The homotopy groups of the spheres $S^n$ (see Wikipedia) vanish for the circle $S^1$ as, naively speaking, there are not higher order holes to be grasped by higher order homotopy groups. This intuitions already breaks down for the two sphere $S^2$, e.g. $\pi_3(S^2)$ is nontrivial because of the Hopf fibration. This nontriviality seems to keep on going for all the higher spheres $S^n$. What makes $S^1$ so fundamentally different?

14$\begingroup$ The "universal covering space" operation is simultaneously very geometric and preserves all homotopy groups except $\pi_1$. This transparently sends $S^1$ to the contractible space $\Bbb R$. There are analogues to this for higher homotopy groups, but the 'Whitehead truncation' operations are no longer so geometric; there is no real way to see visually what the Whitehead truncations of higher spheres are, and indeed they are not contractible. $\endgroup$– mmeJul 20, 2019 at 15:39

3$\begingroup$ I think it is a combination of two facts: (1) taking the universal cover is "geometric". In particular if $X$ is a $d$dimensional manifold, so is its universal cover. (2) There aren't that many simply connected 1manifolds. Note that taking higher Whitehead covers does not preserve being a manifold (and in fact tends to send finitedimensional objects to infinitedimensional objects, in some sense) $\endgroup$– Denis NardinJul 20, 2019 at 16:00

1$\begingroup$ I'm not sure what you mean by "this nontriviality seems to keep on going"; for instance, the homotopy group pi_10 S^6 (in the stable range) vanishes. There are, however, infinitely many nontrivial homotopy groups of S^n for n>1. This is a consequence of a result known as the McGibbonNeisendorfer theorem, which states that if X is a simplyconnected finite complex which is not plocally trivial, then pi_n X has ptorsion for infinitely many n. From this point of view, the failure of S^1 to be simplyconnected is one root of the issue. $\endgroup$– skdJul 20, 2019 at 16:00

1$\begingroup$ @DenisNardin actually $n$dimensional torus $(S^1)^n$ has vanishing of higher homotopy groups, and they are nmanifolds. $\endgroup$– user43326Jul 20, 2019 at 16:11

6$\begingroup$ @user43326 I never said there are no aspherical manifolds, there are obviously lots (tori, hyperbolic manifolds,...). I meant to say that if you start with a manifold with nontrivial $\pi_n$ and you take the $n$th Whitehead cover, it will usually not be the homotopy type of a manifold anymore (in fact it will usually tend to become "infinitedimensional" in some sense). $\endgroup$– Denis NardinJul 20, 2019 at 16:14
6 Answers
So far the discusion mostly focused on a geometric explanation, I'd like to mention the algebraic one as well:
One way to formulate it involves the delooping machinery: up to delooping, $\mathbb{S}^n$ corresponds to the free group like $E_n$algebra on one generator.
Small recall: the usual delooping machinery say that the looping/delooping construction induce an equivalence between pointed spaces $X$ such that $\pi_k X =0$ for all $k<n$ and group like $E_n$algberas. Through that correspondence $\mathbb{S}^n$ corresponds to the free group like $E_n$algebra on one generators, as the looping/delooping adjunction justs shift the $\pi_n$ you can describe the homotopy group of sphere as shifted homotopy group of these free group like $E_n$algebra. A more explicit way to say all this is that the free group like $E_n$algebra is $\Omega^n \mathbb{S}^n$.
Now, when you construct the free $E_1$algebra, there is not much you can do: you only have one way to multiply elements and the free $E_1$algebra on one generator is just the monoid $\mathbb{N}$ (and $\mathbb{Z}$ for the group like one).
But for $E_n$ you get $n$ "compatible" (in a homotopy theoretic sense) way to multiply the elements and all the higher elements in the homotopy group comes from the interaction (the coherence law, that are given by homotopies) between these multiplications, for example the free $E_2$algebra has all the braid groups appearing as its various $\pi_1$ due to that, and the free group like $E_2$algebra become too complicated to described (well... it is $\Omega^2 \mathbb{S}^2$ ).
So the difference is that for $n=1$ no such interaction is happening because to get interaction you need at least two compatible multiplication.
Alternatively to the delooping machinery, one can (somehow equivalently) think of spaces as $\infty$groupoids and as the $n$sphere as the $\infty$groupoid freely generated by a cell in dimension $n$. The discussion is pretty much the same except that now the $n$ "compatible" multiplication are simply the compositions in direction $k$ for $k$ from $0$ to $n1$.
Edit: Here is how you get a non trivial element of $\pi_3(\mathbb{S}^2)$, in the second perspective. I'm using an unspecified model of weak $\infty$groupoid, and applying freely the operation of strict $\infty$categories to give a feel of how it works, this is not mean't to be formal (but it is formalizable in any algebraic model of weak $\infty$groupoid, or in Hott )
Given a two cells $u$ and $v$ whose source and target is a (weak) identity, the usual Eckman Hilton argument (so the typical example of interaction between $\#_0$ and $\#_1$ as I mentioned above) gives an isomorphism $\theta_{u,v} : u \#_0 v \simeq v \#_0 u$.
If $e$ is the generating 2cell of the 2sphere then this gives an isomorphisms $\theta_{e,e}: e \#_0 e \simeq e \#_0 e $ taking $e^*$ a $0$ inverse of $e$, one has that $e^* \#_0 \theta_{e,e} \# e^*$ is a 3cell whose source and target are (up to the coherence isomorphism expressing that $e$ and $e^*$ are inverse) identities, so it gives an element of $\pi_3(\mathbb{S}^2)$, which is nonzero by a universality argument. I'm convince it is a generator (so either the Hopf fibration or its opposite depending which way you have rotated the 'EckmanHilton clock') but I don't know how to prove it using only this type of tools.

