I have an exercise in topology and I am struggling with it:

a) Proof that a homeomorphism $S^1 \times S^1\to \mathbb T ^2$ exists.

b) For which $p,q \in \mathbb R$ is the curve $$\gamma(p,q):=\mathbb{R}\to S^1 \times S^1 , t\mapsto (\exp(2* \pi*t*p),\exp(2*\pi*t*q) $$ injective?

c) Draw a sketch of the curve $\gamma(2,3)$ for $r=1$ and $R=2$.

I think the answer for a) is that $$f((a,b),(c,d))=((c+R)*a,(c+R)*b,d*r)$$ is the requested homeomorphism with the inverse $$f^{-1}(a,b,c)= (\frac{a}{\sqrt{a^2+b^2}}, \frac{y}{\sqrt{a^2+b^2}},\frac{\sqrt{a^2+b^2}-R}{r},\frac{c}{r})$$.

My Problem is b) and c). In b) I think that I have to choose $p$ and $q$ so that the $\exp$ is not periodic, is this right? If it is right how can I find this $p$ and $q$ and if it is nit true, what I have to do? And at the moment im completley confused which effect the parameter have for the sketch in c).

Thank you :)

Definition: $\mathbb T^2$ is the surface of revolution generated by revolving the circle $\{(x,0,z):(x-R)^2+z^2=r^2\}$ around the $z$-axis.

  • $\begingroup$ Your question is not complete unless you tell us what definition of $\mathbb T^2$ you are using. This is important not just to understand your equations, which seem somewhat random, because also because one of the most common definitions, $\mathbb T^2$ is defined to be $S^1 \times S^1$. $\endgroup$ – Lee Mosher Nov 4 '19 at 22:44
  • $\begingroup$ I added the definition. $\endgroup$ – alpaka123 Nov 4 '19 at 23:03

For b. As for $q=0$ or $p=0$ is clear that the map is not injective, we may assume $pq\neq 0$. Note that if $\gamma$ is not inyective there should exist $t_1,t_2$ such that (I added an $i$ that I think you forgot).

$$ (e^{2\pi it_1 p},e^{2\pi it_1 q})=(e^{2\pi it_2 p},e^{2\pi it_2 q})\quad \Rightarrow \quad e^{2\pi i(t_1-t_2) p}=1=e^{2\pi i(t_1-t_2) q} $$

Therefore $(t_1-t_2) p=n\in \mathbb Z$ and $(t_1-t_2) q=m\in \mathbb Z$. Therefore $p/q=(n/m)$ is rational. Which proves that $\gamma$ is NOT injective if $p/q$ is rational. I leave to you the proof of the reverse implication but the final result is $\gamma$ is injective if and only if $p/q$ is irrational.

That gives a fairly good idea for c. As (2/3) is rational $\gamma$ is not injective. Moreover you have to look for the smallest $t$ such that both $2t$ and $3t$ are integers. 2,3 are co-prime then the smallest such $t$ is 1 and the image $\gamma(2,3)(\mathbb R)$ is equal to $\gamma(2,3)([0,1])$. Lastly noting than $e^{2\pi i 2t},e^{2\pi i 3t}$ covers the circle two and three times for $0\leq t\leq 1$ respectively. you see the plot is a line that winds around one direction two times and around the other three times.

I guess that's enough to draw it but if you have troubles let me know :)


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.