Let $Y\hookrightarrow X$ be a inclusion of a smooth divisor $Y$ (i.e. codimension 1 closed subscheme) in to smooth variety $X$. Let $\mathcal{L}$ be a vector bundle over $X$, we regard it as a locally free $\mathcal{O}_X$-module. Let $\mathcal{S}(\bullet)$ be the symmetric product of quasi-coherent $\mathcal{O}_X$-module. My question: How to compute $$\mathcal{S}(i_*(\mathcal{O}_Y)\otimes\mathcal{L})$$ exactly? I want to know the case $\Delta: \Delta\mathbb{P}^1\hookrightarrow\mathbb{P}^1\times\mathbb{P}^1$ and $\mathcal{L}=\mathcal{O}_{\mathbb{P}^1\times\mathbb{P}^1}(m,n)$ (i.e. the line bundle of $\mathbb{P}^1\times\mathbb{P}^1$), where $\Delta\mathbb{P}^1$ is the diagonal.
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$\begingroup$ What does it mean for you to 'compute' this? It's already a pretty concrete thing, but maybe you're looking for an answer in some specific form? $\endgroup$– R. van Dobben de BruynCommented Nov 1 at 19:24
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$\begingroup$ @R. van Dobben de Bruyn. Yes, I want know the form that is a infinty sum of $O_X$-module $\endgroup$– fool rabbitCommented Nov 1 at 19:56
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3$\begingroup$ Ah, in degree $0$ it will be $\mathcal O_X$, and in positive degree $n$ it will be $i_*i^*\mathscr L^{\otimes n}$. Indeed, what you have to check is that both $i^*$ and $i_*$ preserve tensor products. For $i^*$ this should be clear by definition, and for $i_*$ you use the local computation that $M \otimes_A N = M \otimes_{A/I} N$ if $M$ and $N$ are $A/I$-modules. (I'm assuming you mean that $\mathcal S$ is the symmetric algebra; please clarify if you mean something else.) $\endgroup$– R. van Dobben de BruynCommented Nov 1 at 21:10
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$\begingroup$ @R.vanDobbendeBruyn Thanks for you answer, this is useful for me! $\endgroup$– fool rabbitCommented Nov 1 at 21:53
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