$\begingroup$ Is it actually possible to do some computations with one of these perspectives, ideally the second one? I've thought a bit about it but I don't know how to find the Hopf fibration as a generator of the endomorphism group of the 2identity in a weak 3groupoid with one object freely generated by a 2endomorphism of the 1identity. Maybe its double, via Whitehead products, but even in HoTT to get the generator they seem to construct the fibration and use the long exact sequence, which I think of as a fundamentally geometric argument. $\endgroup$ Jul 20, 2019 at 20:59

4$\begingroup$ @KevinCarlson : I've edited to show how to get a non trivial element in $\pi_3(\mathbb{S^2})$. In theory you can get all elements of $\pi_n(\mathbb{S}^m)$ this way, though it is going to be very non automatic and I have no idea how you could show that you have found all elements, so I wouldn't call that a way to compute $\pi_n(\mathbb{S}^m)$, to me it is more a way of 'naming' its elements. $\endgroup$ Jul 20, 2019 at 21:48

$\begingroup$ @NoahSnyder : I think I'm misunderstanding your comment: isn't "counterclockwise" EckmanHilton the inverse of the clockwise EckmanHilton ? so if you compose them you should get something trivial no ? $\endgroup$ Jul 21, 2019 at 21:36

2$\begingroup$ Though in a sense I think the nonvanishing of $\pi_3(S^2 \vee S^2)$ more directly illustrates why just the group structure fails to capture higher loop spaces. The free grouplike $E_2$ algebra generated by two 2loops has an interesting 3loop given by $xy \rightarrow yx \rightarrow xy$ because there are two distinct proofs of EckmanHilton, the clockwise one and the counterclockwise one. $\endgroup$ Jul 21, 2019 at 22:16

1$\begingroup$ Right, that’s the right thing to do for $S^2$. What I was writing down gives $2 \in \mathbb{Z}$ instead of 1. But note it doesn’t give 0. To work out what a given construction does you should work out the corresponding PontryaginThom diagram. $\endgroup$ Jul 21, 2019 at 22:32
Homotopy groups of spheres correspond to framed submanifolds of Euclidean space through the PontrjaginThom construction. For example, the Hopf map corresponds to a circle in $\mathbb{R}^3$ framed “with a twist”. The homotopy groups of $S^1$ thus correspond to framed codimension one submanifolds. But such are canonically framed and all bound, so there are no interesting/ nontrivial examples.

1$\begingroup$ This is probably wellknown, but could you describe what the ambient manifold is here when you say "submanifolds of Euclidean space"? Is this something like R^\infty with the direct limit topology? $\endgroup$ Jul 25, 2019 at 22:44

2$\begingroup$ Sorry to be brief: if we consider $\pi_n(S^k)$ then the submanifolds in question are codimension $k$ submanifolds of $\mathbb{R}^n$. (The ambient manifold can be taken to be $\mathbb{R}^n$ rather than $S^n$ because the map is based so the preimage of a nonbasepoint will be disjoint from the basepoint of $S^n$.) $\endgroup$ Jul 26, 2019 at 17:07

1$\begingroup$ This framed cobordism perspective highlights the fact that the only ingredient that is interesting in cobordism of codimension one submanifolds is separation. So the homotopy groups of $S^1$ are different from the homotopy groups of $S^k$ with $k>1$ in much the same way that $S^0$ is different from $S^k$ for $k>0$, i.e. $S^0$ is not connected. $\endgroup$ May 6, 2021 at 21:51
In a sense, failure of the the higher homotopy groups of $S^n$ to be trivial, $n>1$, is due to them not representing singular cohomology. If the higher homotopy groups were trivial, all spheres would be EilenbergMacLane spaces and would represent cohomology. For most spheres, this failure to represent cohomology can be seen because they are not loop spaces which in turn is because they are not groups.

