# Formal completion of a quotient stack


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.

• 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). – Sam Gunningham Sep 25 '20 at 13:11
• 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 $Spf(\hat{O}(X))/G$? – Sam Gunningham Sep 25 '20 at 13:14
• @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. – Adrien Sep 25 '20 at 15:15
• 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 ? – Adrien Sep 25 '20 at 15:18

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

• 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 ! – Adrien Sep 25 '20 at 16:45