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I am doing the exercise a) II.5.16 on the tensor operations on sheaves from Hartshorne. This exercise asks the reader to prove that, if $\mathscr{F}$ is locally free of rank n, then $T^r( \mathscr{F})$,$S^r(\mathscr{F})$, and $\bigwedge^r(\mathscr{F})$ are also locally free, of ranks $n^r$,$C^{n-1}_{n+r-1}$, and $C^n_r$ respectively. Here is my attempt trying to show that $T^r(\mathscr{F})$ is locally free of rank $n^r$. But I do not know how to proceed:
Since $T^r(\mathscr{F})$ is the sheafification of the presheaf $\mathscr{P}$ sending $U$ to $$\mathscr{F}(U)\otimes_{\mathscr{O}_{X}(U)}\cdots\otimes_{\mathscr{O}_{X}(U)}\mathscr{F}(U)$$ They should share the same stalk at $x\in X$. And because $\mathscr{F}|_U\cong \mathscr{O}_U^{\oplus n}$ and, taking stalk is a left adjoint funcor hence commutes with colimits, we have $$\mathscr{F}_x\cong \bigoplus_r \mathscr{O}_{X,x}$$ Hence, if we consider $\mathscr{P}_x$, as tensor commutes with taking stalks, we would have $$\mathscr{P}_x\cong T^r(\mathscr{F})_x\cong\bigoplus_{n^r}\mathscr{O}_{X,x}$$ Now I wish to lift it as in exercise II.5.7. Even though in this exercise, $\mathscr{F}$ is locally free of finite rank, hence is coherent. However, in 5.7 $X$ is noetherian while we do not have that condition here. I think this approach is not really working and, unfortunately this exercise seems to be considered trivial in many places so I cannot find a proper proof.
Any help is appreciated! Thanks in advance.

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    $\begingroup$ The simplest approach here is using the internal logic of sheaves (though this requires some background knowledge). These statements are just internalisations of the corresponding statements for free modules of rank $n$, all of which can be proved purely constructively. $\endgroup$ Jun 25, 2023 at 10:25
  • $\begingroup$ @MarkSaving This sounds interesting, but do you mind giving some references? $\endgroup$
    – Mizutsuki
    Jun 26, 2023 at 4:42
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    $\begingroup$ @Mizutsuki -- the most accessible reference is almost certainly Ingo Blechschmidt's PhD thesis $\endgroup$ Jun 26, 2023 at 6:57

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If $\mathscr{F} \cong \mathcal{O}_X^{\oplus n}$ is free, then the presheaf $\mathscr{P} \cong \mathcal{O}_X^{\oplus n^r}$ is also free, and hence is already a sheaf. For $U \subset X$ open, $$T^r(\mathscr{F})|_{U} = T^r(\mathscr{F}|_{U}),$$ so the result follows.

EDIT: I'll elaborate a bit on the isomorphism $\mathscr{P} \cong \mathcal{O}_X^{\oplus n^r}$. Let $e_1, \dots, e_n$ be a basis for $\mathscr{F}(X)$, so that $e_{i_1} \otimes \cdots \otimes e_{i_r}$ form a basis for $\mathscr{P}(X)$. These global sections define an isomorphism $\mathcal{O}_X^{\oplus n^r} \to \mathscr{P}$, because for $U \subset X$, $$e_{i_1}|_U \otimes \cdots \otimes e_{i_r}|_{U} = (e_{i_1} \otimes \cdots \otimes e_{i_r})|_{U}$$ form a basis for $\mathscr{P}(U)$. Indeed, $e_i|_{U}$ form a basis for $\mathscr{F}(U)$.

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  • $\begingroup$ Waow, thanks for this detailed answer. This exercise does not really seem so trivial to me. I think here we are using many claims, for example restriction commutes with sheafification, tensor commutes with restriction. Again, thanks a lot! $\endgroup$
    – Mizutsuki
    Jun 26, 2023 at 4:42
  • $\begingroup$ @Mizutsuki No problem! I agree; I think you pointed out the main gaps. There may be an easier way to do this, as Mark Saving suggested, but I felt that this was more consistent with the material in Hartshorne. $\endgroup$
    – Daniel
    Jun 26, 2023 at 13:43

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