Canonical Line Bundle

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

In §Linear Systems, we saw that a (basepoint-free) complete linear system of a line bundle can be used to embed a variety into projective space, and if this defines a closed embedding, we call such a line bundle very ample.

Despite the considerable influence that line bundles exert on our geometry, we have not yet properly examined how to define line bundles on an arbitrary variety in general. If \(X\) is a smooth variety, we can consider the cotangent bundle \(\Omega_X^1\) defined on it using §Line Bundles and Vector Bundles, ⁋Example 24, and by taking its top exterior power we obtain the canonical bundle \(\omega_X\). The goal of this post is to examine this bundle \(\omega_X\).

Vector Bundles and Quasi-Coherent Sheaves

As mentioned above, to define \(\omega_X\) we start from the cotangent bundle \(\Omega_X^1\). We have already seen that this is the bundle of differential forms on \(X\). Let us verify that this aligns with differentiation in the algebraic setting. ([Commutative Algebra] §Differentials, ⁋Definition 3) To do this, we need to examine the process of transporting an \(A\)-module \(M\) to a vector bundle over \(X\) when an affine variety \(X\) with its coordinate ring \(A\) and an \(A\)-module \(M\) are given.

Our basic philosophy is that using §Affine Varieties, we can translate a homomorphism between coordinate rings into a morphism between varieties in the opposite direction, and thus obtain a bundle defined on \(X\). However, the problem is that \(M\) is not a ring: multiplication is not defined on \(M\). Yet according to §Tensor Algebras, ⁋Definition 5, we can consider the symmetric algebra \(\S(M)\), which forcibly defines a (commutative) multiplication on \(M\).

However, applying this directly is problematic. Recall that our goal is to view \(M\) as a vector bundle defined on \(X\). Roughly speaking, our goal is to attach \(M\) nicely above each point of \(X\), but according to §Affine Varieties, if \(M\) appears in the world of varieties (as a fiber), the coordinate ring defining it must be its coordinate functions. That is, we must use \(M^\vee\) instead of \(M\), and therefore we consider \(\S_A(M^\vee)\) rather than \(\S_A(M)\). Then this is an \(A\)-algebra, and thus we obtain a function \(A\rightarrow \S_A(M^\vee)\) between coordinate rings; applying §Affine Varieties to this yields a morphism from some variety \(V(M)\) to \(X\).

Let us verify that this morphism actually carries a vector bundle structure over \(X\). A point \(x\in X\) corresponds to the maximal ideal \(\mathfrak{m}_x\) of the coordinate ring \(A\), and therefore the points of the set-theoretic fiber \(V(M)_x = \pi^{-1}(x)\) above \(x\) in \(V(M) \to X\) are the maximal ideals of \(\S_A(M^\vee)\) containing \(\mathfrak{m}_x\cdot \S_A(M^\vee)\).

To obtain the coordinate ring defining this fiber algebraically, we must first think about what functions are defined on it. Now, the functions contained in the maximal ideal \(\mathfrak{m}_x\) defining \(x\in X\) all vanish at \(x\), so it is reasonable to think of the functions defined on this fiber as \(A/\mathfrak{m}_x\)-valued functions. We call this field \(\kappa(x)=A/\mathfrak{m}_x\) the residue field at \(x\), and therefore in general \(\kappa(x)\) is an algebraic extension of \(\mathbb{K}\). (§Nullstellensatz, ⁋Theorem 4) Since we usually consider the case where \(\mathbb{K}\) is an algebraically closed field, we may simply think of \(\kappa(x)\) as \(\mathbb{K}\).

