SURFACES NOTES Ravi Vakil Taught a Course On

SURFACES NOTES

AARON LANDESMAN

1. INTRODUCTION Ravi Vakil taught a course on algebraic surfaces at Stanford in Winter 2019. These are my “live-TeXed“ notes from the course. Conventions are as follows: Each lecture gets its own “chapter,” and appears in the table of contents with the date. Of course, these notes are not a faithful representation of the course, either in the mathematics itself or in the quotes, jokes, and philosophical musings; in particular, the errors are my fault. By the same token, any virtues in the notes are to be credited to the lecturer and not the scribe.1 Please email suggestions to [email protected]

1This introduction has been adapted from Akhil Matthew’s introduction to his notes, with his permission. 1 2 AARON LANDESMAN

2. 1/6/20 This class will vaguely follow Badescu’s book on algebraic surfaces. We’ll roughly linearly move through the book. One thing Ravi is interested in is understanding why proper surfaces are projective. Ravi is also interested in understanding the classification of algebraic surfaces in light of the minimal model program. We may have student lectures on topics like (1) Tsen’s theorem, (2) resolution of singularities for surfaces (in char 0) (3) bend and break. Our assumptions throughout the class is that we’ll be working over a field k, and consider separated finite type schemes over k. We’d like to have as few black boxes as possible. The first few weeks will focus on (1) Snappers theorem (with intersection theory in the proper, not necessarily projective case). (2) Nakai-Moishezon (without projective hypotheses) (3) Zariski’s theorem that all proper smooth surfaces are projective. For references, we’ll follow Badescu, but also see Kollar’s´ appendix to rational curves on algebraic varieties and Kleiman’s “Towards a numerical criterion of ampleness” and maybe FGA explained.

2.1. Snapper’s theorem. Let F be a coherent sheaf on a proper scheme X (over a ﬁeld k).

Theorem 2.1 (Snapper’s theorem). Given L1, ... , Lt line bundles on X, the Euler characteristic

n1 nt χ(F ⊗ L1 ⊗ · · · ⊗ Lt ) t is a polynomial in n1,..., nt from Z → Z of total degree at most dim Supp F . Remark 2.2. The key ideas of the proof is Devissage (a kind of Noetherian induction). Suppose some of the coherent sheaves on X are “good,” meaning they satisfy the theorem statement. Here, “good” sheaves play well in exact sequences (so if two out of three of them are good, then the third is) and for all irreducible subvarieties V then OV is good. We want to show all sheaves are good. The proof is now by induction on the dimension and length of the support of the coherent sheaf. One can reduce to the case that the scheme V is irreducible. One can then further reduce to the case V again using induction on the dimension and the length at the generic point of the support. SURFACES NOTES 3

Finally, we arrive at the last step. We need to show that if F is scheme the- oretically supported on an integral subvariety V ⊂ X of rank r at the generic point. We want to reduce to a rank r − 1 sheaf. This can be accomplished in the projective case by sufﬁciently twisting up until we get a section. For the reduction to the proper case, we can use Chow’s lemma.

n1 nt Proof. We say a sheaf F is good if χ(F ⊗ L1 ⊗ · · · ⊗ Lt ) is a polynomial of degree at most dim Supp F . n1 n2 Suppose we are given F , L1, ... , Lt on X. Let χ(F ⊗ L1 ⊗ · · · ⊗ Lt ) is a polynomial of degree ≤ dim Supp F . When the dimension of the support of F is 0, we are certainly good. Also, n1 nt OV is good for V irreducible. On V, χ L1 ⊗ · · · ⊗ Lt is a polynomial. This can be seen in the projective case because every L is a difference of two very ample line bundles, and we can reduce to the very ample case. We assume V is integral. We can next reduce to the case V is normal by replacing V by its normalization. (This is not working strictly in the context of subvarieties.) The reason is that χ(G ) for G a sheaf on the normalization ν : Ve → V then χ(ν∗G ) = χ(G ). So, now we may assume V is normal. We may take a rational section of L1. Given a rational section of L1 we get a Weil divisor D+ − D− on V. So L1 = O(D+ − D−). The plan is to reduce our statement to questions on D+ and D−. Example 2.3 (Differences of Weil divisors which are not Cartier may be Cartier). If we take a cone and the difference of two lines through the cone point, each line individually is a Weil divisor that is not Cartier, but the difference is Cartier. ∨ Note that L1 6= O(D+) ⊗ O(D−) in general but L1 ⊗ O(D−) = O(D+) and ID = O(−D) so L1 ⊗ ID+ ' ID− . We have two exact sequences (2.1) n1 nt n1 nt n1 nt 0 ID+ ⊗ L1 ⊗ · · · ⊗ Lt L1 ⊗ · · · ⊗ Lt ⊗L1 ⊗ · · · ⊗ Lt 0

(2.2)

n1−1 nt n1−1 nt n1−1 nt 0 ID− ⊗ L1 ⊗ · · · ⊗ Lt L1 ⊗ · · · ⊗ Lt OD− ⊗ L1 ⊗ · · · ⊗ Lt 0 We now see the ﬁrst two terms of the above exact sequences agree. The third terms are supported on lower dimensions, so are polynomials. There- fore, the difference of the two middle terms have Euler characteristic is 4 AARON LANDESMAN

a polynomial. So the Euler characteristic of the middle sheaf in the ﬁrst sequence is a polynomial of the desired degree. Remark 2.4. What happened in Snapper’s theorem that was different in the proper vs. projective case? We had to deal with subvarieties V that were proper, but not necessarily projective. Here we needed Chow’s lemma to make Devissage work. We also need to use this trick that line bundles are differences of divisors, and then induct on the support dimension using the above exact sequences.

