In Faltin-Metropolis-Ross-Rota's [FMRR] paper [*The Real Numbers as a Wreath Product*](https://doi.org/10.1016/0001-8708(75)90115-2) [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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Lamplighter_group#Generalizations) $\mathbb{Z}/(b) \,\wr\, \mathbb{Z}$.



The real numbers as a wreath product?
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 $n=3$.


