Let $M$ be a compact Riemannian manifold with vanishing curvature of Levi-Civita connection. Such manifolds were classified by Bieberbach; sometimes they are called Bieberbach manifolds. According to this classification, a Bieberbach manifold is a quotient of a torus by a finite group freely acting on it by isometries. Crystallographic group is a fundamental group of a Bieberbach manifold. There is an exact sequence
$$ 0 \rightarrow {\mathbb Z}^n \rightarrow G \rightarrow L\rightarrow 0,$$ 
where $G$ is a crystallographic group, ${\mathbb Z}^n$ a group of translations in $G$,
and $L$ the holonomy group of the Levi-Civita connection. The group $L$
is naturally embedded to the group $GL(T\_xM)$ of automorphisms of the tangent space of the Bieberbach manifold.


Could $T\_x M$ be irreducible as a representation of $L$?
I have checked for dimension $\leq 5$, and $T\_xM$
is never irreducible for dimension $\leq 5$.



Flat manifolds and irreducible representations
Here is the solution (with thanks to Misha Kapovich 
who pointed this out).


Gerhard Hiss, Andrzej Szczepanski, ["On torsion free crystallographic groups"](http://www.sciencedirect.com/science/article/pii/0022404991900476),
Journal of Pure and Applied Algebra
Volume 74, Issue 1, 10 September 1991, Pages 39-56


Abstract:
The torsion free crystallographic groups arise as fundamental groups of compact flat Riemannian manifolds. Let R be a crystallographic group with point group G and translation group T. In this paper we consider the G-module T⊗z, for which we prove: If R is torsion free, then G does not act irreducibly on T⊗z. A proof of this theorem for solvable groups G was first given by G. Cliff. The theorem proves a conjecture made by the second author. The proof of the theorem uses the classification of the finite simple groups.