Definition 2.5. We now define (L1 ··· Lt · F ) to be the coefficient of n1 ··· nt n1 nt in Snapper’s polynomial χ(F ⊗ L1 ⊗ · · · ⊗ Lt ). 3. 1/8/20 3.1. Review. Let’s begin by reviewing what we did last class. If X is normal, every line bundle defines a Weil divisor, and so every line bundle L can be written as D+ − D−, a difference of two effective divisors. Here are some facts on ampleness which we need. Here is the cohomological criterion for ampleness. Proposition 3.1. Assume X is proper. Then, if for all F is coherent on X, h1(F ⊗ L ⊗n) = 0 for i > 0 if n 0. Exercise 3.2. A line bundle L on a proper algebraic scheme X over a field k is ample if and only if L restricted to each irreducible component of Xred is ample. Hint: Use Devissage. Call a sheaf “good” if it satisfies the above cohomological criterion on all components. Then filter the sheaf by sheaves supported on each component. Another thing to remember is Lemma 3.3. If f : X → Y is finite, then L is ample on Y if and only if f ∗L is ample. We’ll also need Snapper’s theorem from last time, which says that χ(F ⊗ ⊗n1 nt L1 ⊗ · · · ⊗ Lt ) is a polynomial in n1, ... , nt of degree at most dim Supp F . We also defined an intersection product (L1 ··· Lt · F ) as the coefficient of n1 ··· nt in the above polynomial defined by the Euler characteristic of ⊗n1 nt χ(F ⊗ L1 ⊗ · · · ⊗ Lt ). Lemma 3.4. We have ∨ ∨ ∨ ∨ (L1 ··· Lt · F ) = χ(F ) − χ(F ⊗ L1 ) · · · − χ(F ⊗ Lt ) + χ(F ⊗ L1 ⊗ L2 )) ··· and this is symmetric and multilinear. SURFACES NOTES 5

Proof. The key trick for multilinearity is that (L · L 0 · F ) = (L · F ) + (L 0 · F ) − ((L ⊗ L 0) · F ), but we can assume the dimension of the support of F is 1 (by sticking in more line bundles) so that the left hand side is 0. 3.2. More useful facts. Suppose π : Y → X is surjective and generically ﬁnite. We have the pullback formula: ∗ ∗ deg π (L1 ··· Lt) = (π L1 ··· π Lt) . There is similarly a projection formula. 3.3. Nakai Moishezon. Here is the main result: Theorem 3.5 (Nakai Moishezon). Suppose X is proper and L is a line bundle on X. Then, L is ample if and only if for all integral subvarieties V ⊂ X, we have    

L ··· L  = L ··· L ·OV > 0. | {z } | {z } dim V V dim V X Proof. One implication is easy, so we will just focus on showing that if the restriction of L to each subvariety has positive self intersection, then L is ample. We can first reduce to the case X is integral, using Devissage, as we showed in an exercise above. Our goal is to show L satisfies the cohomological criterion for ampleness. It suffices to show L is ample on all integral subvarieties V ⊂ X. We can assume by induction on dimension that L is ample on all integral subvarieties V ⊂ X. Our goal is now to show the following. Suppose X is integral, L a line bundle, (L dim X) > 0 and L ample on all proper subvarieties, and F is coherent on X. We want hi(X, L ⊗n) = 0 for i > 0 and n 0. We can replace X by its normalization. Write L = O(D) and D = D+ − D− and O(−D−) = O(−D+) ⊗ O(D). We then have the same two short exact sequences as last class.

(3.1) 0 ID+ (mD) OX(mD) OD+ (mD) 0 (3.2)

0 O((m − 1)D)ID OX((m − 1)D) OD− ((m − 1)D) 0 For m 0 and i ≥ 2, we have an isomorphism i i H (OX(mD)) ' H (OX((m − 1)D)). ( ( )) Ddim X dim X We also know χ O mD is a polynomial in m with leading term (dim X)! m . 6 AARON LANDESMAN

One can see this using that if we consider χ(L n1 ··· L nt ) as a polynomial dim X of degree at most t, with coefficient of n! ··· nt given by D . We also have ! that χ(O(mD)) is a polynomial in m = n1 + ··· + nt, so the coefficient t times the coefficient of n1 ··· nk, using the binomial theorem for expanding t (n1 + ··· + nk) . Actually, we’ll revise the proof, and won’t demonstrate the cohomological criterion, but just produce the map to projective space. We know that for m 0 we have h0(mD) 0 (since hi for i ≥ 2 vanish as we have shown above). Hence, we may assume D is effective. We then have an exact sequence

(3.3) 0 O((m − 1)D) O(mD) OD(mD) 0. For m 0 using induction as D is smaller dimensional, we have

(3.4) H1(O(m − 1)D) H1(O(mD)) 0 0

so H1 eventually stabilizes, and hence h1 stabilizes as m → ∞. This means we have a surjection

0 0 (3.5) H (O(mD) H (OD(mD)) 0 0 We basically know some multiple of D is generated by global sections, which will give us a map to projective space. This proof will be continued next time.

