The real numbers as a wreath product?

Question

In Faltin-Metropolis-Ross-Rota's [FMRR] paper The Real Numbers as a Wreath Product [Adv. Math. 16(3), 278-304 (1975)], the real numbers are constructed as a quotient of a certain subset of the ring of formal Laurent series $\mathbb{Z}((T))$, emphasizing the digit-expansion aspect of elements of $\mathbb{R}$ by formalizing the idea of "infinite carrying". The paper by itself is not hard to follow, and my question is actually more about the title of the paper, for even after reading it completely, I did not understand what exactly was being referred to as a wreath product. (The term wreath product is mentioned exactly once in the entire paper: in the title!)

Here's a (very) quick summary of the construction:

Let $b\geq 2$ be a natural number (base). Write $\mathbf{C}_b \subseteq \mathbb{Z}((T))$ for the ring of convergent sequences: $$ \mathbf{C}_b = \left\{ \sum_{n\in\mathbb{Z}} a_n T^n \in \mathbb{Z}((T)) ~\bigg|~ |a_n| = o(b^n) \right\}. $$ Write $\mathfrak{K}_b := 1 - bT$ for the carry constant. Then, denote the ring of bounded sequences and the carry ideal by, resp., $$ R_b := \left\{ \sum_{j\in\mathbb{Z}} a_j T^j \in \mathbb{Z}((T)) ~\bigg|~ \sum_{j=1}^{n} |a_j| b^{n-j} = O(b^n) \right\}, \quad I_b := (\mathfrak{K}_b \mathbf{C}_b) \cap R_b, $$ so that $I_b \subseteq R_b \subseteq \mathbf{C}_b$. Then:

Theorem (FMRR). $R_b/I_b \simeq \mathbb{R}$ as ordered topological fields, for every $b \geq 2$.

Remark. The big-$O$ notation is just a shorthand for the appropriate condition (which, although similar, is defined solely in terms of $\mathbb{Z}$). Both the order and the topology turn out to be the "obvious ones" (of course); while a topology may be defined in terms of the magnitude of $\sum_j |\alpha_j - \beta_j| b^{n-j}$ as $n$ grows, the order depends on a canonical representation of strings (referred to as clearing in [FMRR]).

My question is: what exactly in this construction is being referred to as a wreath product? I see that $\prod_{j\in\mathbb{Z}} \mathbb{Z}/(b)$ could be acting as a basis for some potential wreath product, but I see no obvious group action that makes it substantially different from the generalized Lamplighter group $\mathbb{Z}/(b) \,\wr\, \mathbb{Z}$.

Answer 1

Consider the finite analogue where we model $\mathbb{Z}/b^n \mathbb{Z}$ as the ring $R = \mathbb{Z}[T]/\langle T^n \rangle$ modulo the ideal generated by $1-bT$. Note that $1$ in $R$ corresponds to $b^{n-1}$. Reversing sequences, so we get the 'little-endian' representation, we take the abelian group $\mathbb{Z}^n$ and quotient by the subgroup generated by all

$$ (0,\ldots, 0, -b, 1, 0, \ldots, 0) .$$

A set of representatives for the cosets is the $b$-ary words of length $n$. These are in bijection with the leaves of the rooted $b$-ary tree with $n$ levels. The automorphism group of the tree is the iterated wreath product $S_b \wr \ldots \wr S_b$, with $n$ factors. Since addition respects congruence modulo powers of $b$, addition of a constant is an automorphism of the tree. This already gives some connection.

To bring in carries, consider the binary case. Now addition of $1$ is the automorphism which swaps the two halves of the tree, and then continues on the (new) right-hand subtree, by swapping its two halves, and then continues working further down the tree. (For instance, after the first swap, an even numbered leaf $2m$ is replaced with the odd numbered leaf $2m+1$, after the second, an odd numbered leaf $4m+1$ is replaced with $4m+2$, and so on.) I think this repeated swapping is essentially the carry relation.

To make this explicit take $b=2$ and $n=3$. Then the leaves of the tree are labelled $0,4,2,6,1,5,3,7$, corresponding to the binary words

$$000, 001, 010, 011, 100, 101, 110, 111.$$

Addition of $1$ is the permutation $(0,1)(4,5)(2,3)(6,7)$ followed by $(0,2)(4,6)$ followed by $(0,4)$. (Confusingly the elements in these permutation come from successive left sub-trees, because the labels change as we apply the automorphism.) Chasing $3$, we get $3 \mapsto 2 \mapsto 0 \mapsto 4$. This corresponds to

$$110 \mapsto 210 \mapsto 020 \mapsto 001$$

in the carries model. For $7$ the carries model requires four steps

$$111 \mapsto 211 \mapsto 021 \mapsto 002 \mapsto 000$$

rather than the three predicted by the permutations, but this should be expected because we truncated the elegant infinite construction in the paper at $n=3$.