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

  • 2
    $\begingroup$ (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. $\endgroup$
    – YCor
    Sep 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.) $\endgroup$ Sep 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. $\endgroup$
    – YCor
    Sep 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 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 $Z$. (...) $\endgroup$
    – YCor
    Sep 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. $\endgroup$
    – YCor
    Sep 22, 2016 at 16:15

1 Answer 1


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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.