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In our study of automorphism groups of transcendental field extensions, we have encountered the situation where we have a group $F$ together with an endomorphism $\alpha \colon F \to F$, resulting in a group $G$ which is the projective limit of infinitely many copies of $F$ (indexed by the natural numbers) with respect to $\alpha$:

$$ G = \{ (a_0, a_1, a_2, \dots) \mid a_i \in F, \alpha(a_{n+1}) = a_n \text{ for all } n \geq 0 \} . $$

Does this construction have a name?

We are applying this specifically in the case where $F$ is the multiplicative group of $\mathbb{C}$ (or any other suitable field) and $\alpha \colon a \mapsto a^p$ for some fixed prime $p$. In this case, our group $G$ contains the group $\mathbb{Q}_p$ of $p$-adic numbers (w.r.t. addition), but it is (much?) larger. Is this group a "known group"?

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    $\begingroup$ In your example $F$ is isomorphic to $\mathbf{Q}^{(c)}\times\mathbf{Q}/\mathbf{Z}$ as abstract group. Unless you take into consideration the topology of the group $\mathbf{C}^*$, in which case it's more interesting: $\mathbf{C}^*\simeq\mathbf{R}\times\mathbf{R}/\mathbf{Z}$. The resulting group is then $G=\mathbf{R}\times S_p$ (, $S_p$ = $p$-solenoid). This is the Pontryagin dual of $\mathbf{R}\times\mathbf{Z}[1/p]$. The dense embedding $\mathbf{Z}[1/p]\to\mathbf{Q}_p$ inducing a dense embedding $\mathbf{Q}_p\to S_p$. $\endgroup$
    – YCor
    Commented May 25, 2021 at 16:53
  • $\begingroup$ @YCor Thanks — but what is $\mathbf{Q}^{(c)}$? Could you expand (and perhaps post this as an answer)? We do not take the topology of $\mathbf{C}^*$ into account, but I don't see how it plays a role: the resulting group is independent of the choice of the field provided it has $p^n$-th roots for all $n \geq 0$ and characteristic not equal to $p$. Could you explain what you mean? Thanks! $\endgroup$
    – Tom De Medts
    Commented May 26, 2021 at 7:24
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    $\begingroup$ $G^{(X)}$ means the set of finite support functions $X\to G$. Thus $Q^{(c)}$ means "a vector space over $Q$ of dimension $c:=2^{\aleph_0}$". My point is that in your example there's a topology that makes things more interesting, and the embedding of $\mathbf{Q}_p$ is indeed continuous and natural. As a plain abstract group, an embedding of $\mathbf{Q}_p$ is just the same as $\mathbf{R}$. $\endgroup$
    – YCor
    Commented May 26, 2021 at 7:40
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    $\begingroup$ If $K$ is an algebraically closed field of characteristic zero then $K^*$ is (non-canonically) isomorphic to $\mathbf{Q}^{|K|}\oplus\mathbf{Q}/\mathbf{Z}$, and hence to $\mathbf{Q}^{|K|}\oplus\bigoplus_{q\neq p}C_{q^\infty}\oplus C_{p^\infty}$. $\endgroup$
    – YCor
    Commented May 26, 2021 at 8:21
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    $\begingroup$ The underlying abstract group of the projective limit is the same when $F$ is viewed as discrete or topology (said otherwise: forgetting the topology and taking the projective limit commute, when the projective limit is taken within topological groups). As discrete group $F=H\oplus C_{p^\infty}$ where $H$ is the direct sum of my previous comment. In $H$ multiplication by $p$ is bijective. So as discrete group $G=H\oplus\mathbf{Q}_p$. Actually if we view $H$ as discrete group but consider the proj. limit as top group, this gives $G\simeq H\oplus\mathbf{Q}_p$ as topological group ($H$ discrete). $\endgroup$
    – YCor
    Commented May 26, 2021 at 9:50

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