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?
No there are none (apart from the trivial rep.). This follows not because of the metric but because the transformations on Euclidean space still form a non-compact group, like Poincaré.
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
Repka, J. and de Guise, H., 1999. Some finite dimensional indecomposable representations of E (2). Journal of Mathematical Physics, 40(11), pp.6087-6109
(available here) and follow up papers mostly by A. Douglas.
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.
