Does there exist finite dimensional irreducible representation of Euclidean or Poincare group in which translation and rotation both act nontrivially?

Question

Does there exist any finite dimensional irreducible rep. of Euclidean or Poincare group in which translation and rotation both act nontrivially?

Let me firstly clarify my question. For example, we obviously have a faithful rep. of Poincare group, $\begin{pmatrix} \Lambda & x \\ 0 & 1 \end{pmatrix}$, where $\Lambda$ is Lorentz transformation and $x$ is translation. But this is a reducible and indecomposable representation. We can always define an irreducible rep. of Poincare group by

$$f:\begin{pmatrix} \Lambda & x \\ 0 & 1 \end{pmatrix}\rightarrow D_{(i,j)}(\Lambda)$$ where $D_{(i,j)}(\Lambda)$ is some irreducible rep. of Lorentz group. But in this case, the translation acts trivially.

Because Euclidean and Poincare group are noncompact, their unitary and irreducible rep. must be infinite dimensional. However it doesn't say we don't have finite dimensional and irreducible but not unitary rep. of these groups. The above is an example. So I want to know whether there is other finite dimensional irreducible rep. of Euclidean and Poincare group in which rotation and translation both act nontrivially? If it doesn't exist, how to prove or tell me the name of theorem.

Answer 1

The answer is "No".

You don't specify which kind of representation you have in mind. I assume these are finite dimensional complex representations. Thus you ask about (continuous) homomorphisms $G\to \text{GL}_n(\mathbb{C})$. In the groups $G$ you consider you have a non trivial normal nilpotent subgroup $N$, the group of translations (aka the unipotent radical). Let $Z$ be the center of $N$. It is still normal in $G$. In your case $Z=N$. Decompose $V=\mathbb{C}^n$ to a direct sum of $Z$-eigenspaces. Each of these is $G$-invariant. Thus, if $V$ is $G$-irreducible then each element of $Z$ must act by a scalar multiplication on $V$. In particular, in the image of $G$, $Z$ must be central. In your specific groups the only normal subgroups are $\{e\},G$ and $N$ ($=Z$ ), and $N$ is not central in $G$. So $N$ must be in the kernel of the representation.