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This question is a sequel to An infinite profinite group such that any $p$-adic representation has finite image .

Fix a prime $ p $. We call an infinite profinite group $G$ a Boston group (with respect to $ p $) if every continuous homomorphism $ G\to {\rm GL}_{n}(\overline{\mathbb{F}_{p}((t))}) $ has finite image for any positive integer $n$. Again, the motivation for defining such a group comes from the following example:

Let $ K $ be a number field, $ S $ a finite set of primes of $ K $ not containing any primes above $ p $ and $ G_{K,S} $ the Galois group of the maximal extension of $ K $ unramified outside $ S $. Nigel Boston conjectured that $ G_{K,S} $ is a Boston group (with respect to $p$), cf. Conjecture 2 in [Nigel Boston, Some cases of the Fontaine-Mazur conjecture. II, J. Number Theory 75 (1999), no. 2, 161–169.]

My question is that: can someone give more examples of Boston groups? For example, is $ {\rm SL}_{n}(\mathbb{Z}_{p}) $ a Boston group?

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    $\begingroup$ Yes, because for $n\ge 3$, the group $\mathrm{SL}_n(\mathbf{Z})$ has no faithful finite-dim representation over any field of finite characteristic. This is because there is no distorted infinite cyclic subgroup in these groups. This also apply to many other groups for the same reason, e.g., the profinite or pro-$p$-completion of the integral Heisenberg group. This is also true for $n=2$, arguing with $\mathrm{SL}_2(\mathbf{Z}[1/\ell])$ for any choice of prime $\ell\neq p$. $\endgroup$
    – YCor
    Commented Jun 11, 2022 at 22:16
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    $\begingroup$ I think the quotient of the free rank $n$ pro-$p$ group by $n$ elements chosen at random should have this property with probability 1, but I don't know if this is a theorem. $\endgroup$
    – Will Sawin
    Commented Jun 11, 2022 at 23:54
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    $\begingroup$ For every prime $p$ and every $n\ge 2$, the free pro-$p$-group on $n$ generators is not, as topological group, linear over any local field (Barnea-Larsen J. Algebra 1999 DOI link) $\endgroup$
    – YCor
    Commented Jun 12, 2022 at 8:43

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