In the following everything is over some field $k$.

Let $G$ be a discrete group. We write $G^{\text{alg}}$ for its pro-algebraic completion. The latter is a pro-affine pro-algebraic group which arises as the Tannakian dual to the category $\mathrm{Rep}_{\text{fd}}(G)$ of finite-dimensional $G$-representations. By definition there is an equivalence of categories $$ \mathrm{Rep}_{\text{fd}}(G)\simeq \mathrm{Rep}_{\text{fd}}(G^\text{alg}). $$ Under what conditions does this lift to an equivalence of categories of possibly-infinite dimensional representations?

  • $\begingroup$ For the usual story, I think you want finite dim representations such that $G^{alg}\to Gl_n(k)$ is a morphism of group schemes. I sort of doubt there is a useful statement for infinite dim reps, unless $G$ is finite of corse. $\endgroup$ Aug 6, 2020 at 6:47
  • $\begingroup$ Is it possible that finiteness is necessary? Certainly Rep(G) (infinite dim) recovers G, and Rep(G^alg) recovers G^alg (it's just the ind-completion I think), so if the two are equivalent doesn't that say G and G^alg are iso as group schemes? $\endgroup$ Aug 6, 2020 at 6:58
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    $\begingroup$ I agree $Rep(G{alg})$ should be the ind-completion, What concerns me is that this may not contain the regular representation of G, when it’s infinite. $\endgroup$ Aug 6, 2020 at 7:33
  • $\begingroup$ Right, but the finite dimensional representations are already enough to recover G^alg, by definition, so what's the issue? Maybe I'm missing something obvious $\endgroup$ Aug 6, 2020 at 7:38
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    $\begingroup$ If the definition of infinite-dimensional representation of $G^{alg}$ is what I think it is, then it sounds to me like the answer is "iff $G$ is finite" precisely for the reason Donu points at: the regular representation of $G$ won't be a representation of $G^{alg}$ because it will fail to be a filtered colimit of finite-dimensional representations. Right? $\endgroup$ Aug 6, 2020 at 7:59


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