Discrete complex reflection groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 12 May 2014

Notes and References

This is an excerpt of the lecture notes Discrete complex reflection groups by V.L. Popov. Lectures delivered at the Mathematical Institute, Rijksuniversiteit Utrecht, October 1980.

Several auxiliary results and the classification of irreducible infinite noncrystallographic complex $r -groups$

Now we move on to proofs. The greater part of these proofs relies heavily on the fact that for an infinite $r -group$ $W$ the subgroup $Tran W$ is sufficiently "massive".

The subgroup of translations

In order to clarify the last statement we need the following principal fact:

Theorem. Let $W \subset A (E)$ be an infinite $r -group.$ Then $Tran W \neq 0 .$

Proof.

Assume that $Tran W = 0 .$ Then $Lin : W \to U (V)$ is a monomorphism.

It follows from the discreteness of $W$ that ${(\overline{Lin W})}^{0}$ is a torus (here bar denotes the closure and $0$ denotes the connected component of unity). (See [Bou1972], Ch. III §4, ex 13a.)

Set $S = Lin W \cap {(\overline{Lin W})}^{0} .$

Then $Lin W ⊳ S$ and $[Lin W : S] < \infty .$ Let also $V_{0} = {v \in V | {(\overline{Lin W})}^{0} v = v}$ and let $V_{1}$ be the subspace of $V$ determined by $V = V_{0} \oplus V_{1}, V_{0} ⊥ V_{1} .$ The subspaces $V_{0}$ and $V_{1}$ are $S -invariant.$ Definitely, $V_{1} \neq 0$ (indeed, if not, then $S = 1,$ and hence $| Lin W | < \infty$ which is a contradiction with $| Lin W | = | W |).$

The set ${P \in {(\overline{Lin W})}^{0} | H_{P} = V_{0}}$ is dense in ${(\overline{Lin W})}^{0} .$ Therefore ${P \in S | H_{P} = V_{0}}$ is dense in ${(\overline{Lin W})}^{0} .$ Hence ${P \in S | H_{P} = V_{0}} \neq 0$ (because $S$ is dense in ${(\overline{Lin W})}^{0}).$

We claim that there exists a point $a \in E$ with $γ (a) - a \in V_{0}$ for every $γ \in W .$

Consider an arbitrary point $b \in E$ and an operator $P_{0} \in S$ such that $H_{P_{0}} = V_{0} .$ Take $γ_{0} \in W$ with $Lin γ_{0} = P_{0} .$ We have $γ_{0} (b) - b = t_{0} + t_{1}$ where $t_{0} \in V_{0},$ $t_{1} \in V_{1} .$ But the restriction of $1 - P_{0}$ to $V_{1}$ is nondegenerate. Hence there exists a vector $t \in V_{1}$ such that $(1 - P_{0}) t = t_{1} .$ So, we have $γ_{0} (b + t) - (b + t) = (γ_{0} (b) + P_{0} t) - (b + t) = (γ_{0} (b) - b) + (P_{0} - 1) t = t_{0} + t_{1} - t_{1} = t_{0} \in V_{0} .$ Put $a = b + t .$ We have $γ_{0} (a) - a \in V_{0} .$ We shall prove that $a$ is a point as wanted.

Let us first check that $γ (a) - a \in V_{0}$ if $γ \in W,$ $Lin γ \in S .$ Write $P = Lin γ,$ $v_{0} = γ_{0} (a) - a \in V_{0}$ and $v = γ (a) - a .$ We need to prove that $v \in V_{0} .$ But $S$ is commutative, so $P P_{0} = P_{0} P .$ Hence $γ γ_{0} = γ_{0} γ .$ We have now: $\begin{matrix} κ_{a} (γ_{0}) = (P_{0}, v_{0}) = κ_{a} (γ γ_{0} γ^{- 1}) = (P, v) (P_{0}, v_{0}) (P^{- 1}, - P^{- 1} v) \\ = (P P_{0} P^{- 1}, - P P_{0} P^{- 1} v + P v_{0} + v) = (P_{0}, - P_{0} v + P v_{0} + v) . \end{matrix}$ So, $- P_{0} v + P v_{0} + v = v_{0},$ i.e. $(1 - P_{0}) v = (1 - P) v_{0} .$ But $P \in S,$ hence $(1 - P) v_{0} = 0 .$ It follows from $Ker (1 - P_{0}) = V_{0}$ that $v \in V_{0} .$ This establishes the claim.

