When one is given a partition $\lambda=(\lambda_1,...,\lambda_r)$ and a locally free sheaf $\mathcal{E}$ on for example a Grassmannian variety one can apply the Schur-functor $\Sigma^{\lambda}(\mathcal{E})$ for some partition $\lambda$. Now take an invertible sheaf $\mathcal{L}$ and my question is: What is $\Sigma^{\lambda}(\mathcal{E}\otimes \mathcal{L})$ in terms of $\Sigma^{\lambda}(\mathcal{E})$ and $\mathcal{L}$ or $\Sigma^{\lambda}(\mathcal{L})$? Is ther a formula for computing Schur-functor of tensorproduct?
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4$\begingroup$ $\Sigma^\lambda(E\otimes L) = \Sigma^\lambda E \otimes L^{\sum\lambda_i}$. $\endgroup$– SashaCommented Dec 29, 2013 at 15:27
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$\begingroup$ Ah I see. In the article of Kapranov on the derived category of Grassmannians do we have $\mathrm{Hom}(\Sigma^{\alpha}(\mathcal{S}),\Sigma^{\beta}(\mathcal{S}))\simeq \Sigma^{(\beta-\alpha)}(V)$, when $\beta_i-\alpha_i\geq 0$ and $\mathrm{Grass}(r,V)$? $\endgroup$– AleksaCommented Dec 29, 2013 at 15:54
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$\begingroup$ Not quite. The correct formula is $Hom(\Sigma^\alpha (S),\Sigma^\beta(S)) = \Sigma^{\beta/\alpha}V^*$, where $S$ is the DUAL tautological bundle and $\Sigma^{\beta/\alpha}$ is the skew Schur functor. E.g. if $\beta = (2,1)$ and $\alpha = (1,0)$ then the RHS is $\Sigma^{(2,1)/(1,0)}V^* = V^*\otimes V^*$. $\endgroup$– SashaCommented Dec 29, 2013 at 16:58
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$\begingroup$ So what do we have for the skew Schur functor in the case $\Sigma^{\beta/\alpha}(\mathcal{E}\otimes \mathcal{L})$ when $\mathcal{L}$ is invertible? $\endgroup$– AleksaCommented Dec 29, 2013 at 17:12
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1$\begingroup$ Twist by $\sum (\beta_i - \alpha_i)$. $\endgroup$– SashaCommented Dec 29, 2013 at 17:21
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