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.


1 Answer 1


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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