User:TakuyaMurata/Analysis on complex manifolds

Sheaf theory

We say ${\mathcal {F}}$ is a pre-sheaf on a topological space $X$ if

• (i) ${\mathcal {F}}(U)$ is an abelian group for every open subset $U\subset X$
• (ii) For each inclusion $U\hookrightarrow V$, we have the group morphism $\rho _{V,U}:{\mathcal {F}}(V)\to {\mathcal {F}}(U)$ such that

  $\rho _{U,U}$ is the identity and $\rho _{W,U}=\rho _{V,U}\circ \rho _{W,V}$ for any inclusion $U\hookrightarrow V\hookrightarrow W$

A pre-sheaf is called a sheaf if the following "gluing axiom" holds:

For each open subset $U$ and its open cover $U_{j}$, if $f_{j}\in {\mathcal {F}}(U_{j})$ are such that $f_{j}=f_{k}$ in $U_{j}\cap U_{k}$, then there exists a unique $f\in {\mathcal {F}}(U)$ such that $f|_{U_{j}}=f_{j}$ for all $j$.

Note that the uniqueness implies that if $f,g\in {\mathcal {F}}(U)$ and $f|_{U_{j}}=g|_{U_{j}}$ for all $j$, then $f=g$. In particular, $f|_{U_{j}}=0$ for all $j$ implies $f=0$.

4 Example: Let $G$ be a topological group (e.g., $\mathbf {R}$). Let ${\mathcal {F}}(U)$ be the set of all continuous maps from open subsets $U\subset X$ to $G$. Then ${\mathcal {F}}$ forms a sheaf. In particular, suppose the topology for $G$ is discrete. Then ${\mathcal {F}}$ is called a constant sheaf.

Given sheaves ${\mathcal {F}}$ and ${\mathcal {G}}$, a sheaf morphism $\phi :{\mathcal {F}}\to {\mathcal {G}}$ is a collection of group morphisms $\phi _{U}:{\mathcal {F}}(U)\to {\mathcal {G}}(U)$ satisfying: for every open subset $U\subset V$,

$\phi _{U}\circ \rho _{V,U}=\rho _{V,U}\circ \phi _{V}$

where the first $\rho _{V,U}$ is one that comes with ${\mathcal {F}}$ and the second ${\mathcal {G}}$.

Define $(\operatorname {ker} \phi )(U)=\operatorname {ker} \phi _{U}$ for each open subset $U$. $\operatorname {ker} \phi$ is then a sheaf. In fact, suppose $f_{j}\in \operatorname {ker} \phi _{U_{j}}$. Then there is $f\in {\mathcal {F}}(U)$ such that $f|_{U_{j}}=f_{j}$. But since

$(\phi _{U}f)|_{U_{j}}=\phi _{U_{j}}(f|_{U_{j}})=\phi _{U_{j}}f_{j}=0$

for all $j$, we have $\phi _{U}f=0$. Unfortunately, $\operatorname {im} \phi$ does not turn out to be a sheaf if it is defined in the same way. We thus define $(\operatorname {im} \phi )(U)$ to be the set of all $f\in {\mathcal {G}}(U)$ such that there is an open cover $U_{j}$ of $U$ such that $f|_{U_{j}}$ is in the image of $\phi _{U_{j}}$. This is a sheaf. In fact, as before, let $f\in {\mathcal {G}}(U)$ be such that $f|_{U_{j}}\in \operatorname {im} \phi _{U_{j}}$. Then we have an open cover of $U$ such that $f$ restricted to each member $V$ of the cover is in the image of $\phi _{V}$.

Let ${\mathcal {F}}^{0},{\mathcal {F}}^{1},{\mathcal {F}}^{2}$ be sheaves on the same topological space.

A sheaf ${\mathcal {F}}$ on $X$ is said to be flabby if $\rho _{X,U}:{\mathcal {F}}(X)\to {\mathcal {F}}(U)$ is surjective. Let ${\mathcal {F}}_{p}=\lim _{U\ni p}{\mathcal {F}}(U)$, and, for each $f\in {\mathcal {F}}(U)$, define $\operatorname {supp} f=\{x\in U|f|_{p}\neq 0\}$. $\operatorname {supp} f$ is closed since $f|_{p}=0$ implies $p$ has a neighborhood of $U$ such that $f|_{q}=0$ for every $q\in U$. Define $\operatorname {Supp} {\mathcal {F}}=\{x\in X|{\mathcal {F}}_{x}\neq 0\}$. In particular, if $i:Z\hookrightarrow X$ is a closed subset and $\operatorname {Supp} {\mathcal {F}}\subset Z$, then the natural map ${\mathcal {F}}\to i_{*}i^{-1}{\mathcal {F}}$ is an isomorphism.

