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The universal cover of the torus $T$ is the complex plane $\mathbb{C}$. If $p: \mathbb{C} \to T$ is the covering map, why is $p$ doubly periodic?

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    Is it because we see $\mathbb{C}$ as $\mathbb{R} \times \mathbb{R}$, and so we have a period in the $y$-direction and in the $x$-direction?2012-06-10
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    Well, the periods don't have to be in the horizontal and vertical direction: they can be any two linearly independent vectors over $\mathbb R$. But the real question is: why must any covering map be periodic at all? In the real-smooth ($C^{\infty}$) setting there are many covering maps $p\colon \mathbb R^2\to\mathbb T^2$ that are not periodic. But complex-analytic maps are much more rigid...2012-06-10
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    Oh, so this really *is* coming from the complex nature of this universal cover? But how?2012-06-10
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    Universal covering map is unique up to an automorphism of the covering space. A complex-analytic automorphism of $\mathbb C$ is of the form $az+b$, hence preserves double-periodicity (though it may change the periods themselves). So, as long as we have one doubly-periodic cover such as the quotient map, we know that all covering maps are doubly-periodic.2012-06-10

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Since $\mathbb C$ is simply connected, it is a universal covering space. Any two covering maps $\mathbb C\to\mathbb T$ are related by an automorphism of $\mathbb C$. Such automorphisms are linear, therefore preserve periodicity. So if one covering map is doubly periodic, all of them are. To get one such map, use the quotient map that comes from the definition of torus as the quotient of $\mathbb R^2$ by a lattice.