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Toen-Vaquié construct a category of schemes relative to some complete cocomplete closed symmetric monoidal category $C$. Affine schemes correspond by definition 1:1 to commutative monoid objects in $C$. For $C=\mathsf{Ab}$ we get the usual category of schemes, where affine schemes correspond to commutative rings. For $C=\mathsf{Set}$ one gets one of the various definitions of schemes over $\mathbb{F}_1$. One of the drawbacks of this quite general theory is that schemes are defined via their functors on commutative monoid objects, without any geometric incarnation.

Question. What happens when $C$ is the category of graded abelian groups (equipped either with the usual symmetry, or with the twisted symmetry)? Here affine schemes correspond to (graded) commutative rings. Is there any connection with the usual Proj construction? Is there any more geometric interpretation of these schemes? For example one might hope for a fully faithful functor into the category of locally ringed spaces. Is this category of schemes something new at all?

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    Aren't these just schemes with a $G_m$ action, or equivalently schemes over $BG_m$? More generally if you have a Tannakian reconstruction identifying $C$ with $QCoh(X)$ for some geometric stack $X$, isn't $C$-geometry just geometry over $X$? – David Ben-Zvi Mar 14 '13 at 01:10
  • This sounds quite reasonable, thank you. I would appreciate it if you expand this to an answer. – Martin Brandenburg Mar 14 '13 at 01:22
  • If these are schemes with a $\mathbb G_m$-action, then GIT modding out by the $\mathbb G_m$ action should get you the Proj, no? – Will Sawin Mar 14 '13 at 01:58
  • You could have a look at Michael Temkin's paper "On local properties of non-Archimedean spaces II" Isr. J. of Math. 140 (2004), 1-27 (see http://www.math.huji.ac.il/~temkin/papers/Local_Properties_II.pdf), especially the first section. – Jérôme Poineau Mar 14 '13 at 07:09
  • "Connection with Proj": I believe the natural functor you get would be homogeneous Spec. From there you can recover Proj. – Dylan Wilson Mar 14 '13 at 12:32
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    And you're dealing with locally graded-ringed spaces. These are spaces with a sheaf of graded rings so that the stalks are graded local in the sense that they have one homogeneous maximal ideal. – Dylan Wilson Mar 14 '13 at 12:34
  • @Dylan: This sounds very natural. But what are the affine graded schemes in this picture? Perhaps you can elaborate on this in an answer? Thank you. – Martin Brandenburg Mar 16 '13 at 20:31

1 Answers1

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It is my impression (without being very careful about it) is that the C-geometry (for C the category of graded abelian groups) is the same as geometry over $BG_m$, i.e., geometry of schemes with an action of the multiplicative group. This should follow from the identification of $C$ with representations of $G_m$. (In particular there is a close relation with Proj - except we don't throw away the "irrelevant ideal", we just consider the stacky quotient of a homogeneous variety in affine space by $G_m$.)

More generally, suppose your category $C$ is identified by a generalized Tannakian reconstruction theorem with $QCoh(X)$, the (complete cocomplete closed symmetric monoidal) category of quasicoherent sheaves on $X$, a quasicompact stack with affine diagonal. Then $C$-schemes should be identified with schemes over $X$, so that $C$-geometry just means geometry relative to the base $X$. Thus for example $C$ could be $R-mod$ for a commutative ring $R$ ($X=Spec(R)$), or $Rep(G)$ for an affine group scheme $G$ ($X=BG$).

David Ben-Zvi
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    Hi David. While certainly a morally correct answer to the original question, I wonder if this is literally correct. It is my impression (without being too careful about it!) that what T-V give you is the G_m-schemes that can be made by gluing affine G_m-schemes together along open affine G_m-subschemes, and repeating this. But there exist G_m-schemes that don't have open covers by affine G_m-schemes. For instance, take P^1 with its usual G_m action and glue 0 and infinity together (transversely, say). Then the only G_m-stable open neighborhood of 0 is the whole variety, which is not affine. – JBorger Mar 14 '13 at 08:37
  • Hi James - excellent point! Idle question: is the problem with the word "scheme"? i.e. if we replace the Zariski topology with the etale (or flat) topology (which I think T-V define in this generality), then things should get better? In other words, shouldn't algebraic spaces or stacks over C again be just spaces over $X$? – David Ben-Zvi Mar 14 '13 at 14:58
  • My little counterexample can be circumvented by working in the etale topology. P^1 with two points glued has an etale double-cover formed by gluing two copies of P^1 together by 0 in each is glued to infinity in the other. The G_m action lifts, and this double cover has a cover by affine G_m-schemes (delete the singular points). I don't know if it's always possible to get around such problems by working with the etale or flat topologies. – JBorger Mar 21 '13 at 04:18
  • It kind of reminds me of looking at G-equivariant derived categories. You have the option of taking the derived category of G-equivariant objects, or of adding some G-structure to the ordinary derived category. My understanding (you would know better) is that the latter is preferred. Perhaps the two phenomena are particular instances of a general categorical principle. – JBorger Mar 21 '13 at 04:21
  • If G is an affine group scheme (as in our current setting), then there's no difference between categories of G-equivariants (as module categories for Rep(G) ) or categories with G-action (if by derived category we mean say infinity-category) - this is a theorem of Gaitsgory in the derived setting. It's a form of saying that BG is "1-affine", or affine as far as quasicoherent sheaves of categories are concerned. – David Ben-Zvi Mar 22 '13 at 01:31
  • Seems very likely to me that the assertions are true in the flat (and hence probably etale) topology -- basically because BG is affine (in the usual sense) flat-locally, though not Zariski-locally.. – David Ben-Zvi Mar 22 '13 at 01:34
  • Interesting! . – JBorger Mar 25 '13 at 07:26