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I am reading de la Harpe's book "Topics in Geometric Group Theory". On page 145, there is a theorem: Let $V$ be a complete $n$-Riemannian manifold with sectional curvature satisfying $K\ge 0$. Then there exist a finite normal subgroup $F_1$ of the fundamental group $\pi_1(V)$ and a short exact sequence of the form: $$ 1 \to \mathbb Z^k \to \pi_1(V)/F_1 \to F_2 \to 1 $$ where $k\le n$ and where $F_2$ is a finite group.

I have two questions:

1) why the author put $F_1$ in such a special position, why not absorb $F_1$ into $F_2$, since they are basically all finite group.

2) In other papers/books, peopole called a discrete, torsion free and cocompact subgroup of $O(n)\ltimes \mathbb R^n $ A Bieberbach group. So why "torsion free" is important?

Thank you!

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In 1) you can indeed absorb $F_1$ into $F_2$, see Theorem 2.1 in Wilking's preprint wwwmath.uni-muenster.de/sfb/about/publ/wilking3.ps. For 2) the torsion-free case is easier and cleaner, and also these are the groups of compact flat manifolds, which matter for geometers. –  Igor Belegradek Sep 5 '12 at 14:43
@Igor, Thanks. Is it true that the groups mentioned in Wilking's paper Thm2.1 is $\mathbb Z^n\ltimes F$ where $F$ is a finite group. The difference between the Flat Bieberbach group and the one in Wilking's list is for Bieberbach group $F$ can be considered as a subgroup of $O(n)$. However for non-negatively curved case, it is a subgroup of $GL(n, \mathbb Z)$? –  John B Sep 5 '12 at 18:22
The author's name is "de la Harpe". –  Igor Rivin Sep 5 '12 at 18:39
@MG: The groups in Thm 2.1 need not be semidirect products, eg the fundamental group of a flat manifold is torsion-free, so it cannot contain nontrivial finite subgroups. Your other question seems misstated for any finite subgroup of $GL_n(\mathbb R)$ preserves an inner product, so it lies in a conjugate of $O(n)$. In any case a finitely generated group is crystallographic iff it is virtually abelian and has no normal finite subgroups. Thus torsion-free crystallographic groups (also known as Bieberbach groups) are precisely the torsion-free virtually-$\mathbb Z^n$, see rmk 2.5(2) in Wilking. –  Igor Belegradek Sep 5 '12 at 18:47
That you can absorb $F_1$ into $F_2$ follows from the fact that polycyclic-by-finite groups are residually finite (I think this is due to Malcev). Actually, here you reduce to the finite-by-abelian case, which easily boils down to showing that every f.g. 2-nilpotent group is residually finite; this is a simple exercise. –  Yves Cornulier Sep 5 '12 at 22:36
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1 Answer

up vote 3 down vote accepted

There is a ``quasi-isometric rigidity'' context in which a finite kernel arises naturally. I am not sure how general this is so let me state it in the narrower context of constant sectional curvature equal to zero.

Theorem (quasi-isometric extension of the Bieberbach Theorem): If $\Gamma$ is an abstract group quasi-isometric to Euclidean $n$-space then there exists a finite normal subgroup $F_1$ of $\Gamma$ such that $\Gamma / F_1$ is a Bieberbach group (of the general kind, allowing torsion), and there is a short exact sequence of the form $1 \to \mathbb{Z}^n \to \Gamma / F_1 \to F_2 \to 1$ where $F_2$ is a finite group. So the group $\Gamma/F_1$ is the orbifold fundamental group of a compact $n$-dimensional Euclidean orbifold.

This is a consequence of Gromov's polynomial growth theorem. I don't know whether this is what de la Harpe had in mind, but maybe.

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A finite kernel also arises if you characterize discrete (not necessarily cocompact) groups of isometries of Euclidean spaces. –  Yves Cornulier Sep 5 '12 at 22:39
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