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I know that the second homotopy group of $S^2$ is $\mathbb{Z}$. I also know that a simple representative of the class of maps that generates this group is the identity map on the sphere, $i:S^2\rightarrow S^2, p \rightarrow p$ In other words, this map has degree 1. My question is, are there simple representatives of the other elements of this group, and if so what are they? For instance, what is an example of map $S^2 \rightarrow S^2$ with degree 2? What about for arbitrary degree $n\in \mathbb{Z}$?

Thanks in advance!

Tom
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3 Answers3

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If $n$ is an integer, the complex power map $z \mapsto z^{n}$ extends to a map of the Riemann sphere $\mathbf{C} \cup \{\infty\}$ having degree $n$. (It's understood that if $n = 0$, then $z^{n} = 0$ for all $z$, while if $n > 0$ then $\infty^{n} = \infty$, and if $n < 0$, then $\infty^{n} = 0$ and $0^{n} = \infty$. These conventions do not constitute an unqualified endorsement of the equations "$1/0 = \infty$", "$1/\infty = 0$", or "$\infty^{0} = 1$".)

Visually, it may help to restrict the power mapping to the unit circle ("wind the circle around itself $n$ times"), then to suspend.

Andrew D. Hwang
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The suspension of the map $z\to z^n$ of $\mathbb{S}^1$ is a map of $\mathbb{S}^2$ of degree $n$. Geometrically, this wraps each latitudinal line of the sphere around itself $n$ times and maps the north pole to itself and the south pole to itself. Analytically, this is the extension of the complex function $z\to z^n$ to the Riemann sphere $\mathbb{S}^2$.

Hope this helps!

Amitesh Datta
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One way to do this is to take the degree $n$ map $f : S^1 \to S^1$ given by $e^{iz} \mapsto e^{inz}$ and extending this to the map $g = f\times I : S^1 \times I \to S^1 \times I$ given by $(e^{iz}, t) = (e^{inz}, t)$, compose with an appropriate homeomorphism $S^1 \times I \cong S^2 - \{(0, 0, 1), (0, 0, -1)\}$ and finally extending continuously to the poles. An explicit formula can be a bit wacky.

This is essentially the construction of the suspension map $Sf : S^2 \to S^2$. This works because, by Hurewicz theorem, $\pi_2(S^2) \cong H_2(S^2)$. We know that $H_i(SX) = H_{i+1}(X)$, which implies that every element in $\pi_2(S^2)$ is obtained from suspending maps that generate $H_1(S^1) \cong \pi_1(S^1)$.

Balarka Sen
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