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Let $S_1$ be the unit sphere in $\mathbb{R}^2$ and define the following map:

$$f_k: S^1 \mapsto S^1: (\cos(2\pi t), \sin(2\pi t))\mapsto (\cos(2k\pi t), \sin(2k\pi t)).$$

We are asked to calculate the degree of this map. De degree of $f$ is defined as follows: If $f: S^1 \mapsto S^1$, then we can take a look at its induced map: $f_*:H_1(S^1) \mapsto H_1(S^1)$. Since this is a map from $\mathbb{Z}$ to $\mathbb{Z}$, we can see this map as $\alpha \mapsto k\alpha$. In this case, $k$ is seen as the degree of the map $f$. I suppose the degree of this map equals $k$, but I have no clue on how to do this. Does someone know how to proceed?

If my guess is right, we could use this result to define a map $f: S^n \mapsto S^n$ with arbitrary degree. Since $H_n(S^n)=\mathbb{Z}$, we can use the suspension map of $f_k$, which has the same degree.

Kevin.S
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emiel647
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    What is your definition of degree? – Paul Frost Jan 06 '22 at 10:14
  • Try to first answer the easier question: Regard $f_{k}$ as a loop in $S^{1}$. Looking at the identity as the generator of $\pi_{1} (S^{1} ) \cong \mathbb{Z}$, what is the class $[f_{k}]$? How does $\pi_{1} (S^{1})$ relate to $H_{1} (S^{1}) $? – Tom Ariel Jan 06 '22 at 10:27
  • @Tom Ariel, If we consider $f_k$ as a loop, isn't $[f_k]$ equal to $\mathbb{Z}$ since every loop starting at $(1,0)$ is homotopic to any other loop starting at the same point? Furthermore, I do not see how the fundamental group and the first homology group are related. I know that they are both equal to $\mathbb{Z}, but I do not see any relation between the two. – emiel647 Jan 06 '22 at 11:10
  • How do you define degree in this line? This result can be used to define a map $f\colon S^n \to S^n$ with arbitrary degree* since the suspension map has the same degree.* – Sumanta Jan 06 '22 at 11:16
  • And how did you calculate $\pi_1(S^1)$? Via the covering projection $p : \mathbb R \to S^1$? – Paul Frost Jan 06 '22 at 12:55
  • As a note: How to find the degree of a map $f\colon M\to N$ between two closed connected orientable topological manifolds of dimension $n$? Suppose we have a closed $n$-ball $B$ embedded in $N$ with $f^{-1}(B)=\bigsqcup_{i=1}^n B_i$ where $B_i$ are closed $n$-balls embedded in $M$ such that each $f\big|B_i\to B$ is a homeomorphism. So, there exists an integer $\varepsilon_i\in {+1,-1}$ such that $H_n(f\big|B_i)\colon H_n(B_i,\partial B_i)\cong\Bbb Z\to \Bbb Z\cong H_n(B,\partial B)$ is multiplication by $\varepsilon_i$. Then $$\deg(f)=\sum_{i=1}^n\varepsilon_i.$$ – Sumanta Jan 07 '22 at 06:31

1 Answers1

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Triangulate $S^1$ as a $k$-gon, with edges $e_1,\cdots,e_k$ where $e_i$ is the arc: $$\left\{\left(\cos(2\pi t),\sin(2\pi t)\right)|\,\, t\in \left[ \frac{i-1}k,\frac ik\right]\right\}.$$ Then $(f_k)_*$ maps: $$e_i\mapsto \sum_{j=1}^k e_j,$$ for all $i$.

The generator of $H_1(S^1)$ is $\mu=\sum_{i=1}^k e_i$. Thus $$ (f_k)_*(\mu)=\sum_{i=1}^k (f_K)_*e_i=\sum_{i=1}^k \mu=k\mu, $$ as required.

tkf
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  • How do you know that $\mu$ is the generator of $H_1(S^1)$? – emiel647 Jan 07 '22 at 13:27
  • The definition of homology is kernel of outgoing boundary operator, modulo image of incoming boundary operator. In this case there is no incoming operator to worry about, so we need only find the kernel of the boundary operator. The boundary of an edge from vertex $u$ to vertex $v$ is $v-u$. Thus a given linear combination of the $e_i$ will lie in the kernel if and only if the coefficient on each $e_i$ equals the coefficient on $e_{i+1}$ (indicices taken cyclically modulo $k$). This is if and only if we have a multiple of $\mu$. – tkf Jan 07 '22 at 19:04
  • @tfk Thanks, this is clear to me now. However, I am looking for an answer to the problem which does not use triangulation, since it is not a part of our course. Do you think the map you defined can be viewed in an other way? For example with paths or another construction? – emiel647 Jan 07 '22 at 23:37
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    If you are familiar with the Hurewicz homomorphism, you can use the method provided by @TomAriel. A third way is to use De Rham cohomology: $$(f_k)^*dt=kdt.$$ – tkf Jan 08 '22 at 00:50