By Karel Dekimpe

Ranging from uncomplicated wisdom of nilpotent (Lie) teams, an algebraic conception of almost-Bieberbach teams, the elemental teams of infra-nilmanifolds, is constructed. those are a average generalization of the well-known Bieberbach teams and lots of effects approximately traditional Bieberbach teams end up to generalize to the almost-Bieberbach teams. furthermore, utilizing affine representations, specific cohomology computations could be performed, or leading to a class of the almost-Bieberbach teams in low dimensions. the idea that of a polynomial constitution, another for the affine buildings that usually fail, is brought.

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**Extra resources for Almost-Bieberbach Groups: Affine and Polynomial Structures**

**Example text**

Case 2: r ( N ' ) = 1 In this situation N ~ is a torsion free nilpotent subgroup, containing N as a subgroup of finite index. e. N C_ N ~ C G). Therefore, conjugating N in E with an element of N ~ is exactly conjugating with an element of G, from which it follows t h a t ~ is not injective. This l e m m a offers an alternative way to see that the translational subgroup of an AC-group is m a x i m a l nilpotent. Indeed, let us investigate the map ~ : F --~ Out (G) defined above. Let ~ E E / ( E N G), then x = ( g , ~ ) , for some g E G and (~ E Aut (G).

R = 1. e. f(g, 1) = f(1, g) = 1) satisfying a cocycletype condition 1. Vg, h E F r162 = #(f(g, h))r 2. Vg, h, k E F f(g, h)f(gh, k) = r The extension E compatible with r N • F with multiplication: ( f ( h , k)) f(g, hk). determined by f is then obtained as Vn, m e N, Vg, h ~ F (n, g)"(~Z) ('~, h) = ( n r h), gh). Let us denote this particular extension by E(r If the set E x t r N) is not e m p t y it is in bijective correspondence to H2(F, Z(N)), where the F - m o d u l e structure of Z ( N ) is induced by r Let us now notice the following property.

We claim that the map k o j : E -~ G>~Aut (G) = Aft(G) is the desired embedding. As the restriction of k o j to N is the canonical embedding of N into its Mal'cev completion, the only thing left to show is that k o j is injective. But as the kernel of the map k o j and N only have the neutral element in common, this kernel has to be a finite normal subgroup of E and hence trivial. At this point we remark that it is not true that F will, in general, be isomorphic to the holonomy group of the almost-crystallographic group k o j ( E ) .