1 Classification of Surfaces

1.1 The Intersection Form

Given two distinct irreducible curves, C and D with local defining equations f and g on a surface X wedefine their intersection at a point x as follows

mx(C ∩D) = dimCOX,x/(f, g).

Their total intersection is defined to be

C.D =∑

x∈C∩Dmx(C ∩D).

We define the sheafOC∩D = OX/(OX(−C) +OX(−D))

whose stalk at a point x is OX,x/(f, g) and is thus a skyscraper sheaf. Furthermore, we see that

C.D = h0(OC∩D).

Lemma 1.1.1. Define

L1.L2 = χ(OX)− χ(L−11 )− χ(L−12 ) + χ(L−11 ⊗ L−12 ).

Then OX(C).OX(D) = C.D and this intersection is bilinear.

Proof. Consider the exact sequence

0→ OX(−C −D)(g,f)−−−→ OX(−C)⊕OX(−D)

(f,−g)−−−−→ OX → OC∩D → 0.

To prove exactness look at the stalks. Note that OX is regular, hence a UFD, so the middle joint is exact. Theadditivity of Euler characteristic then implies the second part of the Lemma. For the first part, we must showthe following sublemma: For smooth C, we have degL

∣∣C

= L.OX(C). This follows from Riemann-Roch,additivity of the Euler characteristic, and the exactness of the sequence

0→ OX(−C)→ OX → OC → 0.

and its tensor with L.Now, consider

s(L1, L2, L3) =L1.(L2 ⊗ L3)− L1.L2 − L1.L3

=− χ(OX)

+ χ(L−11 ) + χ(L−12 ) + χ(L−13 )

− χ(L−11 ⊗ L−12 )− χ(L−11 ⊗ L

−13 )− χ(L−12 ⊗ L

−13 )

+−χ(L−11 ⊗ L−12 ⊗ L

−13 )

This is symmetric in the Li. Furthermore, it vanishes when L1 = OX(C) for a smooth curve C by thesublemma. So by symmetry, it vanishes whenever any of the Li = OX(C). Say L2 = OX(D) for somedivisor D. We can express D as A−B for A and B very ample divisors. By Bertini’s theorem, A and B arerepresented by smooth curves. Then,

0 = s(L1, L2, B) = L1.A− L1.L2 − L1.B

so we have L1.L2 = L1.A−L1.B = degL1

∣∣A− degL1

∣∣B

. This expression is bilinear in L1 as degree adds ona smooth curve.

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Theorem 1.1.1 (The Riemann-Roch Theorem).

χ(D) = χ(OX) +1

2(D.D −D.K).

Proof. This follows from the definition and Serre duality, which states that hi(D) = h2−i(K −D).

Corollary 1.1.1 (The Genus/Adjunction Formula). For an irreducible curve C, the genus g(C) := h1(C,OC)is given by

g(C) = 1 +1

2(C.C + C.K).

Proof. This follows from the exactness of the usual sequence associated to C, along with the fact thath0(C) = 1. We can in fact strengthen this result when C is smooth to the adjunction formula

(C +K)∣∣C

= ωC .

To prove it, we take the normal sequence

0→ TC → TX∣∣C→ NC → 0.

Dualizing, we get the exact sequence

0→ OX(−C)∣∣C→ Ω1(X)

∣∣C→ ωC → 0

Taking determinants, we get the result.

We can get awesome results like Bezout’s theorem very quickly: Every degree d curve is linearly equivalentto dL for a line L. If D and E are curves of degree d and e, then

D.E = dL.eL = deL.L = de.

By GAGA analytic and algebraic Pic are the same, so we can use the exponential exact sequence 0 →Z→ O → O∗ → 0. The long exact sequence gives

· · · → H1(Z)→ H1(O)→ Pic(X)c1−→ H2(Z)→ H2(O)→ · · · .

By Hodge theory, the kernel of the last map is H2(Z)∩H1,1(X) := Num(X). The intersection form dependsonly on the image of an element of Pic under c1, and is equal to the usual cap product there.

First note that [C] ∩ [D] = C.D for transverse curves C and D, since holomorphic curves intersectpositively. Thus, since we can replace C with A − B, we get that the intersection form and cap productagree. Note that c1(L) is the curvature form associated to L and its integral on an orientable surface isequal to the degree of L restricted to that surface. Hence 〈c1(OX(C)), D〉 = degOX(C)

∣∣D

= C.D for everyorientable submanifold D. Thus, by Poincare duality and representability of H2(X,Z) by submanifolds, weget c1(OX(C)) = [C], as desired.

1.2 Birationality

To blow up a point p on a surface X, take a chart U with coordinates x, y centered at p. The blow up isgotten by taking the subset U × P1 given by xt − ys = 0 and glueing it to the rest of X. When x = y = 0we get a copy of P1, called the exceptional divisor. Away from p, the projection ε to U is an isomorphism.The strict transform C of a curve C is defined as the Zariski closure of the inverse image away from p.

If a curve C passes through p with multiplicity m, its defining equation has lowest order term degree mat p. So f(x, y) = a0y

m + · · · , and choosing x, y, we may assume a0 6= 0. Then, a pair of coordinates on thechart s 6= 0 of the blow up is x, t with xt = y. In these coordinates, the equation becomes

f(x, tx) = xm(a0tm + a1t

m−1 + · · · ) + xm+1(b0tm+1 + b1t

m + · · · ) + · · · .

So, the strict transform is defined by the equation x−mf(x, tx), and hence intersects the exceptional divisorx = 0 at m points. Furthermore, the pullback ε∗(C) = C +mE.

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Note that ε∗D.E = 0 and ε∗D.ε∗D′ = D.D′ for all D,D′ ∈ Pic(X) because we can always replace D andD′ by divisors that don’t go through p. Since ε∗C.E = C.E + mE.E = m + mE.E = 0, we conclude thatE.E = −1. Since KX = ε∗KX + nE for some n, we can use the genus formula on E to conclude that n = 1.

Furthermore, it is clear that Pic(X) = ε∗Pic(X)⊕ZE. If ε∗D+nE = 0 then n = 0 by dotting with E, andD = 0 by pushing forward (or using Hartog’s lemma).

Theorem 1.2.1 (Birational Factorization). A birational map φ : X 99K Y factors as a sequence of blow-upsfollowed by a sequence of blow-downs.

Proof. Step 0. The map φ and its inverse are defined on all but a finite set of points. To see this, embedY into Pn such that it lies in no hyperplane. We can reduce to P1 by taking pairs of coordinates on Pn, asthe values on the pairs of coordinates determines the map into Pn. But a function to P1 is undefined onlyat where the poles and zeroes intersect, which is codimension two. A rational map to Pn contained in nohyperplane corresponds to a linear system with no fixed part, by simply taking φ∗O(1).

Step 1. The first step is to eliminate indeterminacy. Consider the linear system |D| associated to φ.Then, blow up a fixed point of |D|, to get the linear system |ε∗D| on the blow up. The fixed part of |ε∗D| isthen its previous fixed points away from p plus kE for some positive k. Subtracting off the fixed part kE,we get a new linear system |D1| which satisfies D2

1 = D2− k2 and defines the same embedding away from p.Continuing inductively, we get a linear system such that D2

n = 0, that is to say, there are no fixed points.Step 2. Now that indeterminacy is eliminated, we wish to show that a birational morphism is a sequence

of blow-downs. Nonsingularity of the target is essential, as this is false otherwise. The first part we mustprove is Zariski’s Main Theorem, which is just a fancy way of saying that if the inverse of a birational mapis undefined, then the inverse image is a connected subvariety of positive dimension. Though, I think this istotally obvious with a little bit of topology, for completeness sake, here is the algebraic proof for surfaces.

Step 2a. Let φ : X → Y be a birational morphism, with X possibly singular. Suppose that φ−1 : Y 99KX is undefined at p. Then, pick an open set U containing p, assume φ−1 : U 99K An, and write

φ−1 = (g1, . . . , gn)

with gi rational functions. As φ−1 is undefined, some gi must be undefined at p, say it is g1. Write g1 = u/vwith u, v ∈ OY,p coprime and v(p) = 0. Then, since x1 is regular function on φ−1(U), and x1 = u

v φ, weconclude that

V := V(v φ) ⊂ V(u φ)

on φ−1(U). Furthermore, φ(V ) ⊂ V(u, v). Hence by relative primality of u and v, we conclude that φ(V ) isa finite set in U . Shrinking U if necessary, we there is a curve V such that φ(V ) = p.

S

Xφ -

π1

Y

π2

-

Step 2b. Next, we show a similar property, but for birational maps, not morphisms. Take the closureof the graph of φ : X 99K Y , and denote it by S, possibly a singular surface. If φ−1 is undefined at p, theprojection map π2 : S → Y is too. Thus, there is a curve C such that π2(C) = p, by Step 2a. Since Cis a curve, π1(C) is a curve D (C can’t be contracted to a point in both projections). This curve satisfiesφ(D) = p.

Y

Xφ -

s

Y

ε-

Step 2c. We can now prove the universal property of blow-ups, which is that if φ−1 is undefined at p,then φ factors through the blow-up at p. Now, suppose that s−1 is undefined at some point q ∈ X. Also,

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φ(q) = p, as ε−1 is defined away from p. By Step 2b, there is a curve C ⊂ Y such that s(C) = q, and soε(C) = p. Thus, C = E. Now, consider OX,q. We claim that there is a local coordinate y at p such thatφ∗y ∈ m2

q. This is true because both local coordinates on Y pull back to functions which vanish on the curveφ−1(p), and so some linear combination of them vanishes to order at least 2 at q. Now, let e ∈ E be one ofthe many points where s is defined. Then,

s∗φ∗y = ε∗x ∈ m2e.

