Let $F$ be a locally free sheaf on some variety $X$. Does something like $\mathrm{Sym}^d(F)^\vee\simeq\mathrm{Sym}^d(F^\vee)$ hold? I can't find anything like this in the literature, and I also can't find the existence of a canonical isomorphism in the vector space ($X=\mathrm{Spec}(\mathbb{C})$) setting.
1 Answer
Yes, this claim is true, but only in characteristic zero. Since tensors and duals commute for finitely generated projective modules, they commute for locally free sheaves of finite rank, and so we have $(F^\vee)^{\otimes d}\cong (F^{\otimes d})^\vee$, where the isomorphism is given by sending $\phi_1\otimes\cdots\otimes\phi_d \mapsto (f_1\otimes\cdots\otimes f_d \mapsto \phi_1(f_1)\cdots\phi_d(f_d))$. It remains to show that the symmetric bits you care about are sent to eachother under this isomorphism. But this can be shown using the averaging operator which sends an element $a_1\otimes\cdots\otimes a_d$ to $\frac{1}{d!}\sum_{\sigma\in S^d} a_{s^{-1}(1)}\otimes\cdots\otimes a_{s^{-1}(d)}$ - everything in the image of this endomorphism of $A^{\otimes d}$ will actually be in $Sym^d(A)$ and you can verify what you need.
This averaging operator is what makes things break in characteristic $p$, where you will need to use a divided power algebra instead, see for example https://mathoverflow.net/questions/34452/symmetric-powers-and-duals-of-vector-bundles-in-char-p.
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$\begingroup$ Thanks a lot! very instructive answer $\endgroup$– BernieCommented Jun 23, 2018 at 13:06