Timeline for What is the geometric meaning of $H^2(X, \mathscr{O}_X)$?
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22 events
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| Jan 3, 2022 at 22:51 | comment | added | user40276 | @Z.M I think we should discuss this elsewhere. I guess we are spamming the thread. Can you create a chat room? I don't know how to do that... For now, I say the following 1) The cotangent complex is almost always perfect 2) Deloopings exist in the general setting of $\infty$-topoi for any internal group | |
| Dec 28, 2021 at 16:14 | comment | added | Z. M | @user40276 Basically, I did not believe that the formal neighborhood of $BG$ is determined by the formal neighborhood of $G$ at origin. Is the assertion about $B^nG$ true when the cotangent complex of $G$ is not bounded, or even not of finite Tor-amplitude (I am thinking about the base being an affine scheme like $\operatorname{Spec}(\mathbb Z)$, not necessarily a field of char 0), and you seem to indicate that $B^nG$ makes sense even when $G$ is not commutative (or "$E_n$")? | |
| Dec 27, 2021 at 22:17 | comment | added | AmorFati | @SándorKovács It just goes to show how important the Hodge decomposition is. Alternatively, how deluded we are by the Hodge decomposition :) | |
| Dec 27, 2021 at 22:15 | comment | added | user40276 | @Z.M Yes. For any say affine algebraic $G$, the cotangent complex of $B^nG$ is the shifted dual Lie algebra with the coadjoint representation (notice that qcoh modules over $BG$ are exactly$G$-representations). I don't get exactly what you mention to be counter-intuitive. What's in being taken is the formal neighborhood of $B^2\mathbb{G}_m$ around the identity, so there's only the identity and a bunch of infinitesimal stuff. A formal group over say $Spec (\mathbb{C})$ has only one closed point. | |
| Dec 27, 2021 at 20:10 | comment | added | Sándor Kovács | ps: @AmorFati, so perhaps it is a bad case... :) | |
| Dec 27, 2021 at 20:09 | comment | added | Sándor Kovács | @AmorFati, uh, ok, I should read the question more carefully. :) | |
| Dec 27, 2021 at 19:52 | comment | added | AmorFati | @SándorKovács Dear Sándor, thank you for your comment. For a non-Kähler complex manifold, one could certainly have $H^1(X, \mathbb{Z})=0$ without $H^1(X, \mathscr{O}_X)=0$. I'm primarily interested in the non-Kähler case. | |
| Dec 27, 2021 at 16:48 | comment | added | Z. M | @user40276 Is something like the tangent of $B^nG$ at origin is the $n$-th shift of the tangent of $G$ at the origin for abelian group scheme $G$ (I guess that this is how $\mathbb G_a$ and $\mathbb G_m$ relate)? If so, is there any obvious reason? (I asked about tangent because I find it counterintuitive that the tangent of $G$ is specific at a chosen point $e$, although in the abelian case, the conjugation action is trivial, but in the formulation of the tangent $BG$ at the origin, there seems no preferred choice like that of $e$). | |
| Dec 27, 2021 at 0:34 | comment | added | Sándor Kovács | @AmorFati: I don't think that can happen. If $H^1(X,\mathscr O_X)$ embeds into $H^1(X, \mathscr O_X^*)$, then $H^1(X,\mathbb Z)=0$, but then $H^1(X,\mathbb C)=0$, so $H^1(X,\mathscr O_X)=0$. | |
| Dec 26, 2021 at 20:08 | comment | added | AmorFati | @JLA Thank you for this comment, I will look into gerbes and bundle gerbes. | |
| Dec 26, 2021 at 20:08 | comment | added | AmorFati | @PiotrAchinger Thank you for these comments, what if $H^1(X, \mathscr{O}_X) \simeq H^1(X, \mathscr{O}_X^{\ast})$, but is non-zero; is this a bad case? | |
| Dec 26, 2021 at 19:21 | comment | added | Piotr Achinger | @Z.M No derived stuff necessary! Simply consider the functor associating to an Artinian local complex algebra $A$ the kernel of $H^2(X_A, O^*)\to H^2(X, O^*)$. In good cases (e.g. $H^1(X, O_X)=0$) this is representable by a formal group called by Artin and Mazur the formal Brauer group of $X$. Its tangent space is $H^2(X, O_X)$. | |
| Dec 26, 2021 at 19:07 | comment | added | user40276 | @Z.M Take the identity in $B^2\mathbb{G}_m$ as a $2$-group and look at the formal neighbourhood around it. That's the formal thing. Truncate it by killing the squares in the horizontal direction. That's the tangent space. Maybe it's a good idea to take a look at the usual definitions of a formal moduli problem. See, for instance, Lurie's ICM math.ias.edu/~lurie/papers/moduli.pdf . If you're unfamiliar with the basics of $\infty$/derived stuff, it's a good idea to start with the cotangent complex. See also ncatlab.org/nlab/show/Brauer+stack . | |
| Dec 26, 2021 at 17:50 | comment | added | Z. M | @PiotrAchinger Sorry for my ignorance: what do you mean by "the tangent space of the (formal) Brauer group"? I imagine that it might be related to the stack $B^2\mathbb G_m$, but I don't see what kind of tangent should we consider? | |
| Dec 26, 2021 at 8:35 | comment | added | Piotr Achinger | For the follow up question I encourage you to take a look at the cohomology sequence from the exponential sequence. | |
| Dec 26, 2021 at 5:52 | comment | added | Josh Lackman | It classifies gerbes with band $\mathbb{C}$ (also look at bundle gerbes). | |
| Dec 26, 2021 at 1:43 | history | edited | LSpice | CC BY-SA 4.0 |
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| Dec 26, 2021 at 0:23 | history | edited | AmorFati | CC BY-SA 4.0 |
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| Dec 25, 2021 at 22:04 | comment | added | AmorFati | For others that may be interested: A nice reference for the geometric side looks like Schröer's Topological Methods for complex-analytic Brauer groups -- arxiv.org/pdf/math/0405223.pdf It is very well-written from what I have seen so far. | |
| Dec 25, 2021 at 21:42 | comment | added | AmorFati | @PiotrAchinger Dear Piotr, thank you for this! I was not aware of Brauer groups at all and have begun looking into them. Do you have a reference suggestion for using Brauer groups to study the geometry of a complex manifold? I'm admittedly far from knowledgable about number theory and my understanding of commutative algebra is only at a working level (although, this is a great excuse to learn more!) :) | |
| Dec 25, 2021 at 20:24 | comment | added | Piotr Achinger | It is the tangent space of the (formal) Brauer group. It contains obstruction classes to deforming line bundles on $X$ to infinitesimal deformations. | |
| Dec 25, 2021 at 20:13 | history | asked | AmorFati | CC BY-SA 4.0 |