From the above discussion, we now know that we must consider the collection \(\S_A(M^\vee)\otimes_A\kappa(x)\) of \(\kappa(x)\)-valued functions of \(\S_A(M^\vee)\). Then from the fact that symmetric algebra commutes with tensor product, we obtain the following equation

\[\S_A(M^\vee)\otimes_A\kappa(x)=\S_{\kappa(x)} (M^\vee\otimes_A\kappa(x))=\S_{\kappa(x)}(M_x^\vee)\]

where \(M_x = M \otimes_A \kappa(x)\) on the right-hand side, and the tensor products appearing here can be thought of roughly as viewing all objects as \(\mathbb{K}\)-vector spaces (more precisely, as \(\kappa(x)\)-vector spaces), just as explained above.

Now \(\S_{\kappa(x)}(M_x^\vee)\) is a polynomial algebra with coefficients in \(\kappa(x)\) having each element of \(M_x^\vee\) as a linear form, and then the fiber of \(V(M)_x\) consists of points having the elements of \(M_x^\vee\) as coordinate functions, and therefore these points can be thought of as the double dual of \(M_x\). Now if \(M\) is a finitely generated \(A\)-module, the canonical isomorphism \(M_x\cong M_x^{\vee\vee}\) exists, and from this we can understand each fiber \(V(M)_x\) as the \(\kappa(x)\)-vector space \(M_x\).

Through additional calculation we can verify that this data satisfies the locally trivial condition, and thus we can confirm that when a finitely generated \(A\)-module \(M\) is given, we can think of it as a vector bundle defined on \(X\).

On the other hand, showing that this vector bundle is locally trivial is essentially the same as examining it in the language of sheaves. We have not used the language of sheaves much for geometric intuition, but roughly speaking it is as follows.

As above, when an affine variety \(X\) with its coordinate ring \(A\) and an \(A\)-module \(M\) are given, we define the sheaf \(\widetilde{M}\) on \(X\) by the formula

\[\widetilde{M}(U)=M\otimes_A \mathcal{O}_X(U)\]

This is basically the same context as introducing \(\kappa(x)\) above, and although we did not compute it in detail, when showing the local triviality of \(V(M)\) we would have used the structure sheaf to perform base change in this manner. Then in particular for the whole of \(X\) we have

\[\widetilde{M}(X)=M\otimes_A A=M\]

so that \(M\) becomes the global section space of \(\widetilde{M}\).

These two definitions are merely expressing the same object in different geometric languages. That is, for an affine variety \(X\) with its coordinate ring \(A\) and a finitely generated \(A\)-module \(M\), the étale space of \(\widetilde{M}\) is precisely \(V(M)\), and the section sheaf of \(V(M)\) is \(\widetilde{M}\).

In general, compared to concrete geometric language, the advantage of sheaf language is that it can be applied in more general cases. For example, we define the following. (§Basic Notions, ⁋Definition 8)

Definition 1 An \(\mathcal{O}_X\)-module \(\mathcal{F}\) on a general variety \(X\) is called a quasi-coherent sheaf if there exist an affine open cover \(\{U_i\}\) of \(X\) and \(A_i=\mathcal{O}_X(U_i)\)-modules \(M_i\) for each such that \(\mathcal{F}\vert_{U_i}\cong \widetilde{M_i}\). If each \(M_i\) is a finitely generated \(A_i\)-module, we call \(\mathcal{F}\) a coherent sheaf.

In general, when dealing with quasi-coherent sheaves, we must be careful because different \(M\)’s may be attached on each affine cover, but if we restrict to the affine case, \(M\mapsto \widetilde{M}\) defines a categorical equivalence from \(\lMod{A}\) to \(\QCoh(X)\). This can be verified by checking that for any quasi-coherent sheaf \(\mathcal{F}\), \(\widetilde{\Gamma(X,\mathcal{F})}\) recovers \(\mathcal{F}\) itself. That is, our slogan is: in the affine case, a quasi-coherent sheaf is an \(A\)-module, and a coherent sheaf is a finite rank \(A\)-module.