Now we can prove $γ (a) - a \in V_{0}$ for arbitrary $γ \in W .$ Indeed, write $P = Lin γ,$ $v = γ (a) - a,$ as before. We have: $κ_{a} (γ γ_{0} γ^{- 1}) = (P P_{0} P^{- 1}, - P P_{0} P^{- 1} v + P v_{0} + v) .$ It follows from $S ⊲ Lin W$ that $V_{0}$ is $P -invariant.$ Therefore we have $P v_{0} \in V_{0} .$ Moreover, if $γ' = γ γ_{0} γ^{- 1},$ then $Lin γ' \in S$ and, by the claim, $γ' (a) - a = - P P_{0} P^{- 1} v + P v_{0} + v \in V_{0} .$ Therefore, $- P P_{0} P^{- 1} v + v \in V_{0}$ and hence $- P_{0} P^{- 1} v + P^{- 1} v \in P^{- 1} V_{0} = V_{0},$ i.e. $(1 - P_{0}) P^{- 1} v \in V_{0} .$ But the image of $1 - P_{0}$ is $V_{1},$ hence $(1 - P_{0}) P^{- 1} v = 0 .$ Therefore $P^{- 1} v \in V_{0},$ i.e. $v \in P V_{0} = V_{0}$ and we are done.

Now we consider two subgroups of $W :$ $W',$ resp. $W ″$ is the subgroup generated by those reflections $γ$ for which $H_{γ} ∋ a,$ resp. $H_{γ} ∌ a .$

The subgroup $W'$ is finite. Indeed, identifying $A (E)$ and $G L (V) V$ by means of $κ_{a},$ we have $κ_{a} (γ) = (R, 0) \in U (V)$ for any reflection $γ \in W'$ and hence for any $γ \in W' .$ So $κ_{a} (W')$ is a discrete subgroup of a compact group $U (V),$ hence finite.

We claim now that $W ″$ is infinite. Before proving this, we shall show how to finish the proof of the theorem if the statement holds. Thus, assume $W ″$ is an infinite $r -group.$

Note that $Tran W ″ = 0 .$ Using the above arguments and constructions we obtain a torus ${(\overline{Lin W ″})}^{0},$ a subgroup $S ″ = Lin W ″ \cap {(\overline{Lin W ″})}^{0}$ and a decomposition $V = V_{0}^{″} \oplus V_{1}^{″},$ $V_{0}^{″} = {v \in V | {(\overline{Lin W ″})}^{0} v = v},$ $V_{0}^{″} ⊥ V_{1}^{″} .$ Also, we have $V_{1}^{″} \neq 0$ and ${P \in S ″ | H_{P} = V_{0}^{″}} \neq \emptyset .$ But $W ″ \subset W,$ therefore $Lin W ″ \subset Lin W,$ whence ${(\overline{Lin W ″})}^{0} \subset {(\overline{Lin W})}^{0} .$ So $S ″ \subset S$ and $V_{0}^{″} \supset V_{0} .$ It follows now that $γ (a) - a \in V_{0}^{″}$ and $γ (a) - a \neq 0$ for every $γ \in W ″ .$

