Let $k$ be an algebraically closed field of positive characteristic. Let $G$ be an affine reductive group acting on a smooth projective variety $X$. Let $E$ be a vector bundle on $X$ such that the action of $G$ lifts to an action on $E$. I am looking for a criterion which tells me when $E$ descends to a bundle on the GIT quotient $X//G$.

For example, in characteristic 0, one has the following criterion, which is Theorem 6.8 in notes by Drezet which can be found here: https://hal.archives-ouvertes.fr/hal-00742479/document

(*) $E$ descends to a bundle on $X//G$ iff for every closed point $x \in X$, such that the orbit $G.x$ is closed, the stabilizer $G_x$ acts trivially on the fiber $E_x$.

Is this result true in positive characteristic?

(Clarification on notation. Here $E_x:=E\otimes_{\mathcal O_X}{\mathcal O}_{X,x}/{\mathfrak m}_x$ )

EDIT: If $G$ is not linearly reductive, but is smooth, connected and reductive, then is (*) true in positive characteristic? In the answer of Jason, $G$ is not connected. In the counterexample in Alper's article, mentioned in the comment by Niels, $G$ is not smooth.

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    $\begingroup$ If $G$ is linearly reductive this is a particular case of results of Alper, Good moduli spaces for Artin stacks, see notably §1.3 and §10 in loc. cit. I guess it was known earlier, though. $\endgroup$ Commented Feb 23, 2019 at 13:44
  • 3
    $\begingroup$ I support Matthieu's statement - but watch out J.Alper, Adequate moduli spaces and geometrically reductive group schemes Example 5.6.1 for a counter-example for a general group scheme. $\endgroup$
    – Niels
    Commented Feb 23, 2019 at 14:23
  • $\begingroup$ @Matthieu: Thanks for a very precise reference. $\endgroup$
    – user45758
    Commented Feb 24, 2019 at 5:30
  • $\begingroup$ @Niels: Thanks. $\endgroup$
    – user45758
    Commented Feb 24, 2019 at 5:30

2 Answers 2


I have one question / comment. The notation $E_x$ is used in some sources to mean the stalk of a sheaf $E$ at a point $x$, considered as a module over the local ring $(\mathcal{O}_{X,x},\mathfrak{m}_x).$ However, by the context here, I am guessing that $E_x$ is intended to mean the "fiber" $E_x/\mathfrak{m}_x E_x$. Otherwise I am a bit confused what is meant by the action of $G_x$ on $E_x$, since this is not a finite dimensional $k$-linear representation.

Question. To the OP: is this what you meant?

In the following, I assume that the relevant $k$-linear $G_x$-represenation is the fiber $E_x/\mathfrak{m}_xE_x$ rather than the stalk $E_x$. Triviality of this finite dimensional $k$-linear representation of $G_x$ is a necessary condition for descent. This necessary condition is also sufficient when the smooth, affine group $k$-scheme is linearly reductive.

Nonetheless, there exists an action of a reductive (smooth) group $k$-schemes that is not linearly reductive satisfying the necessary condition yet failing descent. In other words, in the (geometrically) reductive case, the necessary condition is not sufficient. The following example is roughly the same as in my solution to an earlier MathOverflow question: Quotient of a smooth curve by a finite group and differentials

Let $k$ be an algebraically closed field of characteristic $p$. Denote by $\mathbb{P}^1_k$ the $k$-scheme $\text{Proj}\ k[s,t]$ together with the tautologial invertible quotient, $$(s,t):\mathcal{O}_{\mathbb{P}^1}^{\oplus 2}\to \mathcal{O}_{\mathbb{P}^1}(1).$$ By the universal property of projective space, for every $k$-scheme $Z$ and for every quotient invertible $\mathcal{O}_Z$-module, $$(\sigma,\tau):\mathcal{O}_Z^{\oplus 2}\twoheadrightarrow \mathcal{L},$$ there exists a unique $k$-morphism and $\mathcal{O}_Z$-module isomorphism, $$\phi:Z\to \mathbb{P}^1_k, \alpha:\phi^*\mathcal{O}(1) \xrightarrow{\cong} \mathcal{L},$$ such that $(\sigma,\tau)$ equals $\phi^*(s,t)\circ \alpha$.