4. 1/10/20 If X is normal, the Picard group is a subgroup of the class group. A Q- Cartier divisor only makes sense when X is normal (so that we have the above injection). A Q-divisor means a Weil divisor with Q coefﬁcients, i.e., an element of Cl(X) ⊗Z Q. Last time, we were working out the proof of Nakai-Moishezon for proper surfaces. The normality hypotheses are so that we can work with Q-Cartier divisors, and so that we can work with valuations along codimension 1 points. Recall the statement from last time: Theorem 4.1 (Nakai-Moishezon). Let L be a line bundle on a proper scheme X. Then, L is ample if and only if for all integral subschemes Y ⊂ X, L dim Y · Y > 0. Proof. We will sloppily use O(D) in place of L . So far, we have reduced to the case X is integral, using that to check if an invertible sheaf is ample, we SURFACES NOTES 7

can check this on each irreducible component. We want to show the direction proving that L is ample. Using induction, we saw we could assume that L is ample on every scheme of dimension smaller than X. Last time we saw that hi(X, L n) stabilizes in n for ﬁxed i ≥ 2 using two short exact sequences. So, we showed h0 − h1 tends to ∞ which means h0 tends to inﬁnity. This proves D is effective. Using that D is effective, we shoed h1 also stabilizes. The reason is that we take the long exact sequence associated to

(4.1) 0 O((m − 1)D) O(mD) O(mD)|D 0. Here, D is a fixed effective divisor, not just a divisor class. Because D is 1 smaller dimensional, we know H (D, O(mD)|D⊗) = 0 for m 0. So this implies H1(X, L m) stabilizes, because there are surjections H1(X, (m − 1)D) → H1(X, mD). We have (L ··· L ) mdim X | {z } X χ(L m) = dim X . (dim X)! We now arrive at the new part of the proof. Lemma 4.2. We have O(mD) is base point free for m 0. Proof. We’ll separately deal with the cases that a point lies on D or not on D. It’s certainly base point free away from D because there is a section vanishing exactly on D. 0 0 We know H (X, O(mD)) → H (D, O(mD)|D) is surjective for m 0 1 because H stabilizes. Since O(mD)|D is very ample on D, it is base point free, so we can then lift this to a section of O(mD) on X, and hence this is base point free. Hence, we get a map X → PN by O(mD). We claim this is finite onto the image. We need to show the map has finite fibers (since it is automatically proper). Here we are also using Chow’s lemma. If the fibers are not finite, the fiber includes a curve C. The restriction of O(1) to C would be 0 if C were contained in a fiber, by the projection formula. This contradicts our assumption. Therefore, there can be no curves contained in fibers. Ampleness is now a numerical property in terms of how a line bundle intersects all subvarieties. Assuming X is normal, we can define ampleness for Q-Cartier divisors. The following is an atypical definition. 8 AARON LANDESMAN

Deﬁnition 4.3. A line bundle L on X is nef if for all integral subvarieties Y ⊂ X if (L ··· L ·Y) ≥ 0. | {z } dim Y Remark 4.4. If L is ample, it is certainly nef. Further, L is nef on X if and only if L is nef on each component. Lemma 4.5. If L is nef and H is ample, then L ⊗ H is ample. Said another way, “nef + ample = ample.” Proof. This follows from Nakai-Moishezon by expanding

((L + H )dim Y · Y) > 0 using the binomial theorem. One has to reduce to the very ample case to argue the cross terms are positive. Above, we are using raising to the dim Y power to indicate the self intersection dim Y times (and not the dim Yth tensor power). It turns out that a sum of nef divisors is again nef, but we haven’t yet showed that. Kleiman showed that the ample cone is the interior of the nef cone.

Deﬁnition 4.6. Let L be a line bundle on X. We deﬁne nef≥r if for all integral Y ⊂ X with dim Y ≤ r, we have L dim YY ≥ 0.

Of course being nef on X is the same as being nef≤dim X.

Lemma 4.7. Suppose X is projective. We have nef≤r = nef≤r+1 for r ≥ 1. Corollary 4.8. We have, L is nef if and only if its restriction to every curve is positive.

Proof. This follows from the above lemma.

Remark 4.9. If L ∈ nef≤r, and Y is of dimension d, and s ≤ r and H is ample, then

s d−s (L · H )Y ≥ 0. This holds again because we can reduce to the case H is very ample, and then slice by members in the divisor class of H and use the hypothesis that L is nef. SURFACES NOTES 9

Proof of Lemma 4.7. We can easily reduce to the following special case: Sup- dim X pose X has dimension r and is integral, and L ∈ nef≤r−1. We want L ≥ 0. If H is ample (which exists since we are assuming X is projective) consider D + tH with H = O(H) and t ∈ Q. Consider P(t) = (D + tH)r. If r D < 0, then there is some t0 > 0 such that P(t) > 0 for t > t0 and P(t0) = 0. We have P(t) = (D + tH)r−1 D + t(D + tH)r−1 · H

− Since D + tH is ample, the ﬁrst term (D + tH)r 1 D ≥ 0. Also, when we expand (D + tH)r−1, we see that each monomial in the expansion is positive. r r Further, t H is positive because t0 > 0. Therefore, this second term is strictly positive, and cannot be 0. Hence, P(t) > 0. Corollary 4.10. A sum of two nef divisors is nef. Proof. We just need to check the intersection with every curve is nonnegative, and this is true for each constituent, so it is true for the sum.

