Faithful finite-dimensional unitary representations Is there any characterization of the non-compact connected Lie groups that possess faithful finite-dimensional unitary representations?
 A: Proposition. Equivalences ($G$ connected Lie group):

*

*(i) $G$ has a faithful finite-dimensional continuous unitary representation;


*(ii) $G$ is locally isomorphic to some compact Lie group;


*(iii) $G$ is direct product of some Euclidean group (=$\mathbf{R}^d$ for some $d$) with a compact Lie group.
$\bullet$ Indeed (iii) clearly implies (i).
$\bullet$ Suppose (ii), let $G$ be such a group. Write it as quotient of $V\times K$ with $V$ Euclidean group and $K$ compact semisimple, by a discrete central subgroup $Z$. Let $Z_1$ be the intersection of $Z$ with $V$, and $Z_2$ the projection of $Z$ on $V$. Since the kernel of the projection $Z\to Z_2$ is contained in the center of $K$ which is finite, $Z_1$ has finite index in $Z_2$, and hence both have the same span, say $W$, and let $M$ be a supplement subspace of $W$ in $V$. Then $G$ is direct product of $M$ and the compact group $(W\times K)/Z$. So (iii) holds.
$\bullet$ Now suppose (i).  Let $\mathfrak{g}$ be the Lie algebra, which thus embeds in $\mathfrak{u}(n)$ for some $n$, and $\mathfrak{r}$ the solvable radical of $\mathfrak{g}$.
It follows that for every $g\in \mathfrak{g}$, $\mathrm{ad}(g)$ is $\mathbf{C}$-diagonalizable. Hence $\mathfrak{g}$ has no subalgebra isomorphic to the non-abelian 2-dimensional Lie algebra. This already shows that $G$ modulo its solvable radical $R$ is compact [which is enough if you assume beforehand that $G$ is semisimple].
We can assume that $G$ is mapped injectively into $\mathrm{U}(n)$. Let $R$ be the solvable radical of $G$. Then its action can be written as a sum of irreducibles, which by Lie's theorem are 1-dimensional. Each of the eigenspaces determines a weight: a continuous homomorphism $\chi: R\to\mathbf{U}$. Hence $\mathbf{C}^n=\bigoplus_{\chi\in\Phi} V_\chi$, where $V_\chi=\{v:\forall r\in R:r.v=\chi(r)v\}$ and $\Phi=\{\chi:V_\chi\neq 0\}$. Note that faithfulness then already implies that $R$ is abelian.
The group $G$ permutes the weight spaces $V_\chi$, and hence by connectedness, preserves each $V_\chi$. Consider the homomorphism $d:g\mapsto (\det(g|_{V_\chi}))_{\chi\in\Phi}$, $G\to\mathbf{U}$. In restriction to $R$, $d$ has finite kernel. So the diagonal map $d'$ from $G$ into $(G/
R)\times G/\mathrm{Ker}(d)$ has finite kernel; it is readily seen to be surjective. This yields (ii).
(Sorry I should have posted this earlier and not just comment.)