From this perspective, a vector bundle can be thought of as a very special case of a (quasi-)coherent sheaf. Or conversely, when thinking of these (quasi-)coherent sheaves, we may think of them as very general forms of vector bundles. Specifically, a coherent sheaf can be thought of as an extension of the category of (finite rank) vector bundles to make it closed under the operations of an abelian category, that is, kernel, image, cokernel, etc.; intuitively, it can be thought of as a vector bundle whose fiber dimension may vary from point to point. A quasi-coherent sheaf is what remains after removing the finite rank condition from this.

However, one somewhat cautionary point from the above intuition is that §Basic Notions, ⁋Proposition 9 does not exactly match the geometric situation. For example, any coherent sheaf on a smooth variety has a locally free resolution of finite length, but this is not the case on a singular variety.

Canonical Bundle

We are now ready to define the canonical bundle. To do this, we must first introduce the cotangent bundle on a variety; the following definition was already examined in §Line Bundles and Vector Bundles, ⁋Example 24, but we introduce it again for completeness.

Definition 2 The cotangent bundle \(\Omega_X^1\) of a smooth variety \(X\) is the dual vector bundle of the tangent bundle \(\mathcal{T}_X\).

Then the construction we examined in the previous section is for the following.

Proposition 3 For an affine variety \(X\) and its coordinate ring \(A\), \(\Omega_X^1\) is the vector bundle corresponding to \(\widetilde{\Omega}_{A/\mathbb{K}}\). ([Commutative Algebra] §Differentials, ⁋Definition 3)

Proof

For this, it will be convenient to rewrite the previously defined tangent bundle and cotangent bundle in the language of sheaves. First, let us define the tangent sheaf \(\mathcal{T}_X\). For an open subset \(U\) of \(X\), we call the sheaf defined by \(\mathcal{T}_X(U) = \Der_\mathbb{K}(\mathcal{O}_X(U),\mathcal{O}_X(U))\), the collection of \(\mathbb{K}\)-derivations on \(\mathcal{O}_X(U)\), the tangent sheaf.

Our main tool is the universal property of Kähler differentials. ([Commutative Algebra] §Differentials, ⁋Lemma 2) That is, for any \(A\)-module \(N\), we use the natural isomorphism

\[\Der_\mathbb{K}(A,N)\cong\Hom_A(\Omega_{A/\mathbb{K}},N)\]

Then by the fact that \(\widetilde{(-)}\) is a categorical equivalence and the above natural isomorphism, we have

\[\Hom_{\mathcal{O}_X}(\widetilde{\Omega_{A/\mathbb{K}}},\widetilde{N})\cong\Hom_A(\Omega_{A/\mathbb{K}},N)\cong\Der_\mathbb{K}(A,N)\]

Moreover, if we think of the sheaf of derivations, the last term \(\Der_\mathbb{K}(A,N)\) is again \(\Der_\mathbb{K}(\mathcal{O}_X, \widetilde{N})\), so the following formula

\[\Hom_{\mathcal{O}_X}(\widetilde{\Omega_{A/\mathbb{K}}}, \widetilde{N})\cong \Der_\mathbb{K}(\mathcal{O}_X, \widetilde{N})\]

holds. In particular, for the case \(N=A\), that is, \(\widetilde{N}=\mathcal{O}_X\), we have

\[\Hom_{\mathcal{O}_X}(\widetilde{\Omega_{A/\mathbb{K}}}, \mathcal{O}_X)\cong\Der_\mathbb{K}(\mathcal{O}_X, \mathcal{O}_X)\cong \mathcal{T}_X\]

On the other hand, from the fact that \(\Omega_{A/\mathbb{K}}\) is a finitely generated projective \(A\)-module, we know that \(\widetilde{\Omega_{A/\mathbb{K}}^\vee}\cong \widetilde{\Omega_{A/\mathbb{K}}}^\vee\), and therefore

\[\widetilde{\Omega_{A/\mathbb{K}}}^\vee\cong \widetilde{\Omega_{A/\mathbb{K}}^\vee}\cong \widetilde{\Der_\mathbb{K}(A,A)}\cong \mathcal{T}_X\]

so the desired claim holds.