5. 1/17/20 Recall the statement of Kleiman’s criterion: Let X be projective. We use 1 N1(X) as curves up to numerical equivalence and N (X) denote divisors up to numerical equivalence. We consider these as vector spaces over Q. We let NE1 denote the classes of effective curves, and NE1 denotes its closure in the associated real vector space. (1) Then, a Q Cartier Q-divisor D is ample if and only if D · Z > 0 for every Z ∈ NE1(X). (2) For any ample H, for any k ∈ R, the set Z ∈ NE(X) : H · Z ≤ k

is compact. Hence its intersection with N1(X)Z is a ﬁnite set. ∨ Remark 5.1. Said another way, Kleiman’s criterion implies NE1(X) is the nef cone whose interior is the ample cone. 5.1. Finishing the Proof of Kleiman’s criterion. From last time, we prove the ﬁrst part, but we still need to prove Z ∈ NE(X) : H · Z ≤ k

is compact. We have N1(X)R and inside we have NE1(X). Choose a basis for 1 N consisting of ample divisors. Let D1, ... , Dn denote such a basis. Then, 10 AARON LANDESMAN

1 there is some ﬁxed m so that H ± m Di is still ample. Then mH ± Di is ample. So, (mH ± Di) · Z ≥ 0. This implies

km ≥ m(H · Z) ≥ |Di · Z|. Therefore, we have bounded the set we are seeking of Z with H · Z ≤ k in a box as above, since we have bounded each of the coordinates. This shows it is compact. Proposition 5.2. Let C be a smooth proper curve of genus at least 2. Then Aut(C) is ﬁnite.

Proof. We know KC is ample and numerically equivalent to 2g − 2 points. Consider the diagonal ∆ ⊂ C × C. The degree of K∆ is 2g − 2. On the other 2 hand, by adjunction it is (KC×C + ∆) · ∆ = 2(2g − 2) + ∆ , which imp lie 2 ∆ = −(2g − 2). This uses that KC×C is the pullback of two copies of KC, each of which has degree 2g − 2. Now, suppose we have an automorphism σ : C → C. Consider the graph. This has self intersection −(2g − 2). Since KC×C is ample. Hence, there are only ﬁnitely many numerical equivalence classes of curves Z with K · Z = 2g − 2. Note that two different automorphisms correspond to different numerical equivalence classes because they intersect each other positively, while they intersect themselves negatively (with degree 2g − 2. Further, the self intersection of the graph of an automorphism agrees with the self intersection of ∆ because there is an automorphism of the surface given by id ×σ taking one curve to the other, which preserves intersection numbers. Note that we were using g ≥ 2 to ensure 2g − 2 > 0. Remark 5.3. If X is a variety with ample canonical bundle, then it is true that X has ﬁnitely many automorphisms. 5.2. Behavior of cones under maps. For X an arbitrary irreducible projec- 1 tive variety, consider N1(X) which is dual to N (X) and N1(X)R which is 1 dual to N (X)R. We also have the ample cone and effective cone inside of these. We would like to understand how this interacts with maps between varieties. Given X → Z we can factor this through a surjective map π : X → Y. For surfaces, most X have a unique smooth minimal model. The sort of things that can go wrong occur when X is a ruled surface, it can have P2 and P1 × P1 as a minimal model. What goes wrong is that when you take the canonical map it not ample, and quite negative, and maps the surface to a curve of lower dimension. In higher dimensions, we might hope for the statement that when the canonical bundle ample, there will be a unique smooth minimal model. SURFACES NOTES 11

Theorem 5.4. Let X be an arbitrary projective irreducible variety and a map π : X → Y, with Y normal. Let H be an ample divisor on Y. Then, π∗H restricts to 0 on any curve which is contracted. There is a cone in NE1(X) ⊂ N1(X). The ∗ pullback π H defines a functional on NE1(X), and defines a face (possibly of lower dimension) of NE1(X) determines π. We’ll come back to proving this theorem in several classes. What gets contracted tends to be rational curves. Theorem 5.5 (Zariski). Every smooth proper irreducible surface over k is projective. 5.3. Idea of proof. The idea of the proof is as follows: We want to construct an ample divisor. Thanks to Nakai-Moishezon, we need to check it is nonnegative on every curve and has positive self intersection. Start with an affine open U. The complement of U has pure codimension 1, so it is a union of curves. The ample divisor we’ll come up with is some positive combination of these curves. We need to show it meets every curve positively, and itself non-negatively. This is true if the curve we are meeting meet the affine open. So, we just need to cook up the multiples of the curves in the complement of U so that the intersection of this with each such curve is positive.

