## 17.15 Bilinear maps

Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$, $\mathcal{G}$, and $\mathcal{H}$ be $\mathcal{O}_ X$-modules. A *bilinear map* $f : \mathcal{F} \times \mathcal{G} \to \mathcal{H}$ of sheaves of $\mathcal{O}_ X$-modules is a map of sheaves of sets as indicated such that for every open $U \subset X$ the induced map

is an $\mathcal{O}_ X(U)$-bilinear map of modules. Equivalently you can ask certain diagrams of maps of sheaves of sets commute, immitating the usual axioms for bilinear maps of modules. For example, the axiom $f(x + y, z) = f(x, z) + f(y, z)$ is represented by the commutativity of the diagram

Another characterization is this: if $f : \mathcal{F} \times \mathcal{G} \to \mathcal{H}$ is a map of sheaves of sets and it induces a bilinar map of modules on stalks for all points of $X$, then $f$ is a bilinear map of sheaves of modules. This is true as you can test whether local sections are equal by checking on stalks.

Let $\text{Mor}( - , - )$ denote morphisms in the category of sheaves of sets on $X$. Another characterization of a bilinear map is this: a map of sheaves of sets $f : \mathcal{F} \times \mathcal{G} \to \mathcal{H}$ is bilinear if given any sheaf of sets $\mathcal{S}$ the rule

is a bilinear map of modules over the ring $\text{Mor}(\mathcal{S}, \mathcal{O}_ X)$. We don't usually take this point of view as it is easier to think about sets of local sections and it is clearly equivalent.

Finally, here is yet another way to say the definition: $\mathcal{O}_ X$ is a ring object in the category of sheaves of sets and $\mathcal{F}$, $\mathcal{G}$, $\mathcal{H}$ are module objects over this ring. Then a bilinear map can be defined for module objects over a ring object in any category. To formulate what is a ring object and what is a module object over a ring object, and what is a bilinear map of such in a category it is pleasant (but not strictly necessary) to assume the category has finite products; and this is true for the category of sheaves of sets.

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## Comments (2)

Comment #6624 by John Baez on

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