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I'm reading the Huybrechts's Complex Geometry, p.71, Prop.2.2.17 and there is some point that makes me somewhat confused :

enter image description here enter image description here

I can't understand the underlined statement. How can we use Example 2.2.4, viii) for obtaining the Adjunction formula? Here Example 2.2.4, viii) states that "for a short exact sequence of holomorphic vector bundles $ 0 \to E \to F\to G\to 0$, $\operatorname{det}(F) \cong \operatorname{det}(E)\otimes \operatorname{det}(G)$."

As the above image (there $\mathcal{T}_Y$, $\mathcal{T}_X$ are holomorphic tangent bundle of $Y$, $X$ respectively and $\mathcal{T}_{X}|_{Y}$ is the restriction of $\mathcal{T}_X$ to $Y$), we have the normal bundle sequence :

$$0\to \mathcal{T}_Y \to \mathcal{T}_X|_Y\to \mathcal{N}_{Y/X}\to0$$

As in the above image, in Lemma 2.2.15, he states that "If $Y\subset X$ is a complex submanifold, then there is a canonical injection $\mathcal{T}_Y \subseteq \mathcal{T}_{X}|_{Y}$"

How can we use the viii) of Example 2.2.4 ? And $\mathcal{T}_Y \subseteq \mathcal{T}_{X}|_{Y}$ is correct form? If we can use viii) of Example 2.2.4, then from the form of statement of Proposition 2.2.17, a priori(?), it looks that $\mathcal{T}_{X}|_Y \subseteq \mathcal{T}_Y$ is more correct form.

Is it true? Anyone helps?

Plantation
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1 Answers1

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Recall that the canonical bundle is the determinant bundle of the cotangent bundle. From the short exact sequence

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

we obtain the dualised short exact sequence

$$0 \to \mathcal{N}_{Y/X}^* \to \mathcal{T}_X|_Y^* \to \mathcal{T}_Y^* \to 0.$$

Evoking $\mathrm{viii})$ of Example $2.2.4$ we have

$$\det(\mathcal{T}_X|_Y^*) \cong \det(\mathcal{N}_{Y/X}^*)\otimes\det(\mathcal{T}_Y^*) \cong \det(\mathcal{N}_{Y/X})^*\otimes K_Y.$$

On the other hand,

$$\det(\mathcal{T}_X|_Y^*) \cong \det(\mathcal{T}_X^*|_Y) \cong \det(\mathcal{T}_X^*)|_Y \cong K_X|_Y$$

so $K_X|_Y \cong \det(\mathcal{N}_{Y/X})^*\otimes K_Y$. Tensoring both sides by $\det(\mathcal{N}_{Y/X})$ gives the adjunction formula.