Fibre of a sheaf of $\mathcal{O}_X$-modules.

Let $\mathcal{F}$ be a locally free sheaf of modules of rank $n$ over some complex manifold $X$.

To $\mathcal{F}$ we can associate a vector bundle call, $\pi: V \to X$.

The fiber of the sheaf $\mathcal{F}$ is $\mathcal{F}(x) := \mathcal{F}_x \otimes_{\mathcal{O}_x} k(x)$ where $k(x)$ is the residue field for some point $x \in X$. Note that this is definitely a vector space over $k(x)$.

Is $\mathcal{F}(x) \cong \pi^{-1}(x)$?

It seems like a basic question but I cannot find a straightforward answer.

Using the correspondence of between locally free sheaves of modules and vector bundles $\pi^{-1}(x) = \mathcal{F}_x / m_x \mathcal{F}_x$. We have the following:

$$\mathcal{F}_x / m_x \mathcal{F}_x \cong \mathcal{F}_x \otimes_{\mathcal{O}_x} \mathcal{O}_x/m_x' \mathcal{O}_x \cong \mathcal{F}_x / m'_x \mathcal{F}_x$$.

Then this boils down to show that the maximal ideal's, $m'_x \subset \mathcal{O}_x$, acts on $\mathcal{F}_x$ as $m'_x \mathcal{F}_x= m_x \mathcal{F}_x$

Solution 1:

Yes, this is correct.

The Serre-Swan theorem gives an equivalence of categories between vector bundles and locally-free $\mathcal{O}_X$-modules of finite rank.

Under this equivalence, the pullback of vector bundles corresponds to the pullback of sheaves of $\mathcal{O}_X$-modules. In particular, for $x: \bullet \hookrightarrow X$ the inclusion of a point, the pullback functor on vector bundles $$x^*: \mathsf{Vect}(X) \to \mathsf{Vect}(\bullet) \cong k(x)\mathsf{Mod}$$ gives you the fibre of a vector bundle, and the pullback functor on sheaves of modules over locally ringed spaces $$x^*: \mathcal{O}_X\mathsf{Mod} \to k(x) \mathsf{Mod}$$ given by $\mathcal{F} \mapsto k(x) \otimes_{x^{-1} \mathcal{O}_X} \mathcal{F}$ gives you the fibre of the corresponding sheaf. With respect to the Serre-Swan equivalence, $x^* \mathcal{V}$ corresponds to $x^*V$.