# A question regarding germs, sheaves and stalks

As far as I know, a germ is a local representation of an analytic function and a sheaf is a collection of germs. We're also told that there is a function which maps a given germ to the point for which it is locally defined. We then define the stalk of $$z\in\mathbb{C}$$ to be the inverse of such a function (at $$z$$).

My question, how is this (i.e., the inverse) well-defined? Does a sheaf relate to one function ('equivalency'), or are there not multiple functions defined to a single point?

• The inverse image, not the inverse. The stalk at $z$ is $f^{-1}(\{z\}) =$ the set of all germs locally defined at $z$. Mar 20, 2021 at 4:14
• Thanks Alex @AlexKruckman
– Seth
Mar 20, 2021 at 5:10

Question: "My question, how is this (i.e., the inverse) well-defined? Does a sheaf relate to one function ('equivalency'), or are there not multiple functions defined to a single point?"

Answer: If $$(X, \mathcal{O})$$ is a complex manifold and $$x\in X$$ is a point, you define the stalk of the sheaf $$\mathcal{O}$$ at $$x$$ (denoted $$\mathcal{O}_x$$) as follows:

$$\mathcal{O}_x:=\{(U,s):x\in U, s\in \mathcal{O}(U)\}/\cong$$

where two pairs $$(U,s),(V,t)$$ are equivalent iff $$s_{U\cap V}=t_{U \cap V}$$. With this definition it follows $$\mathcal{O}_x$$ is a local ring with a maximal ideal $$\mathfrak{m}_x$$, consisting of equivalence classes of pairs $$(U,s)$$ with the property that there is an open set $$x\in V \subseteq U$$ with $$s_V=0$$ in $$\mathcal{O}(V)$$.

Note: Here $$\mathcal{O}(U)$$ is the ring of complex holomorphic functions on the open set $$U \subseteq X$$. For each open subset $$V \subseteq U$$ you get trivially a restriction map

$$\rho_{UV}: \mathcal{O}(U) \rightarrow \mathcal{O}(V)$$

and you may verify that $$\mathcal{O}$$ is a sheaf of rings on $$X$$.

You get an "evaluation map"

$$ev_x: \mathcal{O}_x \rightarrow \mathcal{O}_x/\mathfrak{m}_x \cong \mathbb{C}$$

defined by $$ev_x(U,s):=\overline{(U,s)} \in \mathcal{O}_x/\mathfrak{m}_x$$. Since $$ev_x$$ is surjective with kernel $$\mathfrak{m}_x$$, it follows $$\mathfrak{m}_x$$ is a maximal ideal.

If the germ $$(U,s)$$ has non-zero value at $$x$$ under the evaluation map, it follows there is an open neigborhood $$x\in V \subseteq U$$ where $$s$$ is non-zero, hence on $$V$$ it follows $$s$$ has an inverse $$t$$. It follows $$(U,s)$$ is a unit in $$\mathcal{O}_x$$ with multiplicative inverse $$(V,t)$$. Hence any element $$(U,s)\in \mathcal{O}_x- \mathfrak{m}_x$$ is a unit and it follows $$\mathcal{O}_x$$ is a local ring.

The ring $$\mathcal{O}_x$$ is a Noetherian regular local ring in the sense of "commutative algebra" and for this reason people use methods from commutative algebra in the study of these rings.

https://en.wikipedia.org/wiki/Sheaf_(mathematics)

https://en.wikipedia.org/wiki/Germ_(mathematics)