Exercise 13, § 4

Equip $\mathbf{R}^n$ with the usual scalar product $((\xi_i), (\eta_i)) \mapsto \sum_{i=1}^n \xi_i \eta_i$. Let $I(n)$ be the group of affine isometrical transformations of $\mathbf{R}^n$. It is canonically isomorphic to the Lie subgroup of $\mathbf{GL}(n+1, \mathbf{R})$ formed by the matrices

where $U \in \mathbf{O}(n)$ and $x$ is an arbitrary element of $\mathbf{R}^n$ (matrix of type $(n, 1)$). One can identify $I(n)$ with the semidirect product of $\mathbf{O}(n)$ by $\mathbf{R}^n$ defined by the canonical injection of $\mathbf{O}(n)$ into $\mathbf{GL}(n, \mathbf{R})$ (the matrix $S$ being then denoted $(U, x)$). Let $p : I(n) \to \mathbf{O}(n)$ be the canonical surjection, with kernel $\mathbf{R}^n$.

a) Let $\Gamma$ be a discrete subgroup of $I(n)$. Consider the closure $\overline{p(\Gamma)}$ of $p(\Gamma)$ in $\mathbf{O}(n)$. Now prove that its identity component is commutative. (Let $V$ be a neighbourhood of $I$ in $\mathbf{O}(n)$ having the properties of exerc. 11 d), and such moreover that $\|U - I\| \leq \frac{1}{4}$ for every $U \in V$ (the norm on $\mathrm{End}(\mathbf{R}^n)$ being deduced from the euclidean norm of $\mathbf{R}^n$). Argue by contradiction by supposing that there exist $S_1 = (U_1, x_1), S_2 = (U_2, x_2)$ such that $U_1, U_2$ belong to $V$ and do not commute. Now prove that one can write $(S_1, S_2) = ((U_1, U_2), y)$ with

Define then by recurrence the $S_k$ by putting $S_k = (S_1, S_{k-1})$ for $k \geq 3$. Taking account of the choice of $V$, now prove that this sequence has an infinite number of distinct terms and is bounded in $I(n)$, which is absurd.)

b) One says that $\Gamma$ is crystallographic if $I(n)/\Gamma$ is compact. Now prove that, if this is so, for every $x \in \mathbf{R}^n$, the affine subspace $L$ of $\mathbf{R}^n$ generated by $\Gamma x$ is equal to $\mathbf{R}^n$. (In the contrary case, for every $y \in \mathbf{R}^n$, all the points of $\Gamma y$ are at the same distance from $L$, and, since this distance can be arbitrarily large, $I(n)/\Gamma$ is not compact.)

c) If $\Gamma$ is commutative crystallographic, we have $\Gamma \subset \mathbf{R}^n$. (If $S = (U, x) \in \Gamma$ is such that the vector subspace $V \subset \mathbf{R}^n$ formed by the points invariant under $U$ is not equal to $\mathbf{R}^n$, and if we denote by $V^\perp$ the orthogonal subspace to $V$, then, for every $S' = (U', x') \in \Gamma$, $U'$ leaves $V$ and $V^\perp$ stable. On the other hand, $U | V^\perp$ having no nonzero invariant point, show that this permits one to suppose, on changing the origin, that $x \in V$. Using b), there exists $S' = (U', x') \in \Gamma$ such that the orthogonal projection $y'$ of $x'$ on $V^\perp$ is $\neq 0$. By calculating $Sx'$ by the formula $S = S'S{S'}^{-1}$, deduce that $U.y' = y'$ contrary to the definition of $V$.)

d) If $\Gamma$ is crystallographic, $\Gamma \cap \mathbf{R}^n$ is a commutative free group of rank $n$, and $\Gamma/(\Gamma \cap \mathbf{R}^n)$ is a finite group (Bieberbach's theorem). (Let $W$ be the vector subspace of $\mathbf{R}^n$ generated by $\Gamma \cap \mathbf{R}^n$. The compact group $\overline{p(\Gamma)}$ has only a finite number of connected components. If $W = \{0\}$, $\Gamma$ contains, according to a), a commutative normal subgroup of finite index $\Gamma_1$; $\Gamma_1$ is then crystallographic, which leads to a contradiction with c). Hence $W \neq \{0\}$. Show that $p(\Gamma)$ leaves $W$ stable and that $p(\Gamma) | W$ is finite: otherwise, there would exist a sequence $(S_m)$ in $\Gamma$ such that if $S_m = (U_m, x_m)$, the $U_m$ are pairwise distinct and tend to $I$; form then the $(I, a_j)S_m(I, a_j)^{-1}S_m^{-1}$ where $(a_j)$ is a basis of $\Gamma \cap \mathbf{R}^n$, and obtain a contradiction with the assumption that $\Gamma$ is discrete. Finally, in order to see that $W = \mathbf{R}^n$, show that otherwise the action of $\Gamma$ on $\mathbf{R}^n/W$ would be that of a crystallographic group containing no translation $\neq 0$.)