler Lawson Apr 27 '10 at 12:57
One reason could be is that the category of finite sets, and the symmetric groups are complicated. See question mathoverflow.net/questions/76541 – Spice the Bird Feb 15 '12 at 1:46

This question can be answered at two levels. I'm going to take the easy one. I really hope that someone more advanced than me in the lore of algebraic topology can pick up the hard one.

The easy answer is that if the homotopy groups of spheres weren't so complicated then we wouldn't be talking about the homotopy groups of spheres so much.

Let me expand on that by an analogy. A penknife is a useful tool. One can do a lot with a penknife, but there's a lot of things that it's not that good at: getting corks out of bottles, descaling fish, sawing small bits of wood, getting annoying bits of food out from between your teeth ... I mean, I know that you can do a lot of those things with a penknife if it's all you've got, but it's not the best way to achieve those ends. Now a Swiss army knife is much better at doing all those. The latest probably also have inbuild GPS! But Swiss army knives are quite complicated gadgets. So when you say, "Why are Swiss army knives so complicated?" then the easy answer is that if they weren't, we wouldn't be using them so much and we would have found something else that was complicated to use instead.

In slightly less prosaic language, the fact that the homotopy groups of spheres are so complicated is what makes algebraic topology actually useful. We want to build complicated objects out of simple ones. What could be simpler than spheres? But to get something complicated, there has to be a source of complexity (I'm speaking very informally here) otherwise there would be no real hope of algebraic topology ever helping with other things. I mean that we know that general stuff in mathematics is quite complicated, so we're going to need some complicated tools to study it. If the homotopy groups of spheres were simple, then algebraic topology wouldn't be half so useful as it is; and if that were the case then there wouldn't be so many algebraic topologists around and your friend probably wouldn't have heard of the homotopy groups of spheres.

Let me finish with an attempt to clarify what I think is the hard part of this question to answer. That is, "Why spheres?". We accept as given, as I've argued above, that we need a complicated theory to study complicated objects; but the methods of algebraic topology are to probe the complicated objects by simple ones and so, hopefully, for any specific question to get rid of all unnecessary complexity and be able to see clearly the structure required for that specific question (I think that the proof of the Kevaire invariant problem is an example of what I mean here). So we need a good source of "simple objects" to probe with. Now these "simple objects" are those that look simple when we look at them with the tools of algebraic topology. So spheres are simple because they have very simple cohomology.

But we can probe something in two ways: we can either throw mud at it and see what sticks (that's homotopy), or we can take pictures of it and see what it looks like from different angles and with different lighting conditions (that's cohomology). As I've argued, the theory needs to have some complexity somewhere, so it's to be expected that the objects that are simple with respect to one method will look complicated when viewed at from the other. So spheres have complicated homotopy because they have simple cohomology. In contrast, the Eilenberg-Mac Lane spaces have complicated cohomology because they have simple homotopy.

But still, "Why spheres?". I mean, no-one ever asks, "Why do the Eilenberg-Mac Lane spaces have complicated cohomology?". I guess that's because no-one outside algebraic topology ever meets Eilenberg-Mac Lane spaces and so they aren't common objects across all of mathematics. So of course they have complicated cohomology because they are some weird tool that algebraic topologists have constructed and who knows what secret rites were used to do it?

So maybe I do have an answer to my "hard part" of this question: it's historical. In the early days of algebraic topology, the pioneering homotopy theorists got the idea of studying a space by throwing mud at it and seeing what stuck. As this was a new thing to try, they looked for the simplest thing that they could find: spheres. Then they found that they had a useful theory that had enough complexity to study spaces, and this was evidenced by the complexity of the homotopy groups of spheres. Had the homotopy groups of spheres been simple, algebraic topology wouldn't have gotten off the groups and, as I said, your friend would probably never have heard of it or them.

So, in summary, my answer is: something powerful enough to study a space by being thrown at it is going to have some complexity somewhere; spheres were the first thing that people tried, and they proved to be sufficient. (One could continue this by asking: why were spheres enough? But the answer is the same: if they weren't, we would have gone further. Spheres aren't enough to study everything, but they are enough to study most things that people are interested in.)

