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

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.

  • 3
    $\begingroup$ 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). $\endgroup$ Sep 25, 2020 at 13:11
  • 2
    $\begingroup$ 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 $Spf(\hat{O}(X))/G$? $\endgroup$ 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
    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
    Sep 25, 2020 at 15:18

1 Answer 1


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

  • 1
    $\begingroup$ 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
    Sep 25, 2020 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.