5.4. Hodge Index Theorem. Here is a nice linear algebra fact Ravi really likes, but didn’t know before. This is in the lead up to the Hodge index theorem. If q is a nondegenerate real quadratic form which has a signature with ±1’s. The 1’s correspond to the largest positive deﬁnite subspace. Here is another nice fact. Lemma 5.6. If we have a Q-vector space V with a symmetric quadratic form q. Suppose ei span V and Bq(ei, ej) ≥ 0 for Bq(a, b) = q(a + b) − q(a) − q(b). Then for z = ∑i aiei with ai > 0 and Bq(z, ei) = 0, then x · x ≤ 0 for all x ∈ V. Further, the set of x with x · x = 0 is a linear subspace equal to the radical of q. Further, the dimension of the radical is the number of connected components of the graph formed by joining ei to ej if Bq(ei, ej) > 0. It seems this statement may have a typo because Ravi wants the radical to be trivial when there is a unique connected component.

6. 9/22/20 When we have a proper map X → Y, there are some curves which are contracted. For example, it could be a blow up, or it could be a sort of ﬁb-ration. When one contracts curves, we’ll have some negativity, i.e., on 12 AARON LANDESMAN

a surface the curves may have negative self intersection. And conversely, when one has this negativity, one will be able to contract the curves. We’ll now return to the hodge index theorem. The following proof is due to Grothendieck. Lemma 6.1. Suppose D, E are two divisor classes on X with E effective. Then, dim h0(D + E) ≥ dim h0(D).

Proof. Given any element of h0(D), just add E. Proposition 6.2 (Hodge Index Theorem). Let X be a smooth surface with D2 > 0. Let H be ample. Then, exactly one of these holds. Either (1)D · H > 0 and dim |nD| → ∞ as n → ∞ (2)D · H < 0 and dim |nD| → ∞ as n → ∞. Proof. By Riemann Roch for surfaces h0(nD) + h0(K − nD) ≥ h0(nD) − h1(nD) + h2(nD) = n2D2/2 + an + b for suitable a and b depending on nD and X. We can’t have both h0(nD) → ∞ and h0(K − nD) → ∞ since by the above lemma this would imply h0(K) ≥ h0(nD). Similarly, the same holds as n → −∞. Then, as n → ∞, we can’t have both h0(nD) → ∞ and h0(−nD) → ∞. And similarly as n → ∞, we can’t have both h0(K + nD) → ∞ and h0(K − nD) → ∞. Altogether, Riemann Roch as above shows as n → ±∞, we know one of h0(nD) → ∞ or h0(K − nD) → ∞, but the same one cannot go to infinity by the above observations. So, if h0(nD) → ∞ as n → ∞ we must have h0(K − nD) → ∞ as n → −∞. This forces D · H > 0 since D is effective. Which gives the first case. Otherwise h0(nD) → ∞ as n → −∞, in which case h0(K − nD) → ∞ as n → ∞ and we find D · H < 0 since −D is effective. Corollary 6.3. If D · H = 0 in the context of the hodge index theorem, then D2 ≤ 0. Proof. This follows from the hodge index theorem, it is essentially the con- verse. Theorem 6.4. Let D and H be divisors on a smooth projective surface X with H ample and D · H = 0. Then D2 = 0 if and only if D ≡ 0. Proof. We know D2 ≤ 0 by the hodge index theorem. We have DH = 0 and D2 = 0. Whet if D 6≡ 0? Then, there is some E with which D · E 6= 0, because 1 the pairing on N1 and N is nondegenerate. SURFACES NOTES 13

Set F = E + αH and note that D · F = D · E = 0 because D · H = 0. We have H · F = H · E + αH · H Choose α so that H · F = 0. We have now that tH is ample, D · H = 0, D2 = 0, D · F 6= 0, H · F = 0. 2 2 2 2 Now, replace D by Dn := nD + F. Then, Dn = n D + 2nDF + F = 2 2nDF + F and note that D · F 6= 0. But Dn · H = 0. By choosing n very 2 positive or negative, we know Dn > 0 1 Exercise 6.5. To make the above proof complete, check that N1(X) = N (X).

Corollary 6.6. The signature of the quadratic form on N1(X)R given by intersecting classes is (1, dim N1(X)R − 1) where the 1 corresponds to the ample class. Proof. This follows from the above theorem because the perpendicular space to the ample class H is negative deﬁnite. The previous hodge index theorem showed it was semideﬁnite. 2 Exercise 6.7. Let X be a projective smooth surface D1, D2 with D1 > 0. Show 2 2 2 D1 D2 ≤ (D1D2) . 6.1. A Proposition from linear algebra. Proposition 6.8. Let B be a symmetric bilinear form on a Q vector space M. Let ei span M. Assume ei · ej ≥ 0 for i 6= j. Suppose Z = ∑i aiei with ai > 0. Then, x · x ≤ 0 for all x ∈ M. and the dimension of the null space is the number of connected components of the adjacency graph of the ei (where ei is connected to ej if ei · ej 6= 0).

We should think of ei’s as irreducible components of ﬁbers of a surface ﬁbered over a curve.

