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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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    $\begingroup$ 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. $\endgroup$ – Matt Apr 27 '10 at 6:26
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    $\begingroup$ I think mathematical objects should be complicated until proven simple. $\endgroup$ – Qiaochu Yuan Apr 27 '10 at 6:35
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    $\begingroup$ 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? $\endgroup$ – Martin Brandenburg Apr 27 '10 at 8:08
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    $\begingroup$ @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. $\endgroup$ – Tyler Lawson Apr 27 '10 at 12:57
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    $\begingroup$ One reason could be is that the category of finite sets, and the symmetric groups are complicated. See question mathoverflow.net/questions/76541 $\endgroup$ – Spice the Bird Feb 15 '12 at 1:46
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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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    $\begingroup$ 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. $\endgroup$ – Thomas Kragh Apr 27 '10 at 7:29
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    $\begingroup$ If you don't think that Eilenberg-MacLane spaces have complicated cohomology then go ahead and compute the integral Steenrod algebra. I dare you. $\endgroup$ – Jeffrey Giansiracusa Apr 27 '10 at 9:26
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    $\begingroup$ @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. $\endgroup$ – Tyler Lawson Apr 27 '10 at 14:28
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    $\begingroup$ 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! =) $\endgroup$ – Tom Church Apr 28 '10 at 8:28
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    $\begingroup$ I'm afraid I don't understand this answer at all. In order to be useful, things need to be sufficiently simple, not sufficiently complicated. And indeed the parts of algebraic topology which are most applicable in other areas of mathematics are all much simpler and better understood: e.g. (i) singular co/homology, (ii) fundamental groups, (iii) fiber bundles and characteristic classes, etc. Have the homotopy groups of spheres ever been applied to anything, including in algebraic topology itself? $\endgroup$ – Pete L. Clark Apr 28 '10 at 8:50
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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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    $\begingroup$ 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. $\endgroup$ – Sean Tilson Apr 28 '10 at 22:55
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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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    $\begingroup$ By the way, what's the nowadays status about calculations of homotopy groups of spheres? Can we really know ALL of them? Thanks. $\endgroup$ – user1832 Apr 27 '10 at 13:55
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    $\begingroup$ @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. $\endgroup$ – Tyler Lawson Apr 27 '10 at 14:27
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    $\begingroup$ @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.) $\endgroup$ – Torsten Ekedahl Apr 28 '10 at 8:31
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    $\begingroup$ 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 $\endgroup$ – Lennart Meier Apr 28 '10 at 9:51
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    $\begingroup$ @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). $\endgroup$ – Torsten Ekedahl Apr 29 '10 at 20:23
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Another perspective of the question is from the point of view of computational complexity. According to the this paper (https://arxiv.org/pdf/1304.7705.pdf), for a simply connected finite simplicial complex $X$, computing $\pi_n (X)$ is a computationally very hard problem with respect to the parameter $n$ ($W[1]$-hard to be concrete).

It has been known for a long time that higher homotopy groups are algorithmically computable (the first results go back to Brown), and later it was shown (https://arxiv.org/pdf/1211.3093.pdf) that there is an algorithm which for a finite simply connected simplicial complex $X$ and $n \geq 2$ computes $\pi_n (X)$ in time, polynomial in the size of $X$. However, in this polynomial result $n$ is taken as part of the input. The result I stated above is that if you vary $n$, then the complexity of the problem explodes. I find this point of view particularly nice because it gives some concrete feeling for how complicated the computation of higher homotopy groups is (including the homotopy groups of spheres). On the other hand, one may point out that many results have been obtained regardless of the complexity but the sophistication of the tools required to obtain those results is yet another instance of this complexity.

Hope that is useful.

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Another kind of answer involves the EHP sequence. If $\pi_n^k=\pi_n(S^k)/\text{odd torsion}$, then there are exact sequences $$ \pi_{n+2}^{2k+1}\xrightarrow{P}\pi_n^k \xrightarrow{E} \pi_{n+1}^{k+1} \xrightarrow{H} \pi_{n+1}^{2k+1} \xrightarrow{P} \pi_{n-1}^k. $$ (These are homomorphisms of abelian groups, except that when $n=k=0$ the map $H$ is the self-map of $\pi_1^1=\mathbb{Z}$ given by $m\mapsto m(m-1)/2$.) We also have the "boundary conditions" that $\pi_n^k=0$ for $n<k$, or for $n>1$ when $k=1$, and the fact that $HPE^2$ is multiplication by $2$ on the group $\pi_{4n-1}^{4n-1}=\mathbb{Z}$. This provides a rather intricate pattern of connections between all the groups. You can try to write down a system of groups and homomorphisms satisfying these conditions, without worrying about whether they are actually the same as the real homotopy groups of spheres, and you will find that it quickly gets very complicated.

There are two partial solutions that can be written down explicitly. The groups $\mathbb{Q}\otimes\pi^n_k$ are known: there is a copy of $\mathbb{Q}$ for $(n,k)=(i,i)$ or $(4i-1,2i)$, and everything else is zero. It is easy to write down the maps $E$, $H$ and $P$ in this context and to check that everything is exact. There is a version of the $\Lambda$ algebra that is a relatively simple structure with properties similar to those specified above, but it has $P=0$ and each group $\Lambda_n^k$ is an infinitely generated $\mathbb{Z}/2$-module. (The $\Lambda$-algebra also has a differential, and it can be regarded as the $E_1$ page of an unstable Adams spectral sequence converging to $\pi_*^*$.)

I do not know any direct construction of a system of finitely generated abelian groups and homomorphisms with the required properties, and I do not think that anyone else does either. My guess is that any such system is of similar complexity to the homotopy groups of spheres.

Here is another interesting fact that sheds some light on this circle of ideas. Suppose we have systems $A_*^*$ and $B_*^*$ as above, and a morphism $f\colon A\to B$ that is compatible with $E$, $H$ and $P$, such that $f\colon A_1^1\to B_1^1$ is an isomorphism. One can then show that $f$ is an isomorphism in all bidegrees. This is another illustration of how tightly everything is linked together by the EHP sequence.

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I might be misinterpreting your friend's question, but if your friend knows very little about the subject and is expressing surprise that something so seemingly simple turns out to be so complicated, then maybe you should concentrate on explaining the Hopf fibration to him or her. There is an interesting visualization in Thurston and Levy's book Three-Dimensional Geometry and Topology. I think that anyone who successfully goes through the exercise of visualizing the Hopf fibration will be disabused of any preconception that the homotopy groups "should" be simple.

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  • $\begingroup$ I think this gets it exactly right. Before you learn about the Hopf map it's quite reasonable to think that homotopy groups of spheres might be easy (after all, $\pi_k(S^n)$ for $k \leq n$ is easy), but once you learn about the Hopf map your intuition should switch "if we have this construction why shouldn't there be zillions of other similar constructions"? (This is especially true if you think of the Hopf map as coming from the Eckman-Hilton argument a la Pontryagin rather than as involving a division algebra argument.) $\endgroup$ – Noah Snyder Sep 20 '18 at 16:56

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