I don't have much background in Lie theory (especially about different kinds of Lie algebras), hence I'm struggling a bit with the following:
Every connected Lie group $G$ with semi-simple Lie algebra $\mathfrak g$ is unimodular.
We know the modular function $\Delta_G\colon G\to \Bbb{R}_{>0}$ is continuous (as for any lc group), thus a smooth Lie group homomorphism. Using the exponential map $\Delta_G \circ \exp_G = \exp_{\Bbb{R}} \circ\, d\Delta_G$ and thus it would suffice to show that $\ker(d\Delta_G) = \mathfrak g$. I guess now there is a pretty basic argument to see this (i mean, $\Bbb R$ is a pretty basic Lie algebra), but it's a bit late and I'm not too familiar with the definitions. Maybe you could help me finish this argument?
Best Answer
The point is just that a semisimple Lie algebra is also a perfect Lie algebra, meaning its abelianization $\mathfrak{g}/[\mathfrak{g}, \mathfrak{g}]$ is trivial. So $\mathfrak{g}$ does not admit any nontrivial homomorphisms to an abelian Lie algebra, meaning $G$ does not admit any nontrivial homomorphisms to a connected abelian Lie group.