Proof. We can begin by replacing ei by aiei to assume z = ∑i ei. Write x = ∑i ciei. Then, 2 2 2 2 x · x = ∑ ci ei · ei + 2 ∑ cicj(ei · ej) ≤ ∑ ci ei · ej + ∑(ci + cj )(eiej) i i

Remark 6.9. In fact, the proof shows that the only possible dependence relation between the ei is that ∑i ei = 0 when the graph is connected. 7. 1/27/20 Ample implies nef and big, and today we will discuss the notion of bigness more. Nefness and bigness play better under birational transformations. Proposition 7.1. Let D be a line bundle on a projective n-dimensional X. Suppose (1) for all i, hi(X, mD) = O(mn) (2) if D is nef, then hi(X, mD) = O(mn−1) for i > 0. In particular, h0(X, mD) = n Dn n−1 m n! + O(m ). Proof. We’ll start with the ﬁrst part. Write D = E1 − E2. We have exact sequences

(7.1) 0 OX(mD − E1) OX(mD) OE1 (mD) 0 (7.2)

0 OX((m − 1)D − E2) OX((m − 1)D) OE2 ((m − 1)D) 0 The left terms are the same and the right terms are O(mn−1) by induction, so we get 0 i n−1 h (OX(mD)) ≤ h (OX((m − 1)D)) + O(m ).

i m0 i i n−1 Therefore, h (OX(m0D)) = ∑m=0 h (OX(mD)) − h (OX(m − 1)D) = m0O(m0 ) = n O(m0 ). Part (b) holds by the inductive hypothesis for i ≥ 2 by the same long exact sequences. We now know h0(X, mD) − h1(X, mD) = mnDn/n! + O(mn−1) and Dn ≥ 0 for D nef. If h0(X, mD) = 0 for all m > 0 we get Dn = 0 and so h1(X, mD) = O(mn−1). If h0(X, mD) > 0 for m 0 we have

(7.3) 0 OX((m − m0)D) OX(mD) OE(mD) 0 We have 1 1 n−2 h (OX(mD)) ≤ h (OX((m − m0)D)) + O(m ) 1 n−1 and by induction this means h (OX(mD)) = O(m ). SURFACES NOTES 15

Deﬁnition 7.2. A line bundle D on a variety X is big if

h0(X, mD) > lim sup n 0. m→∞ m Remark 7.3. A line bundle is nef and big then it is in particular nef, so Dn > 0, by the above proposition. Also, we have Kodaira vanishes for ample line bundles. There is a gener- alization called Kawamata-Viehweg vanishing for nef and big line bundles.

Remark 7.4. The sum of two nef and big divisors is again nef and big. We have already seen it is nef. We need to check the sum of two big divisors is n big, assuming they are nef. Just expand (D1 + D2) and we know each term n in the expansion is non-negative because they are nef and then D1 is strictly n positive, so (D1 + D2) is strictly positive. Lemma 7.5. Suppose X is projective. If D is nef and big, then there is an effective E such that D − tE is ample for 0 < t ≤ 1.

Proof. Take H ample. We claim

h0(X, mD − H) > 0 for m 0.

We have

(7.4) 0 OX(mD − H) OX(mD) OH(mD) 0

n−1 0 n For m 0, we know OH(mD) is O(m ) while h (OX(mD)) = O(m ), so 0 0 h (OX(mD − H)) is effective. This shows there is some E with E = mD − H. 0 1 Let mE = E which is an effective Q-divisor. Then, D − E = m H, which is ample. We will now show D − tE is ample for 0 < t ≤ 1. Indeed, D − tE = t m H + (1 − t)D which is the sum of an ample and a nef. Here, (Q-)ample means that some multiple is very ample. Remark 7.6. Something we might want to say that if D is nef and big, then there exists m > 0 so that O(mD) gives a rational map X → Pr with image of dimension n. We take R := ⊕H0(X, mD), and take proj. Suppose this is ﬁnitely generated (which is hard). This gives a map to projective space 0( ( ) ⊗ → ( ) corresponding to the surjection H O mD Ok OX O mD given by evaluation of sections and using bigness tells us the dimension, because it has on the order of mn sections. 16 AARON LANDESMAN

8. 1/29/20 Last time we discussed how, given a surface fibered over a curve with geometrically connected fibers, any fiber has intersection form which is nonpositive definite, and has only a single 0 eigenvalue (and all other eigenvalues strictly negative). We won’t prove the following, but it is useful to know. Theorem 8.1 (Du Val). Let f : X → Y be a map of surfaces with X smooth Y normal and f is proper birational which is an isomorphism away from y ∈ Y. Suppose the preimage of y as a set is a union of curves E1 ∪ · · · ∪ En. Then, Ei · Ej is negative definite. Remark 8.2. The vague idea is that we can add in an additional divisor which intersects every component quite positively, and then the resulting matrix satisfies the linear algebra lemma we saw last time. The flavor of the general statement is that given anything negative you should be able to contract it. Example 8.3. Here is Nagata’s example of a non-contractible negative curve. We’ll work over C. We’ll start with a cubic curve E in P2 embedded by some line bundle 3p. Choose a point x0 which is not torsion in the group law, if p corresponds to the identity on the elliptic curve. There is no curve 2 in P which meets E only in the point x0, since x0 is not torsion. 2 Now, blow up P at x0. Then, blow up the point over X0 in the proper transform of E in the blow up. The self intersection is now 8. Then repeat this 9 more times, until the self intersections of C is −1. Suppose we could contract C to obtain some variety Y. Let y be the image of the contraction. Choose a curve in Y missing y. The preimage in the successive blow ups is some curve C. Then, let D be the image of C in P2. In this case, D can only meet E at the point x0. But we assumed x0 was not torsion, which means there is no D meeting E only at x0. Indeed, because O(d)|E = NO(x) we obtain Nx = 3dp, which means x − p is 3d torsion.

