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For some $m \in \mathbb{Z}$, consider the map $$f: S^1 \rightarrow S^1: (\cos 2\pi t, \sin 2 \pi t) \mapsto (\cos 2\pi m t, \sin 2 \pi m t).$$

I want to compute the degree of $f$. My intuition tells me that it should be simply $m$, and that it can be solved with the Mayer-Vietoris sequence. However, I am struggling to find the solution.

For clarity the degree of a map $f: S^n \rightarrow S^n$ is defined as the integer $k$ such that the homomorphism induced on $H_n (S^n)$ induced by $f$ acts on generators $\alpha$ of $H_n (S^n)$ like $$f_*(\alpha) = m \alpha.$$

Geigercounter
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  • See here – A. Goodier Dec 25 '23 at 12:12
  • I don't understand the answer given there and I want to prove it using the ideas I have written above (if this approach is correct) – Geigercounter Dec 25 '23 at 12:17
  • If $m\neq 2$, then Mayer-Vietoris will probably not help you, since $f$ will not preserve the covering. What other tools do you have available? Do you know that $H_1$ is the abelianization of $\pi_1$? – Vincent Boelens Dec 26 '23 at 18:24
  • I don't understand why M-V won't work, could you give some more detail? – Geigercounter Dec 26 '23 at 19:08
  • Well I guess you want to use the upper hemisphere $S^1_+$ and the lower hemisphere $S^1_-$, but $f(S^1_+)=S^1=f(S^1_-)$, so you won't get a map of exact sequences after applying M-V. The same thing presumably also happens if you use another cover, since a contractible subset of $S^1$ is homeomorphic to an interval. – Vincent Boelens Dec 27 '23 at 13:13
  • So you only know singular homology, excision and M-V? Not even cellular homology? – Vincent Boelens Dec 28 '23 at 11:50

1 Answers1

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You said that you are working with Vick's "Homology Theory". In Chapter 1 he introduces singular homology theory and studies its properties.

In the example on p. 22 / 23 he proves that $H_1(S^1) \approx \mathbb Z$ and shows that the $1$-cycle $c + d$ with singular $1$-simplices $c , d : \sigma_1 \to S^1$ as depicted in Fig. 1.6 represents a generator $\omega$ of this group.

For the sake of convenience let $h : I = [0,1] \to \sigma_1, h(t) = (1-t)(1,0) + t(0,1)$, denote the obvious homeomorphism and let $$\bar c : I \to S^1, \bar c(t) = e^{\pi i t}, \\\bar d : I \to S^1, \bar d(t) = e^{\pi i t + \pi} .$$ Then $c = \bar c \circ h^{-1}, d = \bar d \circ h^{-1}$. Note that we regard $S^1$ as the subset of $\mathbb C$ consisting of all $z$ with $\lvert z \rvert = 1$.

Let us next prove the following

Lemma. Let $\bar u , \bar v : I \to X$ be two paths in a space $X$ such that $u(1) = v(0)$ and $$\bar w = \bar u * \bar v : I \to X, t \mapsto \begin{cases} \bar u(2t) & t \le 1/2 \\ \bar v(2t-1) & t \ge 1/2 \end{cases}$$ be the concatenated path. Let $u = \bar u \circ h^{-1}, v = \bar v \circ h^{-1}, w = \bar w \circ h^{-1}$ be the associated singular $1$-simplices. Then the $1$-chain $u + v - w$ (which is actually a $1$-cycle) is the boundary of a singular $2$-simplex in $X$.

Proof. The idea is this: Consider the standard $2$-simplex $\sigma_2$ with vertices $v_0 = (1,0,0), v_1 = (0,1,0), v_2 = (0,0,1)$. Map the line segment $[v_0,v_1]$ by $u$ to $X$, $[v_1,v_2]$ by $v$ to $X$ and $[v_0,v_2]$ by $w$ to $X$. You get a map from the boundary of $\sigma_2$ to $X$. It is easy to see that this map extends to $\sigma_2$ which proves the lemma.

The lemma shows that the $1$-cycles $c + d$ and $e = \bar e \circ h^{-1}$ with

$$\bar e : I \to S^1, \bar e(t) = e^{2\pi i t}$$ represent the same element in $H_1(S^1)$. Actually both represent the generator $\omega$.

For $k = 1, \ldots, m$ define paths $$\bar p_k : I \to S^1, \bar p_k(t) = e^{\frac{2\pi }{m} i t + \frac{2\pi (k-1)}{m} i} .$$

The induced singular $1$-simplices $p_k = \bar p_k \circ h^{-1}$ give a $1$-cycle

$$\gamma = p_1 + \ldots + p_m .$$

Applying the lemma inductively, we see that $\gamma$ and $e$ represent the same element in $H_1(S^1)$.

Using complex numbers, the map $f$ is given by $$f(z) = z^m .$$ We have $$f_*(\gamma) = f \circ p_1 + \ldots + f \circ p_m.$$ An easy computation shows that $$f \circ p_k = e, $$ thus $$f_*(\gamma) = me .$$ Going to homology we get $$f_*(\omega) = f_*([\gamma]) = [f_*(\gamma)] = [me] = m[e] = m\omega .$$ This shows that $f$ has degree $m$.

Paul Frost
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  • @Geigercounter See my update. – Paul Frost Dec 29 '23 at 10:11
  • Thanks! This is quite clear – Geigercounter Dec 29 '23 at 11:28
  • I don't see why you don't just consider the $1$-simplices instead of introducing $h$ etc. For example, it seems to me like the map $c: \sigma_1 \rightarrow S^1: (t_0,t_1) \mapsto (\cos \pi t_1, \sin \pi t_1)$ does the trick too (with similar for the other maps). Or am I missing something important by doing this? – Geigercounter Jan 14 '24 at 18:34
  • @Geigercounter You are right, you can write the singular $1$-simplex $e : \sigma_1 \to S^1$ in the form $e(t_0,t_1) = e^{2\pi it_1}$.This is due to the fact that $h(t) = (1-t,t)$. But with this "direct interpretion" the Lemma has to be reformulated by concatenating singular $1$-simplices. Of course this can easily be done, but I think the concatenation of paths is a more standard concept. – Paul Frost Jan 14 '24 at 23:28
  • I have proven the concatenation of singular 1-simplices before, so to me the proof is more streamlined in this way. Also I am confused on why you say 'going to homology' in the last part. Is this precisely because you are working with paths at the start? – Geigercounter Jan 15 '24 at 09:16
  • @Geigercounter $f_*(\gamma) = me$ is an equation on the chain level. Going to homology means that we consider the homology classes. – Paul Frost Jan 15 '24 at 10:26
  • Is this consistent with Vick's notation? On the pages you mentioned it seems like he doesn't write it like this... – Geigercounter Jan 15 '24 at 10:30
  • @Geigercounter Where precisely do you see an inconsistency? – Paul Frost Jan 15 '24 at 11:41
  • In the 'going to homology' part. Vick seems to suggest that $\omega = c+d$, but that is not consistent with your notation as $\omega = [c+d]$ which I agree with – Geigercounter Jan 15 '24 at 11:48
  • @Geigercounter Vick writes "we must represent $\omega$ by the sum of two chains, $c + d$." This may perhaps mislead to understand it as $\omega = c +d$. But $\omega$ is a homology class and $c + d$ is a cycle., so this does not make sense. What he means is that the homology class $\omega$ is represented by the cycle $c + d$, i.e. $\omega = [c + d]$. – Paul Frost Jan 15 '24 at 13:50