Welcome to MathOverflow

Question

1

Suppose $\mathcal{A}$ is a quasi coherent sheaf of algebras over a group scheme $\mathcal{G}$. Suppose it is generated by global section. Then , what can we say about the external tensor product $\mathcal{A}\boxtimes\mathcal{A}$? Will this sheaf also be generated by the tensor product of the global section with itself? Or is it bigger than this?

flag	asked Oct 6 2010 at 11:08 Neha 79●5

Sasha · Accepted Answer · 2010-10-06 13:25:33Z UTC

4

What do you mean by $A\boxtimes A$? If this is the sheaf $p_1^*A\otimes p_2^*A$ on $G\times G$ then certainly it is globally generated by the tensor square of global sections. It follows easily from right-exactness of the pullback and of the tensor product --- let $V$ be the space of global sections, then $V\otimes O_G \to A$ is surjective, hence $V\otimes O_{G\times G} \to p_i^*A$ is surjective, hence $V\otimes V\otimes O_{G\times G} \to A\boxtimes A$ is surjective.

answered Oct 6 2010 at 13:25

Sasha
9,923●1●7●17

Thanks Sasha for the answer. Yes, $A\boxtimes A$ is the sheaf given by $p_1^∗A\otimes p_2^∗A$ 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. – Neha Oct 7 2010 at 16:38

Neha · Answer 2 · 2010-10-07 14:51:51Z UTC

0

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.

answered Oct 7 2010 at 14:51

Neha
79●5

1

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 $V\otimes V\otimes O_{G\times G} \to A\boxtimes A$. – Sasha Oct 7 2010 at 16:41

Thanks Sasha, this is really useful. – Neha Oct 8 2010 at 12:14

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. – Neha Oct 8 2010 at 15:48

External tensor product of sheaves

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.