0
$\begingroup$

In $P_{295}$ of Daniel Huybrechts - Complex Geometry An Introduction, he gives the definition of fiber and the comparison between fiber and stalk as follows:

If $\left(M, \mathcal{R}_{M}\right)$ is a locally ringed space, i.e. the stalks of $\mathcal{R}_{M}$ are all local, then one can define the fibre of any $\mathcal{R}_{M}$ -module $\mathcal{F}$ at a point $x \in M$ as $\mathcal{F}(x):=\mathcal{F} \otimes_{ \mathcal{R}_{\mathcal{M}, z}} \mathcal{R}_{M, x} / \mathfrak{m},$ where $\mathfrak{m} \subset \mathcal{R}_{M, x}$ is the maximal ideal. One should be aware of the essential difference between the stalk $\mathcal{F}_{x}$ and the fibre $\mathcal{F}(x)$ of a sheaf. If $\mathcal{F}$ is the sheaf of sections of a vector bundle $E,$ then its fibre $\mathcal{F}(x)$ is nothing but the fibre $E(x)=\pi^{-1}(x)$ of the vector bundle. Its stalks are much bigger.

My question: why stalks are much bigger than fibers?(Of course, when refering to the holomorphic vector bundle, this claim is very apparently true.)( however according to the definition of the fibre, it equals the stalk tensored by sth. In my opinion, the tensor operation make the fiber more bigger than the stalk.

$\endgroup$
1
  • $\begingroup$ Have you computed an example? Try taking your space to be the projective line and compute the stalk and fiber of the structure sheaf at a point. $\endgroup$
    – KReiser
    Apr 25, 2020 at 10:14

1 Answer 1

3
$\begingroup$

The fibre is a quotient of the stalk and therefore "smaller". In fact, this follows from the isomorphism $$\mathcal F_x\otimes_{\mathcal R_{M,x}}\mathcal R_{M,x}/\mathfrak m \simeq \mathcal F_x/\mathfrak m \mathcal F_x.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .