Exercise 13, § 4

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

where $U \in \mathbf{O}(n)$ and $x$ is an arbitrary element of $\mathbf{R}^n$ (matrix of type $(n, 1)$). $I(n)$ can be identified with the semi-direct 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 $S$ is then denoted by $(U, x)$). 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 $\mathbf{O}(n)$. Show that its identity component is commutative. (Let $V$ be a neighbourhood of $I$ in $\mathbf{O}(n)$ with the properties of Exercise 11 (d) and such that further $\|U - I\| \leq \frac{1}{4}$ for all $U \in V$ (the norm on $\mathrm{End}(\mathbf{R}^n)$ being derived from the Euclidean norm on $\mathbf{R}^n$). Argue by reductio ad absurdum, assuming that there exist $S_1 = (U_1, x_1), S_2 = (U_2, x_2)$ such that $U_1$ and $U_2$ belong to $V$ and do not commute. Show that $(S_1, S_2) = ((U_1, U_2), y)$ with

Then define the $S_k$ inductively by $S_k = (S_1, S_{k-1})$ for $k \geq 3$. Using the choice of $V$, show that this sequence has an infinity of distinct terms and is bounded in $I(n)$, which is absurd.)

(b) $\Gamma$ is called crystallographic if $I(n)/\Gamma$ is compact. Show that if this is so, for all $x \in \mathbf{R}^n$, the affine subspace $L$ of $\mathbf{R}^n$ generated by $\Gamma x$ is equal to $\mathbf{R}^n$. (Otherwise, for all $y \in \mathbf{R}^n$, all the points of $\Gamma y$ are the same distance from $L$ and, as this distance can be arbitrarily large $I(n)/\Gamma$ is not compact.)

(c) If $\Gamma$ is commutative and crystallographic, then $\Gamma \subset \mathbf{R}^n$. (If $S = (U, x) \in \Gamma$ is such that the vector subspace $V \subset \mathbf{R}^n$ consisting of the points invariant under $U$ is not equal to $\mathbf{R}^n$ and $V^\perp$ denotes the orthogonal subspace of $V$, then, for all $S' = (U', x') \in \Gamma$, $U'$ leaves $V$ and $V^\perp$ stable. On the other hand, since $U|V^\perp$ has no non-zero invariant point, show that this allows us to assume, by changing the origin, that $x \in V$. Using (b), there exists $S'' = (U', x') \in \Gamma$ such that the orthogonal projection $y'$ of $x'$ onto $V^\perp$ is $\neq 0$. By evaluating $Sx'$ 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 free commutative group of rank $n$ and $\Gamma/(\Gamma \cap \mathbf{R}^m)$ 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 $\bar{p}(\Gamma)$ has only a finite number of connected components. If $W = \{0\}$, $\Gamma$ contains, by (a), a commutative normal subgroup of finite index $\Gamma_1$; $\Gamma_1$ is then crystallographic, which implies a contradiction with (c). Hence $W \neq \{0\}$. Show that $\bar{p}(\Gamma)$ leaves $W$ stable and that $\bar{p}(\Gamma)|W$ 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 distinct and tend to $I$; then form 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 hypothesis that $\Gamma$ is discrete. Finally, 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 $\neq 0$.)