Timeline for Why is there no Brauer scheme?
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9 events
| when toggle format | what | by | license | comment | |
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| Jun 26, 2017 at 15:55 | comment | added | Benjamin Antieau | This follows fairly easily from the Leray spectral sequence for the morphism $E\rightarrow Spec k$. Look at the pushforward of the sheaf $\mathbb{G}_m$ on $E$. I guess you probably want to look at the étale cohomology of a smooth projective curve over an algebraically closed field of characteristic zero for some input information. | |
| Jun 25, 2017 at 14:50 | comment | added | guest | @Antieau: Thanks for the informative answer; what's a good reference for this? Your first exact sequence with the Brauer groups of $k$ and $E$ and $H^1_{et}(k,E)$ is very intriguing..could you please point me to a proof of this sequence? The first map is clear but what is the second map. | |
| Oct 3, 2016 at 13:02 | comment | added | Benjamin Antieau | @Qixiao If $l/k$ is an extension, meaning that there is a map of fields $k\rightarrow l$, then the induced map of schemes is $X_l\rightarrow X_k$. As for injectivity, think of the $\mathbb{R}$-points of an $\mathbb{R}$-variety sitting inside the $\mathbb{C}$-points. | |
| Oct 2, 2016 at 1:22 | comment | added | user39380 | @BenjaminAntieau I am a little bit confused, suppose $l/k$ is an extension, should the map of Brauer groups be of the reverse direction $\mathrm{Br}(X_l)\to\mathrm{Br}(X_k)$ by pullback? Also can you explain a little bit why injectivity is expected if a functor is representable? Thank you! | |
| Oct 30, 2014 at 14:44 | comment | added | Benjamin Antieau | You're absolutely right. | |
| Oct 30, 2014 at 8:22 | comment | added | Qiaochu Yuan | To say it more explicitly, the topological analogue of $B \text{Pic}(X)$ is not $B^2 G$ but the space of maps from $X$ to $B^2 G$. | |
| Oct 29, 2014 at 16:56 | comment | added | Benjamin Antieau | You're absolutely correct that in topological spaces a K(G,n) has only one homotopy group. But, now the K(G,n)s I'm talking about are étale sheaves of spaces in some sense. They have only a single homotopy sheaf, but the global sections are more complicated. In general, $\pi_i(\Gamma(X,K(G,n)))\cong H^{n-i}(X,G)$ for $0\leq i\leq n$ and $0$ otherwise. | |
| Oct 29, 2014 at 15:54 | comment | added | bananastack | I have question. I don't know too much about topology and higher stacks, but when I see K(G,n) I expect to find only one homotopy group. Instead, you are saying that BPic(X) is more like $K(G_m, \leq 2)$ rather than $K(G_m,2)$. No? Is there a moral explanation of this? | |
| Oct 28, 2014 at 23:08 | history | answered | Benjamin Antieau | CC BY-SA 3.0 |