But a local coordinate on Y pulls back to a local coordinate on E for all but one e ∈ E. Contradiction. Sos−1 is everywhere defined.

Step 3. If the process of applying Step 2c never terminates, then there would be arbitrarily many distinctcurves on X that were contracted to a point in Y . This is impossible, as this number is finite. Hence, everybirational morphism is a sequence of blow-downs. Combined with Step 0, this gives the result.

We get some good results from this. For example, every surface has a birational morphism to a minimalsurface, as the rank of Neron-Severi strictly decreases every time we blow down. Also, blowing up has the“reverse universal property,” which is to say that if a birational morphism contracts the exceptional curve,then if factors through the blow-down. This follows quickly by reducing the target to affine space thenapplying Hartog’s lemma.

Theorem 1.2.2 (Castelnuovo’s Contractibility Criterion). Let E be a curve on X isomorphic to P1 withE2 = −1. Then E is an exceptional curve.

Proof. The idea is to find a linear system on X whose associated map to Pr is the contraction. Consider avery ample divisor H on X such that H1(OX(H)) = 0, and suppose H.E = n. Let H ′ = H + nE. Thereis a map from sections of H + (k − 1)E to sections of H + kE by multiplying by a standard section f ofO(E). Thus, |H ′| is an embedding away from E. Because H ′.E = 0, we conclude that all sections of H ′ areconstant when restricted to E. We claim |H ′| contracts E to a point. We have exact sequences

0→ O(H + (k − 1)E)→ O(H + kE)→ OE(n− k)→ 0

so by induction and the long exact sequence in cohomology, we know H1(H + kE) = 0 for all 0 ≤ k ≤ n.Thus, we can write down the sections of H ′ as

fns0, . . . , fnsm, f

n−1a0,n−1, . . . , fn−1an−1,0, . . . , fa0,1, fa1,0, a0,0

where ai,j generate H0(E,O(i + j)) and si generate H0(OX(H)). In this basis, E maps to the pointp = [0 : · · · : 0 : 1]. Let the image of |H ′| be denoted Y , and the map by φ.

The only thing left to check is that Y is smooth. Let [x0 : · · · : xr] be the coordinates on Pr given bythe basis of H ′ above. In the chart xr = 1, the ideal mp = (x0, . . . , xr−1). Note that p is a smooth point ifmp/m

2p has dimension 2. We claim

x0, . . . , xr−3 ≡ 0 mod m2p.

The divisor of φ∗xi is D + kE, where D does not have E as a component, k ≥ 2, and D.E = k. Since a0,1and a1,0 generate any section of OE(k), we may choose some degree k polynomial Q(xr−2, xr−1) so that

(f−kφ∗Q)∣∣E

= (f−kφ∗x)∣∣E

by letting the zeroes of Q be D ∩E and scaling. Then, the pullback of xi −Q(xr−2, xr−1) vanishes to orderat least k + 1 on E. Continuing inductively, we conclude that there exist Qk so that

xi −∞∑k=2

Qk(xr−2, xr−1) ∈ OY,p

vanishes to infinite order on E when pulled back, thus is exactly zero. Hence xi ∈ m2p and mp/m

2p has

dimension (at most) 2. By standard algebra, the dimension of the completion equals the dimension of theoriginal ring, thus OY,p is regular.

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Let’s define some invariants of a surface. We have the Hodge diamond with entries hp,q(X) = hq(Ωp),the dimension of the Dolbeault cohomology groups. We define q = h1(O) = h0(Ω1) and pg = H2(O) =H0(K). Furthermore, there are the Betti numbers, the topological and holomorphic Euler characteristics, andinvariants of the intersection form on H2(X,Z), such as signature. Both q and pg are birational invariants.We can pull back one- or two-forms to meromorphic differentials, which are in fact holomorphic, as the locusof poles has codimension 1. Finally, there’s Noether’s Formula, which is notoriously evil:

12χ(OX) = K2 + χtop(X).

A later section is devoted to proving it.

1.3 Ruled Surfaces

A surface is ruled if it is birational to C × P1 for some curve C. A surface is geometrically ruled if it has amap to a curve C such that any (scheme-theoretic) fiber is a smooth P1. The main results on ruled surfacesare that the geometrically ruled surfaces are projectivizations of rank 2 vector bundles, and that the minimalruled surfaces are geometrically ruled. The primary tool is the Noether-Enriques theorem.

Theorem 1.3.1 (Noether-Enriques Theorem). Suppose that π : S → C is a map from a surface to a curvesuch that S is smooth on a fibre F and F ∼= P1. Then, there is an open set of C over which S is a trivializableP1 bundle.

Proof. Denote the fiber F . Then, F 2 = 0, as we can replace p by a linearly equivalent disjoint set of points.Since g(F ) = 0 the genus formula implies K.F = −2.

Since F is effective and F 2 ≥ 0, F.E ≥ 0 for all E effective. In particular, K is not effective, i.e.H2(O) = 0. Since K.F = −2, we also know F is not multiple, as the thing it was a multiple of would haveno well-defined genus. Hence, by Poincare duality, there is an H ∈ Pic(S) such that H.F = 1. Consider theexact sequence

0→ O(H + (n− 1)F )→ O(H + nF )→ OF (1)→ 0.

The long exact sequence contains H1(H + (r − 1)F ) → H1(H + rF ) → 0. In particular, h1(H + rF ) ≤h1(H + (r − 1)F ) so these dimensions stabilize for large enough r. Hence, we eventually get

H0(H + rF )→ H0(OF (1))→ 0.

Choose a pencil (two-dimensional linear subsystem) V ⊂ |H + rF | which surjects onto H0(OF (1)). Then,the fixed locus of V does not intersect F , and so must be contained in a finite set of fibres. Throwing theseout, along with any singular fibres, we have that |V | has no fixed points and defines a morphism φ to P1.Set Ct = φ−1(t). Note that Ct is a section for all t: If Ct contains a fibre, Ct ∩ Ct′ 6= 0, contradicting thatthere are no fixed points (as t and t′ span the linear system). Ergo, (π, φ)→ U × P1 is an isomorphism.

We may now easily prove that all geometrically ruled surfaces are projectivizations of rank 2 bundles. Wehave Zariski local trivializations U × P1. Which gives the structure of a PGL2 bundle. The exact sequence

0→ O∗ → GL2(O)→ PGL2(O)→ 0

shows that a PGL2 bundle lifts to a GL2 bundle whenever H2(O∗) = 0. This is true, because C is one-dimensional. Also, on any variety, O∗ has a resolution 1 → O∗ → K∗ → Div → 0 by flasque sheaves. SoHk(O∗) = 0 for all k ≥ 2. Furthermore, this shows that two rank two bundles give the same P1 bundle ifand only if they differ by a line bundle.

Theorem 1.3.2. Every minimal, ruled, nonrational surface is geometrically ruled.

Proof. First, we construct a map to C. By elimination of indeterminacy, we can factor S 99K C × P1 intoblow ups followed by a birational morphism. The exceptional divisor of the last blow up must be contractedto a point, since the base C isn’t rational. Thus, by the “reverse universal property” we didn’t need to dothe last blow up. Inductively continue to get a map S → C with generic fibre P1. Every fibre is connected,

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because all but finitely many are connected. Furthermore, any fibre is irreducible, because if it had twocomponents, then one such component would satisfy C.K < 0 and C2 < 0 (since F 2 = 0 and Ci.Cj > 0 forsome j.) This contradicts minimality. We’ve already ruled out the case that F is multiple. Now, F 2 = 0and F.K = −2 implies F ∼= P1, as desired. Finally, geometrically ruled surfaces are minimal, because anyP1 must be a fibre, so does not have square −1.

S - S′

C?

- C ′?

Given a diagram as above of geometrically ruled surfaces, the fibre of S → C includes into a fibre ofS′ → C ′ because the composition to C ′ must be trivial. Thus, there are induced maps C → C ′ and viceversa, implying that C ∼= C ′. Hence, two geometrically ruled surfaces PC(E) and PC′(E′) are isomorphic if

and only if α : C∼=−→ C ′ and α∗E′ = L⊗ E for some line bundle L.

To any geometrically ruled surface, we can associate a hyperplane class h. Consider the bundle π∗E onPC(E). There is a “tautological” line subbundle of this rank 2 bundle given by associating to each pointin the fibre PEp the line that goes through it. The quotient line bundle is by definition OS(1) = H. Thedivisor H satisfies H.F = 1. Finally, we compute Pic, H2(Z), and K of a ruled surface. First, we claimPic(S) = π∗Pic(C)⊕ ZH. We use asymptotics of Riemann-Roch. But first, given a divisor D modify it bysubtracting multiples of H until D.F = 0. Then, consider D + nF . By Riemann-Roch

h0(D + nF ) + h0(K −D − nF ) = h0(D + nF ) ≥ χ(OS) +1

2

(D2 + 2n−K.D

),

where we again the fact that F.(K −D− nF ) < 0 implies K −D− nF ineffective. In particular, eventuallyD+nF is effective. Since (D+nF ).F = 0 we conclude that D+nF is a sum of fibres, hence D ∈ π∗Pic(C).Furthermore, we know that π∗Pic(C) and H are independent because they intersect differently with F .

