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$\DeclareMathOperator\Spm{Spm}\DeclareMathOperator\SL{SL}\DeclareMathOperator\Hom{Hom}$Let $T$ be the topos of $\Spm\mathbb{Q}_p$-rigid analytic spaces, $G$ an abstract group, and $\Hom(G,\SL_2)$ the sheaf which associates to an affinoid $X\in T$ with Tate algebra $A$ the set of continuous homomorphisms from $G$ to $\SL_2(A)$.

When $G$ is a finite group, generated by $r$ elements, the sheaf $\Hom(G,\SL_2)$ may be represented as a scheme by identifying it as being cut out from $\SL_2^r$ subject to the conditions coming from the relations of $G$.

Is there any obstruction to the representability of the same Hom sheaf when $G$ is infinite? I am particularly interested in the case when $G$ is profinite, and more precisely the absolute Galois group of the rationals.

Thanks in advance!

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  • $\begingroup$ In the profinite case, do you really want homomorphisms $G \to \operatorname{SL}_2(A)$ instead of continuous homomorphisms? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jan 5, 2023 at 21:06
  • $\begingroup$ Yeah, I want a continuous morphism. I will edit my question indeed. Otherwise the same construction works. $\endgroup$
    – kindasorta
    Commented Jan 5, 2023 at 21:27
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    $\begingroup$ If the Hom sheaf is representable for every finite group $G$, then for every profinite group, the sheaf of continuous homomorphisms is representable by the associated ind scheme. $\endgroup$
    – Jason Starr
    Commented Jan 5, 2023 at 23:45
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    $\begingroup$ Could you please elaborate? Perhaps it would be easier to specialize to the case where $G = \mathbb{Z}_p$ under addition. $\endgroup$
    – kindasorta
    Commented Jan 6, 2023 at 0:09
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    $\begingroup$ I missed that you are specifically looking at continuous maps to $\textbf{SL}_2(A)$. I thought that you were talking about the sheaf of groups of $\textbf{SL}_2$ on the category of schemes, with no extra topological / analytic structure. Then "continuous" means "factors through a finite quotient". So the functor is representable by the ind scheme associated to all finite quotients. I do not know the answer for $\textbf{SL}_2(A)$ with its natural topological structure. $\endgroup$
    – Jason Starr
    Commented Jan 6, 2023 at 11:32

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