This result shows that the cotangent bundle is represented by differential \(1\)-forms as we imagine.

Example 4 The cotangent bundle of \(\mathbb{A}^n\) is \(\Omega_{\mathbb{A}^n}^1 \cong \mathcal{O}_{\mathbb{A}^n}^{\oplus n}\). Algebraically, if we fix the coordinate ring \(\mathbb{K}[\x_1, \ldots, \x_n]\) of \(\mathbb{A}^n\), the Kähler differentials of this \(\mathbb{K}\)-algebra are the free module \(\bigoplus_{i=1}^n \mathbb{K}[\x_1, \ldots, \x_n] d\x_i\), so this result aligns well with our intuition.

On the other hand, for any smooth variety \(X\) of dimension \(n\) and its cotangent bundle \(\Omega_X^1\), since each fiber of \(\Omega_X^1\) is \(n\)-dimensional, we know that taking its \(n\)-fold exterior product yields a line bundle.

Definition 5 We define the canonical line bundle \(\omega_X\) of a smooth variety \(X\) of dimension \(n\) as the top exterior power of the cotangent bundle

\[\omega_X = \bigwedge\nolimits^{\!n} \Omega_X^1\]

We call a global section \(s\in \Gamma(X, \omega_X)\) of the canonical bundle \(\omega_X\) a regular \(n\)-form on \(X\). These are the \(n\)-forms of the form \(fd\x_1 \wedge \cdots \wedge d\x_n\) for a regular function \(f\), when we take a trivializing open set \(U\) of \(\omega_X\) and identify it with the cotangent bundle on affine space as in Example 4.

On the other hand, from the correspondence between line bundles and divisor classes, we can define the following.

Definition 6 The divisor class corresponding to the canonical bundle \(\omega_X\) is called the canonical divisor and denoted by \(K_X\). That is, \(\omega_X \cong \mathcal{O}_X(K_X)\).

Since we use §Line Bundles and Vector Bundles, ⁋Proposition 19 for this, note that \(K_X\) is defined only as a divisor class.

The Canonical Bundle of \(\mathbb{P}^n\)

As in previous posts, the most familiar example for us is that of \(\mathbb{P}^n\). Intuitively, if we unpack the quotient

\[\mathbb{P}^n=(\mathbb{A}^{n+1}\setminus\{0\})/\mathbb{K}^\ast\]

defining \(\mathbb{P}^n\), the \(\mathbb{K}^\ast\)-action is the action in the direction radiating from the origin, that is, the direction defined by the Euler vector field, and from the perspective of \(\mathbb{P}^n\) this is merely a trivial line bundle. Then the tangent space of \(\mathbb{P}^n\) corresponds to the remaining part after quotienting the directions of \(\mathbb{A}^{n+1}\), that is, the linear forms, by this trivial line bundle. That is, the following short exact sequence corresponding to the tangent bundle

\[0 \rightarrow \mathcal{O}_{\mathbb{P}^{n}}\rightarrow \mathcal{O}_{\mathbb{P}^n}(1)^{\oplus (n+1)}\rightarrow T_{\mathbb{P}^n}\rightarrow 0\]

exists, and taking the dual of this yields the following.

Proposition 7 (Euler Exact Sequence) There exists an exact sequence of vector bundles on \(\mathbb{P}^n\)

\[0 \rightarrow \Omega_{\mathbb{P}^n}^1 \rightarrow \mathcal{O}_{\mathbb{P}^n}(-1)^{\oplus(n+1)} \rightarrow \mathcal{O}_{\mathbb{P}^n} \rightarrow 0\]

To compute the canonical bundle of \(\mathbb{P}^n\) now, we must take the top exterior power of this exact sequence. More generally, suppose the following short exact sequence

\[0\rightarrow E\rightarrow F\rightarrow L\rightarrow 0\]

is given. Here let \(E\) be a vector bundle of rank \(r\) and \(L\) a vector bundle of rank \(1\). Taking \(\bigwedge\nolimits^{\!r}(-)\) of this sequence, from the fact that determinant is compatible with tensor product we know that

\[\det(F)\cong \det(E)\otimes \det(L)\]

