Definition 17.5.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules.
The support of $\mathcal{F}$ is the set of points $x \in X$ such that $\mathcal{F}_ x \not= 0$.
We denote $\text{Supp}(\mathcal{F})$ the support of $\mathcal{F}$.
Let $s \in \Gamma (X, \mathcal{F})$ be a global section. The support of $s$ is the set of points $x \in X$ such that the image $s_ x \in \mathcal{F}_ x$ of $s$ is not zero.
Of course the support of a local section is then defined also since a local section is a global section of the restriction of $\mathcal{F}$.
Lemma 17.5.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules. Let $U \subset X$ open.
The support of $s \in \mathcal{F}(U)$ is closed in $U$.
The support of $fs$ is contained in the intersections of the supports of $f \in \mathcal{O}_ X(U)$ and $s \in \mathcal{F}(U)$.
The support of $s + s'$ is contained in the union of the supports of $s, s' \in \mathcal{F}(U)$.
The support of $\mathcal{F}$ is the union of the supports of all local sections of $\mathcal{F}$.
If $\varphi : \mathcal{F} \to \mathcal{G}$ is a morphism of $\mathcal{O}_ X$-modules, then the support of $\varphi (s)$ is contained in the support of $s \in \mathcal{F}(U)$.
Proof. This is true because if $s_ x = 0$, then $s$ is zero in an open neighbourhood of $x$ by definition of stalks. Similarly for $f$. Details omitted. $\square$
In general the support of a sheaf of modules is not closed. Namely, the sheaf could be an abelian sheaf on $\mathbf{R}$ (with the usual archimedean topology) which is the direct sum of infinitely many nonzero skyscraper sheaves each supported at a single point $p_ i$ of $\mathbf{R}$. Then the support would be the set of points $p_ i$ which may not be closed.
Another example is to consider the open immersion $j : U = (0 , \infty ) \to \mathbf{R} = X$, and the abelian sheaf $j_!\underline{\mathbf{Z}}_ U$. By Sheaves, Section 6.31 the support of this sheaf is exactly $U$.
Lemma 17.5.3. Let $X$ be a topological space. The support of a sheaf of rings is closed.
Proof. This is true because (according to our conventions) a ring is $0$ if and only if $1 = 0$, and hence the support of a sheaf of rings is the support of the unit section. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)