In particular, denote by $(f,a)$ the pair as above for the following quotient invertible $\mathcal{O}_{\mathbb{P}^1}$-module, $$(s+t,t):\mathcal{O}_{\mathbb{P}^1}^{\oplus 2} \to \mathcal{O}_{\mathbb{P}^1}(1),$$ $$ f:\mathbb{P}^1_k \to \mathbb{P}^1_k, \ \ a:f^*\mathcal{O}_{\mathbb{P}^1}(1) \xrightarrow{\cong} \mathcal{O}_{\mathbb{P}^1}(1).$$ Together with the pullback map on global sections, this pair induces a $k$-linear map of global sections, $$a\circ f^*(-):H^0(\mathbb{P}^1_k,\mathcal{O}_{\mathbb{P}^1}(1)) \to H^0(\mathbb{P}^1_k,\mathcal{O}_{\mathbb{P}^1}(1)), \ \ s \mapsto s+t, \ \ t\mapsto t.$$ This linear map is a $k$-linear isomorphism. Moreover, it has finite order $p$. Thus, $f$ is a finite order automorphism of order dividing $p$, and $a$ is a linearization of the invertible $\mathcal{O}_{\mathbb{P}^1}$-module $\mathcal{O}_{\mathbb{P}^1}(1)$ with respect to the cyclic subgroup $$G=\langle f \rangle \leq \text{Aut}(\mathbb{P}^1_k,\mathcal{O}_{\mathbb{P^1}}(1)) = \textbf{GL}_{2,k}.$$ Denote the induced action on $\mathbb{P}^1_k$ by $G$ as follows, $$\mu:G\times \mathbb{P}^1_k \to \mathbb{P}^1_k, \ \ (f^d,[s,t]) \mapsto [s+dt,t].$$ Note that the action $\mu$ is a free action when restricted to the open subscheme $W:= D_+(t)$. The non-free locus of $f$ equals the $k$-point $x=\text{Zero}(t) = [1,0]$, and this is a fixed point.

One $G$-invariant open neighborhood of $x$ equals $U=D_+(s(s^{p-1}-t^{p-1}))$. The global section $s$ of $\mathcal{O}_{\mathbb{P}^1}(1)$ is a generator on $U$. Thus, there exists a unique section $z$ over $U$ of $\mathcal{O}_{\mathbb{P}^1}(U)$ such that $t$ equals $zs$ as sections over $U$ of $\mathcal{O}_{\mathbb{P}^1}(1)$. The zero scheme of $z$ on $U$ equals the $k$-point $x$. The local ring $\mathcal{O}_{X,x}$ equals the DVR $k[z]_{\langle z \rangle}$.

Since $s$ is a generator of $\mathcal{O}_{\mathbb{P}^1}(1)$ on the $G$-invariant open subscheme $U$, also $f^*s$ is a generator of $f^*\mathcal{O}_{\mathbb{P}^1_k}(1)$. The $\mathcal{O}_{\mathbb{P}^1_k}$-module isomorphism $a$ sends $f^*s$ to $s+t$. For the specified generators $f^*s$ and $s$, the stalk of $a$ at $x$ is equivalent to the following isomorphism of invertible modules over $k[z]_{\langle z \rangle}$, $$k[z]_{\langle z \rangle} \xrightarrow{1+z} k[z]_{\langle z \rangle}.$$ For $E=\mathcal{O}_{\mathbb{P}^1}(1)$, the induced map of the fiber $E_x/\mathfrak{m}_xE_x$ is the identity map. Thus, the stabilizer $G$ of $x$ acts trivially on the fiber of $E$ at the unique non-free point $x$.

On the other hand, without even computing the quotient, we know that $E=\mathcal{O}_{\mathbb{P}^1}(1)$ is not isomorphic to a pullback. Indeed, the quotient map has degree $p$, so that every pullback has degree divisible by $p$. Yet the degree of $\mathcal{O}_{\mathbb{P}^1}(1)$ equals $1$. Therefore it is not a pullback.

In fact, a description of the quotient is given in my previous answer. One $G$-invariant map that is a uniform categorical quotient is the following, $$ q:\mathbb{P}^1_k \to \mathbb{P}^1_k, \ \ b:q^*\mathcal{O}_{\mathbb{P}^1}(1)\xrightarrow{\cong} \mathcal{O}_{\mathbb{P}^1}(p), $$ $$b(q^*s) = s(s^{p-1}-t^{p-1}), \ \ b(q^*t) = t^p.$$ This is a finite, flat $k$-morphism. Thus, if the pullback of a quasi-coherent sheaf is flat, then the quasi-coherent sheaf is also flat. Therefore, any quasi-coherent sheaf that pulls back to an invertible sheaf is itself an invertible sheaf. Every invertible sheaf on $\mathbb{P}^1_k$ is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(d)$ for some integer $d$. Via the isomorphism $b^d$, the pullback $q^*\mathcal{O}_{\mathbb{P}^1}(d)$ is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(pd)$. In particular, since $1$ is not a multiple of $p$, the $G$-linearization $a$ of the invertible sheaf $\mathcal{O}_{\mathbb{P}^1}(1)$ is not a pullback of an invertible sheaf.

