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For $A$ a ring, we know any linear map from $A^{\oplus n} \to A^{\oplus m}$ is given by a $m \times n$ matrix. For $(X,\mathcal{O}_X)$ a locally ringed space (or scheme), is every $\mathcal{O}_X$-module homomorphism $\mathcal{O}_X^{\oplus n} \to \mathcal{O}_X^{\oplus m}$ given by a matrix with coefficients in $\Gamma(X,\mathcal{O}_X)$?

Charles
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  • $\mathcal{O}_X^{\oplus n}$ is a sheaf and not itself a module. However, for each open $U\subset X$, $\mathcal{O}_X^{\oplus n}(U)$ is an $\mathcal{O}_X(U)$-module. Since the map when restricted to $U$ is a map between free modules, you can indeed encode the information about the map in a matrix. – Sergio Da Silva Sep 04 '14 at 16:17
  • I'm not quite sure of the answer, butI think it should be false, as there could be more homomorphims of sheaves rather than matrices with coefficients in the global sections. For example, have you thought about what happens if you pick $X=\mathbb P^1$ over a field? Because in that case the global sections are the constants, but $\mathbb P^1$ itself has many automorphisms... – Ferra Sep 04 '14 at 17:16
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    @SergioDaSilva $\mathscr{O}_X$-module means (abelian) sheaf with an $\mathcal{O}_X$-action. This is standard terminology. – Zhen Lin Sep 05 '14 at 19:43
  • In general, for an object $A$ of an abelian category, we have $\mathrm{Hom}(A^n,A^m) \cong M_{m \times n}(\mathrm{End}(A))$. Now apply this to $A=\mathcal{O}_X$. – Martin Brandenburg Sep 07 '14 at 13:04

2 Answers2

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Yes, and the reason is basically the same as for rings: on any ringed space $(X, \mathscr{O}_X)$, the $\mathscr{O}_X$-module $\mathscr{O}_X$ represents the global sections functor on the category of $\mathscr{O}_X$-modules. In other words, there's a natural isomorphism $\mathrm{Hom}(\mathscr{O}_X, \mathscr{F}) \to \Gamma(X, \mathscr{F})$ of abelian groups, given by mapping $\phi \in \mathrm{Hom}(\mathscr{O}_X, \mathscr{F})$ to the image of 1 under $\phi$ on the map $\Gamma(X, \mathscr{O}_X) \to \Gamma(X, \mathscr{F})$ induced by $\phi$ on global sections. I think proving this fact is not too difficult an exercise, but I can provide more details if you let me know.

Having proved this, we have isomorphisms

$\mathrm{Hom}(\mathscr{O}_X^{\oplus n}, \mathscr{F}) \to \mathrm{Hom}(\mathscr{O}_X, \mathscr{F})^{\oplus n} \to \Gamma(X, \mathscr{F})^{\oplus n}$.

The first isomorphism is a consequence of the universal property of direct sums. The second one is obtained by taking a direct sum of isomorphism that we established earlier. If you chase through it, this composite isomorphism is given by mapping $\phi \in \mathrm{Hom}(\mathscr{O}_X^{\oplus n}, \mathscr{F})$ to $(\phi(e_1), \dotsc, \phi(e_n))$, where $e_i$ is the standard basis vector of $\Gamma(X, \mathscr{O}_X^{\oplus n}) = \Gamma(X, \mathscr{O}_X)^{\oplus n}$ and $\phi(e_i)$ is its image under the map induced by $\phi$ on global sections.

Now taking $\mathscr{F} = \mathscr{O}_X^{\oplus m}$ gives you what you want.

chthonian
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I'll answer the $m =n =1$ case and the rest follows because finite direct sums are the same as the corresponding finite direct sum of presheaves of $\mathcal{O}_X$-modules (Lemma 3.2 of modules, Stacks Project). Now for any $U \subseteq X$, consider the diagram

$$\begin{array}{ccc} \mathcal{O}_X(X) & \stackrel{f(X)}{\longrightarrow} & \mathcal{O}_X(X) \\ \downarrow && \downarrow \\ \mathcal{O}_X(U) &\stackrel{f(U)}{\longrightarrow}& \mathcal{O}_X(U)\end{array}$$

Now I claim everything is completely determined by the image of $1 \in \mathcal{O}_X(X)$ under $f(X)$. Indeed, the point is that the restriction $\mathcal{O}_X(X) \to \mathcal{O}_X(U)$ is a ring homomorphism and so $1$ is sent to $1$. Now $f(U)$ being an $\mathcal{O}_X(U)$-module homomorphism implies it is completely determined by the image of $1$. It follows by commutivity of the diagram that $$f(U)(1) = f(X)(1)|_U$$ and so $$\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{O}_X,\mathcal{O}_X) = \Gamma(X,\mathcal{O}_X)$$

as desired.

David Benjamin Lim
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