Remark 8.4. Ravi said he would bet up to $10 that if x0 were torsion in the above example, then you can contract the curve. Here is one of Artin’s contractibility criteria. Theorem 8.5. Suppose X is a smooth projective surface and E is a geometrically connected curve on X with E = E1 ∪ · · · ∪ En. Then, the following are equivalent (1) There exists some f : X → Y with Y normal and f is an isomorphism except in the ﬁber over a single point y whose preimage is E. Additionally, χ(OX) = χ(OY). SURFACES NOTES 17

(2) The intersection matrix (EicE˙ j) is negative deﬁnite and pa(Z) ≤ 0 for each positive Z ⊂ E. Furthermore, in the above setup, f : X → Y in the ﬁrst part is unique. Additionally, the singularity at y is a rational singularity. Remark 8.6. The second condition is eminently checkable since you can take a linear combination of the Ei and compute the genus using adjunction. Example 8.7. Suppose p ∈ X is a rational double point (meaning the singularity is rational, and is a double point, which is something like the tangent cone has degree 2) It is also true that if you have a −1 curve, you can contract it, and the resulting contracted surface is smooth. The smoothness is the hard part, and you can compute the nth formal neighborhood of a point in the contracted surface and its preimage corresponds to powers of the ideal sheaf of the −1 curve. It is also true in the case of rational double points that the preimage of the nth power of the ideal sheaf of the double point is the nth power of the ideal sheaf of the contracted curve. 8.1. Facts about curves which analytically locally embedable in P2. Let E1 ∪ · · · ∪ En be a geometrically connected scheme in a surface X and let = ∈ Z Z ∑ri≥0 riEi with ri . Lemma 8.8. The following are equivalent: 1 (1)H (OZ) = 0 0 0 0 (2) For all Z ⊂ Z, meaning Z = ∑i siEi with 0 ≤ si ≤ ri we have pa(Z ) ≤ 0 (the arithmetic genus pa is at most 0). Proof. Let’s see that (1) =⇒ (2). We have

(8.1) 0 OZ/Z0 OZ OZ0 0. Taking the long exact sequence in cohomology we get 1 1 H (Z, OZ) → H (OZ0 ) → 0 1 0 0 1 which shows H (OZ0 ) = 0. This implies pa(Z ) = 1 − h (OZ0 ) + h (OZ0 ) = 0 1 − h (OZ0 ) ≤ 0. For the other direction, once you’ve chosen Z, one can induct on Z by adding an appropriate Ei. But we omit the proof. Fact 8.9. This is a fairly difﬁcult fact. n There is a map d : PicZ/k(Spec k) → Z sending L to (L · Ei)1≤i≤n. Then, the ﬁrst two conditions in Lemma 8.8 are equivalent to 18 AARON LANDESMAN

(3) the map d has ﬁnite kernel (3’) the map d is an isomorphism n Moreover there is a map Z → PicZ/k(Spec k) is given by sending (ti) to OX(∑i tiEi)|Z. If the above four equivalent conditions (1), (2), (3) and (30) are all true, then H1(O ) = 0 then every line bundle on Z has deg (L) ≥ 0 globally Z Ei 1 0 0 generated and h (L) = 0 Riemann Roch implies h (L) = ∑i ridi + h (OZ). A surface which has resolution satisfying any of the above properties, is said to have rational singularities.

9. 1/31/20 Suppose we start with a map f : X → Y which contracts some conﬁg- −1 uration of curves Ei on X with E := ∪iriEi connected. Suppose f (y) is supported on the Ei and f is proper and an isomorphism away from y. Suppose X and Y are normal. Theorem 9.1. In the above situation, the following are equivalent.

(c) Let Z = ∑i riEi with ri ≥ 0 but not all ri = 0. The genus pa(Z) ≤ 0. 1 (d) Let Z = ∑i riEi with ri ≥ 0 (not all 0). Then H (OZ) = 0. (e) Let Z = ∑i riEi with ri > 0. Then there is a canonical map ⊕n PicZ/k(Spec k) → Z L 7→ deg L , . . . , deg L E1 1 En which is injective and has image of ﬁnite index Equivalently there is a map Zn → Zn sending L := O( r E )| to deg L , and that map ∑i i i E Ei should factor through PicZ/k(Spec k) with ﬁnite kernel. 1 (a)R f∗OX = 0 (b) χ(OX) = χ(OY). Proof. The idea is that c and d imply all components are rational and form a tree, which will imply e. To go back from e to d, we can see the curve must 1 be a tree of genus 0 curves, and H (OZ) is the tangent space to Pic, which will be trivial exactly when the map from Pic to Zn is injective. To connect these to (a) to (b), we know f∗OX = OY by normality of Y, which gives the equivalence by the Leray spectral sequence which degener- i i i ates, so H (Y, OY) ' H (Y, f∗OX) ' H (Y, OY). The equivalence of (a) and 1 (d) follows because R f∗OX Let E = ∑i riEi is supported on the point y and 1 \1 ( is equivalent to H (E, OE) = 0. (R f∗OX)y = limri→∞ H OZ). SURFACES NOTES 19

Theorem 9.2 (Artin). Suppose we have a map f : X → Y as in the previous theorem. The following are equivalent (1) The map f has rational singularities

