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$\newcommand{\Rep}{\operatorname{Rep}}$ $\newcommand{\mo}{\operatorname{-mod}}$ $\renewcommand{\hat}{\widehat}$

I apologize in advance if this is a naive question but my background in algebraic geometry is fairly superficial. I mostly care about global quotients $X/G$ where $X$ is an affine scheme over $\mathbb C$ and $G$ a complex connected affine algebraic (reductive if you like) group. My understanding of those is pretty much limited to the fact that we have an equivalence of symmetric monoidal categories $$QC(X/G)\simeq O(X)\mo_{\Rep G}$$ where $O(X)$ is the algebra of global functions, and $\Rep G$ the category of $O(G)$-comodules.

Let $x \in X$ be a fixed point of the $G$ action. In a nutshell my question is:

What is the correct definition of the formal completion of $X/G$ at $x$ ? In particular what is its category of quasi-coherent sheaves thinking of it as an "ordinary" rather than formal stack (f that makes sense) ?

A basic observation is that $\hat O(X)$, the completion of $O(X)$ by the ideal of functions vanishing at $x$, is not an object in $\Rep G$. Now it seems there are different things one can do:

  1. Look at the category $\hat O(X)\mo_{\Rep \mathfrak g}$, which I guess should be like quasi-coherent sheaves on $\hat X/\hat G$
  2. Think of $\hat O(X)$ as a topological algebra, hence as an object in a certain category of topological $G$-representation (say the pro-completion of the category of finite dimensional $G$-modules).
  3. We can look at the coalgebra $C(X)$ of "distributions supported at $x$", i.e. the coalgebra which satisfies $C(X)^*=\hat O(X)$, which is a a coalgebra in $\Rep G$ so that you take take comodules over it.. This is the idea that formal affine scheme are the same as "cospectrum" of cocommutative coalgebras, and I think the category you get is equivalent to the one in 2 by taking duals.
  4. Although $\hat O(X)$ is not an object in $\Rep G$, it still makes sense to look at modules for this algebra that happens to be in there, i.e. $\hat O(X)\mo_{\Rep G}$ do makes sense.

Is any of those the, or a, correct definition ? Any insight or reference would be much appreciated.

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    $\begingroup$ I haven't thought through your different options yet, but another natural thing to consider is the completion of $O(X)$ internal to the category $Rep(G)$. In some cases, this will just be $O(X)$ again (e.g. if $G$ is reductive and $O(X)$ has finite multiplicity for each irreducible). You could consider this either as a ring object in $Rep(G)$ or as a pro ring object in Rep(G) (in the latter case, you probably end up with the same category as in 2 or 3). $\endgroup$
    – Sam Gunningham
    Commented Sep 25, 2020 at 13:11
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    $\begingroup$ Another vague thought: it seems to me that there is not an algebraic $G$-action on $Spec(\hat{O}(X))$ - though there is one on $Spf(\hat{O}(X))$. So from some perspective perhaps there is no "ordinary" stack, only the formal stack $Spf(\hat{O}(X))/G$? $\endgroup$
    – Sam Gunningham
    Commented Sep 25, 2020 at 13:14
  • $\begingroup$ @SamGunningham Thanks, indeed. I should say, I'm interested in the general picture, but in practice I want to translate say the fact that $\exp$ is a formal isomorphism of stacks $\mathfrak g/G \rightarrow G/G$ into an equivalence of categories of QC sheaves in a "minimal" way, ie I want to keep working with $Rep\ G$ and not $Rep\ \mathfrak g$. I suspect using 2, i.e. thinking of $\exp$ as inducing the PBW $G$-equivariant coalgebra iso $U(\mathfrak g) \cong S(\mathfrak g)$ is the way to go, but I was wondering if I had missed a "better" way. $\endgroup$
    – Adrien
    Commented Sep 25, 2020 at 15:15
  • $\begingroup$ I think it matches what you say in your second comment, but I guess I'm still confused by what it means exactly that the action of $G$ on the formal spectrum is algebraic, unless you're using "the corresponding coalgebra is an object in $Rep\ G$" as a definition of algebraic action in that case ? $\endgroup$
    – Adrien
    Commented Sep 25, 2020 at 15:18

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I will assume $X$ is smooth for simplicity, but it is probably not needed. Given the stack $X/G$, there are two completions one may consider:

  1. Completing along $\mathrm{B}G\rightarrow X/G$ one obtains $\hat{X}/G$.
  2. Completing along $\mathrm{pt}\rightarrow X/G$ one obtains $\hat{X}/\hat{G}$.