Since the image of π∗Pic(C) in H2(Z) is ZF , we conclude that H2(S,Z) = ZF ⊕ ZH. To completeour understanding, we must compute H.H. Then we can know the intersection form. Using the fact that0→ N → E → OS(1)→ 0, can conclude

c2(E) : = N.OS(1) = N−1.OS(1)−1

= χ(OS)− χ(N)− χ(OS(1)) + χ(∧2p∗E)

= χ(OS)− χ(p∗E) + χ(∧2p∗E)

depends only on E. Hence, resolving E on C by 0 → L → E → M → 0 shows c2(p∗E) = p∗L.p∗M = 0.Hence, N.OS(1) = 0 (which is actually obvious if you think about it). Also, N ⊗ OS(1) = p∗(∧2E) Hence,dotting with OS(1) gives H.H = H.p∗(∧2E) = deg(E). Finally, we compute the canonical bundle. LetK ≡ aH + bF . Dotting with F shows a = −2. Then, let a section be equivalent to H + rF . Using the genusformula, r cancels, and we conclude that b = deg(E) + 2g(C)− 2.

1.4 Rational Surfaces

We begin by determining the geometrically ruled surfaces over P1. Theorem 1.3.2 does not apply, so wecannot conclude that these are the minimal ruled surfaces, but we still know that all such surfaces arise fromprojectivizing a rank two bundle on P1. All such bundles are of the form O(n) ⊕ O(m). The main tool isRiemann-Roch theorem for vector bundles on curves.

Theorem 1.4.1. Let V be a holomorphic vector bundle on a genus g curve. Then

χ(V ) = deg(V ) + rank(V )(1− g).

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Proof. We first claim that V can be filtered by line bundles. V (d) has a holomorphic section s for d largeenough. Let D = div(s). Then, there is a map 0→ O(D)→ V (d) injective on fibers: Divide by the canonicalsection of O(D) then multiply by s. So the quotient is a vector bundle. A more highfalutin way of phrasingthis: There is a map from V (d)∗ → O gotten by evaluating on s. The image is the subsheaf O(−D) for someD effective. Since this is a line bundle, the kernel of V (d)∗ → O(−D) is a vector bundle, and we can thendualize the resulting exact sequence. In either case, we may twist back to get an exact sequence

0→ O(D)→ V → E → 0

where D can be effective (trivial) if V has a holomorphic (nonvanishing) section. Furthermore, when h0(V ) ≥rank(V ) either V is the trivial bundle, or r sections in H0(V ) are somewhere linearly dependent, implyingthey have a linear combination which vanishes at some point, and D can be chosen of positive degree.

Now, we simply do induction on the rank of V . Assuming Riemann-Roch for rank less than V , we knowthat χ(V ) = χ(O(D))+χ(E) = deg(D)+1−g+deg(E)+(rank(V )−1)(1−g) = deg(V )+rank(V )(1−g).

Now let E be rank two on P1. Let E(n) have degree d equal to 0 or −1. By Riemann-Roch, h0(E) ≥ 1so 0→ O(n)→ E → O(d− n)→ 0 is exact and n ≥ 0. The extension group is H1(2n− d) = 0 so E splits.Thus, the geometrically ruled surfaces over P1 are Fn = PP1(O ⊕O(n)), called the Hirzebruch surfaces. Bythe previous section, Pic(Fn) = ZH ⊕ ZF , with H2 = n, H.F = 1 and F 2 = 0.

There are two special “curves of interest” on Fn. Recall that isomorphic P1 bundles are arrived at bytensoring with line bundles. Thus,

Fn = PP1(O ⊕O(n)) = PP1(O(−n)⊕O).

There are obvious choices of section over P1; the ones which map into the trivial factor O. Let E0 andE∞ be the image curves. The curve E0 is an example of a so-called zero section, i.e. it is the image of(1, σ) for a section σ ∈ H0(O(n)). All such sections are homologous and E2

0 = n. The curve E∞ is calledthe “section at infinity,” since removing it gives the total space of O(n). It is the zero section in the totalspace of O(−n) and therefore has self-intersection −n. Since E∞.F = 1, we can write E∞ = H + rF . ThenE2∞ = −n implies E∞ = H − nF . In fact E∞ is the unique irreducible curve with negative self-intersection:

Let aH + bF represent a different irreducible curve. Then, (aH + bF ).F ≥ 0 =⇒ a ≥ 0. Furthermore,(aH + bF ).(H − nF ) = b ≥ 0. Thus, (aH + bF )2 = a2n+ 2ab ≥ 0.

Since Fn has a unique irreducible curve of negative self-intersection −n, we know that Fn ∼= Fm iff n = m.Furthermore Fn is minimal for all n 6= 1. Castelnuovo’s Rationality Criterion will imply that P2 and Fn forn 6= 1 are the minimal rational surfaces.

Linear systems on P2 define various interesting embeddings. The image of the linear system of conics inP5 is the Veronese surface V . The image of a line is a conic in V because the hyperplane class is 2l and2l.l = 2. There are no lines on V because H is even. A generic projection of V into P4 is an embedding. Theproof is as follows: Consider the union of all planes containing the conics on V . Every chord is containedin such a plane because any two points are joined by a line in P2, hence by a conic in V . The dimension ofsuch planes is 2, as they correspond to the lines in P2. So the set of such planes is dimension 4, implyingthat the chord variety is dimension 4.

We can similarly take all conics going through a point p. This defines an embedding F1 → P4, becausethere are five degrees of freedom, and only three are necessary to separate tangents on the exceptional divisor.The image is isomorphic to a projection of V from a point on V . The hyperplane section now is 2l −E. Soevery line not going through p is a conic, and every line going through p has strict transform l − E whichdots with H to give 1 so is a line. Thus, the pullback of a line going through p is the union of two lines. Callthis surface S.

Amazingly, S is the intersection of a two-dimensional linear system of quadrics! Any pencil in this linearsystem intersects at the union of a plane and S. A quadric in P4 intersects S at 4l− 2E, or quartics passingthrough p with multiplicity 2. This imposes three conditions on the quartic (as it must pass through, andboth its partials must vanish). Since dim |OP4(2)| = dim |OP2(4)| = 10 there is (at least) a three dimensionalspace of quadrics in P4 that contain S (and so a two-dimensional linear system). The intersection of twoquadrics is a degree 4 surface. Since deg(S) = 3, the intersection of any two quadrics is S ∪P . So the pencilgenerated by these two quadrics vanishes on S ∪P . These are the only quadrics containing S ∪P , since they

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generate the ideal of S ∪ P . Let P be defined by L = M = 0. Then the two quadrics can be expressed asQi = AiL − BiM for Ai, Bi linear forms i = 1, 2. So the determinant Q3 = A1B2 − A2B1 vanishes on S.Furthermore, Q1∩Q2∩Q3 = S. If the intersection contained a line, we could express the Qi in the quadraticterms of three linear forms. The subspace of products of two linear forms would be a closed, four-dimensionalsubspace of this five-dimensional projective space. Thus, since the Qi span a two-dimensional space of thesequadratic forms, their span would intersect the space of products of two linear forms. This would in turnimply that S is contained in a hyperplane, which is impossible.

Enough of this stuff, let’s do cubics! Define the del Pezzo surface Sr as the image of Pr, or P2 blown upat r generic points p1, . . . , pr, embedded via cubics going through those points. The map is well-defined byL’Hopital’s Rule. The projection is into P9−r, as passing through a point reduces the degrees of freedom by1. The hyperplane class is

H = 3l −∑i

Ei = −K.

The degree of this surface is H2 = 9− r. For r ≤ 6, the del Pezzo surface is smooth. It is sufficient to checkfor r = 6. The only non-trivial step is showing that tangents on the exceptional divisor are separated. Thisis accomplished as follows: Let x ∈ E1. Take a conic Qx23 which goes through p1, p4, p5, p6 and x. ThenQx23 ∪L23 and Qx24 ∪L24 are cubics separating the tangents of x. This is because Qx23 ∩Qx24 = 4 by Bezout’stheorem, so they intersect at p1 with multiplicity 2. (If they had the same tangents at x, the would intersectwith multiplicity at least 3).

Hence, we get a smooth cubic surface in P3 isomorphic to P2 blown up at six points. We can analyze itsexceptional divisors as follows: There are six obvious exceptional divisors, Ei. Any line going through twoof the points pi is an exceptional divisor, since self-intersection decreases by m2. Any conic going throughfive of the point is, too, for the same reason. We claim these are the only exceptional divisors. To provethis, note that any irreducible curve note Ei is the strict transform of its image. Suppose then that C passesthrough pi with multiplicity mi. The conditions that its strict transform is exceptional are C2 = −1 andK.C = −1. (Or equivalently H.C = 1, viz., C is a line!) Thus, we are led to the following equations:

r∑i=1

m2i = d2 + 1

r∑i=1

mi = 3d− 1.

For r < 9, we can use inequalities to solve these equations. Using the method of Lagrange multipliers (orsomething easier like AM-QM inequality) we see that the upper-left-hand sum is minimized by mi = 3d−1

r .Plugging it in,

r(d2 + 1) ≥ (3d− 1)2 or (r − 9)d2 + 6d+ r − 1 ≥ 0.

So we have the following table of possibilities:

r = 0, 1 d = 0r = 2, 3, 4 d ≤ 1r = 5, 6 d ≤ 2r = 7 d ≤ 3r = 8 d ≤ 6r ≥ 9 d <∞

As you can see from the table, we’ve already found all exceptional divisors on Pr with r ≤ 6. So the delPezzo cubics have 27 lines on them. In fact, all cubics in P3 are P2 blown up at six points, as we will see later.The number of exceptional divisors on P7 is 7 +

(72

)+(75

)+ #cubics giving rise to exceptional divisors.

Such a cubic must have∑mi = 8 and

∑m2i = 10 which is possible if the cubic passes through 6 points with

multiplicity 1, and 1 point with multiplicity 2, which we will notate 2116. So the number of exceptionaldivisors is 7 + 21 + 21 + 7 = 56.

On P8, quartics giving exceptional divisors satisfy∑mi = 11 and

∑m2i = 17. Only 2315 works.

