16
$\begingroup$

Suppose $S^n$ is an $n$-dimensional sphere.

Definition of the degree of a map: Let $f:S^n \to S^n$ be a continuous map. Then $f$ induces a homomorphism $f_{*}:H_n(S^n) \to H_n(S^n)$ . Considering the fact that $H_n(S^n) = \mathbb {Z}$ , we see that $f_*$ must be of the form $f_*(n)=an$ for some fixed integer $a$. This $a$ is then called the degree of $f$.

Question: For every $k \in {\mathbb Z}$ how does one construct a continuous map $f: S^n \to S^n$ with $\deg(f) = k$?

$\endgroup$
19
$\begingroup$

Here is another solution.

Claim 1: If $f:S^n \to S^n$ has degree $d$ then so does $\Sigma f: S^{n+1} \to S^{n+1}$

Proof: Use the Mayer-Vietrois sequence for $S^{n+1}$. Let $A$ be the complement of the North pole, and $B$ the complement of the South pole. Then $S^n \simeq A \cap B$ and the connecting map $\partial_*$ in the Mayer-Vietrois sequence is an isomorphism. We get the following commutative diagram

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\la}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} H_{n+1}(S^{n+1}) & \ra{\partial_*} & H_n\left(A \cap B\right) & \la{i_*} & H_n(S^n)\\ \da{\Sigma f_*} & & \da{} & & \da{f_*} \\ H_{n+1}(S^{n+1}) & \ra{\partial_*} & H_n\left(A \cap B\right) & \la{i_*} & H_n(S^n)\\ \end{array} $$

in which each horizontal map is an isomorphism. Thus $\Sigma f_* = \partial_*^{-1} i_* f_* i_*^{-1}\partial_*$ and applying homology shows that $\text{deg}(f) = \text{deg}(\Sigma f)$

Thus we are reduced to simply showing that there is a map $f:S^1 \to S^1$ of degree $k$. But this is just the winding number and it is (reasonably well known) that the map $z \mapsto z^k$ (where we view $S^1$ as the unit circle in $\mathbb{C}$) has degree $k$.

Finally I would direct you to have a look at Algebraic Topology by Hatcher:

  • Example 2.31 gives a direct construction of a map of arbitrary degree;
  • Example 2.32 works through the calculation of the map $f(z)=z^k$ proving it has degree $k$; and
  • Prop 2.33 gives another prove of Claim 1 above (which basically takes a different route to the same commutative diagram).
$\endgroup$
  • 2
    $\begingroup$ That is a great solution! $\endgroup$ – Asaf Karagila Feb 28 '12 at 11:06
  • $\begingroup$ @Yoav: Glad to be the one giving, rather than receiving help! $\endgroup$ – Juan S Mar 4 '12 at 22:46
  • $\begingroup$ Juan: Could you please give the definition of $\Sigma f$? $\endgroup$ – Asaf Karagila Mar 6 '12 at 2:27
  • $\begingroup$ @Asaf: This is just the suspension: en.wikipedia.org/wiki/Suspension_%28topology%29 $\endgroup$ – Juan S Mar 6 '12 at 2:44
12
$\begingroup$

Hint: Parameterize $S^n$ as $$g(t_1,\ldots,t_n)=\begin{pmatrix} \cos(t_1)\\ \sin(t_1)\cos(t_2)\\ \vdots\\ \sin(t_1)\cdots\sin(t_{n-1})\cos(t_n)\\ \sin(t_1)\cdots\sin(t_{n}) \end{pmatrix}$$ over $t_1,\ldots,t_{n-1}\in [0,\pi]$ and $t_n\in [0,2\pi)$, and define $f:S^n\to S^n$ by $$f(g(t_1,\ldots,t_{n-1},t_n))=g(t_1,\ldots,t_{n-1},kt_n)$$ which amounts to wrapping the sphere around itself $k$ times in the $x_nx_{n+1}$ plane. One way to show that this is the desired map is to show that it is continuous and then consider one of the simplices perpendiclar to the $x_nx_{n+1}$ plane, when you view the simplicial complex of $S^n$ as a subdivided version of $\partial I^{n+1}$ sitting in $E^{n+1}$ (for example, the top edge of a square, top face of a cube, etc).

$\endgroup$
  • $\begingroup$ I think $\cos(t)$ is suposed to be $\cos(t_1)$, right? $\endgroup$ – Integral Dec 3 '13 at 18:24
  • $\begingroup$ @Integral Yes, thanks $\endgroup$ – Alex Becker Dec 3 '13 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.