Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Please let me refer you to:

Example 4.18. The skew line $f: \mathbb R \to S^1 \times S^1$ $$ f(t) = (e^{it}, e^{i\alpha t}). $$ If $\alpha$ is irrational then the image of $f$ is dense in $S^1 \times S^1$, so if $V$ is an open neighborhood of $f(t)$ in $S^1 \times S^1$, then $$ \overline{V \cap f(\mathbb R)} = V $$ so $V \cap f(\mathbb R) \neq f(U)$.

(Source:, Page $16$, example $4.18$.)

Why is the image of the skew line dense (assuming $\alpha$ is not rational)?

share|cite|improve this question
This is called Kronecker winding on the torus. Have a look here: – Giuseppe Negro Oct 5 '11 at 22:37
up vote 1 down vote accepted

Essentially because, in $S^1$, $e^{it}=e^{i(t+2\pi n)}$ for integer $n$, and the fractional part of $\alpha n$ is dense in $[0,1)$.

share|cite|improve this answer
Can you please explain more? – user17182 Oct 5 '11 at 23:03
@user17182: which bit? – Henry Oct 5 '11 at 23:34
why the result follows from what you wrote? I don't see it – user17182 Oct 6 '11 at 2:26
@user17182: If you are looking close to $\left(e^{i2\pi x},e^{i2\pi y}\right)$, consider $\left(e^{i2\pi (x+n)},e^{i2\pi (\alpha x+\alpha n)}\right)$ for integer $n$. – Henry Oct 6 '11 at 6:15

Hint: It is enough to show that the intersection with $\{1\}\times S^1$ is dense in $S^1$. Because $\alpha$ is irrational, this intersection is infinite, so by compactness of $S^1$ it contains points arbitrarily close together. Take two such points and subtract their parameter values ...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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