Are there infinite abelian real reflection groups?

Question

Lately I have been studying reflection groups, and there is a particular example of a complex reflection group that has been very good for guiding my intuition. I would like to know if there is an analogue over the real numbers. To keep this post self-contained, let me start by stating my definition of a reflection group.

Definition: Let $k = \mathbb{R}$ or $\mathbb{C}$, and let $V$ be a finite-dimensional vector space over $k$, equipped with an inner product. A reflection is an operator $s \in \text{GL}(V)$ such that

1. $s(\alpha)= \zeta \alpha$ for some nonzero vector $\alpha \in V$ and nontrivial root of unity $\zeta \neq 1$, and
2. $s$ fixes the codimension 1 orthogonal complement of the line $k\alpha$.

A reflection group is a subgroup $G \subseteq \text{GL}(V)$ generated by reflections.

At least in my experience, many counterexamples when studying reflection groups can be found by looking at either infinite or abelian reflection groups, so it seems natural to ask if there is a reflection group which is both. We can construct such a complex reflection group as follows:

Over $\mathbb{C}$: For each integer $n \geq 2$, pick a primitive $n^{th}$ root of unity $\zeta_n \in \mathbb{C}$. We can then consider the rank 1 reflection group $W = \langle \zeta_n : n \geq 2 \rangle$ which is clearly both infinite and abelian.

This construction doesn't work over $\mathbb{R}$ because the only nontrivial root of unity is $\zeta = -1$. This leads me to wonder:

Question: Do there exist infinite abelian real reflection groups?

In case it helps to either construct an example or prove nonexistence, let me give the arguments for or against that arose out of my attempts to answer this question.

Arguments for:

1. The existence of such an example over $\mathbb{C}$ is encouraging.
2. We can construct finite abelian real reflection groups of arbitrarily large order as follows: for each $n \geq 1$, let $V = \mathbb{R}^n$ and let $\{e_i\}_{i=1}^n$ be the standard basis vectors. Then the operators $s_{e_i}$ have order 2, commute and there are no other relations between them, so the reflection group they generate is isomorphic to $(\mathbb{Z}/2)^n$. I would like to do something like "take the limit" of this construction, but my definition requires that $V$ be finite-dimensional.

Arguments against:

Such a group would have to be infinitely-generated (in line with our $\mathbb{C}$ example). If $W$ were a finitely-generated abelian reflection group then, by the standard classification theorem, we would have $$ W \cong \mathbb{Z}^n \oplus \text{torsion}.$$

For $W$ to be infinite, the exponent $n$ needs to be positive. But $\mathbb{Z}$ does not contain any element of order 2 and, in particular, cannot be generated by a reflection.

Answer 1

I think that you can prove by induction on the dimension of $V$ that there is no infinite Abelian (finite-dimensional) real reflection group $G$. Suppose that $G \subseteq {\rm GL}(V)$ is such an Abelian infinite real reflection group with ${\rm dim}_{\mathbb{R}}(V)$ minimal. Then $V$ is certainly not $1$-dimensional. Let $t$ be one of the generating reflections of $G$. Then $V = U_{t} \oplus W_{t}$ is a $G$-invariant direct sum decomposition of $V$, where $U_{t}$ is the (one-dimensional) $-1$-eigenspace of $t$ on $V$ and $W_{t}$ is the $1$-eigenspace of $t$.

Furthermore, every element of $G$ acts as $\pm 1$ on $U_{t}$, and no non-identity of $G$ acts as the identity on both $U_{t}$ and $W_{t}.$ The elements of $G$ which act as $1$ on $U_{t}$ form a subgroup of $G$ of index $2$, and every generating reflection of $G$ (other than $t$) acts as a reflection on $W_{t}.$ It follows that $G/\langle t \rangle$ is isomorphic to an infinite Abelian reflection subgroup of ${\rm GL}(W_{t})$, contrary to the minimality of the dimension of $V$.