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Let $$1 \rightarrow (\mathbb{C}^{\times})^r \rightarrow G \rightarrow V/L \rightarrow 0$$

be an extension of complex lie groups, where $L$ is a lattice in a complex vector space $V$.

Is $G$ necessarily commutative?

safety stegosaurus
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1 Answers1

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$G$ acts on $(\mathbb{C}^\times)^r$ by conjugation, but $(\mathbb{C}^\times)^r$ is commutative so the action factors through the quotient. However the automorphism group of $(\mathbb{C}^\times)^r$ is discrete (it's $GL_r(\mathbb{Z})$) so the action of $V/L$ must be trivial and therefore the group is commutative.

Nate
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  • Dear @Nate, It seems to me that the argument is incomplete. For example, let $D_8$ be the dihedral group of order $8$, $Z$ be its center, which is of order $2$. Then $D_8/Z$ is of order $4$, so abelian. Then the exact sequence $0\to Z\to D_8\to D_8/Z\to0$ induces a trivial homomorphism $D_8\to Aut(Z)$ by $x\mapsto x\bullet x^{-1}$. However, $D_8$ is not commutative. – Doug Jan 28 '23 at 16:01
  • The result is indeed true, see Lemma 2.10 of https://arxiv.org/pdf/2103.07015.pdf – Doug Jan 30 '23 at 16:22
  • Yea. I don't know what I was thinking (4 years ago). All that argument really does is show that $(\mathbb{C}^\times)^r$ must be central. – Nate Jan 30 '23 at 23:48