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I really dont think Eilenberg MacLane spaces have complicated cohomology compared to the homotopy groups of spheres. They do play the dual role of spheres for cohomology, but for some reason they are much nicer. Otherwise I fully agree with the answer. – Thomas Kragh Apr 27 '10 at 7:29
If you don't think that Eilenberg-MacLane spaces have complicated cohomology then go ahead and compute the integral Steenrod algebra. I dare you. – Jeffrey Giansiracusa Apr 27 '10 at 9:26
@Jeffrey: On the other hand, Thomas seemed to be making a statement about relative complexity compared to homotopy groups of spheres, which I would say is completely fair. – Tyler Lawson Apr 27 '10 at 14:28
The cohomology of Eilenberg-Mac Lane spaces includes in particular all of group cohomology, and it's fair to say there's just as much unknowable blackness there as in the homotopy groups of spheres. There are even parallels: e.g. lots of people would like to know the cohomology of mapping class groups, we only recently know the stable cohomology, and there's so much unstable cohomology it seems plausible that we may never know what all the cohomology is. I'm sure hundreds of mathematicians could say the same about their own favorite groups. Not all EM spaces are simply-connected! =) – Tom Church Apr 28 '10 at 8:28
@Pete: As far as I understand it, the homotopy groups of spheres are essential in high-dimensional topology. For example, the classification of exotic spheres (i.e. smooth manifolds homeomorphic to the sphere) would (essentially) be known if the homotopy groups of spheres would be known. This is only the tip of the iceberg - in whole of differentiable surgery theory the role of the stable homotopy groups of spheres (aka framed bordism groups) is essential. – Lennart Meier Apr 28 '10 at 9:44

You're going to get many different answers depending on the tastes of the topologist answering...

I like to think about homotopy groups of spheres through framed cobordism. Theories like unoriented and complex cobordism are understandable for a couple reasons. Technically they are calculable because we can understand their cohomology so well over the Steenrod algebra. But morally they are understandable because they are amenable to analysis through characteristic classes. But for framed bordism, the structure group is the trivial group. So either the theory is going to be trivial, or really hard because there are no characteristic classes to use. It turns out that it is the latter.

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this is a very interesting answer, quite a different perspective. I think this also explains why we might care about the homotopy groups of spheres, they contain a lot of geometric information. – Sean Tilson Apr 28 '10 at 22:55

It is not only that homotopy groups of spheres are very complicated, homotopy classes of maps between manifolds tend in general to be very difficult. Of course, there are some exceptions, e.g. maps from spheres into hyperbolic manifolds, but in general there is no reason to expect that it is easy to count the ways one can map a high-dimensional thing into a lower-dimensional thing; at least, after one has seen the Hopf map as an example that this is indeed possible in a non-trivial way.

One reason, one picks usually the homotopy groups of spheres, I think, is that we have a nice infinite family of spheres and that we can build so much out of it. If one wants to compute some (stable) homotopy group of some manifold, I think, the usual try would be to build as a CW-complex out of spheres and use the computation for spheres.

Apart from the argumentation that they are difficult because there is no reason to expect them to be easy: over the last decades, it became more and more apparent that there is rich arithmetic hidden in the (stable) homotopy groups of spheres. The e-invariant and the J-homomorphism link them to the denominators of Bernoulli numbers and the f-invariant (see e. g. Laures and Hornbostel&Naumann) to congruences between modular forms. The last phenomenon has something to do with Topological Modular Forms and the work of Behrens and Lawson on Topological Automorphic Forms gives hope to relate the homotopy groups of spheres even to the arithemetic of automorphic forms.

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By the way, what's the nowadays status about calculations of homotopy groups of spheres? Can we really know ALL of them? Thanks. – user1832 Apr 27 '10 at 13:55
@unknown: There are various computational methods for getting at them, but actual knowledge of all of them (e.g. in a recursive formula) seems currently out of reach. I believe that the first indeterminacy in our knowledge of stable homotopy groups of spheres is around dimension ~mid-50s or so (at the prime 2), although a lot is known beyond that point. Other primes are somewhat easier. – Tyler Lawson Apr 27 '10 at 14:27
@Eric: This is not a correct description of Freyd's conjecture. It says that stable homotopy groups give a full embedding of the finite homotopy category into that of modules over the stable homotopy ring of spheres. The stable homotopy category also is not abelian, it is triangulated. It is even known that the stable homotopy category is not the derived category of a ring. (I think it may be the derived category of a differential graded ring and I even have some vague recollection that this has been proved.) – Torsten Ekedahl Apr 28 '10 at 8:31
Indeed, it has been disproved that the stable homotopy category is equivalent to the derived category of modules over a differential graded ring. See Stefan Schwede's work: math.uni-bonn.de/~schwede/torsion.pdf or math.uni-bonn.de/~schwede/algebraic_topological.pdf – Lennart Meier Apr 28 '10 at 9:51
@Thomas: As I understand it Freyd meant (and I just tried to follow him) the subcategory of the stable category consisting of finite complexes and their shifts (which of course can be defined without embedding it in the larger category). – Torsten Ekedahl Apr 29 '10 at 20:23