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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?)2010-12-27
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    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 (http://en.wikipedia.org/wiki/Quasiperiodic_motion).2010-12-27
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    This might be a dumb question, but what is $\tau^{{\mathbb Z}}$? I haven't seen this notation before, I don't think.2010-12-27
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    Called the *irrational torus* according to wikipedia (http://en.wikipedia.org/w/index.php?title=Diffeology&oldid=325733774#Examples)2010-12-27
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    I'm assuming that $\tau^{\mathbb{Z}}$ refers to the subgroup of integer powers of $\tau$. Writing $\tau=e^{2\pi i\alpha}$, this is equivalent to $\mathbb{R}/(\mathbb{Z}+\alpha\mathbb{Z})$.2010-12-27
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    ...also called an irrational foliation (Kronecker foliation) of the torus: http://en.wikipedia.org/w/index.php?title=Foliation&oldid=399033782#Kronecker_foliation2010-12-28
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    @George: the Kronecker foliation happens in the two-dimensional torus. Here, it seems, we are talking about $\mathbb R/(\mathbb Z+\alpha\mathbb Z)$.2010-12-28
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    @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.2010-12-28
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    @Mariano: Aren't they the same thing? The two dim torus quotiented out by the lines of slope $\alpha$ is $X=\mathbb{R}^2/(\mathbb{Z}^2+\mathbb{R}(\alpha,1))$. The map $\mathbb{R}\to X$ given by $x\mapsto(x,0)$ is onto with kernel $\mathbb{Z}+\alpha\mathbb{Z}$. So $X\cong\mathbb{R}/(\mathbb{Z}+\alpha\mathbb{Z})$. I'm not sure about the topology though.2010-12-28
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    @George: the Kronecker foliation is not the quotient space of the torus by that subgroup, it is *the foliation* by the cosets of that subgroup. The second is an abelian group with an useless topology; the first one is, well, a foliation.2010-12-28

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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.