# When does a group and its pro-algebraic completion have equivalent categories of arbitrary representations?

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?

• 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. – Donu Arapura Aug 6 '20 at 6:47
• 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? – Patrick Elliott Aug 6 '20 at 6:58
• 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. – Donu Arapura Aug 6 '20 at 7:33
• 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 – Patrick Elliott Aug 6 '20 at 7:38
• 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? – Qiaochu Yuan Aug 6 '20 at 7:59