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Mar 31, 2011 at 2:08 vote accept Andreas Holmstrom
Mar 30, 2011 at 20:33 comment added Faisal I don't really know much about abelian varieties, but I can offer a comment that might help. If $G = \mathbb{C}^g/\mathbb{Z}^{2g}$ is a complex torus, then its Lie algebra can be identified with $\mathbb{C}^g$ in such a way that the exponential map may be thought of as the quotient map $\mathbb{C}^g \to G$; in particular, the kernel of exp is $\mathbb{Z}^{2g}$. On the other hand, $G$ is homeomorphic to a product of $2g$ copies of the unit circle, so that $H_1(G,\mathbb{Z}) = \mathbb{Z}^{2g}$.
Mar 30, 2011 at 20:28 answer added mephisto timeline score: 16
Mar 30, 2011 at 19:55 answer added Simon Rose timeline score: 0
Mar 30, 2011 at 19:12 history asked Andreas Holmstrom CC BY-SA 2.5