Let us suppose the the group $G:=\mathbb{Z}/2\mathbb{Z}=(1,i)$ freely act on a smooth projective variety/$k$ $X$ and denote by $Y$ the G.I.T. quotient $X/G$. Let $\pi:X\longrightarrow Y$ the quotinet map. Now take a $G$-linearised coherent sheaf $(\mathcal{F}, \lambda)$, one can construct the sheaf of invariants of $\mathcal{F}$,the sheaf $\pi_\ast(\mathcal{F})^G$, that is a sheaf on $Y$. Then given $\overline{\alpha}=\pi(\alpha)$ a point of $Y$ it is possible to consider the fiber of $\pi_*(\mathcal{F})^G$ at $\overline{\alpha}$, $\pi_\ast(\mathcal{F})^G(\overline{\alpha})$. On the other and one can consider the fiber of $\pi_\ast(\mathcal{F})(\overline\alpha)$ that is (I think) isomorphic to the direct sum $$\mathcal{F}(\alpha)\oplus\mathcal F(i\cdot\alpha)\simeq \mathcal F\otimes(k(\alpha)\oplus k(i\cdot\alpha))$$ Now this vector space admit a $G$-action induced by $\lambda$, given by $\lambda\otimes\sigma$ where sigma is the action on $k(\alpha)\oplus k(i\cdot\alpha)$ given by permutation.

My question is: the fiber of the sheaf of invariants is isomorphic to the invariant subspace of the fiber (in the given action)? I think the answer is no (the first is a subspace of the latter), but I would prefer it to be yes...

If the answer is yes, how do I prove it? could you give me some references?

If the answer is no, could you give me an explicit counterexample?

Thank you very much for the time you dedicated to me and a special thanks to every one who will answer me best regards Stgermain


It all works out as well as you could want in every possible sense because of the freeness of the action. As you know, you can make a cover by $G$-stable affine opens, so the real work is in that case. So we focus on the affine case, and then all hypotheses on the affine can be removed: let $A$ be any ring whatsoever and $G$ a finite group acting freely on $X = $ Spec($A$) in the sense of acting freely on the set of geometric points valued in any algebraically closed field, which forces the strongest sense of acting freely on points valued in any ring at all. Thus, $G \times X := \coprod_{g \in G} X \rightarrow X \times X$ as functors via $(g,x) \mapsto (x, g.x)$ is a subfunctor inclusion and thus an equivalence relation on $X$ in the sense of functors. Now for the real content: that equivalence relation condition, coupled with $G \times X \rightrightarrows X$ being finite locally free (even finite etale) implies that SGA3, Expose V, 4.1(iv) applies, so $X := {\rm{Spec}}(A)$ is a finite etale cover of $Y := {\rm{Spec}}(A^G)$ and the natural map $$\coprod_ {g \in G} X \rightarrow X \times_ Y X$$ via $x_g \mapsto (x, g.x)$ on the $g$th copy of $X$ is an isomorphism. This is a deep result of Grothendieck in such generality. In particular, if $A^G$ is a $B$-algebra and $B \rightarrow B'$ is any map of rings then the natural map $$B' \otimes_B A^G \rightarrow (B' \otimes_B A)^G$$ is an isomorphism.

So the $G$-invariant map $\pi:X \rightarrow Y$ is a $G$-torsor for the etale topology and hence $Y$ is a quotient of $X$ by the $G$-action in all good senses (quotient sheaf, good behavior under base change, universal mapping property, etc.). In particular, by etale descent theory (see the example of "Galois covers" in section 6.2 or so of the book "Neron Models") it follows that $\pi_{\ast}^G$ and $\pi^{\ast}$ induce inverse equivalence between the category of quasi-coherent $O_Y$-modules and $G$-equivariant quasi-coherent $O_X$-modules.

In particular, for $G$-equivariant quasi-coherent $O_X$-modules $F$, the formation of $\pi_{\ast}(F)$ and $\pi_{\ast}(F)^G$ commute with any base change on the quotient (e.g., passing to a fiber there). Hence, the geometric fibers of $\pi_{\ast}(F)$ are the $G$-invariants on the fibers, as you desired, and more vividly the fiber of $\pi_{\ast}(F)$ at a geometric point $y$ of $Y$ is the direct sum of the fibers at the points in the $G$-orbit fiber of geometric points on $X$ over $y$, with $G$ acting simply transitively on this collection of fibers at points of $X_y$.

Edit: As t3suji points out, the situation is much simpler (without needing freeness conditions) if $|G|$ is a unit on the schemes in question.

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    $\begingroup$ On the other hand, one question being asked is simply `is the fiber of the sheaf of invariants equal to the invariants in the fiber of the direct image'? In this form, the answer is still positive (assuming char k and card(G) are coprime) even if the action is not free, and the proof is quite easy. $\endgroup$ – t3suji Jul 8 '10 at 12:58
  • $\begingroup$ Wow! Thank you a lot for your answers/comments! That was really helpful. Now I will try to do my homework and proving the case in which the action is not free.. thank you again best wishes Stgermain $\endgroup$ – Rurik Jul 8 '10 at 14:11
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    $\begingroup$ Dear Stgermain: "unit on a scheme" means "non-vanishing section of the structure sheaf". So an integer $n$ is a unit on a scheme $S$ when char($k(s)$) doesn't divide $n$ for all $s \in S$, or equivalently $S$ is a $\mathbb{Z}[1/n]$-scheme. As for "equivariant culture", by which I assume you mean "general stuff about quotients by group actions", begin by learning about smooth & etale maps (Milne's book on etale cohomology, Chapter 2 of "Neron Models", SGA1...) and then descent theory (early part of Chapter 6 of "Neron Models", SGA1, FGA Explained...) and then read Expose V in SGA3. $\endgroup$ – Boyarsky Jul 9 '10 at 14:06
  • $\begingroup$ I am sorry to bother you again, but there are some things I still do not get: if X is an abelian variety and Y its (singular) kummer variety (char$k\neq 2$) then the action is not free and the quotient map is not flat. If I have a G−sheaf on $X$ then I canconsider the sheaf $\pi_*F^G$ If I take the fiber at a a singular point $\overline{x}$ what do I obtain? I believed to get $\pi_*(F(x))^G$ but I do not think this is the same as $(\pi_*(F)(\overline{x}))^G$ (that is quoting t3suji the 'invariants in the fiber of the direct image'). What I am doing wrong? Thanks again and again sorry! $\endgroup$ – Rurik Jul 14 '10 at 10:10
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    $\begingroup$ Take a point $y\in Y$. The map $\pi:X\to Y$ is finite; consider the scheme-theoretic preimage $\pi^{-1}(x)$. For any $G$-equivariant sheaf $F$ on $X$, the following spaces are identified: - The fiber of the sheaf of invariants $\pi^*(F)^G$ at $y$; - The $G$-invariants in the fiber $\pi_*F(y)$ of $\pi_*F$ at $y$; - The $G$-invariants in the sections $H^0(\pi^{-1}(y),F)$. For the last description, you must take the scheme-theoretic preimage. For instance, even if $\pi^{-1}(y)$ is a single point $x$ with non-trivial scheme structure (which IIUC is what you ask), it is not just $F(x)^G$. $\endgroup$ – t3suji Jul 14 '10 at 13:55

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