Relations between the fiber of a vector bundle and the stalk of the corresponding sheaf

Question

I know that the fiber of a vector bundle and the stalk of the corresponding sheaf are different objects. Shafarevich says that there's a general relation between this fiber and the aforementioned stalk, that's to say

$E_x=(\mathcal{L}_E)_x/\mathcal{M}_x (\mathcal{L}_E)_x$, where $\mathcal{M}_x$ is the maximal ideal of $\mathcal{O}_x$. How can I proof this relation?

Answer 1

For convenience I will drop the subscript $E$ on $\mathcal{L}_E$. So $\mathcal{L}$ is the sheaf of sections of the vector bundle $q: E \to X$, i.e., for any open $U \subset X$, $\mathcal{L}(U) =\{ s : U \to E_U | qs = id\}$. Such a section $s$ assigns to $x \in U$ a value $s(x) \in E_x$. This is exactly the condition that $qs = id$.

The stalk $\mathcal{L}_x$ consists of all pairs $(s, U)$ where $s \in \mathcal{L}(U)$ for $x \in U$ under the equivalence relation $(s,U) \sim (t,V)$ if and only if there exists $W \subset U \cap V$ containing $x$ such that $s|_W = t|_W$. In particular, since all these open sets contain $x$ and the equivalence relation identifies only sections that agree at $x$, it makes sense to talk about the value $\bar{s}(x) \in E_x$ of an equivalence class $\bar{s} \in \mathcal{L}_x$. By definition, $\mathfrak{m}_x$ are the functions in the stalk $\mathcal{O}_x$ that vanish at $x$, so that $\mathfrak{m}_x\mathcal{L}_x$ are the $\bar{s} \in \mathcal{L}_x$ such that $\bar{s}(x) = 0$.

Given all that above, we can now describe an explicit isomorphism $E_x \cong \mathcal{L}_x/\mathfrak{m}_x\mathcal{L}_x$. Evaluation at $x$ gives a homomorphism

$$ ev_x: \mathcal{L}_x \to E_x \enspace \enspace \enspace \bar{s} \mapsto \bar{s}(x). $$

$ev_x$ is surjective. We can see this by taking a trivializing open set $U \ni x$ so that $E_U \cong U \times V$ where $V = E_x$. Then since $E_U \to U$ is a trivial vector bundle, the constant maps $s_v: y \mapsto (y,v)$ for any $v \in V$ are sections and by definition $ev_x(s_v) = v \in E_x$ for any $v \in E_x$.

The kernel of $ev_x$ is the set of $\bar{s}$ such that $ev_x(\bar{s}) = 0$, but this is exactly $\mathfrak{m}_x\mathcal{L}_x$ and so $ev_x$ factors as an isomorphism $\mathcal{L}_x/\mathfrak{m}_x\mathcal{L}_x \cong E_x$.