Quintics satisfy∑mi = 14 and

∑m2i = 26. So 2612 works. Finally sextics must satisfy

∑mi = 17 and

8

∑m2i = 37. So 3127 works. Hence, there are 8+

(82

)+(83

)+2 ·

(82

)+(83

)+(82

)+8 = 240 exceptional curves.

There are infinitely many exceptional curves if we blow up at 9 sufficiently generic points.

Theorem 1.4.2. Every smooth cubic S in P3 is a del Pezzo surface.

Proof. First note that there is at least one line on the cubic, using the Grassmannian argument; G(1, 3)×P19

is a projective variety that maps to lines in P3 via projection onto first coordinate. The projection onto thesecond coordinate maps to the space of cubics. Consider incidence correspondence of pairs X = (l, C) : l ⊂C. It is a closed subvariety because this is a closed condition. The first projection of X has 15 dimensionalfibre, as the space of cubics containing a line is 15 for any line. Hence S has dimension 19. The secondprojection has some finite fibre, as we can explicitly give a cubic that has only finitely many lines on it, e.g.x0x1x2 = x33. Hence by lower-semicontinuity of the fibre dimension, the generic fibre is finite, and the mapis finite. But a finite map between projective varieties is surjective, hence every cubic contains a line.

Consider the pencil of planes Pλ containing our line `. Note that Pλ∩S = `∪Cλ for some conic Cλ. Weclaim that whenever Cλ is singular, it contains two distinct lines. Let the corresponding plane in the pencilbe L = 0. Then if L = 0 contained a double line, say N = 0, the equation of the cubic is LQ+MN2, whereM = 0 is the equation of ` in L = 0. But, computing derivatives shows this is singular at L = N = Q = 0.So there are no double lines.

Now, we can explicitly compute the number of singular conics. If the line is defined by Z = T = 0, thenthe equation of S is AX2 + BXY + CY 2 + 2DX + 2EY + F = 0 where A,B,C,D,E, F are homogenouspolynomials in Z, T . The singular conics are given when the determinant of

∆(Z, T ) =

A B DB C ED E F

vanishes. This is a homogenous quintic, so it has five roots. We need only show that the roots are distinct.Suppose that Z = 0 is a root. We can assume the singular conic is defined by either the equation X2−T 2 = 0or XY = 0 depending on whether the singular point of the conic does or does not lie on `. In the latter case,all coefficients except B must be divisible by Z. Furthermore, since the point X = Y = Z = 0 is smooth,F must only be divisible by Z and not Z2. Hence ∆ can only have a single root at Z = 0. Similarly ifthe conic is X2 − T 2 = 0, then all coefficients except A and F are divisible by Z. Furthermore the pointX = Z = T = 0 is smooth, so C must not be divisible by Z2. As before ∆ only has a single root at Z = 0.Thus, there are five distinct planes with singular conics.

We claim that the five pairs of lines (the singular conics) do not meet each other. They could onlypossibly intersect each other on `. But if three lines go through the same point, they are all contained in thetangent plane at the point, so in fact coplanar.

Finally, to complete the proof, take two disjoint lines ` and `′. Then, we define a rational map φ :`× `′ 99K S by taking the line through the two points and mapping it to the third point of intersection withS. Conversely, given a point in S− `− `′, there is a unique point on ` such that the line through the originalpoint and it intersect `′. Call this map ψ. Clearly φ and ψ are inverse to each other, so S is birationalto P1 × P1. Furthermore, ψ can be extended to ` and `′ by requiring that the line go through the givenpoint twice (that is, be contained in the tangent plane). Hence ψ is a birational morphism to P1 × P1. Bybirational factorization, it must be a sequence of blow downs. We need only count the number of them.

Consider the plane containing ` and two lines d, e. Then, this plane meets `′ at exactly one point. Since` and `′ are disjoint, either d or e intersects `′. It is impossible for both d and e to meet `′, as then the planewould be tangent at that point and contain all of `′. Hence there are exactly five lines contracted by ψ. SoS is P1 × P1 blown up at five points, or P2 blown up at six.

We now prove a wonderful criterion for determining when a surface is rational.

Theorem 1.4.3 (Castelnuovo’s Rationality Criterion). A surface is rational if and only if q = P2 = 0.

Proof. Main Step. We can assume minimality because the quantities in the theorem are birational invari-ants. The idea of the proof is to find a smooth curve C ∼= P1 on the surface whose self-intersection C2 isnon-negative. This proves the theorem: We get an exact sequence

0→ H0(OS)→ H0(OS(C))→ H0(OC(C))→ 0

9

because q = 0. Thus h0(OS(C)) ≥ 2, implying that there is a pencil of curves on S, at least one of which isC. By the Noether-Enriques theorem, S is birational to P1 × P1, and we’re done.

We must work the conditions to find such a curve. Suppose that C is an irreducible curve such thatC.K < 0, and |C +K| = ∅. Then by Riemann-Roch and the genus formula,

h0(−C) + h0(C +K) = 0 ≥ χ(OS) + g(C)− 1 = g(C).

Hence, C is a smooth sphere. Since S is minimal we cannot have C.K = −1 , so in fact C.K ≤ −2, implyingthat C2 ≥ 0. It is sufficient to find an effective divisor D such that D.K < 0 and |D + K| = ∅, since somecomponent will satisfy C.K < 0 and still |C + K| = ∅ will hold. In the arguments below, we say such a D“works.” There are three cases, K2 < 0, K2 = 0, and K2 > 0.

Case 1: Suppose K2 < 0. If suffices to find an effective divisor D with D.K < 0. This is because asbefore, we may choose an irreducible component C with C.K < 0 and by minimality C2 ≥ 0. Hence, C isnef. In particular, since C.(C + nK) is eventually negative, C + nK is eventually not effective. So someC + rK works.

Pick a very ample divisor H. Then, the divisor

D = −(K.K)H + (H.K)K

dots with K to be 0. Note that H.D = (H.K)2 −H2K2 > 0, so −D is ineffective. Thus, by Riemann-Roch

h0(D +K) ≥ 1 +1

2

((D +K)2 − (D +K).K

)= 1 +

1

2D2 = 1− 1

2(K.K)(H.D) ≥ 1.

and (D +K).K = K.K < 0, so D +K works.Case 2: Suppose K2 = 0. Combining P2 = 0 and Riemann-Roch, we have h0(−K) ≥ 1 + K2 = 1.

Choose a very ample H. Then H.K < 0 since −K cannot be the trivial bundle. Hence H+nK is eventuallyineffective, as it eventually dots with H negatively. Thus, there is a D = H + rK that works.

Case 3: Suppose K2 > 0. As before, h0(−K) ≥ 2. Suppose −K is had multiple components, say| −K| = A+ B. Assume without loss of generality A.K < 0. Then, A works, as |A+K| = | − B| = ∅. Soassume that every section of −K is irreducible. Every effective divisor dots with −K positively: |−K| formsa family of irreducible curves on S that goes through every point (a linear combination of two elements of|−K| can always be made to vanish at a given point). Hence every curve will intersect some member of thisfamily. The intersection is positive, as even intersection by containment gives positive intersection. Hence,by the Moishezon-Nakai criterion, −K is ample.

Suppose that rank of Neron-Severi is larger than 1. Choose a nontrivial D with D.K = 0. Then D isineffective since −K.D = 0. But D − nK is effective for sufficiently large n. So a divisor D − rK works.

Finally, assume that rank of Neron-Severi is 1. Pick an ample generator D. Then by Poincare duality,D2 = 1. Furthermore D.K < 0, and if D.K < −1 then D is a smooth sphere with positive self-intersection,so we win. So assume that D.K = −1, that is to say Num(S) = ZK. This contradicts Noether’s formula,of which we now know all the quantities.

Conclusion: The “only if” part of the proof follows from the fact that the quantities are birationalinvariants. So, we’re done.

The CRC furnishes us with the necessary tools to classify the minimal ruled surfaces.

Theorem 1.4.4. The minimal ruled surfaces are P2 and Fn for n = 0 and n ≥ 2.

Proof. Let S be a minimal ruled surface. By the proof of CRC, there is an irreducible curve C with C.K < 0and |C +K| = ∅. Subject to these conditions, minimize the quantity

Q(C) = maximum number of components of a curve in |C|.

(Dotting with a very ample H shows this quantity is always finite!) Suppose that some D ∈ |C| werereducible, D = A + B. Then say A.K < 0 and |A + K| = ∅. So A works. But then Q(A) < Q(C) becauseB+ |A| ⊂ |C|. Hence, every curve in |C| is irreducible. As in the proof of CRC, every curve in |C| is genus 0,hence smooth (by irreducibility). Furthermore, by irreducibility, any pencil in |C| has no fixed component.

10

Taking such a pencil, there are C2 base points. If C2 = 0 then the pencil gives a geometric ruling, so we’redone by classification of geometrically ruled surfaces over P1. If C2 > 0, then from the exact sequence

0→ H0(OS)→ H0(OS(C))→ H0(OC(C))→ 0

we may pick a pencil in H0(OS(C)) generated by C and D such that the roots of the section D on C aredistinct. Then, D and C intersect transversely. Thus, blowing up of the points C ∩ D, this pencil is ageometric ruling. But we know that all geometrically ruled surfaces are Fn. The only non-minimal Fn is F1,and it only has one exceptional divisor. So S is the blow-down of F1, implying S ∼= P2.