Your next question is about quasi-coherent sheaves. I will assume your definition of quasi-coherent sheaves on a prestack is given by the right Kan extension from affines. In particular, $\mathrm{QCoh}$ sends colimits of prestacks to limits of categories.

Let me begin with $\mathrm{QCoh}(\hat{X})$. By definition, $\hat{X}$ is a colimit $\mathrm{colim} X_\alpha$ of affines. So, $\mathrm{QCoh}(\hat{X})=\lim \mathrm{QCoh}(X_\alpha)=\lim \mathrm{Mod}_{\mathcal{O}(X_\alpha)}$. If $\mathcal{O}(\hat{X})$ is the corresponding topological algebra, this limit may be identified with the category of complete $\mathcal{O}(\hat{X})$-modules. Also, since $\mathcal{O}(X_\alpha)$ are finite-dimensional, you can rewrite it as $\lim \mathrm{CoMod}_{\mathcal{O}(X_\alpha)^*}$. So, you can identify this category with the category of comodules over the coalgebra of distributions $\mathrm{Dist}(\hat{X})$. (The inclusion of the structure sheaf $p^*\colon\mathrm{Vect}\rightarrow \mathrm{QCoh}(\hat{X})$ admits a left adjoint $p_+\colon \mathrm{QCoh}(\hat{X})\rightarrow \mathrm{Vect}$ and $\mathrm{Dist}(\hat{X})=p_+\mathcal{O}_X$.)

Next, $\mathrm{QCoh}(\mathrm{B}\hat{G})$. Let $i\colon \mathrm{pt}\rightarrow\mathrm{B}\hat{G}$ be the inclusion of the basepoint. The pullback functor $i^*\colon \mathrm{QCoh}(\mathrm{B}\hat{G})\rightarrow \mathrm{Vect}$ does not have a colimit-preserving right adjoint. Instead, it has a left adjoint. One can show that $i^*\colon \mathrm{QCoh}(\mathrm{B}\hat{G})\rightarrow \mathrm{Vect}$ is monadic and identifies $\mathrm{QCoh}(\mathrm{B}\hat{G})\cong \mathrm{Mod}_{\mathrm{U}\mathfrak{g}}$.

(Here is a quick proof on the derived level. By Theorem 10.1.1 in https://arxiv.org/abs/1108.1738 $\Upsilon\colon\mathrm{QCoh}(\mathrm{B}\hat{G})\rightarrow \mathrm{IndCoh}(\mathrm{B}\hat{G})$ is an equivalence since $G$ is smooth. And $\mathrm{IndCoh}(\mathrm{B}\hat{G})=\mathrm{Mod}_{\mathrm{U}\mathfrak{g}}$ by Proposition 2.4.31 in https://www.math.ias.edu/~lurie/papers/DAG-X.pdf.)

Combining the two equivalences, you get $$\mathrm{QCoh}(\hat{X}/\hat{G}) = \mathrm{CoMod}_{\mathrm{Dist}(\hat{X})}(\mathrm{Mod}_{\mathrm{U}\mathfrak{g}}),\qquad \mathrm{QCoh}(\hat{X}/G) = \mathrm{CoMod}_{\mathrm{Dist}(\hat{X})}(\mathrm{Rep}(G)).$$

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    $\begingroup$ Ha, that makes a lot of sense, I was aware that $QC(B\hat G)$ is $Rep\ \mathfrak g$ but as you correctly guessed really I was confused by how to properly interpret $\hat X/G$ as a completion of $X/G$. Thanks for this perfect-as-usual answer ! $\endgroup$
    – Adrien
    Commented Sep 25, 2020 at 16:45
  • $\begingroup$ Sorry for my ignorance. Do you have a reference of your two equivalences (mainly the second)? I imagine that the completion can be computed in terms of Čech complex ("Adams" completion), but I want a reliable reference for that. $\endgroup$
    – Z. M
    Commented Dec 20, 2022 at 15:22

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