Let us apply this to the Euler exact sequence of Proposition 7 (Euler Exact Sequence). Since \(E=\Omega_{\mathbb{P}^n}^1\) has rank \(n\), \(F=\mathcal{O}_{\mathbb{P}^n}(-1)^{\oplus(n+1)}\) has rank \(n+1\), and \(L=\mathcal{O}_{\mathbb{P}^n}\) has rank \(1\), we have

\[\det(\mathcal{O}_{\mathbb{P}^n}(-1)^{\oplus(n+1)})\cong \det(\Omega_{\mathbb{P}^n}^1)\otimes \det(\mathcal{O}_{\mathbb{P}^n})\]

Now \(\det(\mathcal{O}_{\mathbb{P}^n})\cong \mathcal{O}_{\mathbb{P}^n}\) on the right-hand side, and the left-hand side is \(\mathcal{O}_{\mathbb{P}^n}(-1)^{\otimes(n+1)}\cong \mathcal{O}_{\mathbb{P}^n}(-n-1)\), so we obtain

\[\omega_{\mathbb{P}^n}=\det(\Omega_{\mathbb{P}^n}^1)\cong \mathcal{O}_{\mathbb{P}^n}(-n-1)\]

At this time the canonical divisor is given by \(K_{\mathbb{P}^n}=-(n+1)H\). From this calculation and §Line Bundles and Vector Bundles, ⁋Example 16, we know that \(\omega_{\mathbb{P}^n}\) has no regular sections.

Example 8 We can also verify the above calculation from the perspective of transition functions of \(n\)-forms. Setting the affine coordinates on the standard open cover \(U_i = \{\x_i \neq 0\}\) of \(\mathbb{P}^n\) as \(\y_j^{(i)} = \x_j / \x_i\) (\(j \neq i\)), we can consider the \(n\)-form on \(U_i\)

\[d \y_1^{(i)} \wedge \cdots \wedge \widehat{d \y_i^{(i)}} \wedge \cdots \wedge d \y_n^{(i)}\]

On \(U_i \cap U_j\), since \(\y_k^{(j)} = \x_k / \x_j = (\x_k / \x_i) / (\x_j / \x_i) = \y_k^{(i)} / \y_j^{(i)}\), for \(k \neq i, j\) we have

\[d \y_k^{(j)} = d(\y_k^{(i)} / \y_j^{(i)}) = \frac{\y_j^{(i)} d \y_k^{(i)} - \y_k^{(i)} d \y_j^{(i)}}{(\y_j^{(i)})^2}\]

Therefore the \(n\)-form on \(U_j\) transforms on \(U_i \cap U_j\) as

\[\bigwedge_{k \neq j} d \y_k^{(j)} = (\y_j^{(i)})^{-(n+1)} \cdot \bigwedge_{k \neq i} d \y_k^{(i)}\]

Here since \((\y_j^{(i)})^{-(n+1)} = (\x_j / \x_i)^{-(n+1)}\), we can verify that the transition function is \(g_{ij} = (\x_i / \x_j)^{-(n+1)}\). This matches the transition function of \(\mathcal{O}_{\mathbb{P}^n}(-n-1)\).

Adjunction Formula

In many cases, we are interested in varieties obtained from \(\mathbb{P}^n\) through sufficiently many polynomials. Intuitively, this is obtained by successively considering smooth divisors \(D\) of a smooth variety \(X\). The following adjunction formula tells us how to compute the canonical line bundle of \(D\) from that of \(X\) in such cases.

