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 $\leq 5$.


1 Answer 1


Here is the solution (with thanks to Misha Kapovich who pointed this out).

Gerhard Hiss, Andrzej Szczepanski, "On torsion free crystallographic groups", 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.


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