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Let $\tau \in \mathbb{S}^1$ be such that $\tau$ is not a root of unity. Let $E_\tau$ be the quotient space $S^1/\tau^{\mathbb Z}$. Consider it as a pointed space with basepoint the equivalence class of $1$. This is a path-connected space.

What is the fundamental group of this space?

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I may be exposing my ignorance here, but: why is $E_{\tau}$ called a "noncommutative torus"? Isn't it a commutative topological group? (In particular, because it's a topological group, its $\pi_1$ should be commutative, right?) – Pete L. Clark Dec 27 '10 at 23:18
I don't know why it is called a noncommutative torus, but I recognize it as the space of paths of a dynamical system in quasi-periodic motion ( – George Lowther Dec 27 '10 at 23:40
This might be a dumb question, but what is $\tau^{{\mathbb Z}}$? I haven't seen this notation before, I don't think. – james Dec 27 '10 at 23:50
Called the irrational torus according to wikipedia (…) – George Lowther Dec 27 '10 at 23:57
@Pete: the "usual" non-comm. torus is the crossed product $C^*$-algebra $C(S^1)\rtimes\mathbb Z$ with $\mathbb Z$ acting on $C(S^1)$ by rotations of angle $\tau$ of $S^1$: it is morally, à la Connes, the algebra of functions on the "non-comm. space" $S^1/(\tau)$. Calling the actual quotient space $S^1/(\tau)$ a non-comm. torus is probably motivated by this, but, ultimately, wrong. – Mariano Suárez-Alvarez Dec 28 '10 at 4:31

1 Answer 1

Can you see what the topology of the quotient $S^1/(\tau)$ is? Its open sets "are" the open sets in $S^1$ which are invariant under multiplication by $\tau$.

Once you see what the topology is, your question becomes easy.

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