Noncompact (Lie) group has no faithful, finite dimensional, and unitary representations?

Question

In class, a professor said:

Noncompact (Lie) group has no faithful, finite dimensional, and unitary representations.

1. Does this statement hold for noncompact group that is not a Lie group?

2. Can we loosen one of the three restrictions to make the negative statement to be positive?

• Noncompact (Lie) group has faithful, finite dimensional, but non-unitary representations?

• Noncompact (Lie) group has nonfaithful, finite dimensional, unitary representations?

• Noncompact (Lie) group has faithful, infinite dimensional, unitary representations?

Can you provide examples for each case?

Such as for a Lorentz group $SO(1,d)$?

Answer 1

1. This statement doesn't even hold for all Lie groups. For example, $\mathbb{R}$ is noncompact but has a faithful $2$-dimensional unitary representation given by a pair of rotations with incommensurate angles $t \mapsto \left[ \begin{array}{cc} e^{i \alpha t} & 0 \\ 0 & e^{i \beta t} \end{array} \right], \frac{\alpha}{\beta} \in \mathbb{R} \setminus \mathbb{Q}$. It might be true for semisimple Lie groups or something like that though.

2. The first statement is sometimes true and the other two are always true.

   • $GL_n(\mathbb{R})$ has a faithful $n$-dimensional representation almost by definition. On the other hand, the double cover of $SL_2(\mathbb{R})$ (the metaplectic group $Mp_2(\mathbb{R})$) has no faithful finite-dimensional representations; see this answer.
   • Every group has a trivial finite-dimensional unitary representation. Non-faithful representations are very easy to construct!
   • Every locally compact Hausdorff topological group $G$ (which includes in particular every Lie group) has a faithful infinite-dimensional unitary representation on $L^2(G, \mu)$ where $\mu$ is Haar measure. If $G$ is compact the decomposition of this representation is given by the Peter-Weyl theorem. If $G$ is abelian the decomposition is governed by Pontryagin duality, and in particular when $G = \mathbb{R}^n$ we get the theory of the Fourier transform. In the noncompact nonabelian case things are complicated.

Answer 2

As already mentioned by Qiaochu Yuan, it's false, even fixing "Lie group" to "connected Lie group". However we have:

A connected Lie group has a faithful (continuous) unitary representation if and only if it is locally isomorphic to some compact Lie group.

For the non-unitary, the picture is more complicated and Qiaochu gave some illustrating examples.