Canonical Line Bundle

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

Although line bundles thus exert quite an important influence on our geometry, we have not yet properly examined how to define a line bundle on an arbitrary variety in general. If \(X\) is a smooth variety, we can consider the cotangent bundle \(\Omega_X^1\) defined on it via §Line Bundles and Vector Bundles, ⁋Example 24, and by considering its top exterior power, we can consider 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 defined on \(X\). Let us show that this agrees with differentials in the algebraic setting. ([Commutative Algebra] §Differentials, ⁋Definition 3) For this, given an arbitrary affine variety \(X\) and its coordinate ring \(A\), and an \(A\)-module \(M\), we need to examine the process of transferring \(M\) to a vector bundle on \(X\).

Our basic philosophy is that using §Affine Varieties, we can transfer a homomorphism between coordinate rings to 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. That is, multiplication is not defined on \(M\). However, according to [Multilinear Algebra] §Tensor Algebra, ⁋Definition 5, we can consider the symmetric algebra \(\S(M)\), which forcibly defines a (commutative) multiplication on \(M\).

However, there is a problem in applying this directly. Recall that our goal is to view \(M\) as a vector bundle defined on \(X\). Roughly speaking, our goal is to attach \(M\) nicely over each point of \(X\), and according to §Affine Varieties, if \(M\) appears in the world of varieties (as a fiber), the coordinate ring defining it should 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, so we obtain a map between coordinate rings \(A\rightarrow \S_A(M^\vee)\), and applying §Affine Varieties to this gives a morphism from some variety \(V(M)\) to \(X\).

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

Algebraically, to obtain the coordinate ring defining this fiber, we must first think about what functions are defined on it. However, since all functions contained in the maximal ideal \(\mathfrak{m}_x\) defining \(x\in X\) vanish at \(x\), it is reasonable to regard 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 thus in general \(\kappa(x)\) is an algebraic extension of \(\mathbb{K}\). ([Commutative Algebra] §The Nullstellensatz, ⁋Theorem 4) Since we usually consider the case where \(\mathbb{K}\) is an algebraically closed field, we may simply regard \(\kappa(x)\) as \(\mathbb{K}\).

Now, from the above discussion, we know that we should 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 the symmetric algebra commutes with tensor products, we obtain the identity

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

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

Now \(\S_{\kappa(x)}(M_x^\vee)\) is a polynomial algebra with coefficients in \(\kappa(x)\) and with the elements of \(M_x^\vee\) as degree 1 terms, and then the fiber \(V(M)_x\) consists of the points having these elements of \(M_x^\vee\) as coordinate functions, so these points can be thought of as the double dual of \(M_x\). Now if \(M\) is a finitely generated \(A\)-module, there exists the canonical isomorphism \(M_x\cong M_x^{\vee\vee}\), and from this we can understand each fiber \(V(M)_x\) as the \(\kappa(x)\)-vector space \(M_x\).

If \(M\) is a finitely generated projective, that is, locally free \(A\)-module, then through additional computation we can verify that this data satisfies the locally trivial condition, and thereby we can confirm that we can regard this 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 the sake of geometric intuition, but roughly speaking, it is as follows.

As above, given an affine variety \(X\) and its coordinate ring \(A\), and an \(A\)-module \(M\), we define the sheaf \(\widetilde{M}\) defined on \(X\) on a basic open set \(U=D(f)\) by the formula

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

This is basically in the same vein 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 base-changed using the structure sheaf like this. Then in particular for all of \(X\),

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

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

These two definitions merely express the same object in different geometric languages. That is, for an affine variety \(X\) and its coordinate ring \(A\), and a finitely generated projective \(A\)-module \(M\), the total space corresponding to \(\widetilde{M}\) 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 situations. For example, we define the following. ([Commutative Algebra] §Basic Notions, ⁋Definition 8)

Definition 1 On a general variety \(X\), an \(\mathcal{O}_X\)-module \(\mathcal{F}\) is called a quasi-coherent sheaf if \(X\) has an affine open cover \(\{U_i\}\) and there exists an \(A_i=\mathcal{O}_X(U_i)\)-module \(M_i\) for each coordinate ring such that \(\mathcal{F}\vert_{U_i}\cong \widetilde{M_i}\). If each \(M_i\) is a finitely generated \(A_i\)-module, then \(\mathcal{F}\) is called a coherent sheaf.