For this, recall the ideal sheaf \(\mathcal{I}_D=\mathcal{O}_X(-D)\) satisfying the following short exact sequence for a smooth variety \(X\) and a smooth divisor \(D\):

\[0\rightarrow \mathcal{I}_D\rightarrow \mathcal{O}_X\rightarrow \mathcal{O}_D\rightarrow 0\]

(§Line Bundles and Vector Bundles, ⁋Definition 17) Then from this we can compute that the first-order approximation of \(\mathcal{I}_D\) is given by

\[\mathcal{I}_D/\mathcal{I}_D^2=\mathcal{I}_D\otimes_{\mathcal{O}_X}\mathcal{O}_D=\mathcal{O}_X(-D)\vert_D\]

On the other hand, let us compute the tangent sheaves \(\mathcal{T}_X=\Der(\mathcal{O}_X)\) and \(\mathcal{T}_D=\Der(\mathcal{O}_D)\) of \(X\) and \(D\) respectively. Then there exists a natural inclusion \(\mathcal{T}_D\rightarrow \mathcal{T}_X\vert_D\), and we define its cokernel as the normal sheaf \(\mathcal{N}_{D/X}\). That is, the following short exact sequence

\[0\rightarrow \mathcal{T}_D\rightarrow \mathcal{T}_X\vert_D\rightarrow \mathcal{N}_{D/X}\rightarrow 0\]

exists. Then we can verify that the dual of this normal bundle \(\mathcal{N}_{D/X}\) is precisely \(\mathcal{I}_D/\mathcal{I}_D^2\). For this reason we call this the conormal sheaf, and specifically this is obtained by verifying

\[0 \rightarrow \mathcal{I}_D/\mathcal{I}_D^2\rightarrow \Omega_X^1\lvert D\rightarrow \Omega_D^1\rightarrow 0\]

which corresponds to the dual of the above short exact sequence. Here the first arrow is given by \(f\mapsto df\). Taking the top exterior power of this short exact sequence yields the following.

Proposition 9 (Adjunction Formula) For a smooth divisor \(D\) of a smooth variety \(X\),

\[\omega_D \cong (\omega_X \otimes \mathcal{O}_X(D))\vert_D\]

From this, the claim for the canonical divisor also follows immediately. In any case, the content of this proposition is that if we twist the canonical bundle \(\omega_X\) of the ambient variety \(X\) by the normal bundle \(\mathcal{O}_X(D)\) and then restrict to \(D\), we obtain the canonical bundle \(\omega_D\) of the subvariety \(D\). In simpler terms, differential forms on \(D\) are obtained by adding information about the normal direction to the differential forms of the ambient space.

The following example shows a concrete calculation using this.

Example 10 Let \(C \subset \mathbb{P}^2\) be a smooth curve of degree \(d\). By the adjunction formula,

\[\omega_C \cong \omega_{\mathbb{P}^2}\vert_C \otimes \mathcal{O}_{\mathbb{P}^2}(C)\vert_C \cong \mathcal{O}_{\mathbb{P}^2}(-3)\vert_C \otimes \mathcal{O}_{\mathbb{P}^2}(d)\vert_C \cong \mathcal{O}_C(d-3)\]

Therefore \(K_C \sim (d-3)H\vert_C\), and since the degree of \(H\vert_C\) is \(d\), we have \(\deg K_C = d(d-3)\).

On the other hand, it is well known in classical algebraic geometry that the genus of a plane curve (that is, a projective curve in \(\mathbb{P}^2\)) is given from its degree by

\[g=\frac{(d-1)(d-2)}{2}\]

(Degree-genus formula) From this we can verify that

\[\deg K_C=d(d-3)=(d-1)(d-2)-2=2g-2\]

The degree-genus formula is in fact a special case of the Riemann-Roch theorem that we will examine later, and in that post we will derive both the result \(\deg K_C=2g-2\) of the above calculation and the degree-genus formula.

Canonical Divisor of a Blow-up

We examined the blow-up of \(\mathbb{A}^2\) at a point \(0\) in §Rational Maps, ⁋Example 12. We now generalize this example further and examine how the canonical divisor behaves in this general setting.

