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.

  • $\begingroup$ Despite that my answer below says a clear "no", the answer to the question as asked is "yes", but this is due to bad terminology. 1) "trivial representation" means nothing but the one dimensional representation in which all elements act trivially (not only the translations). 2) even when the translations are in the kernel there are many irreducible representations of the Lorentz group. The term "the irreducible rep" is misleading. You probably mean "the standard rep". Any irreducible rep of the Lorentz group which is not the standard will provide a positive answer to the asked question. $\endgroup$
    – Uri Bader
    Apr 8, 2016 at 9:51
  • $\begingroup$ To clarify: my answer "no" is the correct answer to the question asked at the title, but not to the question expressed in the body, which is slightly different. $\endgroup$
    – Uri Bader
    Apr 8, 2016 at 10:01
  • $\begingroup$ @user89334 Yes, I have redefined the "trivial" in the body. I just want to clarify the motivition of this question. $\endgroup$
    – 346699
    Apr 8, 2016 at 13:42

1 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.


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