Is there any characterization of the noncompact connected Lie groups that possess faithful finitedimensional unitary representations?

2$\begingroup$ (You can remove "noncompact" from the question, no reason to exclude trivial case). The answer is: connected Lie groups that are are locally isomorphic to a compact group, or equivalently admitting a discrete central subgroup such that the quotient is a compact group, or equivalently whose quotient by the center is a semisimple compact Lie group. $\endgroup$– YCorSep 22, 2016 at 15:04

$\begingroup$ This is great! I was not able to find a characterisation. Would you be kind enough to provide me with a reference? Especially to the first condition you mentioned  being locally isomorphic to a compact group. Thanks in advance. (You can also write it as an answer so that I can accept it.) $\endgroup$– William of BaskervilleSep 22, 2016 at 15:08

$\begingroup$ There's a more standard result of the same vein: if a Lie group admits a invariant (definite positive) scalar product on its Lie algebra, then it's locally isomorphic to a compact group (i.e. its Lie algebra is direct product of an abelian one and a semisimple compact one). (Maybe somebody has a reference?) Since this condition passes to closed subgroups, one implication follows. $\endgroup$– YCorSep 22, 2016 at 16:10

$\begingroup$ Conversely if a connected Lie group $G$ has this condition, then it's quotient of a product $A\times H$, with $A$ an abelian connected Lie group, $H$ a semisimple compact connected Lie group, by a finite central subgroup which can be viewed as the antidiagonal of $Z\times Z$ where $Z$ is a finite subgroup of both $A$ and $H$. So representations of $G\times G$ are in 11 correspondence with pairs consisting of a representation of $H$, a representation of $A$ in its centralizer, both coinciding on $Z$. (...) $\endgroup$– YCorSep 22, 2016 at 16:13

$\begingroup$ (...) then fix one faithful unitary representation of $H$ on a f.dim. space $V$. Extend it to a representation of $A$. In an additional orthogonal space $W$, fix a faithful unitary representation of $A/Z$, and extend it to a representation of $G$, trivial on $H$. Then the resulting unitary representation of $G$ on $V\oplus W$ is faithful. $\endgroup$– YCorSep 22, 2016 at 16:15
1 Answer
Proposition. Equivalences ($G$ connected Lie group):
(i) $G$ has a faithful finitedimensional 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 nonabelian 2dimensional 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 1dimensional. 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.)