4$\begingroup$ Well, $S^3$ is a group (yeah, yeah, not a commutative one...) $\endgroup$ Jul 21, 2019 at 21:34

1$\begingroup$ The “most” is meant to apply both to failing to represent cohomology and failing to be a group in that only a couple actually accomplish either. $\endgroup$ Jul 21, 2019 at 22:14
Here's a very simple argument as to why not all of the spheres have higher homotopy groups vanish. Higher homotopy groups vanishing (along with the lower ones) is equivalent to the sphere representing ndimensional cohomology. By Yoneda, natural transformations between functors correspond to maps between representing objects. Hence, natural transformations between nth cohomology with integral coefficients and mth cohomology with integral coefficients correspond to maps $S^n \rightarrow S^m$ if these both represent singular cohomology. However, there are nontrivial cohomology operations (say for example, reduction mod p followed by the integral Bockstein), and so this implies that we have a map between the spheres representing them. However, this would imply that the higher homotopy groups of this sphere do not vanish, and so we get a contradiction.
One explanation follows from the fact that if $X$ is a space and $\tilde X$ is its universal cover, then for $i\geq 2$ we have $\pi_i X \cong \pi_i \tilde X$.
Then you can just observe that the universal cover of $S^1$ is $\mathbb R$ (which is contractible and hence has vanishing higher homotopy groups), while for $n > 1$, the universal cover of $S^n$ is just $S^n$ itself (it is simplyconnected, so you can just take the identity as a covering map).

14$\begingroup$ That's not really an explanation for why the homotopy groups of spheres are nonzero though... All you can conclude here is that $\pi_k(S^n) = \pi_k(S^n)$! $\endgroup$ Jul 20, 2019 at 20:46

17$\begingroup$ @NajibIdrissi It does answer the question in the body: "What makes $S^1$ so fundamentally different"? $\endgroup$– WojowuJul 20, 2019 at 22:07

6$\begingroup$ @NajibIdrissi Modulo Whitehead theorem it explains why at least one higher homotopy group is nonzero. $\endgroup$ Jul 20, 2019 at 22:32

$\begingroup$ @kp9r4d Sorry, I don't really understand what you mean. Which Whitehead theorem? The one about weak equivalences of CW complexes being homotopy equivalence, the one about embeddings...? I don't really see how either one applies. $\endgroup$ Jul 22, 2019 at 12:21

1$\begingroup$ @NajibIdrissi : nonzero homology implies noncontractibility without mentioning homotopy groups or Hurewicz; so I don't see how that's circular. The homology here is not part of the question as all spheres have "the same" homology (the formula is the same) $\endgroup$ Jul 24, 2019 at 16:27
Iʼm no homotopy theorist, but I have a little scraplet of intuition that may be helpful. I havenʼt studied the underlying concepts enough to go into much meaningful detail here. Maybe someone else can expand on this; maybe my thoughts are utterly wrong. But without further ado…
I'm going to compare the 1sphere and the 2sphere by describing them in alternation, one paragraph each.
If you want to trace out a 1sphere, one way to do that is to take a 0sphere (a pair of points), anchor one of the points, and move the other point in a loop, starting and ending at that anchor point. By doing that, you've traced out a path within the 1sphere, and that path is the generator of the space.
Likewise, if you want to trace out a 2sphere, you can take a 1sphere, anchor one of the points, and move the opposite point in a loop, starting and ending at that anchor point. By doing that, you've traced out a homotopy (a 2path) within the 2sphere, and that homotopy is the generator of the space.
Of course, the 1sphere has more loops (paths from the anchor point to itself) than just the generator. There's a trivial loop, and you can also reverse loops and compose them. These are, of course, simply the operations of a group, and the homotopy group $\pi_1(S^1)$ describes how these operations work.
Likewise, the 2sphere has more 2loops (homotopies from the trivial loop on the anchor point, to itself) than just the generator. You have the group operations again, described by the homotopy group $\pi_2(S^2)$.
With the 1sphere, the group operations "tell the whole story". Up to homotopy, there are no more loops than those created by the group operations.
With the 2sphere, the group operations no longer tell the whole story. The generator we identified consists of taking a point and moving it in a circle in a particular direction. The group operations allow us to move in the opposite direction. But it's also possible to move in a perpendicular direction, or sideways, or in any of $S^1$many directions. So in order to tell the whole story, we need additional homotopy groups.
Of course, the question I'm failing to answer is: how, exactly, do additional homotopy groups tell the rest of the story? I don't really know. But hopefully I've given some motivation how, unlike the loop space of $S^1$, the loop space of $S^2$ is itself an interesting space; and the loop space of that space is an interesting space, and so on.