Timeline for External tensor product of sheaves
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
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| Oct 8, 2010 at 15:48 | comment | added | Neha | Then in such a case, when $A$ is generated by global sections, will the following diagram commute(why/why not)? [ \xymatrix { A\boxtimes A(U\times U) \ar@{.>}[rrd]^{\exists !}\\ A(G) \boxtimes A(G) \ar@{^{(}->}[u] \ar@{^{(}->}[rr]^-{} && A(U) \otimes A(U). } ] for any open set $U\in G$, Please suggest. Thanks. | |
| Oct 7, 2010 at 16:41 | comment | added | Sasha | The first step is the application of a pullback --- $p_i^*(V \otimes O_G) = V\otimes p_i^*O_G = V\otimes O_{G\times G}$. The last step can be obtained as follows. First we tensor surjection $V \otimes O_{G\times G} \to p_1^*A$ by $p_2^*A$. We get a surjection $V\otimes p_2^*A \to p_1^*A\otimes p_2^*A = A\boxtimes A$. After that we tensor surjection $V \otimes O_{G\times G} \to p_2^*A$ by $V$. We get $V\otimes V\otimes O_{G\times G} \to V \otimes p_2^*A$. Composing it with the surjection obtained earlier we get the required surjection $V\otimes V\otimes O_{G\times G} \to A\boxtimes A$. | |
| Oct 7, 2010 at 14:51 | comment | added | Neha | Thanks Sasha for the answer. Yes, $A\boxtimes A$ is the sheaf given by $p_1^∗A⊗p_2^∗A$ on $G\times G$. Can you please tell me if $V\otimes O_G\to A$ is surjective, then how and why should $V\otimes O_{G\times G}\to p_{i}^{∗} A$ be surjective? And finally how will we get the last step i.e. why would this map become $V\otimes V\otimes O_{G\times G}\to A\boxtimes A$ is surjective? I am sure it is trivial for you, but I am not able to figure out the correct reason, so please help. | |
| Oct 7, 2010 at 14:42 | vote | accept | Neha | ||
| Oct 6, 2010 at 13:25 | history | answered | Sasha | CC BY-SA 2.5 |