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O'Neill's Elementary Differential Geometry, problem 4.3.11, page 156, (Kindle edition), asks the student to discuss the periodicity of the curve $x(at,bt)$ where x is the parametrization of the torus:

$$x(u,v)=((R+r \cos(u))\cos(v),(R+r \cos(u))\sin(v),r\sin(u))$$

And $a/b$ is a rational number $m/n$. I got as far as convincing myself that $2\pi m/a=2\pi n/b$ is a period, where $m$ and $n$ have no common divisor.

$$x(t)=((R+r\cos(at)\cos(bt),(R+r\cos(at))\sin(bt),r\sin(at))$$

Any period of the curve is an integer multiple of the least period of the third component, $2\pi/a$. Since $R>r$, $R+r\cos(at)>0$ and any period is also an integer multiple of $2\pi/b$. The ratio of these two integers is $a/b=m/n$, so the two integers are $m$ and $n$ for the least period. Does this sound correct?

Problem goes on to ask the student to show that the curve $x$ has no self-crossings. According to the first edition of the book, this means that form $x(t)=x(t+T)$ follows that $T$ is a multiple of the least period. How to show that this is the case?

Problem goes on to ask the student to show that if $a/b$ is irrational then $x(at,bt)$ is one-to-one. I guess the thing to do is to show that if it is not one-to-one then $a/b$ is rational. I have no idea how to do this and would like a hint.

This last case is interesting because $x$ comes arbitrarily close to every point on the torus.

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Your reasoning about period of curve seems correct.

To answer second question, about self-crossing, you should again inspect components of curve, and with similar reasoning deduce that $T$ is multiple of period.

Third part of problem now follows from first two. If $x$ is not one-to-one, than there is some point of self intersection (e.g for some $t$ and $T$ holds $x(t)=x(t+T)$). From second part we conclude that $T$ is multiple of period. Now from first part follows that period is rational.

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  • $\begingroup$ "From second part we conclude that T is multiple of period." I don't get this. Here is my reasoning. Is it correct? The second part is "If a/b is rational and x(t)=x(t+T) then T is a multiple of the period." We have x(t)=x(t+T) but we don't have "a/b is rational." That is what we are trying to prove. $\endgroup$ – Gene Naden Sep 9 '18 at 9:29
  • $\begingroup$ So to prove the second part, I have $sin(at)=sin(a(t+T))$ and $cos(at)=cos(a(t+T))$. From this I would like to conclude $aT=2\pi n$. Is this valid? I drew a unit circle and put in angles, and it seems to be true. $\endgroup$ – Gene Naden Sep 9 '18 at 10:25
  • $\begingroup$ So in proving the second part, which is "a/b rational and x(t+T)=x(t) $\Rightarrow$ T is a multiple of P", I got as far as showing that T is an integer multiple of $2\pi/a$ and an integer multiple of $2\pi/b$, with the ratio of these two integers a/b=m/n. I would like to say that T is a common multiple of $2\pi/a$ and $2\pi/b$, and that P is the LCM of $2\pi/a$ and $2\pi/b$, so T is a multiple of P. But this argument talks about the LCM of numbers that are not necessarily integers, so how can it be valid? $\endgroup$ – Gene Naden Sep 9 '18 at 19:45

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