For a smooth variety \(X\) and a codimension \(r\) smooth subvariety \(Z\), the blow-up of \(X\) along \(Z\) is given by the following birational morphism

\[\pi:\widetilde{X}\rightarrow X;\]

here \(\pi\) is an isomorphism away from the fiber of \(Z\), and the fiber of \(Z\) is given by the exceptional divisor

\[E=\mathbb{P}(\mathcal{N}_{Z/X})\]

Here \(\mathcal{N}_{Z/X}\) is the normal bundle of \(Z\) in \(X\), and \(\mathbb{P}(\mathcal{N}_{Z/X})\) means the projective bundle obtained by projectivizing the vector space corresponding to the fiber at each point of \(Z\). This is the analogue of what was done in §Rational Maps, ⁋Example 12, where the exceptional divisor corresponding to the fiber of a point \(0\) was attached by projectivizing the directions coming into this point from outside it.

Now restricting \(\pi:\widetilde{X}\rightarrow X\) to \(E\), let us consider

\[\pi\vert_E: E\rightarrow Z\]

Thinking of the tangent bundle of \(E\), we can divide it into the horizontal direction \((\pi\vert_E)^\ast T_Z\) coming from the base space \(Z\) and the relative tangent bundle direction \(T_{E/Z}\) perpendicular to it. That is, we obtain the following short exact sequence

\[0 \to T_{E/Z} \to T_E \to \pi_E^\ast T_Z \to 0\tag{$\ast$}\]

On the other hand, since \(Z\) was of codimension \(r\) by assumption, \(\mathcal{N}_{Z/X}\) is of rank \(r\) and therefore each fiber of \(E\) is \(\mathbb{P}^{r-1}\). We generalize Proposition 7 (Euler Exact Sequence) to obtain the following.

Proposition 11 (Relative Euler sequence) For any vector bundle \(V\rightarrow B\) and projectivized vector bundle \(\pi: \mathbb{P}(V)\rightarrow B\), the following short exact sequence

\[0\rightarrow \mathcal{O}\rightarrow \pi^\ast V\otimes \mathcal{O}(1)\rightarrow T_{\mathbb{P}(V)/B}\rightarrow 0\]

exists.

Proof

By definition, each point \([v]\in \mathbb{P}(V)\) is a one-dimensional subspace on a fiber of \(V\). Now, just as in \(\mathbb{P}^n\), \(\mathcal{O}_{\mathbb{P}(V)}(-1)\) is the line bundle collecting these lines, and from this we obtain the tautological exact sequence

\[0\rightarrow \mathcal{O}(-1)\rightarrow \pi^\ast V\rightarrow \mathcal{Q}\rightarrow 0\]

and since tensoring a line bundle with an exact sequence preserves exactness, we have

\[0 \rightarrow \mathcal{O}\rightarrow \mathcal{O}(1)\otimes \pi^\ast V \rightarrow \mathcal{Q}\otimes\mathcal{O}(1)\rightarrow 0\]

Now examining the part

\[\mathcal{Q}\otimes \mathcal{O}(1)\cong \Hom(\mathcal{O}(-1),\mathcal{Q})\]

since we already saw in the proof of Proposition 7 (Euler Exact Sequence) that seeing how a line of \(\mathbb{P}^n\) goes to some quotient is precisely the tangent space, this also yields

\[\Hom(\mathcal{O}(-1),\mathcal{Q})\cong T_{\mathbb{P}(V)/B}\]

and from this we obtain the relative Euler sequence

\[0\rightarrow \mathcal{O}\rightarrow \pi^\ast V\otimes \mathcal{O}(1)\rightarrow T_{\mathbb{P}(V)/B}\rightarrow 0\]

Therefore, returning to our situation, we obtain the following exact sequence

\[0\rightarrow \mathcal{O}_E\rightarrow (\pi\vert_E)^\ast \mathcal{N}_{Z/X}\otimes \mathcal{O}_E(1)\rightarrow T_{E/Z}\rightarrow 0\tag{$\ast\ast$}\]

Proposition 12 (Canonical bundle of a blow-up) For the blow-up \(\pi: \widetilde{X} \to X\) of a smooth variety \(X\) along a smooth subvariety \(Z\) of codimension \(r\), letting \(E\) be the exceptional divisor,

\[K_{\widetilde{X}} = \pi^\ast K_X + (r-1)E\]

holds.

