Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

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

1 Answer 1

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 http://www.math.upenn.edu/~wziller/math650/LieGroupsReps.pdf.

share|improve this answer
    
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 at 22:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.