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I'm trying to translate the Proj construction as a kind of quotient by a $\mathbb{G}_m$ action. Here's what I have so far:

Let $X=Spec\,A$ be an affine scheme (after this case is setteled I imagine it would be relatively easy to generalize to relatively affine schemes and get global Proj). Let $\mathbb{G}_m = Spec\,\mathbb Z[x,x^{-1}]$ be the multiplicative group.

The data of an action $\rho :X \times \mathbb{G_m} \to X$ is equivalently a homomorphism $\rho^{\flat}: A \to A[x,x^{-1}]$ which is equivalently a grading on $A$ given by $A=\bigoplus_{d \in \mathbb{Z}} A_d = \bigoplus_{d \in \mathbb{Z}} (\rho^{\flat})^{-1}(Ax^d)$.

This is very similar to a decomposition into irreducible subrepresentations only it doesn't look like a representation but rather a "co-representation".

How should I think of this "co-representation?" Is the theory of comodules over a hopf algebra just "dual" in some sense to the that of modules over a hopf algebra? (meaning everything is practically the same upto flipping arrows).

As far as I understand at this point we'd like to consider the free locus of the action on $X$. This is reasonable (coming from differential geometry where free actions by compact groups always exist) although I wonder if this technique works for any action.

Question 1 : Does the quotient of a scheme by a free action of a group scheme always exist? (as a scheme).

Interpreting the "free locus" of the action in terms of algebra is pretty confusing for me. Having knowledge of the solution these are supposed to be prime submodules of $A$ (prime ideals) which contain the irrelevant ideal $A_+$. The irrelevant ideal corresponds geometrically to the maximal irreducible subset of $X$ on which $\mathbb{G}_m$ acts trivially. So primes containing it correspond to irreducible sets contained in it which we should throw away as well. Is this true/enough?

Having thrown out the problematic points we get a free action on $X\text{ \ }X_p$ which we can now hopefully quotient by the action and get a $\mathbb{G}_m$ torsor. $SpecA \text{ \ } Spec{A_0} \to Proj A$. The fact that orbits of the action are in 1-1 with homogeneous prime ideals (not containing the irrelevant ideal) is a consequence of the fact that the orbit of every point in the free locus is itself a point (irreducible subscheme).

Having a torsor we have an equivalence of categories between $\mathbb{G}_m$-equivariant sheaves on $SpecA \text{ \ } Spec{A_0}$ and sheaves on $Proj A$

Question 2: Is this correct?

Question 3: What's a good reference for equivariant sheaves, quotients by group schemes and quotient stacks in the context of geometric representation theory?

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    $\begingroup$ The answer to question 1 is no: mathoverflow.net/questions/1558/…. Your construction of Proj(R) is not quite right in general: Proj(R) is the G_m-quotient of Spec(R) minus the vertex of the cone. The action of G_m on this is only free if R is generated in degree 1 over R_0. $\endgroup$
    – Marc Hoyois
    Commented May 9, 2016 at 14:44
  • $\begingroup$ @MarcHoyois What's the geometrical interpretation of "finitely generated in degree 1" what does it mean in terms of the action of $\mathbb{G}_m$? $\endgroup$
    – Saal Hardali
    Commented May 9, 2016 at 19:15
  • $\begingroup$ You might also want to take a look Cox's presentation of (normal) toric varieties as quotients of an open subset of an affine space. $\endgroup$
    – Avi Steiner
    Commented May 9, 2016 at 22:08
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    $\begingroup$ @SaalHardali Suppose A is generated over A_0 by homogeneous elements f_i. Then {A[1/f_i]} is a G_m-invariant open cover of the complement of the cone point. So the action is (scheme-theoretically) free iff, for each i, the map A[1/f_i] ⊗ A[1/f_i] → A[1/f_i][t^{±1}], a ⊗ b ↦ abt^{deg(a)} is surjective. An obvious sufficient condition is that A[1/f_i] contain an invertible element of degree 1. As Friedrich pointed out below, this is much weaker than being generated in degree ±1 (though it is equivalent for polynomial rings). $\endgroup$
    – Marc Hoyois
    Commented May 10, 2016 at 22:41
  • $\begingroup$ @MarcHoyois This may be a very stupid question, but wouldn't you want your map (action, identity) to be epi instead of surjective? $\endgroup$
    – user40276
    Commented May 26, 2016 at 7:02

1 Answer 1

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First of all, the ideal $I$ corresponding to the fixed point set is generated by all $A_d$ with $d\ne0$. Thus $I=\bigoplus_d I_d$ with $I_d=A_d$ for $d\ne0$ and $I_0=\sum_{d\ne0}A_dA_{-d}\subseteq A_0$.

Secondly, as already stated, the answer to Q1 in this generality is "no" but, under the given circumstances, the answer is actually "yes": If $\mathbb G_m$ (or any torus for that matter) acts locally freely (i.e., finite isotropy subgroups are ok) on a quasiaffine variety $X$ then the quotient exists as a scheme. This follows from the fact that this is true for affine varieties and $X$ can be covered by invariant affine open subsets.

BUT: The quotient will not be separated, in general. Take for example for $X$ the affine plane with $(tx,t^{-1}y)$-action and remove the origin. Then the action is free and the quotient is the affine line with a double point coming from the two components of $\{xy=0\}$. To get a separated quotient one has to throw out something, i.e., pass to an open subset. There are many choices and selecting one is the main point of GIT. There are even in general open subsets not coming from GIT. These yield quotients which are not quasi-projective. There are some papers of Białynicki-Birula and Swiecicka on this topic.

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  • $\begingroup$ Thanks! According to Marc Hoyois we need generation in degree 1 here. What does generation in degree 1 over $A_0$ has to do with the existence of the quotient? $\endgroup$
    – Saal Hardali
    Commented May 9, 2016 at 18:43
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    $\begingroup$ In the positively graded case, generation in degree 1 ensures that the action is free outside the fixed point set. It is not necessary, though. In the general $\mathbb Z$-graded case that is a bit more complicated to formulate. Anyway, freeness of the action is not required for the quotient to exist. It suffices that all isotropy groups are finite. Freeness is necessary, though, for the variety to be a principal bundle over its quotient. $\endgroup$
    – Friedrich Knop
    Commented May 9, 2016 at 19:27
  • $\begingroup$ Thanks! I think what I'm missing though is how to translate from property of action on $Spec A$ to property of coaction on $A$. In particular I don't understand why in the non negatively graded case generation in degree $1$ is equivalent to freenes of the action on $Spec A \text{ \ } Spec A_0$ is there a book/notes/article where I can find careful discussion of how to pass from properties of action on affine scheme to properties of the comodule over the corresponding hopf algebra? $\endgroup$
    – Saal Hardali
    Commented May 9, 2016 at 20:44
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    $\begingroup$ As I said, it is not equivalent. The condition is just necessary. In ensures that the variety can be embedded into a vector space and there the freeness of the action is clear. As a counterexample take $\tilde A=A_0\oplus\bigoplus_{d\ge d_0}A_d$ with $d_0\ge2$ fixed. Then $Spec A\to Spex\tilde A$ is an isomorphism outside the fixed point set. So the action is still free. $\endgroup$
    – Friedrich Knop
    Commented May 10, 2016 at 4:08

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