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Edit on Nov. 20, 2023. This question is answered below in the case that $0<r_i<1$. And indeed it is shown in the answers to not be an interesting question in that case. So please take all $r_i=1$ in what follows. I still think it is an interesting question in the case that all $r_i=1$, so I will leave it up to see if anyone has an answer in that case. The reason I was arguing against the answer below for the case $0<r_i<1$ is that for any $1>s>r>0$ the "wrong way" maps $\mathbb{Z}_p[[x]]\to \mathbb{Z}_p<\frac{x}{r}> \to \mathbb{Z}_p<\frac{x}{s}>$ are not bounded and I was thinking that meant they were not continuous, but I was mistaken: that is an implication that works for real or complex Banach spaces not here. It is just that there are many Banach norms on the ring of formal power series $\mathbb{Z}_p[[x]]$ that are topologically equivalent to the product topology, yet the maps between them are only bounded in one direction. These norms are also not equivalent to $\mathbb{Z}_p[[x]]$ considered as a product in the Ind-Ban category, but again that doesn't matter for the topology. Please do not write any other comments or answers about the case that all $r_i<1$. You can read the question by replacing $R<\frac{x_1}{r_1}, \dots \frac{x_k}{r_k}>$ by $R<x_1, \dots ,x_k>$. I left the original unedited question below.

Let $K$ be an algebraic number field and let $G$ be either the Galois group of the maximal algebraic extension of $K$ in a separable algebraic closure $\overline{K}$ unramified outside a finite sent of non-archimedean places of $K$, or the Galois group of a the completion of $K$ at some place. Let $R$ be some simple non-archimedean field or ring such as $\mathbb{F}_p$, $\mathbb{Z}_p$, or $\mathbb{Q}_p$. Is anything known about lifting continuous homomorphisms $G \to \operatorname{GL}_n(R[[x_1, \dots, x_k]])$ to continuous homomorphisms $G \to \operatorname{GL}_n(R\langle\frac{x_1}{r_1}, \dots, \frac{x_k}{r_k}\rangle)$? In other words, does there exist positive real numbers $r_1, \dots, r_k$ such that that there is a lift in general? If not, is there some criteria known for their existence? The notation $R\langle\frac{x_1}{r_1}, \dots, \frac{x_k}{r_k}\rangle$ refers to a Tate algebra.

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  • $\begingroup$ $R[[x]]$ refers to the formal power series equipped with the product topology as a countable product of the Banach ring $R$. This notation does not refer to an open disk. By $R<\frac{x}{r}>$ I mean those formal power series $\sum a_i x^i$ such that $\lim |a_i|r^i = 0$ equipped with the norm $|\sum a_i x^i|= \sup |a_i|r^i$ and I think of this as representing functions on a closed disk of radius $r$. $\endgroup$
    – anon
    Commented Nov 20, 2023 at 9:02
  • $\begingroup$ The product topology on $R[[x]]$ is the same as the projective limit topology on the system $R[x]/x^n$, where the finite $R$ modules $R[x]/x^n \cong R^n$ inherit their topology from $R$. Thanks for the thought provoking answers that have been given so far. $\endgroup$
    – anon
    Commented Nov 20, 2023 at 9:40
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    $\begingroup$ This question is now so strewn with wreckage from previous iterations as to be incomprehensible. Please delete it, and open a new one if you still have any actual questions to ask. $\endgroup$ Commented Nov 20, 2023 at 15:58
  • $\begingroup$ This question still makes essentially no sense. $\endgroup$ Commented Nov 21, 2023 at 12:29

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I don't really understand the setup of this question: what is $R\langle x_1/r_1, \dots \rangle$ supposed to mean if $r_i$ is a real number?

That said, if $R = \mathbb{Z}_p$, then $\mathbb{Z}_p[[x_1, \dots, x_k]]$ actually embeds (continuously) into $\mathbb{Z}_p\langle x_1 / p^{a_1}, \dots, x_k / p^{a_k}\rangle$ for any strictly positive $a_1, \dots, a_k$, so you can just map your Galois represetnation along. Geometrically, this is just saying that an open polydisc of radius 1 contains a closed polydisc of any smaller radius. This is such a trivial remark that I don't think it can be the question you meant to ask; but if so, perhaps you could clarify the intended question.

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    $\begingroup$ Ah. I see. So you really want to study the representations up to equivalence. $\endgroup$ Commented Nov 15, 2023 at 21:26
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    $\begingroup$ But $\mathbb{Z}_p[[x]]$ is really the ring of bounded-by-1 functions converging on the open unit disc. Any such series converges on $|x|_p<1$! So, perhaps your perception of the problem you are studying is not quite accurate. $\endgroup$
    – tkr
    Commented Nov 16, 2023 at 4:00
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    $\begingroup$ This is not such a contrast to Mazur's setup as you seem to think, since $\mathbb{Z}_p[[X]]$ embeds in the ring you are calling $\mathbb{Z}_p\langle X / r\rangle$ for any $r < 1$. $\endgroup$ Commented Nov 16, 2023 at 8:49
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    $\begingroup$ Surely any representation into $GL_n(\mathbb{Q}_p[[X]])$ can be conjugated into $GL_n(\mathbb{Z}_p[[X]])$? $\endgroup$ Commented Nov 16, 2023 at 15:56
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    $\begingroup$ That is because $G$ is compact, right? $\endgroup$ Commented Nov 16, 2023 at 16:30

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