I was trying to show that on a ringed space $(X, \mathcal{O})$, that $T^n(\mathscr{F})$ is free of finite rank if $\mathscr{F}$ is a free $\mathcal{O}$-module of finite rank. We recall that $T^r(\mathscr{F})$ is the sheafification of the presheaf $t^r(\mathscr{F})$ defined by $U \mapsto \mathscr{F}(U)^{\otimes r }$ for open sets $U$.
For simplicity assume that $\mathscr{F} = \mathcal{O}^{\oplus n}$. Then, if $U \subseteq X$ is open and $e_1, \dots, e_n$ is the canonical basis of $\mathcal{O}(U)^{\oplus n}$, then the usual argument for modules shows that $t^n( \mathscr{F})(U)$ has a basis given by $e_{j_1} \otimes e_{j_2} \otimes \cdots \otimes e_{j_r}$ for all $r$-tuples formed from $\{1, \dots, n\}$. This yields a canonical isomorphism $t^n(\mathscr{F})(U) \to \mathcal{O}(U)^{\oplus n^r}$. Since basis elements restrict to basis elements, this map is compatible with restrictions to form an isomorphism $t^n(\mathscr{F}) \to \mathcal{O}^{\oplus n^r}$.
This would then imply that $t^n(\mathscr{F})$ is already a sheaf, which seems off since the tensor product(as presheaves) of sheaves is not in general a sheaf.
Is it always a sheaf in the free situation?
Thanks!
Best Answer
What you want to show is that $F$ and $G$ are two (quasi-?)coherent sheaves of $\mathcal O_X$-modules, then for each open affine $U \subset X$ we have $$(F \otimes_{\mathcal O_X} G)(U) = F(U) \otimes_{\mathcal O(U)} G(U).$$
In general, you will certainly have to sheafify. For example, $$\dim \Gamma(\mathbb P^n, \mathcal O(1)) = n+1$$ and $\mathcal O(1)^{\otimes 2} = \mathcal O(2)$, but $$\dim \Gamma(\mathbb P^n, \mathcal O(2)) = \binom{n+2}{n} = \frac{(n+2)(n+1)}{2}\neq (n+1)^2.$$