1.5 Using (and Abusing) the Albanese Map

The Albanese mapAlb : X → Ω1(X)∗/H1(X,Z) = Alb(X)

is defined by

x 7→(∫ x

x0

ω1, . . . ,

∫ x

xo

ωq

)which is well defined only because we modded out by H1(X,Z). Note that by Poincare duality,

Alb(X) = H2,1(X)/H3(X,Z),

which is the (algebraizable) intermediate Jacobian J2(X). The nice thing about the Albanese map is thatis satisfies a universal property: Any map φ : X → T to a complex torus factors through Alb. Let’sprove that. By composing with a translation on T we may as well assume φ(x0) = 0. Consider the mapφ∗ : Ω1(T ) → Ω1(X). The dual of this map is from Ω1(X)∗ → Ω1(T )∗. We claim this descends to a mapm : Alb(X)→ Alb(T ) = T where the latter isomorphism is

∫ p07→ p. Given a loop γ ∈ H1(X,Z), the change

of variables formula states ∫γ

φ∗ω =

∫φ∗γ

ω.

Thus, the dual of φ∗ descends to m : Alb(X) → Alb(T ). We need now only check the following diagramcommutes:

Xφ - T

Alb(X)

Alb?

m- Alb(T )

∫ ∗0 ?

Following all the maps and isomorphisms,

(m Alb(x))(ω) =

(m

∫ x

x0

)ω =

∫ x

x0

φ∗ω =

∫ φ(x)

0

ω =

((∫ ∗0

φ)

(x)

)ω.

If two linear forms agree on every ω, they are equal in the dual space, so

m Alb =

∫ ∗0

φ

as desired. We will be especially interested in the case when the image of Alb is a curve C. In this case, Cis smooth, Jac(C) ∼= Alb(X), and the fibers of Alb are connected.

Alb(X)m - T

X

φ-Alb

Jac C

n

C

i

6

NC

ψ

6

Alb -α-

11

Let C be the normalization of C, and N the normalization map. May assume that α has connected fibers bypossibly replacing α with its Stein factorization. Normality of X induces α. Since the lower, left, and uppertriangles commute, defining ψ = m i N makes the triangle (φ, ψ, α) commute. Hence, the commutativityof the rightmost triangle implies that Jac C satisfies the universal property for Alb(X). By uniquenessAlb(X) = Jac C. Furthermore, the image of X under α is C, which is smooth, and α has connected fibers.We call α the Albanese fibration. Note that g(C) = q.

Also, im(Alb) generates Alb(X) as we could otherwise take the sub-torus generated by the image. Supposepg = 0 and q ≥ 1. Then, the image of Alb is necessarily a curve. If q = 1 we’re already done. So assumeq > 1. If dim(imAlb) = 2 then Alb is etale some point p, so we can construct a holomorphic two-form thatdoes not vanish at Alb(p). Its pullback is nonzero holomorphic, contradicting pg = 0.

Theorem 1.5.1. Non-ruled minimal surfaces have unique minimal models.

Proof. Minimize the number of blow ups needed to eliminate indeterminacy from S 99K S′, with S non-ruled.Let the last exceptional divisor be E. Then E is not contracted to a point, since then the map would needfewer blow-ups to eliminate indeterminacy. Let the image of E in S′ be C. Then C must have been on apoint blown up at least once, or C would be an exceptional divisor of S′. Thus, the strict transform of C(that is, E) has larger intersection with K than C. Hence C.K ≤ −2, implying by the genus formula C2 ≥ 0.Furthermore, C is irreducible. Hence Pn = 0 for all n. If q = 0, we have a contradiction by Castelnuovo’stheorem. If q ≥ 1. Then C must lie in a fiber of the Albanese. Hence C2 ≤ 0, and therefore C2 = 0,C.K = −2. This implies that F = nC, and since F has nonnegative genus, F = C. So by the Noether-Enriques theorem, we have a contradiction again. Thus, no blow-ups are needed to eliminate indeterminacy,and all minimal non-ruled surfaces birational to each other are isomorphic.

Theorem 1.5.2. Let X be a surface. Then P12 = 0 ⇐⇒ κ = −∞ ⇐⇒ X is ruled.

Proof. This is the most difficult and key theorem in the classification. We may assume X is minimal, as allparts of the implication are birationally invariant. The “only if” parts are easy. The fiber F of a ruling isnef, and F.K = −2. So F.nK < 0 for all n > 0, implying that κ = −∞. The implication P12 = 0 =⇒ ruledis what follows. If q = 0, then CRC applies, so X is rational.

So suppose we have pg = 0 and q ≥ 1. Let X → B be the Albanese fibration. There is an easy and ahard case. By Noether’s formula

12− 12q = K2 + (2− 4q + b2)

so K2 = 10 − 8q − b2. In particular, K2 < 0 unless q = 1 and b2 = 2. It is impossible that b2 < 2, as thefiber of the Albanese fibration and an ample divisor must be different classes.

Case 1: Suppose K2 < 0. By Case 1 of the CRC, there are irreducible curves C such that C.K < 0 and|C +K| = ∅. By Riemann-Roch,

h0(−C) + h0(C +K) = 0 ≥ g(C)− q

hence q ≥ g(C). If q > g(C), then C is contained in a fiber F and so C2 ≤ 0. Since C.K < 0, minimalityimplies C2 = 0. By connectedness of the fibre, F = nC. Since F has nonnegative genus, F = C. Thus,F ∼= P1 so by Noether-Enriques, X is ruled.

So assume that q = g(C) and C is not contained in a fiber. Then C maps to B. If q > 1 then this map isan isomorphism, and if q = 1 it is an etale cover. In either case C is smooth. Now, we claim that there is infact such an irreducible curve C with C.K < −1. Furnish an effective divisor D such that D.K < −n and|D + K| = ∅. This is possible by multiplying C by an integer and adding multiples of K until adjunctionterminates. We suppose in contradiction of the claim that D = C1 +C2 with Ci.K = −1 and Ci irreducible.Then, as before

0 = h1(D) + g(D)− q = h1(D) + g(C1) + g(C2) + C1.C2 − 1− q = h1(D) + C1.C2 + (q − 1).

If C1 = C2 then g(C1) = q implies C1.C2 > 0, a contradiction. So suppose Ci are distinct. Then, C1.C2 = 0,q = 1, and h1(D) = 0. Hence, there is an exact sequence

0→ OX(−D)→ OX → OC1 ⊕OC2 → 0.

12

The long exact sequence gives h0(OX) = h0(OC1) + h0(OC2), a contradiction. Thus, we conclude that thereis an irreducible curve C with C.K < −1 and |C +K| = ∅.

Applying Riemann-Roch to C, we get

h0(C) ≥ 1− q +1

2(C2 − C.K) = g(C)− q − C.K ≥ 2.

In particular C moves. Since C is a section for q > 1, as it moves in a smooth fiber it cuts out P1. Ifq = 1, then since the map is etale, and the number of inverse images of a point is always equal to C.F . Inparticular, as C moves in a smooth fiber, it cuts out an unramified cover of P1. But since the fibers of Albare connected, it in fact just cuts out P1. So in either case, by Noether-Enriques, X is ruled.

Case 2: Suppose K2 = 0. Then pg = 0, q = 1, and b2 = 2. As a preface, the structure of the proofis as follows: We assume that the fibers of Alb have genus greater than zero, as otherwise X is ruled overan elliptic curve. In this remaining case we prove P12 6= 0. More exactly that P4 6= 0 or P6 6= 0, which isslightly stronger.

We begin with a topological lemma. Let C be an irreducible curve. Then 2χ(OC) ≤ χtop(C) with equalityif and only if C is smooth. Equality holds for smooth curves by Riemann-Roch and Serre duality. We haveexact sequences

0→ CC → N∗CC → F→ 0

0→ OC → N∗OC → G→ 0

for skyscraper sheaves F → G. The map is injective since a section coming from N∗CC and OC must be inCC . (We could also note that F counts only number of inverse images, but G counts them with multiplicity.)Thus

χtop(C) = χtop(C) + h0(F)

χ(OC) = χ(OC) + h0(G)

Therefore 2(χ(OC) + h0(G))− (χtop(C) + h0(F)) = 0. So 2χ(OC)− χtop(C) = h0(F)− 2h0(G) < 0.And another topological lemma. Suppose we have a fibration X → B. Then we claim that

χtop(X) = χtop(B)χtop(F ) +∑

χtop(Fs)− χtop(F )

where Fs are the singular fibers and F is a nonsingular fiber. Reworking the righthand side, letting U be theopen subset of B with nonsingular fibers, we get χtop(U)χtop(F ) +

∑χtop(Fs). Whence the formula follows

from the long exact sequence of a closed subset, combined with the fact the the Euler characteristic of asmooth fibration is the product of the Euler characteristics.

Let F be a fiber of Alb. We claim that if g(F ) > 1 then all fibers are smooth, and if g(F ) = 1 then theonly singular fibers are multiple fibers. Suppose we have F = A + B. Let H be ample. Then since b2 = 2,we know that A ≡ aF + bH for a, b ∈ Q. Since A.F = 0, we have b = 0. Therefore A.B = 0. Hence A andB cannot contain distinct curves. From this we conclude F is irreducible. Say F = nC. Then, by our firsttopological lemma,

χtop(C) ≥ 2χ(OC) = −C2 − C.K =1

nF.K =

1

nχtop(F ).

By our second topological lemma,

0 = χtop(X) =∑

χtop(Fs)− χtop(F ) ≥∑(

1− 1

n

)χtop(F ).

Equality is necessarily the case, as the RHS is nonnegative. Thus C is smooth. Furthermore, there are nomultiple fibers when χtop(F ) > 0, i.e. g(F ) ≥ 2.

Let’s examine the case g(F ) = 1 more closely. Note that K.F = 0 so K ≡ aF for some rational numbera. If a > 0 then κ 6= −∞. (In fact κ = 1.) If a < 0 then H.K < 0. So adjunction terminates for H, and weget that X is ruled. Thus, the only case we need consider is K ≡ 0. Choose generators C and H of NS(X),with C ≡ aF for some a ∈ Q, and C.H > 0. By Poincare duality, we must have H.C = 1. By Riemann-Roch

h0(E) + h0(K − E) ≥ 1

2E2.