Let $γ \in W ″$ be a reflection. Then $κ_{a} (γ) = (Lin γ, γ (a) - a),$ and in view of $γ (a) - a \neq 0,$ we have: (root line of $Lin γ) = ℂ (γ (a) - a) \subset V_{0}^{″} .$ Therefore $H_{Lin γ} \supset V_{1}$ and $Lin γ$ acts on $V_{1}$ trivially. But $W ″$ is generated by reflections; therefore $Lin W ″$ acts trivially on $V_{1}$ which contradicts the existence of $P \in S ″ \subset Lin W ″$ such that $H_{P} = V_{0}^{″}$ and $V_{1}^{″} \neq 0 .$

It remains to show that $| W ″ | = \infty .$ Let us assume that this is not the case. Then $W ″$ has a fixed point $b \in E .$ By construction, $b \neq a .$ We shall show that $W'$ and $W ″$ commute. This then yields that $W = W' W ″$ is a finite group (for $W'$ and $W ″$ clearly generate $W),$ which is a contradiction.

Let $γ' \in W'$ and $γ ″ \in W ″$ be reflections. We have $a \in H_{γ'},$ $b \in H_{γ ″} .$ $\begin{matrix} \end{matrix}$

But $γ' γ ″ {γ'}^{- 1}$ is a reflection with mirror $γ' (H_{γ ″}) .$

Hence $γ' γ ″ {γ'}^{- 1}$ is either in $W'$ or in $W ″ .$ If this element is in $W',$ then $γ' (H_{γ ″}) ∋ a,$ i.e., $H_{γ ″} ∋ {γ'}^{- 1} (a) = a,$ which is a contradiction. Therefore, $γ' (H_{γ ″}) ∋ b$ and $H_{γ ″} ∋ {γ'}^{- 1} (b) .$ We also have $b \in H_{γ ″}$ and $b \neq {γ'}^{- 1} (b)$ (for otherwise $b \in H_{γ'}$ which is absurd).

Consequently, there is a unique line on $b$ and ${γ'}^{- 1} (b) .$ This line lies in $H_{γ'}$ and is orthogonal to $H_{{γ'}^{- 1}} = H_{γ'} .$ Hence $H_{γ ″} ⊥ H_{γ'},$ i.e. $γ' γ ″ = γ ″ γ' .$

$□$

It follows from this theorem that $Tran W$ is "big enough":

Theorem. Let $W$ be an infinite irreducible $r -group.$ Then $T = Tran W$ is a lattice of rank $n$ if $k = ℝ$ and of rank $n$ or $2 n$ if $k = ℂ$ (here $n = {dim}_{k} E).$

Proof.

Let $k = ℝ .$ Then $ℝ T$ is an invariant subspace of the $Lin W -module$ $W .$ This subspace is nontrivial according to the previous theorem. Therefore $ℝ T = V,$ because of irreducibility (see the theorem in 1.4), and the assertion follows from the equality $rk T = dim ℝ T .$

Let $k = ℂ .$ As above, we have $0 \neq ℂ T = ℝ T + i ℝ T = V .$ Hence $2 n = {dim}_{ℝ} V \leq {dim}_{ℝ} ℝ T + {dim}_{ℝ} i ℝ T = 2 rk T .$ Therefore $rk T \geq n .$ But $ℝ T \cap i ℝ T$ is a $Lin W -invariant$ complex subspace of $V$ and hence either $ℝ T \cap i ℝ T = 0$ or $ℝ T \cap i ℝ T = V,$ i.e. $ℝ T = V .$ If $rk T = dim ℝ T > n$ then $ℝ T \cap i ℝ T \neq 0,$ hence $ℝ T = V,$ i.e. $rk T = 2 n .$

$□$

Corollary.

1)	$\| Lin W \| < \infty;$
2)	if $k = ℂ$ then $W$ is a crystallographic group iff $rk T = 2 n .$

Proof.

1) follows from the fact that $Lin W$ is contained in a compact group and has an invariant lattice of a full rank.