When dealing with quasi-coherent sheaves in general, one must be careful because a different \(M\) may be attached to each affine cover, but when restricted 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 that in the affine case, a quasi-coherent sheaf is an \(A\)-module, and a coherent sheaf is a finitely generated \(A\)-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, one may regard 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 so that it is closed under the operations of an abelian category, namely kernels, images, cokernels, etc., and intuitively it can be thought of as a vector bundle whose fiber dimension may vary from point to point. A quasi-coherent sheaf removes even the finite rank condition from this.

Since the coordinate ring of a variety is Noetherian, by [Commutative Algebra] §Basic Notions, ⁋Proposition 9 the coherent condition and the finitely generated condition coincide even on a singular variety, and thus the above slogan holds just as well in geometric situations. However, one somewhat cautionary point from the above intuition is that how close a coherent sheaf is to a vector bundle depends on the geometry of \(X\). Any coherent sheaf defined 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. For this, we must first introduce the cotangent bundle on a variety; the following definition is one we already saw in §Line Bundles and Vector Bundles, ⁋Example 24, but we introduce it again for completeness.

Definition 2 For a smooth variety \(X\), the cotangent bundle \(\Omega_X^1\) 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 a smooth 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 of \(X\), \(U\), we call the sheaf defined by setting \(\mathcal{T}_X(U)\) to be the collection on \(\mathcal{O}_X(U)\) of \(\mathbb{K}\)-derivations \(\Der_\mathbb{K}(\mathcal{O}_X(U),\mathcal{O}_X(U))\) the tangent sheaf.

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

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

holds. Moreover, if we consider the sheaf of derivations, the last term \(\Der_\mathbb{K}(A,N)\) is again \(\Der_\mathbb{K}(\mathcal{O}_X, \widetilde{N})\), so the equation

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

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

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

holds. 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, since

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

the desired claim holds.

This result shows that the cotangent bundle is represented by differential \(1\)-forms, just 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 of \(\mathbb{A}^n\), \(\mathbb{K}[\x_1, \ldots, \x_n]\), then the Kähler differentials of this \(\mathbb{K}\)-algebra form the free module \(\bigoplus_{i=1}^n \mathbb{K}[\x_1, \ldots, \x_n] \dd{\x_i}\), 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. (§Line Bundles and Vector Bundles, ⁋Example 24)

Definition 5 For a smooth variety \(X\) of dimension \(n\), the canonical line bundle \(\omega_X\) is defined as the top exterior power of the cotangent bundle

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

We call a global section of the canonical bundle \(\omega_X\), \(s\in \Gamma(X, \omega_X)\), on \(X\) a regular \(n\)-form. These are, if we choose a trivializing open set of \(\omega_X\), \(U\), and identify it with the cotangent bundle over affine space as in Example 4, sections that appear for a regular function \(f\) in the form \(f\dd{\x_1} \wedge \cdots \wedge \dd{\x_n}\) as \(n\)-forms.

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.

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 defining \(\mathbb{P}^n\),

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

the \(\mathbb{K}^\times\)-action is an 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, there exists the following short exact sequence corresponding to the tangent bundle

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

and taking the dual of this yields the following.

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

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

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

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

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

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

Now 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

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

On the right-hand side, \(\det(\mathcal{O}_{\mathbb{P}^n})\cong \mathcal{O}_{\mathbb{P}^n}\), 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

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

Here, the canonical divisor is given by \(K_{\mathbb{P}^n}=-(n+1)H\). From this computation 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 computation from the perspective of transition functions of \(n\)-forms. On the standard open cover \(U_i = \{\x_i \neq 0\}\) of \(\mathbb{P}^n\), setting affine coordinates \(\y_j^{(i)} = \x_j / \x_i\) (\(j \neq i\)), we can consider the \(n\)-form on \(U_i\)

\[\dd{\y_0}^{(i)} \wedge \cdots \wedge \widehat{\dd{\y_i}^{(i)}} \wedge \cdots \wedge \dd{\y_n}^{(i)}\]

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

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

On the other hand, the case \(k=i\) is not included in the above computation, and in this case \(\y_i^{(j)} = \x_i / \x_j = 1/t\), so

\[\dd{\y_i}^{(j)} = -t^{-2}\dd{t}\]

Now, if we compute the \(n\)-form \(\bigwedge_{k \neq j} \dd{\y_k}^{(j)}\) on \(U_j\), this exterior product contains the factor with \(k=i\), and since this is a multiple of \(\dd{t}\), the terms with \(\dd{t}\) in the remaining factors all cancel, leaving only \(t^{-1}\dd{\y_k}^{(i)}\). Thus from the \(n-1\) factors with \(k \neq i, j\) we get \(t^{-(n-1)}\), and from the factor with \(k=i\) we get \(-t^{-2}\dd{t} = -t^{-2}\dd{\y_j}^{(i)}\); rearranging these, the \(n\)-form on \(U_j\) transforms on \(U_i \cap U_j\) as

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

Here \((-1)^{i+j}\) is the sign arising from gathering the sign of the factor with \(k=i\) and rearranging the factors in order, and since this can be absorbed by multiplying the \(n\)-form on each \(U_i\) by \((-1)^i\), it is safe to ignore.

