Let, $X$ be a smooth projective surface and $C \subset X$ be a an irreducible curve over an algebraically closed field $\mathbb K$ of characteristic $0$.Let $j:C \to X$ be the inclusion , $A \in Pic(C)$ and $V \subset H^0(C,A)$ be a subspace.
Then my question is that : What is the meaning of the notation " $V \otimes \mathcal {O}_X $ " ?
What I can guess is that may be $V$ stands for the pushforward (on $X$) Of the corresponding constant sheaf associated to a vector space $V$(on $C$).Is this the actual interpretation? Please correct me if I am wrong.
In case the above interpretation is correct ,then is it also the case that pushforward of that constant sheaf and the resultant tensor product both are vector bundles(I mean locally free sheaf)?
Any help from anyone is welcome.
I'm not sure what the context is, but if you let $\{e_i\}_{i=1\ldots n}$ be a basis for $V$, I believe that it should mean that $$ V \otimes_{\mathbb K} \mathcal O_X = \bigoplus_{i=1}^n e_i\otimes \mathcal O_X \cong \mathcal O_X^n. $$ (Think of the $e_i$'s as just labels). For example, given a vector space map $\phi:V\to W$ given by $\phi(e_i)=\sum a_{ij}f_j$, where $\{f_j\}$ is a basis of $W$; and a map of vector bundles $\psi:\mathcal O_X\to \mathcal E$, there is a map $\phi\otimes \psi:V \otimes_{\mathbb K}\mathcal O_X\to W \otimes_{\mathbb K}\mathcal E$ given by $$ (\phi \otimes \psi)(e_i\otimes m) = \phi(e_i)\otimes\psi(m) =\sum_j f_j\otimes a_{ij}\psi(m). $$ You can check that this doesn't depend on the choice of bases.