$□$

Some auxiliary results

Let $K \subset G L (V)$ be a finite $r -group$ and $ℋ$ be the set of mirrors of all reflections from $K .$ Let $H \in ℋ .$

Then it is easy to see that the subgroup of $K$ generated by all reflections $R \in K$ with $H_{R} = H,$ is a cyclic group. Let $m (H)$ be the order of this group.

Theorem. Let ${R_{j}}_{j \in J}$ be a generating system of reflections of $K$ such that the order of $R_{j}$ is equal to $m (H_{R_{j}})$ for every $j \in J .$ Then each reflection $R \in K$ is conjugate to $R_{j}$ for certain $j$ and $l_{j} .$

Proof.

Let $𝒪$ be the $K -orbit$ of $H_{R}$ in $ℋ$ (it follows from $P H_{R} = H_{P R P^{- 1}}$ that $K$ acts on $ℋ).$ Let $χ_{𝒪}$ be the product of linear equations of the mirrors in $𝒪,$ normalized by the condition $χ_{𝒪} (1) = 1 .$ Then $χ_{𝒪}$ is a character of $K,$ i.e. a homomorphism of $K$ into the multiplicative group of $k .$ The group $χ_{𝒪} (K)$ is generated by $χ_{𝒪} (R_{j}),$ $j \in J .$ We have also $χ_{𝒪} (R) \neq 1 .$ Therefore there exists a $j \in J$ with $χ_{𝒪} (R_{j}) \neq 1 .$ But $χ_{𝒪} (R_{j}) \neq 1$ iff $𝒪$ is the orbit of $H_{R_{j}},$ see [Spr1977]. Therefore $g H_{R} = H_{R_{j}}$ for a certain $g \in K$ and the statement follows.

$□$

Theorem. Let $W$ be an $r -group.$ Then for any reflection $R \in Lin W$ there exists a reflection $γ \in W$ with $Lin γ = R .$

Proof.

Let ${ρ_{j}}_{j \in J}$ be the set of all reflections of $W .$ Then ${R_{j} = Lin ρ_{j}}_{j \in J}$ is a generating system of reflections of $Lin W .$ Let $𝒪$ be the $Lin W -orbit$ of $H_{R}$ in $ℋ .$ Then there exists a number $l$ with $H_{R_{l}} \in 𝒪$ and $P_{l} \in Lin W$ such that $P_{l} H_{R_{l}} = H_{P_{l} R_{l} P_{l}^{- 1}} = H_{R} .$ Let $π_{l} \in W$ be such that $Lin π_{l} = P_{l} .$ We have:
$Lin π_{l} ρ_{l} π_{l}^{- 1} = P_{l} R_{l} P_{l}^{- 1}$ and $π_{l} ρ_{l} π_{l}^{- 1}$ is a reflect~on, too. Therefore $χ_{𝒪} (Lin W)$ is generated by $χ_{𝒪} (R_{j})$ where $j \in J' = {j \in J | H_{R_{j}} = H_{R}} .$ We have $χ_{𝒪} (R) = χ_{𝒪} (R_{j_{1}}) \dots χ_{𝒪} (R_{j_{s}}) = χ_{𝒪} (R_{j_{1}} \dots R_{j_{s}})$ for certain $j_{1}, \dots, j_{s} \in J' .$ But $R_{j_{1}}, \dots, R_{j_{s}}$ are the reflections whose mirrors are $H_{R} .$ It follows now (see [Spr1977]) that $R = R_{j_{1}} \dots R_{j_{s}} .$ Let us consider the element $ρ = ρ_{j_{1}} \dots ρ_{j_{s}} .$ We have $Lin ρ = R$ and it easily follows from the parallelity of the mirrors $H_{ρ_{j_{1}}}, \dots, H_{ρ_{j_{s}}}$ that $ρ$ is a reflection.

