Geometric interpretation of the isomorphism $\mathcal{N}_{Y/X} \cong \mathcal{O}_X(Y) \vert_Y$

Question

Let $X$ be a smooth variety / manifold over $\mathbb{C}$ of dimension $n$ and suppose that $Y \subset X$ is a smooth $n-1$-dimensional subvariety. The normal bundle $\mathcal{N}_{Y/X}$ comes from exact sequence

$$ 0 \to \mathcal{T}_Y \to \mathcal{T}_X \vert_{Y} \to \mathcal{N}_{Y/X} \to 0 $$

and it is known that as sheaf $\mathcal{N}_{Y/X}$ is isomorphic to the restriction of the invertible sheaf $\mathcal{O}_X(Y)$ to $Y$. That's more or less a consequence of identification of the conormal bundle $\mathcal{N}_{Y/X}^*$ with the locally free sheaf $\mathcal{I}_{Y/X}/\mathcal{I}_{Y/X}^2 = \mathcal{I}_{Y/X} \otimes \mathcal{O}_Y$ where $\mathcal{I}_{Y/X}$ is the ideal sheaf of $Y$. For a proof, see e.g. 1.4.2 in the book 3264 and All That by Eisenbud & Harris.

Even though the proof is formally clear I'm missing the clear geometric picture of the isomorphism $\mathcal{N}_{Y/X} \cong \mathcal{O}_X(Y) \vert_Y$ and how to think about it as map of vecor bundles. Does this isomorphism have a concrete geometric interpreteation or does it exist only on abstract level? Is it possible to write down an explicit isomorphism $\mathcal{N}_{Y/X} \to \mathcal{O}_X(Y) \vert_Y$ in terms of classical bundle map if we use the correspondence priciple and treat locally free sheaves as vector bundles?

The latter means that we can associate to the free sheaf $\mathcal{O}_X^n $ the trivial bundle $X \times \mathbb{C}^n$ and to the local sections $\mathcal{O}_X^n(U) $ to sections $U \to X \times \mathbb{C}^n$ which correspond to polynomial maps $U \to \mathbb{C}^n$.

Here a guess: we can find a global section $s: X \to \mathcal{O}_X(Y)$ such that at every local trivialisation $U \subset X$ with $\mathcal{O}_X(Y) \vert _U= U \times \mathbb{C}$ the restricted section $s \vert_U $ equals $u \mapsto (u, \tilde{s}(u))$ with $\tilde{s}(u)=0$ iff $u \in U \cap Y$.

The tangent bundle $\mathcal{T}_X $ restricted to $U$ can be identified with $ U \times (\bigoplus_{i=1}^n \mathbb{C} \cdot \frac{\partial}{\partial t_i})$.
Then it looks reasonable to try to define it locally at $U \cap Y$ as the map

$$\mathcal{T}_X \vert _{U \cap Y} \to (U \cap Y) \times \mathbb{C}, \ \ \frac{\partial}{\partial t_i} \mapsto ds(\frac{\partial}{\partial t_i})$$

where $ds$ is differential of $s: U \to \mathbb{C}$ and $\frac{\partial}{\partial t_i} \in \mathcal{T}_X \vert _{U \cap Y} = (U \cap Y) \times (\bigoplus_{i=1}^n \mathbb{C} \cdot \frac{\partial}{\partial t_i})$. By the choice of $s$ the kernel of this map is exactly $ \mathcal{T}_Y \vert _U= \mathcal{T}_{Y \cap U}$.

Does this local map glue to global map $\mathcal{T}_X \vert _{Y} \to \mathcal{O}_X(Y) \vert _Y$? Also, is this the most natural way to "capture the geometric picture" behind the isomorphism $\mathcal{N}_{Y/X} \cong \mathcal{O}_X(Y) \vert_Y$ or is there a more "natural" way to think about it?

Answer 1

Let $ X $ be a $ C^{\infty} $-manifold (or a complex manifold) and $ E $ a smooth (or holomorphic) vector bundle of real (or complex) rank r on $ X $. Suppose $ s $ is a global section of $ E $ and the zero locus $ Z = Z(s) \subset X $ is smooth of codimension $ r $ in $ X $. Then the normal bundle of $ Z $ in $ X $ is just $ E $ restricted to $ X $.

Proof: Let $ A^1_X $ be the sheaf of smooth 1-forms on $ X $and take any connection $ \nabla : E \rightarrow A^1_X \otimes E $ on $ E $. There is a map of bundles $ TX \rightarrow E $ defined by $ v \in TX(U) \rightarrow \nabla_v (s) \in E(U) $ where $ v $ is any local smooth (or holomorphic) vector field on $ U \subset X $ open. Restricting to $ Z $ gives a map of bundles $ ds: TX\mid_Z \rightarrow E \mid_Z $ on Z. The last map is independent of the connection chosen: for if $ \nabla' $ is a second connection, the difference $ \nabla - \nabla' $ is $ \mathcal{C}^{\infty}_X $-linear, i.e. is a $ \mathcal{E}nd (E) $-valued 1-form, and hence vanishes when evaluated at any tangent vector field of $ X $ lying on $ Z $ because $ s $ vanishes on $ Z $. The map also vanishes at any tangent vector field on $ Z $ because $ s $ is constant equal to $ 0 $ on $ Z $.

So $ ds $ factors through $ TZ $ and the normal bundle $ \frac{TX \mid_Z}{TZ} $ is a sub-bundle of $ E \mid_Z $. Because both bundles have rank $ r $, the normal bundle is equal to $ E \mid_Z $.