# Are there finite-dimensional unitary irreducible representations in Euclidean space?

One can show that there are no unitary finite-dimensional irreducible representations of the Poincaré group. The reason is that the generators for boosts are antihermitian.

Since boosts are basically time-space rotations, this led to me wondering whether for Euclidean space, where we don't have minus signs in the metric, we can have finite-dimensional unitary irreps like we can for space-space (read: usual) rotations?

• You should probably add "no finite dimensional unitary". Otherwise, this statement is wrong (as one can see at the post. Jul 2, 2020 at 11:22
• Oh sure, thanks for noticing! Jul 2, 2020 at 11:23

The best you can do apparently is to have indecomposable representations, and that’s a mess because that representation theory is “wild”. The case of $$E(2)$$ is somewhat tractable and discussed at some length in
Recall that (roughly speaking) fully reducible representations can be brought to full block diagonal form: $$T\to \left(\begin{array}{cc} T_1 &\boldsymbol{0}\\ \boldsymbol{0} &T_2 \end{array}\right)\, .$$ Indecomposables can only be made partially block diagonal: $$A\to \left(\begin{array}{cc} A_1 &A_{12}\\ \boldsymbol{0} &A_2 \end{array}\right)\, .$$ For irreducibles one cannot make a $$\boldsymbol{0}$$ block appear anywhere.
In fact, the "natural" representation of the Euclidean group $$E(n)$$ is precisely by a indecomposable matrix: \begin{align} T\to \left(\begin{array}{cc} R&t\\ 0&1 \end{array}\right) \end{align} where $$R\in O(n)$$ is an $$n\times n$$ matrix, and $$t$$ is a column vector of $$n$$ entries giving the translation part of the group action.