$□$

Semidirect products

Let $W$ be a subgroup of $A (E) .$ We have $0 \to Tran W ↪ W \to Lin W \to 1 .$ When is $W$ a semidirect product of $Lin W$ and $Tran W ?$ We have the following criterion:

Theorem. Let $| Lin W | < \infty .$ Then:

a)	$W$ is a semidirect product iff there exists a point $a \in E$ such that $Lin$ induces an isomorphism of the stabilizer $W_{a}$ of $a$ with $Lin W .$
b)	For every finite group $K \subset U (V)$ and every $k -invariant$ subgroup $T$ of $V$ there exists a unique group $W \subset A (E)$ (up to equivalence) such that $Lin W = K,$ $Tran W = T$ and $W$ is a semidirect product of $Lin W$ and $Tran W .$

Proof.

Proof is left to the reader.

$□$

The point a from part a) of this theorem is called a special point of $W',$ see [Bou1968].

Using this theorem we can clarify the structure of infinite $r -groups$ in a number of important cases:

Theorem. Let $W \subset A (E)$ be a group generated by reflections. Assume that $| Lin W | < \infty$ and that $Lin W$ is an essential group (i.e. ${v \in V | (Lin W) v = v} = {0})$ generated by $n = {dim}_{k} E$ reflections. Then $W$ is a semidirect product of $Lin W$ and $Tran W .$

Proof.

Let $R_{1}, \dots, R_{n}$ be a generating system of reflections of $Lin W .$ Then there exists a reflection $γ_{j} \in W$ such that $Lin γ_{j} = R_{j},$ $j = 1, \dots, n,$ see Section 3.2. We have $\cap_{j = 1}^{n} H_{R_{j}} = 0$ because $Lin W$ is essential. Therefore $\cap_{j = 1}^{n} H_{γ_{j}} \neq \emptyset;$ more precisely, this intersection is a single point $a \in E .$

We have now that $Lin : W_{a} \to Lin W$ is a surjective map, hence an a isomorphism. Thus we are done in view of the theorem above.

$□$

The conditions of this theorem are always fulfilled if $k = ℝ$ and $W$ is an infinite irreducible $r -group.$ In general this is not the case if $k = ℂ,$ though "in most cases", it is.

Classification of the irreducible infinite complex noncrystallographic $r -groups.$

Let $k = ℂ$ and let $W$ be a group as in the title. Set $T = Tran W .$ Then $rk T = n = {dim}_{ℂ} E,$ by 3.1.

$ℝ T$ is a $Lin W -invariant$ $ℝ -submodule$ of $V .$ Let us consider the restriction of $⟨ | ⟩$ to $ℝ T .$ We claim that this restriction has only real values. Indeed, $Re ⟨ | ⟩$ defines an euclidean structure on $ℝ T$ such that $Lin W$ is orthogonal with respect to this structure. But $ℂ (ℝ T) = V$ because of irreducibility. Hence, there exists a canonical extension of $Re ⟨ | ⟩,$ determined up to a hermitian $Lin W -invariant$ scalar product, say $(|),$ on $V .$ Thus we have two $Lin -invariant$ hermitian structures, $⟨ | ⟩$ and $(|),$ on $V .$ They are proportional because of irreducibility of $Lin W : ⟨ | ⟩ = λ (|),$ $λ \in ℂ .$ Taking restrictions to $ℝ T,$ we have: $λ = 1 .$ In other words, $ℝ T$ is a real form of $V$ and the restriction of the action of $Lin W$ to $ℝ T$ gives an irreducible real finite $r -group$ (and $Lin W$ itself is the complexification of this group). It follows from the classification that this group is generated by $n$ reflections, see 1.5. Therefore $Lin W$ is generated by $n$ reflections, too. It follows from the theorem above (section 3.3) that $W$ is a semidirect product. Let $a \in E$ be a special point of $W .$ Then $a + ℝ T$ is a real form of $E .$ It is clear that $a + ℝ T$ is $W -invariant.$ The restriction of $W$ to $a + ℝ T$ is a real form of $W .$ Therefore, this restriction is an affine Weyl group and $W$ is its complexification. This completes the proof of the theorem in Section 2.2.

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