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My question is in the title, do you know if SL(2,C) is an unimodular group or not and how to prove it ?

Thank you.

théo jamin
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1 Answers1

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I found the answer in the article of JÖRG WINKELMANN : "Complex analytic geometry of complex parallelizable manifolds" :

Recall that a locally compact topological group $G$ is called unimodular iff a left-invariant Haar measure is also right-invariant.

Assume $G$ is a locally compact topological group.

Page $5$, J. Winkelmann proved that :

  • Let $\Gamma$ be a cocompact discrete subgroup of $G$. Then $G/\Gamma$ admits a $G$ invariant probability measure.

  • Let $\Gamma$ be a discrete subgroup of $G$. Assume that there exists a $G$-invariant probability measure on $G/\Gamma$ then $G$ is unimodular.

Finally, the existence of lattices in $SL_2(\mathbb{C})$ is given by the existence of compact $3$-dimensional hyperbolic manifold and thus we know that $SL_2(\mathbb{C})$ is unimodular.

I hope I did not make mistakes.

théo jamin
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    This works but is unnecessarily complicated. Instead, use the fact that $G=SL(2,C)$ has no nontrivial characters. Then equip $G$ with a left-invariant volume form and prove that it is also right-invariant. – Moishe Kohan Feb 01 '19 at 18:41
  • ...or you could use a hardly modified version of this answer: https://math.stackexchange.com/a/2820058/101420 – Vincent Jun 13 '19 at 10:54