13

Set D = H − ( 12 (H.H) − 1)C. Then D2 = 2 and RR implies D is effective. (−D is ineffective because it

dots with C negatively).The genus formula gives g(D) = 2. So D has exactly one component A which maps to B. If D = A+mC,

then since A2 ≥ 0, we must have m = 1 and A2 = 0. Hence g(A) = 1 and the map A → B is etale. Thisis impossible unless there are no multiple fibers. So assume that D is irreducible. Also, D is smooth, asotherwise its normalization is an elliptic curve, which when we composed with the map down to B would failto be etale. Thus, the ramification type of D → B must be 2, 2 or 3. Furthermore, whenever D passesthrough a multiple fiber with multiplicity m, it must ramify to order that m divides. Hence, the possiblesets of multiple fibers are 2, 2, 2 or 3. In any case, we can base change X to a genus two curve B′

that ramifies over B at exactly its multiple fibers.This base change X ′ is etale over X so we still have K ≡ 0. Base change again to B′′ which etale covers

B′ and is genus 3. We get X ′′ such that K ≡ 0, and χtop(X′′) = 0. Hence χ(OX′′) = 0 by Noether’s formula.

On the other hand, we can pull back the three one-forms from B′′, so q(X ′′) ≥ 3. This is impossible becauseit implies χ(OX′′) < 0. Excuse the speed. The moral of the story is this: We may assume that every fiber issmooth and nonmultiple, even in the case g(F ) = 1. Furthermore, even if there are multiple fibers then 4Khas a section (since the base change to X ′ was at most degree 4).

Given a smooth fibration X → B of genus g curves, there exists a so-called “period map” from

B → Sp2g(Z)\Hg

which associates to each point the period matrix of its fiber. The map is holomorphic, and Hg is biholomorphicto a bounded domain in Cg(g+1)/2. In our case, B is an elliptic curve. Then, we get a map between theuniversal covers C → Hg, which must be constant by Liouville’s theorem. Combining with the local Torellitheorem, we conclude that every fiber is isomorphic. We call such a family isotrivial. Such a family is(analytically) locally trivial. So there is a monodromy map π1(B) = Z⊕ Z→ Aut(F ). Whenever g(F ) > 1,this automorphism group is finite. Hence, we can base change to X ′ = B′ × F .

Suppose that g(F ) = 1. It is easy to show that for any elliptic curve, the automorphism group is theextension of a cyclic rotation group of order 2, 3, 4, or 6 by a translation group. Therefore we can basechange to a surface X ′ which is topologically a torus, such that the degree of the resulting etale coverdivides 4 or 6. This implies K ≡ 0, so κ(X ′) = 0. The result of the following section shows that X ′ is anabelian variety. Then, we have a trace map Ω1(X ′) → Ω1(X) inverse to pullback. So, we have a splittingB = Alb(X)→ Alb(X ′) = X ′. Thus X ′ = B′ × F , as desired. Suppose the monodromy for X is

ρg(b) = ag(b).f + tg(b)

where ag ∈ Aut0(F ) and tg is a translation. Amazingly, by a little trickery, another etale cover gets rid ofthe translations! Using Riemann-Roch on sums of fibers, there is a morphism ρ : B → F such that

ρ(gb)− ag.ρ(b) = ntg(b).

Then, we may take the automorphism u of B × F , defined by u : (b, f) → (b, f − ρ(b)). The group G canthus be chosen to act independently on B′ and F in all cases.

At this point, let us note we already have κ 6= −∞, since there is an etale cover X ′ = B′ × F . More,we can explicitly compute that P4 6= 0 or P6 6= 0. There is an isomorphism between G-invariant forms onB′ × F and forms on X. Thus, we get

0 = H0(KX) = H0(B′,KB′)G ⊗H0(F,KF )G = g(F/G).

Hence we may assume F/G ∼= P1. Then, since all pluricanonical forms on B′ are translation invariant, wealso get

Pk(S) = H0(F,K⊗kF )G = H0(F/G,O(n))

where n = −2k +∑⌊

k(

1− 1eP

)⌋. Using standard Riemann-Hurwitz inequality-style arguments, we get

P4 6= 0 or P6 6= 0. Thus P12 6= 0 in any of these cases. Furthermore Pn →∞ unless X is bielliptic.Side note: We in fact neglected to show in the case with multiple fibers where K ≡ aF , a > 0 that

P12 6= 0. There will be an ADDENDUM. All cases with multiple fibers follow relatively easily from thecanonical bundle formula.

14

1.6 Surfaces with κ = 0

We begin by classifying such surfaces. Note that if κ = 0, 1 then K2 = 0. This is because asymptotics onRiemann-Roch applied to −nK shows that κ = 2 when K2 > 0 (and X is not rational). If is in fact truethat K ≡ 0 when κ = 0! First, we note that the set of n such that Pn = 1 are the positive multiples of someinteger. (By the abundance theorem, this number is 1, 2, 3, 4, or 6.) This is because a section in mK tothe nth power must equal a section in nK to the mth power. So there is combination which is a section ingcd(m,n)K.

Applying Noether’s formula gives 10− 8q + 10pg = h1,1. And pg = 0, 1. Hence there are five cases:

(Case I) pg = 0, q = 0. Called Enriques. 2K = 0.

(Case II) pg = 0, q = 1. Called bielliptic or hyperelliptic. 4K = 0 or 6K = 0.

(Case III) pg = 1, q = 0. Called K3. K = 0.

(Case IV) pg = 1, q = 1. Impossible.

(Case V) pg = 1, q = 2. Called Abelian. K = 0.

Case IV: Impossible. For every divisor, there is a torus of divisors numerically equivalent to it. We claimthat every divisor D numerically equivalent to K is effective. This is because h0(D) ≥ 1 by Riemann-Roch.Choose E so that 2K = D+E. Then D and E are both effective. But since 2K has only one section, therecan only be finitely many expressions of 2K as a sum of effective divisors. Contradiction.

Case II: To avoid circularity, we must prove Case V independently of this case. The abundance theoremclassified these surfaces as smooth elliptic fibrations over an elliptic curve with either a degree 4 or 6 coverequal to the product of two elliptic curves, where the group acts via translations on the base and nontrivialautomorphisms on the fiber. Here either 4K or 6K is trivial. We can explicitly write down all such examplesE × F/G. Firstly, the group G is a subgroup of the translations of E, so is abelian and generated by twoelements. Secondly, the group G must act nontrivially on F . There are only seven possibilities (we write theaction on F after the group is given):

1. F arbitrary, G = Z/2Z, where x 7→ −x

2. F arbitrary, G = Z/2Z⊕ Z/2Z, where x 7→ −x and x 7→ x+ ε

3. F = C/Z[i], G = Z/4Z, where x 7→ ix

4. F = C/Z[i], G = Z/4Z⊕ Z/2Z, where x 7→ ix and x 7→ x+ ε

5. F = C/Z[ρ], G = Z/3Z, where x 7→ ρx

6. F = C/Z[ρ], G = Z/3Z⊕ Z/3Z, where x 7→ ρx and x 7→ 1−ρ3

7. F = C/Z[ρ], G = Z/6Z, where x 7→ −ρx.

These are easily computed to be the only possibilities because the “special” automorphisms of above ellipticcurves commute with very few translation groups.

Case III: Applying Riemann-Roch to 2K, we get h0(2K) + h0(−K) ≥ 2. Hence h0(−K) = 1. Sincealso h0(K) = 1, we get K = 0.

Case I: By CRC, we must have P2 6= 0. By our gcd observation, since K has no sections, 3K has nosections. Thus, h0(3K) + h0(−2K) ≥ 1 implies h0(−2K) ≥ 1. Similarly to case III, 2K = 0. In fact, thesection of 2K gives rise to a multivalued section of K, which describes a covering map. In particular, X isdouble-covered by a surface with K = 0, which must be K3, since the topological Euler characteristic of anEnriques surface is 12, so its cover has nonzero Euler characteristic.

Case V: Proving that such a surface is abelian is tricky. The key is the Albanese map, which we wouldlike to show is an isomorphism. First, we make two long overdue observations: (i) the intersection matrixof a connected fiber is negative semi-definite with kernel spanned by F , and (ii) if a map between surfacescontracts some curve, then any linear combination of those curves has negative self-intersection. This firstis easy linear algebra, and the second follows from considering the pullback of a hyperplane passing through

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p. Suppose that K 6= 0. Then K =∑niCi. By minimality Ci.K ≥ 0. But K2 = 0, so in fact Ci.K = 0.

Then, either C2i = −2 and Ci intersects some other component of K OR C2

i = 0 and Ci does not intersectany other Cj . Thus, K is the disjoint union of (1) genus one curves, and (2) intersecting smooth rationalcurves, and each of these components has self-intersection 0.

Suppose that the image of Albanese is a curve B. It is smooth of genus 2, so all disjoint curves D of Klie in fibers Fi. Since the fibers of Albanese are connected, by observation (ii), D ≡ aF for a ∈ Q+ becauseD2 = 0. Hence, K ≡ aF for a ∈ Q+. This contradicts κ = 0. Now assume that Alb surjects. By observation(ii), every component of K is not contracted to a point. Since there are no rational curves on Alb(S), weconclude that K has no rational components, only elliptic curves. Furthermore, these elliptic curves mustmap to smooth elliptic curves in Alb(S). Composing with a translation, we can always move such a curveto an Abelian subvariety. Passing to the quotient contracts this smooth elliptic curve, so as before, we haveK ⊃ aF and |nK| → ∞.

