Morphisms of a simple sheaf over an algebra to its double dual - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T20:50:32Z http://mathoverflow.net/feeds/question/39620 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39620/morphisms-of-a-simple-sheaf-over-an-algebra-to-its-double-dual Morphisms of a simple sheaf over an algebra to its double dual TonyS 2010-09-22T14:58:07Z 2010-09-22T15:29:27Z <p>Given a smooth and projective surface $S$ over an algebraically closed field $k$ and a sheaf of Azumaya algebras $R$, i.e. $R$ is a locally free $O_S$-module of finite rank. Let $M$ be a coherent and torsion free $O_S$-module, which is also a left $R$-module, such that generically $M_\eta$ is a simple $R_\eta$-module. Then we have $Hom_R(M,M)=k$.</p> <p>Now $M^*:=Hom_{O_S}(M,O_S)$ is a right $R$-module and $M^{**}$ is a left $R$-module. We have the canonical map $\iota: M \rightarrow M^{**}$.</p> <p>Is it true that $Hom_R(M,M^{**})$ just consists of the muliples of $\iota$, i.e. is it a one dimensional $k$-vector space?</p> <p>I tried to use the sequence $0\rightarrow M\rightarrow M^{**} \rightarrow Q\rightarrow 0$. Since $M$ is torsion free $Q$ has support in codimension 2. Then apply $Hom_R(M, - )$, which is left exact, so we get, with $Hom_R(M,M)=k$: $0\rightarrow k\rightarrow Hom_R(M,M^{**}) \rightarrow Hom_R(M,Q)$. But here i am stuck.</p> <p>Or is this assertion wrong, i.e. are there more morphisms? If it is right, can it be generalized to a bigger class of algebras $R$?</p> http://mathoverflow.net/questions/39620/morphisms-of-a-simple-sheaf-over-an-algebra-to-its-double-dual/39625#39625 Answer by Torsten Ekedahl for Morphisms of a simple sheaf over an algebra to its double dual Torsten Ekedahl 2010-09-22T15:29:27Z 2010-09-22T15:29:27Z <p>Any $R$-homomorphism (in fact any $\mathcal O_S$-homomorphism) <code>$M \to M^{**}$</code> extends to a morphism <code>$M^{**}\to M^{**}$</code> (as $M$ is locally free in codimension $1$ and <code>$M^{**}$</code> is the maximal extension from outside codimension $2$. This gives what you want. as <code>$Hom_R(M^{**},M^{**})=k$</code> for the same reason as it is true of $M$.</p>