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I have been studying the Hopf fibration which is an example of a map from $S^3$ to $S^2$. It is a member of $\pi_3(S^2)$ and shows that this group is non-trivial. It can be shown using a long exact sequence applied to the Hopf fibration that $\pi_3(S^2) \cong \mathbb{Z}$.

However, the set of notes I am using to learn this material (here) then claims that the Hopf map is a generator of this group. Furthermore, the Hopf invariant, which is a sort of linking number of the preimages of distinct points can be used to define this isomorphism with $(\mathbb{Z},+)$.

I fail to see how the hopf map generates $\pi_3(S^2)$. Specifically, I would like to know how you can compose the hopf map to generate maps with higher hopf invariant that make up this integer-like group structure. It would be really helpful to explicitly see this in action (i.e. pointers on how to writ this out using elementary algebra).

Note: I have a Chemistry background so my understanding of Mathematics is quite basic though I am willing to learn.

Paul Frost
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Jack Holmes
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  • "shows that this group is not null-homotopic" -- that doesn't really make sense. – John Hughes Oct 26 '19 at 00:01
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    You can think of a $3$-sphere as a cube with boundary identified to a point. You can compose two maps of the $3$-sphere by gluing two cubes along one of their boundaries. When you look at the Hopf invariant, it obviously adds, since it is the linking number of the preimage in one cube plus the linking number of the preimage in the other cube, – Cheerful Parsnip Oct 26 '19 at 00:03
  • @JohnHughes Corrected. – Paul Frost Oct 26 '19 at 00:03
  • @CheerfulParsnip I think I understand what you mean but it would help a lot to know how to write this composition out explicitly. E.g. the Hopf map can be written as $(x_1,x_2,x_3,x_4) \rightarrow (2(x_1x_2+x_3x_4), 2(x_1x_4-x_2x_3), (x_1^2 + x_3^2 - x_2^2 - x_4^2))$. How would I compose this algebraically to form a map with Hopf invariant $2$? – Jack Holmes Oct 26 '19 at 00:10
  • What is your question: Whether the Hopf-map $h : S^3 \to S^2$ generates $\pi_3(S^2)$ or how the multiples $n [h] \in \pi_3(S^2)$ look like? – Paul Frost Oct 26 '19 at 23:10
  • @PaulFrost The latter. – Jack Holmes Oct 28 '19 at 04:11

2 Answers2

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Suppose we write $S^3$ as the set of all pairs $(z, w)$ of complex numbers with $|z|^2 + |w|^2 = 1$, and write $S^2$ as the complex plane plus a point at infinity. The the Hopf map is just $$ (z, w) \mapsto \frac{z}{w}. $$ (The formula you gave is, I believe, pretty much the same as this, but after stereographic projection from the plane to 3-space.

See below for some rethinking of this paragraph

To get a degree-two map, we can simply take a degree two map from $S^2$ to $S^2$ and compose. In particular, we can define $$ H_2(z, w)= \left(\frac{z}{w}\right)^2 $$ and more generally, define $$ H_k(z, w)= \left(\frac{z}{w}\right)^k $$ for every integer $k$. The map $k \mapsto H_k$ defines an injective (use the Hopf invariant!) map from $\Bbb Z$ to $\pi_3(S^2)$. And $H_1 = H$ is a generator, because the Hopf invariant for that map is a $1$, which is a unit.

rethinking Connor Malin's remark, and Trebor's later remark made me go back and rethink this a bit.

Certainly another way to do things is to consider $S^3$ not only as pairs of complex numbers, but also as quaternions (the pair $(z, w)$ corresponds to the quaternion $z + wj$, where $j$ is the second of the three "imaginary generators" of the quaternions).

Now we can define a map $P_k$ from $S^3$ to $S^3$ of degree $k$ via $q \mapsto q^k$, and then compose the Hopf map with this, i.e., define $L_k(q) = H(P_k(q))$. I think that this map is somewhat more certain to represent the element $k \in \pi_3(S^2)$ than is $H_k$, but I confess I'm not up to writing out the details.

If the two maps really DO both represent the element $2\in \pi_3(S^2)$, then it should be possible to see a homotopy between them, but I'm definitely too tired to consider that.

John Hughes
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    You should be careful about the distinction between precomposing and postcomposing with degree m maps of spheres in this context. Given a map in $\pi_k (S^n)$, it is not generally true that precomposing by a degree m map is the same as postcomposing. In particular, for the Hopf map postcomposition by a map of degree -1 has no effect even though the Hopf map is not its own inverse. Probably safer to just precompose by a degree n map, as that is how addition in the homotopy groups is defined. – Connor Malin Oct 26 '19 at 02:22
  • Though it always agrees after a suspension. – Connor Malin Oct 26 '19 at 02:23
  • Sorry for taking a while to thank you. I have been familiarising myself with the consequences of your answer. Thanks again! – Jack Holmes Oct 30 '19 at 01:19
  • The Hopf invariant of $H_k$ is $k^2$, if my calculations are correct. So in fact this is not an injective map. – Trebor Dec 13 '25 at 21:48
  • Thanks, Trebor. You may well be right. Although I wonder how that could be consistent with Conner's remark about suspensions. – John Hughes Dec 14 '25 at 00:12
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The Hopf map is the projection map of a bundle $ S^1 \rightarrow S^3 \rightarrow S^2$. From this fiber bundle we get a long exact sequence. This sequence splits up into exact sequences $0 \rightarrow \pi_n (S^3) \rightarrow \pi_n (S^2) \rightarrow 0$ for $n>2$ because the higher homotopy groups of a circle vanish. In particular, this isomorphism is coming from post composition with the Hopf map. Specializing to $n=3$, this isomorphism must take any generator of $\pi_3 (S^3)$ to a generator of $\pi_3 (S^2)$. Since a generator of the first is the identity, the Hopf map must generate $\pi_3 (S^2)$.

Connor Malin
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    The question explicitly says that OP knows all this, but wants an explicit description of the multiples of the Hopf map. Your answer seems ... baffling, as if you hadn't read the question. – John Hughes Oct 26 '19 at 01:41
  • If you read the question carefully, he says that he knows this argument proves there is an isomorphism. He then says he’d like an argument as to why the Hopf map generates the group. This is what I addressed in my answer. Hope that debaffles you. – Connor Malin Oct 26 '19 at 02:14
  • Thanks, Connor. – John Hughes Dec 14 '25 at 00:10