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How can I find the deck transformations of $p:\mathbb R\to S^1$, where $p(t)=e^{2\pi ti}$? I tried in this way:

Let $\phi:\mathbb R\to \mathbb R$ be a deck transformation, so $p(\phi(t))=p(t)$ for all $t\in \mathbb R$.


$e^{2\pi i\phi(t)}=e^{2\pi it} \implies cos(2\pi\phi(t))+isin(2\pi\phi(t))= cos(2\pi t)+isin(2\pi t)$

$\implies cos(2\pi\phi(t))=cos(2\pi t)$ and $sin(2\pi\phi(t))=sin(2\pi t)$

Then $\phi(t)=kt$, for all $k\in \mathbb Z$. Since $\phi$ is a covering automorphism, then the automorphism group of the covering is $G(\mathbb R)=\mathbb Z$.

My approach is correct?


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Instead of $\phi(t)=kt$, it should be $\phi(t)=t+k$, because $cos(2\pi t)$ is equal to $\cos(2\pi t+2\pi k)$ but not to $\cos(2\pi kt)$ (and similarly for $\sin$). – Andreas Blass Jan 12 '13 at 20:16
@AndreasBlass yes, of course. Thank you for your comment. – user42912 Jan 12 '13 at 22:54
up vote 3 down vote accepted

To find Deck transformations, you only have to know that $\cos(x_1)=\cos(x_0)$ and $\sin(x_1)=\sin(x_0)$ is equivalent to $x_1=x_0+ 2k \pi$ for some $k \in \mathbb{Z}$. So your transformations are $G=\{ \phi_k : t \mapsto t+k \ | \ k \in \mathbb{Z} \}$ and you can verify that $\phi_k \mapsto k$ is an isomorphism between the groups $G$ and $\mathbb{Z}$.

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