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could any one give me a hint how to show a complex torus has topological genus one by constructing an explicit homeomorphism to $S^1\times S^1$? Complex Torus: $\mathbb{C}/L$, where $L=\{\mathbb{Z}\omega_1+\mathbb{Z}\omega_2\}$. Thank you.

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What definition of "complex torus" are you using? –  Chris Eagle Jul 25 '12 at 8:50
    
@ChrisEagle, edited –  El Angel Exterminador Jul 25 '12 at 9:00
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For your definition, this is more or less trivial. Choose coordinate axes on $\mathbb{C}$ parallel to $\omega_1$ and $\omega_2$, and then there is an obvious map $\mathbb{C} / L \to S^1 \times S^1$. –  Zhen Lin Jul 25 '12 at 9:02
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@Zhen: that can probably be an answer. –  Willie Wong Jul 25 '12 at 9:16
    
well, let $(x_1,y_1)+L=z_1+L$ be a point in $\mathbb{C}/L$, then the map is $(x_1,y_1)+\mathbb{Z}e_1+\mathbb{Z}e_2\mapsto (x_1/||x_1||,e_1)\times(y_1/||y_1||,e_2)$?where $e_1=(1,0),e_2=(0,1)$ –  El Angel Exterminador Jul 25 '12 at 9:26

2 Answers 2

up vote 3 down vote accepted

For convenience, I will identify $S^1$ with the topological group $\mathbb{R} / \mathbb{Z}$. Let $\Lambda$ be the lattice $\mathbb{Z} \omega_1 + \mathbb{Z} \omega_2$. Since $\omega_1$ and $\omega_2$ are not parallel, there exists a choice of (real!) coordinates $(x, y)$ on $\mathbb{C}$ such that $\omega_1 = (1, 0)$ and $\omega_2 = (0, 1)$. It is clear that there is a continuous (indeed, smooth) group homomorphism $\mathbb{C} \to S^1 \times S^1$ given by the formula $$(x, y) \mapsto (x + \mathbb{Z}, y + \mathbb{Z})$$ and $\Lambda$ lies in the kernel of this homomorphism, so it factors as a homomorphism $\mathbb{C} / \Lambda \to S^1 \times S^1$. It is not hard to check that this is bijective. On the other hand, $\mathbb{C} / \Lambda$ is compact and $S^1 \times S^1$ is Hausdorff, so this is not just a continuous bijection but a homeomorphism.

There is a much more exciting answer involving Weierstrass $\wp$-functions, but for that one should take a different definition of "genus one" and/or "complex torus"...

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From a topological viewpoint $\Bbb C/L$ is indistinguishable from $\Bbb R^2/\Bbb Z^2$. But then it should be clear that $$ \frac{\Bbb R^2}{\Bbb Z^2}\simeq\frac{\Bbb R}{\Bbb Z}\times\frac{\Bbb R}{\Bbb Z} \simeq S^1\times S^1. $$

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sweet answer... –  El Angel Exterminador Jul 25 '12 at 9:49

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