Faithful representations of non-amenable Lie groups [duplicate]

Question

Can a non-amenable connected Lie group have a faithful finite-dimensional unitary representation?

Answer 1

You have an extension $1\rightarrow R\rightarrow G\rightarrow S\rightarrow 1$ where $R$ is a solvable subgroup and $S$ semisimple by using the Levi decomposition.

Let $f:G\rightarrow U(n)$ be a faithful representation. The restriction of $f$ to $S$ is also faithful. Consider the Iwasawa decomposition of $S=KAN$ where $K$ is compact, $A$ abelian and $N$ nilpotent. Remark that $AN$ is solvable and preserves the hermitian product of $C^n$. This implies that $AN$ is abelian: to see this: remark that there exists a vector $u$ of $C^n$ such that $u$ is an eigenvector for every $g\in AN$, this implies that $AN$ preserves the orthogonal subspace of $Cu$ and you can conclude by using a recursive argument.

The Iwasawa decomposition of $S$ is obtained from the Cartan decomposition of the Lie algebra ${\cal S}=p+l=p+a+n$ of $S$ which verifies $[p,p]\subset l, [l,l]\subset p, [l,p]\subset l$, and $a\subset l$, if $l$ is abelian, we deduce that $l=a$ is normal in ${\cal S}$ so $AN=A$ is a normal subgroup, this implies that there exists an exact sequence $1\rightarrow A\rightarrow S\rightarrow K\rightarrow 1$, since $K$ and $A$ are amenable, we deduce that $S$ is amenable. There exists also an exact sequence $1\rightarrow R\rightarrow G\rightarrow S\rightarrow 1$. Since $R$ and $S$ are amenable, we deduce that $G$ is amenable. Contradiction. Thus such a faithful representation does not exist.