Hence, we may assume that K is trivial. Suppose again that the image of Alb is a curve of genus 2.There is an unramified of this genus 2 curve by a genus 3 curve. The pullback of K is again trivial, andthe Euler characteristic is still 0. But this is a contradiction to Noether’s formula, as we can pull back 3one-forms from the new base.

Thus, the only possible case is when K is trivial and Alb surjects. Take a basis η1, η2 of one-forms onAlb(X). Then Alb∗η1∧Alb∗η2 generically does not vanish, since Alb is generically etale. Since K is trivial, itin fact vanishes nowhere, and one sees this implies that in fact Alb is etale. Since the only covers of Abelianvarieties are Abelian varieties, we’re done.

Let us note that the assumption κ = 0 was unnecessarily strong. In Case III, we only needed that P2 = 1.In Case I, we needed that P2 = 1 and P3 = 0, or equivalently P6 = 1, because then CRC still applies to showthat P2 = 1. Then, if P3 were 1, we’d also have pg = 1, a contradiction. In Case II, if we assume P12 = 1,we conclude that X is bielliptic by a more careful examination of the Riemann-Hurwitz calculation, and ourproof that P12 6= 0. Case IV is still impossible if P2 = 1. The Case V follows from the fact that when P2 = 1,we can construct an elliptic fibration of elliptic curves and apply the theory of pg = 0, q = 1. More detailslater.

Let’s start on the EXAMPLES!Abelian surfaces. Riemann’s bilinear relations Ω = Ωt and im Ω > 0 give a condition for algebraizing

a torus. An example of an Abelian surface is the Jacobian of a genus two curve.K3 surfaces. By the adjunction formula, the only complete intersections with trivial canonical bundle

are S2,2,2 in P5, S2,3 in P4, and S4 in P3. By the Lefschetz hyperplane theorem, they’re all simply connected,so q = 0. Easier, one can simply examine the long exact sequence to prove the hyperplane theorem in thecase hi(O).

Another construction is as follows: Take an Abelian surface A. Let τ be the involution sending x 7→ −x.Blow up at the 16 fixed points of τ , and then mod out by τ . The resulting surface S is nonsingular: Thelocal coordinates on the blow up are x, t where t = y/x. So, τx = −x and τt = t. Thus (x2, t) are localcoordinates on S. Take a two form dx ∧ dy. It is invariant under τ , so it gives a meromorphic two form onS. Its pullback to the blow-up is xdx ∧ dt = d(x2) ∧ dt. Thus it descends to a non-vanishing holomorphictwo-form on S. So KS = 0. Finally, note that S is not an Abelian surface, because it has no one-forms: Ifit did, it would pull back to a τ -invariant one-form, which would descend to A. This is impossible since theone-forms on A are spanned by dx and dy.

There are plenty more K3 surfaces. In fact, one can construct a degree 2g − 2 K3 surface in Pg for allg ≥ 3. Suppose that S is a K3 surface with a smooth genus g curve C. Then, C2 = 2g− 2 since C

∣∣C

= KC .By the exact sequence

0→ H0(S,OS)→ H0(S,OS(C))→ H0(C,KC)→ 0

we get h0(C) = g + 1. Furthermore, the system |C| is without base points for g(C) ≥ 1 because it restrictsto |KC |, which has no base points on C. When g(C) = 2 the map φ defined by |C| is a degree 2 map toP2 whose branch locus is a sextic. Since C2 = 2, we know that φ is degree 2. Let the branch locus be ∆.Let C = φ−1(`). Then, C → ` is branched at ` ∩ ∆ and so `.∆ = 6, whence n = 6. So we have anotherconstruction of a K3 surface as the double cover P2 branched over a smooth sextic.

When g ≥ 3, either φ is a birational morphism or a 2-to-1 morphism depending on whether or not Cis hyperelliptic. To see why, note that φ restricts to the canonical map on any smooth curve in |C|, so iseither an isomorphism or a double cover. From this we see that all generic curves in |C| are of the same

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type (depending on the degree of the morphism). Finally, note that when C is hyperelliptic, the hyperplanesections of the image of φ are the images of curves in |C| so are rational curves. Also, the image is a degreeg − 1 surface. This implies that the image of φ is rational.

In the non-hyperelliptic case:When g = 3, we have that φ(S) is degree 2g − 2 = 4, so is a quartic surface.When g = 4, we have that φ(S) is degree 6. Note that if |D| has a smooth representative, then h1(−D) =

h1(D) = 0 because of the exact sequence. Hence h0(2C) = 14 and h0(3C) = 29. Since h0(OP4(2)) = 15,φ(S) lies in a quadric. Also, h0(OP4(3)) = 35 so φ(S) also lies in a cubic! By degree reasons, φ(S) must bethe intersection of these two.

When g = 5, we have that φ(S) is degree 8. Generically, it is the intersection of three quadrics.Enriques surfaces. These are little bit difficult to give examples of. We use the fact that the intersection

of three quadrics in P5 is a K3 surface S. Suppose the three quadrics are of the form

Qi(x0, x1, x2) +Q′i(x3, x4, x5)

for i = 1, 2, 3 where the Qi and Q′i don’t intersect each other in P2 (which is generically the case). This K3has an involution

τ : (x0, x1, x2, x3, x4, x5) 7→ (x0, x1, x2,−x3,−x4,−x5)

whose fixed locus is the union of two planes x0 = x1 = x2 = 0 and x3 = x4 = x5 = 0. These two planes donot intersect S because we chose the quadrics to not intersect on them. Hence τ has no fixed points on S,and therefore S/τ is an Enriques surface. Apparently a generic Enriques surface arises in this manner.

1.7 Surfaces with κ > 0.

The case κ = 1 is relatively well classified because elliptic fibrations are well-understood, though it isdifficult. The case κ = 2 is called general type and little is known about these surfaces, but we do have theCastelnuovo-de Franchis theorem.

We claim that if X is a non-ruled surface and K2 > 0, then φnK is eventually a birational map to itsimage. To see why, first note that −nK has no sections, because some nK has a nontrivial section (and thusH.K > 0). By Riemann-Roch, h0(nK)→∞ quadratically, so κ = 2. Furthermore, h0(nK −H)→∞ too.Choose E ∈ |nK − H|. Then |nK| separates points and tangents away from E. Thus, φnK is birational.Conversely, if K2 > 0 and S is irrational, then since K2 = 8− 8g for a (minimal) ruled surface, we concludethat S is not ruled, and therefore h0(nK) grows quadratically. Combining with previous results gives thefollowing table:

κ = −∞ P12 = 0 K2 = sign(χ(B)) Ruled or rationalκ = 0 P12 = 1 K2 = 0 Enriques, abelian, bielliptic, or K3κ = 1 P12 > 1 K2 = 0 Elliptic fibration (none of the above)κ = 2 P12 > 1 K2 > 0 General type

Furthermore, when K2 = 0 if we write nK = Z +M where Z is the fixed part and M is mobile, then

K2 = Z2 = M2 = K.M = Z.M = K.Z = 0.

Note that M is nef, since it is mobile. If Z0 ⊂ Z satisfies Z0.K < 0 then Z0.Z < 0 =⇒ Z20 < 0, which

contradicts minimality. Thus, Z.K ≥ 0. But also M.K ≥ 0. Because K2 = 0, we get Z.K = M.K = 0.Dotting the defining equation with M shows that M.Z = M.M = 0. Finally, squaring the defining equationgives Z2 = 0.

In the case κ = 1 eventually |nK| has a mobile part. By the lemma, |M | defines a morphism to Pn, asM2 = 0. Furthermore, this condition says that the sections |M | are fibers of the mapping. (The map mustbe to a curve anyways since κ = 1.) Normalize the image and Stein factorize. Since g(M) = 1, all surfaceswith κ = 1 are elliptic fibrations.

There are lots of surfaces with κ = 1, for example the E × F with E an elliptic curve and g(F ) ≥ 2, orquotients of these. Other examples are a divisor of OB(D)⊗OP2(3) in B × P2, where |D| is base-point freeand deg(D) > 2− 2g(D). One can check that projection onto B is an elliptic fibration.

Let us prove one theorem that actually says something about surfaces of general type:

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Theorem 1.7.1. Let S be non-ruled. Then χ(OS) ≥ 0, and if S is of general type, χ(OS) > 0.

Proof. The value of K2, along with χtop(S) ≥ 0 implies this theorem. Suppose our surface has a fibration.Let Fs =

∑niCi be a singular fiber. Then,

χtop(S) = χtop(B)χtop(F ) + χtop

(∑Ci

)− χtop(F )

≥ χtop(B)χtop(F ) + 2χ(O∑Ci

)− 2χ(OF )

= χtop(B)χtop(F )−(∑

Ci

)2−(∑

Ci.K)

+ F.K

≥ χtop(B)χtop(F ) +∑

(ni − 1)Ci.K

≥ χtop(B)χtop(F ).

where the second-to-last inequality follows from the fact that the intersection form on the curves of a fiber isnegative semi-definite. Note that Ci.K = −2 and C2

i = 0 is impossible because then Fs = P1, contradictingthat Fs is singular. Since C2

i ≤ 0, we have by minimality that Ci.K ≥ 0, whence the last inequality. Evenif non-minimal we can reduce to relatively minimal, and the Euler characteristic increases with blow ups.