Proof

First, by the adjunction formula on \(\widetilde{X}\),

\[K_E = (K_{\widetilde{X}} + E)\vert_E\]

Now since \(\widetilde{X} \setminus E\) and \(X \setminus Z\) are isomorphic, the difference between \(K_{\widetilde{X}}\) and \(\pi^\ast K_X\) can only occur on \(E\). Therefore there exists an integer \(a\) such that we can write

\[K_{\widetilde{X}} = \pi^\ast K_X + aE\]

and we must show that \(a=r-1\).

To do this, restricting the above formula to \(E\) gives

\[K_{\widetilde{X}}\vert_E = (\pi^\ast K_X)\vert_E + aE\vert_E\]

Here \((\pi^\ast K_X)\vert_E = \pi_E^\ast(K_X\vert_Z)\), and by the adjunction formula for \(Z \subset X\),

\[K_X\vert_Z = K_Z \otimes \det(\mathcal{N}_{Z/X})^{-1}\]

so we obtain

\[(\pi^\ast K_X)\vert_E = \pi_E^\ast K_Z \otimes \pi_E^\ast \det(\mathcal{N}_{Z/X})^{-1}\]

Also since \(E\vert_E = \mathcal{O}_E(-1)\),

\[K_{\widetilde{X}}\vert_E = \pi_E^\ast K_Z \otimes \pi_E^\ast \det(\mathcal{N}_{Z/X})^{-1} \otimes \mathcal{O}_E(-a)\]

Therefore

\[(K_{\widetilde{X}} + E)\vert_E = \pi_E^\ast K_Z \otimes \pi_E^\ast \det(\mathcal{N}_{Z/X})^{-1} \otimes \mathcal{O}_E(-a-1)\]

On the other hand, from (\(\ast\)) and (\(\ast\ast\)), taking determinants we obtain the following two formulas

\[\det(T_E) = \det(T_{E/Z}) \otimes \pi_E^\ast \det(T_Z),\qquad \det(T_{E/Z}) = \pi_E^\ast \det(\mathcal{N}_{Z/X}) \otimes \mathcal{O}_E(r)\]

From the second formula we can explicitly obtain the relative canonical bundle as

\[\omega_{E/Z} = \det(T_{E/Z})^{-1} = \pi_E^\ast \det(\mathcal{N}_{Z/X})^{-1} \otimes \mathcal{O}_E(-r)\]

and substituting this again into the first formula, we know that

\[K_E = \det(T_E)^{-1} = \omega_{E/Z} \otimes \pi_E^\ast K_Z = \pi_E^\ast K_Z \otimes \pi_E^\ast \det(\mathcal{N}_{Z/X})^{-1} \otimes \mathcal{O}_E(-r)\]

Comparing these two expressions now yields \(-a-1 = -r\), that is, \(a = r-1\).

Let us examine the following concretely.

Example 13 (Blow-up of \(\mathbb{A}^2\) at a point) Consider the blow-up of \(X = \mathbb{A}^2\) at the origin \(Z = \{0\}\). Since \(K_{\mathbb{A}^2} = 0\) and the codimension of \(Z\) is \(r = 2\), by Proposition 12 (Canonical bundle of a blow-up),

\[K_{\widetilde{\mathbb{A}^2}} = E\]

Since \(K_{\widetilde{\mathbb{A}^2}} = E\), we can see that the exceptional divisor \(E\) plays the role of the canonical divisor.

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References

[Har] R. Hartshorne, Algebraic Geometry, Graduate Texts in Mathematics, Springer, 1977. (II.8. Differentials; III.7. The Dualizing Sheaf)