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?


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Neha
    Oct 7, 2010 at 14:51
  • $\begingroup$ 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 $V\otimes V\otimes O_{G\times G} \to A\boxtimes A$. $\endgroup$
    – Sasha
    Oct 7, 2010 at 16:41
  • $\begingroup$ 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. $\endgroup$
    – Neha
    Oct 8, 2010 at 15:48

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