Do we have $(\mathcal L\otimes_{\mathcal O_X}\mathcal L)(X)=\mathcal L(X)\otimes_{\mathcal O_X(X)}\mathcal L(X)$?

Question

Let $X$ be a scheme and $\mathcal L$ an invertible sheaf on $X$, do we have $(\mathcal L\otimes_{\mathcal O_X}\mathcal L)(X)=\mathcal L(X)\otimes_{\mathcal O_X(X)}\mathcal L(X)$?

Answer 1

No. Let $X = \Bbb P^1_k$ and $\mathcal{L}=\mathcal{O}(1)$. Then $\mathcal{L}^{\otimes 2} = \mathcal{O}(2)$, and the $\dim_k\mathcal{O}(2)(X)=3$. On the other hand, $\dim_k \mathcal{O}(1)(X) = 2$, so $\dim_k \mathcal{O}(1)(X)\otimes_k \mathcal{O}(1)(X) = 4$, so these vector spaces cannot be isomorphic, let alone equal.

On the other hand, there is always a canonical map $\mathcal{F}(U)\otimes \mathcal{G}(U)\to (\mathcal{F}\otimes\mathcal{G})(U)$ for any sheaves $\mathcal{F,G}$ and any open $U$. Not much is known about about how nice this map has to be in general, see Martin Brandenburg's answer at Conditions such that taking global sections of line bundles commutes with tensor product? for some additional details.