Then the \(n\)-forms on \(U_i\) have the frame \(\alpha_i = \bigwedge_{k \neq i} \dd{\y_k}^{(i)}\), and the above computation says that the relation between the two frames is \(\alpha_j = c_{ij}\alpha_i\) with \(c_{ij} = (\y_j^{(i)})^{-(n+1)} = (\x_j / \x_i)^{-(n+1)}\). Now writing a section as \(s = f_i\alpha_i = f_j\alpha_j\), we have \(f_i = c_{ij}f_j\), so by the convention of §Line Bundles and Vector Bundles, ⁋Example 12, the transition function between the trivialized functions is its inverse \(g_{ij} = c_{ij}^{-1} = (\x_i / \x_j)^{-(n+1)}\). 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.

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

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

(§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, there exists the following short exact sequence

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

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 concretely this is obtained by verifying the dual of the above short exact sequence,

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

where the first arrow is given by \(f\mapsto \dd{f}\). Taking the top exterior power of this short exact sequence yields the following.

Proposition 9 (Adjunction Formula) For a smooth variety \(X\) and a smooth divisor \(D\) on it,

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

holds.

From this, the claim for the canonical divisor also follows immediately. In any case, the claim contained in this proposition is that if, for the ambient variety \(X\), one twists the canonical bundle \(\omega_X\) by the line bundle \(\mathcal{O}_X(D)\) and then restricts it to \(D\), the subvariety \(D\)’s canonical bundle \(\omega_D\) is obtained. In simple terms, differential forms on \(D\) are obtained by adding normal direction information to the differential forms of the ambient space.

The following example shows a concrete computation using this.

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

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

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

On the other hand, in classical algebraic geometry it is well known 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

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

holds.

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

Canonical Divisor of a Blow-up

In §Rational Maps, ⁋Example 12, we examined the blow-up of \(\mathbb{A}^2\) at a point \(0\). 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 along \(Z\) of \(X\) is given by the birational morphism

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

where \(\pi\) is an isomorphism away from the fiber over \(Z\), and the fiber over \(Z\) is the exceptional divisor

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

Here \(\mathcal{N}_{Z/X}\) is the normal bundle in \(X\) of \(Z\), 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 §Rational Maps, ⁋Example 12, where the exceptional divisor corresponding to the fiber over 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

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

For notational convenience, we abbreviate this as \(\pi_E\). Now thinking of the tangent bundle of \(E\), we can divide it into the horizontal direction coming from the base space \(Z\), namely \((\pi\vert_E)^\ast T_Z\), and the relative tangent bundle direction \(T_{E/Z}\) perpendicular to it. That is, we obtain the following short exact sequence

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

On the other hand, by assumption \(Z\) had codimension \(r\), so \(\mathcal{N}_{Z/X}\) has rank \(r\) and thus each fiber of \(E\) is \(\mathbb{P}^{r-1}\). Generalizing Proposition 7 (Euler Exact Sequence), we obtain the following.

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

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

Proof

By definition, each point \([v]\in \mathbb{P}(V)\) is a one-dimensional subspace of 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 an exact sequence with a line bundle preserves exactness,

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

holds. But if we examine the part

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

since we have already seen in the proof of Proposition 7 (Euler Exact Sequence) that viewing how a line in \(\mathbb{P}^n\) maps to a 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

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

Thus, 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 \(\pi: \widetilde{X} \rightarrow X\) 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 \(\widetilde{X}\), we have

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

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 \(E\). Thus there exists some integer \(a\) such that we can write

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

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

To this end, restricting the above identity 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 \(Z \subseteq X\),

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

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,

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

Meanwhile, taking determinants from (\(\ast\)) and (\(\ast\ast\)), we obtain the following two identities:

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

Now from the second identity, 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 back into the first identity, we know that

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

Now comparing these two expressions, we obtain \(-a-1 = -r\), that is, \(a = r-1\).

Concretely, let us examine the following.

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),

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

That is, we see that the exceptional divisor \(E\) plays the role of the canonical divisor.

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References

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