Please note: effective descent does hold for the finite flat morphism $q$. This "failure" of descent is due to the fact that the following natural $k$-morphism is not an isomorphism, $$G\times \mathbb{P}^1_k \xrightarrow{(\mu,\text{pr}_2)} \mathbb{P}^1_k \times_{q,\mathbb{P}^1,q}\mathbb{P}^1_k.$$ Indeed, the fiber product $\mathbb{P}^1_k\times_{q,\mathbb{P}^1,q}\mathbb{P}^1_k$ is not normal, and $(\mu,\text{pr}_2)$ equals the normalization.

  • $\begingroup$ I have clarified the notation in the question. Thanks a lot for your detailed explanation. $\endgroup$
    – user45758
    Commented Feb 24, 2019 at 5:22
  • $\begingroup$ In your counterexample, $G$ is not connected. Do you know if the descent criterion holds if $G$ is assumed to be connected and reductive? $\endgroup$
    – user45758
    Commented Feb 28, 2019 at 18:35
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    $\begingroup$ @user45758. For any example of an action $\mu:G\times X \to X$ as above and a $G$-linearized $\mathcal{O}_X$-module $(\mathcal{L},\psi)$, there is a "trick" to get a similar example with a connected group. Let $\rho:G\to \textbf{GL}_n$ be a faithful representation. Now define $X'$ to be the free quotient of $\textbf{GL}_n \times X$ by the "diagonal" action of $G$, i.e., $g\bullet(M,x) = (M\rho(g^{-1}),\mu(g,x))$. The left action of $\textbf{GL}_n$ on itself induces a left $\textbf{GL}_n$-action on $X'$ with $X'//\textbf{GL}_n = X//G$. $\endgroup$ Commented Mar 1, 2019 at 9:21
  • $\begingroup$ Thanks a lot for explaining this. I am adding some details and posting it as an answer. $\endgroup$
    – user45758
    Commented Mar 4, 2019 at 8:13

This answer adds a few details to the comment by Jason Starr. Let $G$, $X$ and $\mathcal L=\mathcal O_{\mathbb{P}^1}(1)$ be as in the answer by Jason Starr. Let $\rho:G\to GL_n$ be a faithful representation. Let $G$ act on $GL_n\times X$ by $$g\bullet (M,x):=(M\rho(g^{-1}),\mu(g, x))$$ Lift the action of $G$ on $GL_n\times X$ to $GL_n\times \mathcal L$ in a similar way, that is, $$g\bullet (M,v):=(M\rho(g^{-1}),\psi(g,v))$$ Both actions are free; denote the quotients by $X'$ and $\mathcal L'$.

The left action of $GL_n$ on itself induces a left action of $GL_n$ on $GL_n\times X$ and on $GL_n\times \mathcal L$. This action commutes with the action of $G$, and so we get an action of $GL_n$ on $X'$ and $\mathcal L'$. Let us denote the points of $X'$ by $[A,p]$. If $B\in GL_n(k)$ stabilizes $[A,p]$, then we have $BA=A\rho(g^{-1})$ and $p=\mu(g,p)$ for some $g\in G$. Thus, if $[A,p]$ has nontrivial stabilizer, then $p=x$ and the stabilizer of $[A,x]$ is exactly the subgroup $AGA^{-1}$. It is checked easily that the stabilizer of $[A,x]$ acts trivially on the fiber $\mathcal L'\vert_{[A,x]}$.

There is a commutative diagram $\require{AMScd}$ \begin{CD} X @>a>> X'\\ @VVV @VVV\\ X//G @= X'//GL_n \end{CD} The map $a$ is the composite $X\to GL_n\times X\to X'$, and is given by $p\mapsto [e,p]$.

If the line bundle $\mathcal{L}'$ were to descend to $X'//GL_n$, then this would mean that the pullback of $\mathcal{L}'$ to $X$ descends to $X//G$. But the pullback of $\mathcal{L}'$ to $X$ is $\mathcal{L}$ and we know $\mathcal{L}$ does not descend.


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