4 Theorem Suppose

$0\longrightarrow {\mathcal {F}}^{0}\longrightarrow {\mathcal {F}}^{1}\longrightarrow {\mathcal {F}}^{2}\longrightarrow 0$

is exact. Then, for every open subset $U$

$0\longrightarrow \Gamma _{Z}(U,{\mathcal {F}}^{0})\longrightarrow \Gamma _{Z}(U,{\mathcal {F}}^{1})\longrightarrow \Gamma _{Z}(U,{\mathcal {F}}^{2})$

is exact. Furthermore, $\Gamma _{Z}(U,{\mathcal {F}}^{1})\to \Gamma _{Z}(U,{\mathcal {F}}^{2})$ is surjective if ${\mathcal {F}}^{0}$ is flabby.
Proof: That the kernel of $\operatorname {ker} {\mathcal {F}}^{0}\longrightarrow {\mathcal {F}}^{1}$ is trivial means that $\operatorname {ker} {\mathcal {F}}^{0}(U)\longrightarrow {\mathcal {F}}^{1}(U)$ has trivial kernel for any $U$. Thus the first map is clear. Next, denoting ${\mathcal {F}}^{1}\to {\mathcal {F}}^{2}$ by $d$, suppose $f\in {\mathcal {F}}^{1}(U)$ with $df=0$. Then there exists an open cover $U_{j}$ of $U$ and $u_{j}\in {\mathcal {F}}(U_{j})$ such that $du_{j}=f|_{U_{j}}$. Since $du_{j}=f=du_{k}$ in $U_{j}\cap U_{k}$ and $d_{U_{j}\cap U_{k}}$ is injective by the early part of the proof, we have $u_{j}=u_{k}$ in $U_{j}\cap U_{k}$ and so we get $u\in {\mathcal {F}}(U)$ such that $du=f$. Finally, to show that the last map is surjective, let $f\in {\mathcal {F}}^{2}(U)$, and $\Omega =\{(U,u)|du=f|_{U}\}$. If $\{(U_{j},u_{j})|j\in J\}\subset \Omega$ is totally ordered, then let $U=\cup _{j}U_{j}$. Since $u_{j}$ agree on overlaps by totally ordered-ness, there is $u\in {\mathcal {F}}(U)$ with $u|_{U_{j}}=u_{j}$. Thus, $(U,u)$ is an upper bound of the collection $(U_{j},u_{j})$. By Zorn's Lemma, we then find a maximal element $(U_{0},u_{0})$. We claim $U_{0}=U$. Suppose not. Then there exists $(U_{1},u_{1})$ with $du_{1}=f|_{U_{1}}$. Since $d(u_{0}-u_{1})=0$ in $U_{0}\cap U_{1}$, by the early part of the proof, there exists $a\in {\mathcal {F}}^{0}(U_{0}\cap U_{1})$ with $da=u_{0}-u_{1}$. Then $d(u_{1}+da)=du_{1}=f|_{U_{1}}$ (so $(U_{1},u_{1})\in \Omega$) while $u_{1}+da=u_{0}$ in $U_{0}\cap U_{1}$. This contradicts the maximality of $(U_{0},u_{0})$. Hence, we conclude $U_{0}=U$ and so $du_{0}=f$. $\square$

4 Corollary

$0\longrightarrow {\mathcal {F}}^{0}\longrightarrow {\mathcal {F}}^{1}\longrightarrow {\mathcal {F}}^{2}\longrightarrow 0$

is exact if and only if

$0\longrightarrow {\mathcal {F}}_{p}^{0}\longrightarrow {\mathcal {F}}_{p}^{1}\longrightarrow {\mathcal {F}}_{p}^{2}\longrightarrow 0$

is exact for every $p\in X$.

Suppose $f:X\to Y$ is a continuous map. The sheaf $f_{*}{\mathcal {F}}$ (called the pushforward of ${\mathcal {F}}$ by $f$) is defined by $f_{*}{\mathcal {F}}(U)={\mathcal {F}}(f^{-1}(U))$ for an open subset $U\subset Y$. Suppose $f:Y\to X$ is a continuous map. The sheaf $f^{-1}{\mathcal {F}}$ is then defined by $f^{-1}{\mathcal {F}}(U)=$ the sheafification of the presheaf $U\mapsto \varinjlim _{V\supset f(U)}{\mathcal {F}}(V)$ where $V$ is an open subset of $X$. The two are related in the following way. Let $U\subset X$ be an open subset. Then $f^{-1}f_{*}{\mathcal {F}}(U)$ consists of elements $f$ in ${\mathcal {F}}(f^{-1}(V))$ where $V\supset f(U)$. Since $f^{-1}(V)\supset U$, we find a map

$f^{-1}f_{*}{\mathcal {F}}\to {\mathcal {F}}$

by sending $f$ to $f|_{U}$. The map is well-defined for it doesn't depend on the choice of $V$.