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Let $f: \mathbb{C}^g/L\to\mathbb{C}^{g'}/L'$ be an isogeny of complex tori, i.e. a surjective Lie group morphism with finite kernel.

Is it obvious that $g\ge g'$ ?

It is easy to show that $f$ is induced by a linear map $\mathbb{C}^g\to\mathbb{C}^{g'}$ that is injective, but I cannot see why this map should be an isomorphism. A surjective morphism of Lie group is not necessarily a submersion, right ?

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Do you mean the linear map $\mathbb C^g \to \mathbb C^{g'}$ is surjective? – Eric O. Korman Apr 10 '14 at 19:34
Or is your question if $g = g'$? – Eric O. Korman Apr 10 '14 at 19:39
$g=g'$, $g\ge g'$ or the linear map is surjective are equivalent, since it is injective. – Klaus Apr 10 '14 at 19:49
up vote 2 down vote accepted

In general, any Lie group homomorphism whose kernel is a discrete subgroup of the center is a normal covering. In particular it is a local isomorphism so in your case $g = g'$. For a proof, see proposition 1.19 of

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Thank you! But it seems a bit involved (the hypothesis being way more general than what we need). I wonder if there is a simpler argument. – Klaus Apr 11 '14 at 22:05

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