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Let $G$ be an algebraic torus or an abelian variety over the complex numbers. Then $G(\mathbb{C})$ is a complex Lie group. Is it true that we have the following exact sequence ?

$ 0 \to H_1(G(\mathbb{C}),\mathbb{Z}) \xrightarrow{\alpha} Lie(G(\mathbb{C})) \xrightarrow{\exp} G(\mathbb{C}) \to 0 $

Examples :

1) If $G=\mathbb{C}^{*}$, the exact sequence is just

$ 0 \to \mathbb{Z} \xrightarrow{2\pi i} \mathbb{C} \xrightarrow{\exp} \mathbb{C}^{*} \to 0 $

2) If $G$ is an abelian variety of dimension $d$, then $G(\mathbb{C})=\mathbb{C}^n/\Lambda$ where $\Lambda $ is a maximal lattice of $\mathbb{C}^n $. The exact sequence is

$ 0 \to \Lambda \to \mathbb{C}^n \to \mathbb{C}^n/\Lambda \to 0 $

Could you please also explain to me why the map $\alpha$ and the exponential map are what they are in the above cases ? Moreover what are their descriptions in general ? I think I know the description of the exponential map in general, but I don't understand why it reduces to the quotient map in example 2. Thank you very much.

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Any real or complex Lie group $G$ has a universal cover $\widetilde{G}$ on which the fundamental group $\pi_1(G)$ acts with quotient $G$. This always comes from a short exact sequence

$$1 \to \pi_1(G) \to \widetilde{G} \to G \to 1.$$

Because $G$ is a topological group, the natural map $\pi_1(G) \to H_1(G)$ is an isomorphism, and if $G$ is abelian, then the exponential map $\mathfrak{g} \to \widetilde{G}$ is also an isomorphism.

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  • $\begingroup$ Thank you for your answer. I wonder if you could give me reference for the statement " If $G$ is abelian, $\mathfrak{g}\to \tilde{G}$ is an isomorphism." $\endgroup$ – raynor14 Feb 18 '15 at 0:23
  • $\begingroup$ @raynor: if $G$ is abelian, then the exponential map is a homomorphism, and $\mathfrak{g}$ is simply connected (above my Lie groups are always connected), so it must be the universal cover. $\endgroup$ – Qiaochu Yuan Feb 18 '15 at 8:36

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