# Faithful finite-dimensional unitary representations

Is there any characterization of the non-compact connected Lie groups that possess faithful finite-dimensional unitary representations?

• (You can remove "non-compact" 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.
– YCor
Sep 22, 2016 at 15:04
• 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.) Sep 22, 2016 at 15:08
• 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.
– YCor
Sep 22, 2016 at 16:10
• 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 anti-diagonal of $Z\times Z$ where $Z$ is a finite subgroup of both $A$ and $H$. So representations of $G\times G$ are in 1-1 correspondence with pairs consisting of a representation of $H$, a representation of $A$ in its centralizer, both coinciding on $Z$. (...)
– YCor
Sep 22, 2016 at 16:13
• (...) 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.
– YCor
Sep 22, 2016 at 16:15

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.)