We will give a proof by contradiction. That is, we assume that χtop(S) < 0 and derive a fibration whichcontradicts the assumptions. We have that 2− 4q+ b2 < 0, hence q ≥ 1. Thus, there are etale covers of anydegree. Choose a cover S′ with χtop ≤ −6. Then,

2− 4q + 2pg ≤ −6,

so pg ≤ 2q− 4. Note that etale covers can only possibly increase q, since there is the pullback map. We now

apply a linear algebra fact: If φ :∧2

V →W and dimW ≤ 2 dimV −4, then ther exist linearly independentv1, v2 such that φ(v1 ∧ v2) = 0. This implies that there are two one-forms with ω1 ∧ ω2 = 0. They must bemultiples of one another, since Ω1 is rank 2! Thus, ω1 = gω2 with g ∈ K(S′). Furthermore, dωi = 0 becauseis it a (2,0)-form such that ∫

dωi ∧ dωi =

∫d(ωi ∧ dωi) = 0.

Thus, dg ∧ ω2 = 0. Hence, ω2 = f dg for f ∈ K(S′). Now, dω2 = 0 implies df ∧ dg = 0.Consider the map φ = (f, g) : S → P1 × P1. At any a point in A1 × A1, the two local one-forms dx

and dy pull back to df and dg, whose wedge is zero. Thus, φ is not generically etale. Therefore the imageis a curve C. Possibly replacing with the normalization and Stein factorizing, we may assume a fibrationwith connected fibers S → C. Consider the meromorphic forms x dy and xy dy on C. Their pullbacks aref dg = ω2 and gf dg = ω1. Furthermore,

div(φ∗α) = φ∗(div(α)) +∑

(ni − 1)Ci

so α has no poles (even where the inverse image of a point is a singular fiber). Hence, C has two distinctholomorphic one-forms, implying g(C) ≥ 2. By the inequality

0 > χtop(S′) ≥ χtop(B)χtop(F ),

we see that χtop(F ) > 0. This is of course only possible when F = P1, hence S′ is ruled. But then S is alsoruled, a contradiction.

Corollary 1.7.1. For any nonruled surface S, we have pg > 2q − 4.

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2 Exercises

2.1 Beauville: Chapter II

Exercise 1: Resolve a curve lying in a smooth surface. We have that C is singular if mp(C) > 1 for some

point p ∈ C. Taking the strict transform, C2 = C2 −m2 and C.K = (ε∗C −mE).(ε∗K + E) = C.K + m.Hence g(C) = g(C)−

(m2

). Thus, genus strictly decreases each time we blow up a singular point. Since this

must stabilize, eventually the strict transform is smooth.

Exercise 2: We havem = C.E =

∑mx(C ∩ E) ≥

∑mx(C)

since the intersection with E may not be generic (if it is, then we have equality). Note that in a smallneighborhood,

mp(C ∩ C ′) = ε∗C.ε∗C ′ = (C +mp(C)E).(C ′ +mp(C′)E) = mp(C)mp(C

′) + C.C ′.

Thus, continuing inductively on the proximate points of higher and higher orders, we get

mp(C ∩ C ′) =∑x

mx(C)mx(C ′)

where the sum ranges over all infinitely near points of p, including order zero. Let N be the normalizationof C. By Exercise 1, we get

g(C) = g(N) +∑i

(mi

2

)where i ranges over all infinitely near points.

Exercise 3: Suppose that an irreducible curve C ⊂ S is contracted to a point by S 99K S′. By birationalfactorization, the strict transform of C must be the strict transform of some exceptional divisor. The stricttransformation must necessarily desingularize C. Then,

C2 = C2 −∑

m2i = −1− n =⇒ C2 =

∑m2i − 1− n.

Similarly,

C.K = C.K +∑

mi = −1 + n =⇒ C.K = −∑

mi − 1 + n.

Furthermore, if C is a rational curve such that C2 ≥∑m2i − 1, we may blow up its singularities to get a

smooth rational curve with C2 ≥ −1. Composing with some more blow ups if necessary, we will eventuallybe able to blow down C. Thus C is of the second kind.

2.2 Beauville: Chapter III

Exercise 1: Let F be a fiber of a geometrically ruled surface. Then F 2 = 0. Blowing up F at a point s,we get F 2 = −1, so F is contractible. Once we contract, we again have a geometrically ruled surface. Theresulting birational map S 99K S′ is called an “elementary transformation.”

Exercise 2: This question doesn’t make sense, since E and C(s) as defined are bundles on different spaces.But, given exercise 1, there is a bundle E′ whose projectivization is S′. So we may ask, how are E and E′

related? The answer is that E′ is the subsheaf of E whose sections pass through s. More formally, we havea map E → p∗C(s)→ 0 by sending a section to its coset in the fiber at p. The kernel of this map is E′.

Exercise 3: Consider two minimal ruled surfaces φ : X 99K S. We claim that φ factors as a series ofelementary transformations (and an automorphism first to make sure the restriction to the base is trivial).Note that this gives an alternate proof of the fact that the minimal ruled surfaces are projectivizations ofrank 2 bundles. Let n(φ) be the minimal number of blow-ups necessary to make φ everywhere defined. We

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show that there is an elementary transformation such that n(t φ) < n(φ). As usual, consider the last blowup’s exceptional curve E. Then E must map to a fiber of S. We first claim that some point of E must lie ona contracted curve. Is so, because E2 = −1 while F 2 = 0, so the strict transform of F cannot be isomorphic.Denote the image of this point by q. Then, the morphism X → S factors through S blown up at q by theuniversal property. Then, contract the strict transform of F . Since E is contracted in the new map, by thereverse universal property, we didn’t need the last blow up. Hence n(t φ) < n(φ). Assuming S = C × P1,exercise 2 shows that every minimal ruled surface is the projectivization of a rank 2 vector bundle! Further-more, the birational transformations of ruled surfaces are generated by the elementary transformations.

Exercise 4: Note that every birational automorphism of a ruled surface over C 6= P1 must map fibers tofibers, since the induced birational map P1 → C can be completed and is therefore constant. Hence there is amap Autb(S)→ Aut(C). It is surjective, because by Noether-Enriques, there is an open subset of S isomor-phic to U × P1 on which we can extend an automorphism of the base; this gives the splitting. The kernel isthe maps with are identity on the base. So we need only consider automorphisms preserving the base, whichare birational automorphisms of C × P1 preserving C. Such things are just rational maps C 99K PGL2(O)(or one can think of transition functions U → PGL2(O)), which is PGL2(K). P.S. This is kinda BS-y, notsure how to rigorize, except that we know transition functions are elements of H0(U,PGL2(O)).

Exercise 5: An elementary transformation of Fn gives either Fn−1 of Fn+1, because a fiber intersects Bat 1 point so the elementary transformation at this point decreases the self intersection of B by 1 while theelementary transformation away from this point increases it by 1. Since B is the unique irreducible curve ofnegative self-intersection, and Fn is characterized by this self-intersection, we get the desired result, and amethod of distinguishing what the elementary transformation is.Exercise 6: The surface F1 contained B which is exceptional because B2 = −1 and it is a smooth sphere(one can also just check that B.K = −1.)

Exercise 7: This problem gives an alternate proof of which rational surfaces are minimal. Suppose we havea birational map φ : S 99K Fn. As usual, we consider the last blowup E. Then E is the strict transformof some curve with non-negative self-intersection. As in the proof of exercise 3, we can apply an elementarytransformation so that we didn’t need the last blow-up. Then, we can factor φ with elementary transforma-tions until it is a sequence of blow-ups followed by an automorphism. Since the only Fn with an exceptionalcurve is F1, the only case where we have any blow ups is P2.

Exercise 8: As noted, two complex line bundles on C are smoothly isomorphic if and only if they have thesame degree. We claim that Lp ⊕ Lq ∼= Lr ⊕ Ls smoothly whenever p+ q = r + s. Tensor with a degree −pline bundle to get O ⊕ Lr+s ∼= Lr ⊕ Ls. Note that smoothly Lr ⊕ Ls has a nonvanishing section, as we cantake a section of Lr which vanishes at r points and add little bump functions on Ls whenever it does. Then,we ask what the degree of the quotient bundle is. It must be r + s since degree is well defined.

Exercise 9: Consider the map to P1 × P1 defined by the pencils of planes through ` and `′ respectively.

Exercise 10: Suppose that every point on S ⊂ P3 lies on a line in S.

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3 Hartshorne

Just for fun, what is the canonical bundle on X = Pn? To find it we take the sequence

0→ O → O(1)⊕(n+1) → TX → 0

gotten by including the tautological bundle into Cn+1. Dualizing and taking determinants, we get thatKX = O(−n− 1) as desired.

Let’s prove finite-dimensionality of cohomology for a projective variety X. We consider a coherent sheafF on X. Embed X into Pn, and push the sheaf forward, reducing to the case X = Pn. We claim that F isa quotient of a sum of line bundles.

Consider the affine open Ui. On this open, F restricts to some finite-dimensional M , with generatorsmij . We claim that for xnimij extends to a section on all of X. This is true, if think about it, because inanother chart, the domain of non-definition may be where xi = 0, but then it is a section localized at xi,implying there is some power moving it into a global section. Choosing n large enough, we get that F (n) isgenerated by global sections on all the Ui, hence on all of Pn. So

Or → F (n)→ 0.

Twisting, we get 0 → G → O(−n)r → F → 0. The kernel G is also a coherent sheaf. Taking the longexact sequence, and knowing the cohomology of O(−n), we conclude that the cohomology of F is finite-dimensional whenever in higher degree, the cohomology of G is finite-dimensional. Thus, be descendinginduction, we need only show that some high cohomology of a coherent sheaf is 0. This is true becausecohomology is computed by taking an affine open cover. Hence, in fact all cohomology of coherent sheaveson Pn is finite dimensional, and only appears in degree n or less.

Now, we’d also like to show that cohomology vanishes past the dimension. This is a general fact aboutNoetherian topological spaces of dimension n and sheaves of abelian groups on them.

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