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Suppose we have two Lie groups $G$ and $H$ and a covering map $f:G\rightarrow H$ with a finite number of sheets which is a Lie group homomorphism.

I want to know whether $G$ is unimodular whenever $H$ is unimodular.

I'd like to know this because I am trying to prove that $SL_n(\mathbb C)$ is unimodular. I have managed to come up with a map

$$SL_n(\mathbb C)\times\mathbb C^*\xrightarrow \phi GL_n(\mathbb C),$$

given by $(A,z)\mapsto zA$. Then $\phi$ turns out to be a Lie group homomorphism which is also a covering map of $n$ sheets. I have also showed that $GL_n(\mathbb C)$ is unimodular by defining a measure which is an integral of a complex form(of type $(n^2,n^2)$) of $\mathbb C^{n^2}$. This sort of imitates an argument found in Serge Lang's book $SL_2(\mathbb R)$.

So if this argument using covering maps works, then $SL_n(\mathbb C)\times\mathbb C^*$ being unimodular implies $SL_n(\mathbb C)$ is unimodular.

Is there another way to show $SL_n(\mathbb C)$ is unimodular using $\phi$?

I'd like to have a sort of elementary answer, please, as I don't know that much about Lie groups. I just know what a Lie group is, what a left invariant measure is, and the definition of the Lie algebra.

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1 Answer 1

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Let $G$ be a Lie group and ${\cal G}$ its Lie algebra, $G$ is unimodular if and only if $tr(ad(g))=0$ for every $g\in{\cal G}$ where $tr(ad(g))$ is the trace of $ad(g):{\cal G}\rightarrow {\cal G}$ defined by $ad(g)(x)=[g,x]$. Let $G\rightarrow H$ be a finite covering, the Lie algebras of ${\cal G}$ and ${\cal H}$ are isomorphic, we deduce that $G$ is unimodular if and only if $H$ is unimodular.

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  • $\begingroup$ nice ${}{}{}{}{}$ $\endgroup$ Oct 2, 2018 at 0:35

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