Timeline for Projective limit of copies of same group w.r.t. some fixed endomorphism
Current License: CC BY-SA 4.0
10 events
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| May 27, 2021 at 7:28 | history | edited | Tom De Medts | CC BY-SA 4.0 |
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| May 26, 2021 at 9:50 | comment | added | YCor | 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). | |
| May 26, 2021 at 8:46 | comment | added | Tom De Medts | @YCor Oh, it's only now that I see that you described $F$ and not $G$ as $\mathbf{Q}^{(c)} \times \mathbf{Q}/\mathbf{Z}$. So the question remains what $G$ is (as a topological group) if we equip $F$ with the discrete topology. | |
| May 26, 2021 at 8:21 | comment | added | YCor | 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 $\mathbf{Q}^{|K|}\oplus\bigoplus_{q\neq p}C_{q^\infty}\oplus C_{p^\infty}$. | |
| May 26, 2021 at 8:19 | comment | added | Tom De Medts | (By the way, what I wrote about the group being independent of the choice of the field is not correct; I had the subgroup in mind of elements starting with $a_0=1$, which is independent of the field.) | |
| May 26, 2021 at 8:18 | comment | added | Tom De Medts | @YCor We do want to take the topology into account arising from the projective limit (but with $F$ equipped with the discrete topology), so then $\mathbf{Q}_p$ is not just the same as $\mathbf{R}$. In addition, I don't really see where the factor $\mathbf{Q}/\mathbf{Z}$ comes from. (Notice that we consider a fixed prime $p$). | |
| May 26, 2021 at 7:40 | comment | added | YCor | $G^{(X)}$ 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 $\mathbf{R}$. | |
| May 26, 2021 at 7:24 | comment | added | Tom De Medts | @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! | |
| May 25, 2021 at 16:53 | comment | added | YCor | 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 $\mathbf{Q}_p\to S_p$. | |
| May 25, 2021 at 16:09 | history | asked | Tom De Medts | CC BY-SA 4.0 |