(2) the matrix E · E is negative deﬁnite and p (Z) ≤ for all Z = i j i,j a 0 ∑i riEi and ri ≥ 0 and not all ri are 0. In the above situation, given X the contraction (in the absence of Y) exists and is unique, in the sense that any other Y0 which realizes such a contraction is isomorphic to Y. Proof. We start with a very ample line bundle H on X. We want to tweak H by the divisors of Ei to make the intersection numbers with the Ei to be 0. We have that the map ψ : Z⊕n → Z⊕n ! (ri) 7→ ∑ riEi Ej i with image contained in k(Z⊕n) for some integer k. This is contained in im ψ. So if we replace H by kH, we will have the restriction map lying in the image. Hence, we can arrange H + ∑i riEi = Z with ri > 0. We have Z · Ei = 0 for all i. We will contract this divisor Z. There will also be a section not vanishing at any given point which gives basepoint freeness; we’ll see this by realizing a certain cohomology vanishes. We’ll similarly see it separates points and tangent vectors which gives an isomorphism away from these curves and all the curves get mapped to a point. 10. 2/3/20

Let X be a smooth projective surface over k and let E = E1 ∪ · · · ∪ En. We had Artin’s theorem on contractions: Theorem 10.1 (Artin contraction). Let X be a smooth surface over a ﬁeld k and E = E1 ∪ · · · ∪ En a connected divisor. Then, the following are equivalent: (1) There exists f : X → Y contracting only E and χ(O ) = χ(O ). X Y (2) The intersection matrix Ei · Ej is negative deﬁnite and pa(Z) ≤ 0 for all Z > 0 supported on E. Further, in the above situation, the contraction is unique. Proof. We have a map n ∨ PicX/k(Spec k) → (Z ) . 20 AARON LANDESMAN

There is a composition map

Zn → (Zn)∨ ! ! ! (r1,..., rn) 7→ ∑ riEi · E1,..., ∑ riEi · En . i i

The map (Zn) × (Zn) → Z

∨ being negative deﬁnite implies there exists some m > 0 with m (Zn) ⊂ im φ. Choose H very ample so that Hi(X, O(H)) = 0 for i > 0. We can ﬁnd ~ Z = ∑i riEi such that φ(H) = −φ(Z). Then φ(H + Z) = 0. We also have Z · Ei < 0 since H is very ample.

Lemma 10.2. Suppose Z = ∑i riEi. Suppose the intersection matrix (Ei · Ej) is negative deﬁnite and Ei · Ej ≥ 0 for i 6= j and Z · Ei < 0. Then, ri > 0.

Proof. First, we claim Z ≥ 0, i.e., Z is effective. Write Z = Z1 − Z2. Observe

0 ≥ Z · Z2 = (Z1 − Z2) · Z2 = Z1 · Z2 − Z2 · Z2 ≥ Z1 · Z2 ≥ 0

using that the matrix is negative deﬁnite for the penultimate step and that the Ei’s in the support of Z1 are disjoint from those in the support of Z2. This implies Z2 = 0. Then, Z · Ei < 0 so ri > 0.

Corollary 10.3. There exists an effective Z with Z · Ei < 0.

So, we now have an effective Z with Z · Ei < 0. It follows O(H + Z)|Z = O. Since H + Z is trivial on Z we have

(10.1) 0 OX(H) OX(H + Z) OZ 0.

1 Taking the long exact sequence on cohomology gives h (OX(H)) = 0. So we 0 0 get an exact sequence on H . h (OE) = 1. So, the sections of OX(H + Z) are basepoint free away from E and also basepoint free away from E, using that H is ample. They separate points and tangent vectors away from E, while the map contracts E. We can produce a section which vanishes on E and does not vanish at a speciﬁed other point using the exact sequence. This tells us E maps to points distinct from other points, so only E is contracted. Finally, because pa(Z) ≤ 0 for all Z ≥ 0, we have χ(OX) = χ(OY). SURFACES NOTES 21

10.1. The fundamental cycle. For the rest of today we’re in Artin’s situation. n n ∨ Consider the map Z → (Z ) given by sending a line bundle L = ∑i riEi to its pairing with each of the curves E1,..., En.

Lemma 10.4. There is a smallest effective class Z = ∑i riEi meeting each Ej negatively. This Z is called the fundamental cycle. 0 0 00 00 Proof. Suppose Z = ∑i ri Ei and Z = ∑i ri Ei both meet each of the Ei’s 0 00 negatively. I claim that Z = ∑ min(ri, ri )Ei works too. 0 00 Suppose r1 ≤ r1 . Then, we can compute E1 · Z as 0 0 00 0 0 0 r1E1 · E1 + ∑ min ri, ri E1 · Ei ≤ r1E1 · E1 + ∑ ri E1 · Ei = Z · E1 < 0. i>1 i≥1

One can similarly argue that the intersection with any Ei is non-positive. Soon, we’ll start with a Z which is the fundamental cycle as in the above −1 n lemma. We’ll see for f : X → Y a contraction, O(−nZ) = f (mY). You can then use this information to work out the degree of the singularity, the embedding deformation, and tangent information, etc. You can work out the degree of the singularity by looking at sections of O(−nZ) as n grows, and so you get a formula in terms of the geometry of Z. You can find when the singularity is a double point, which are classified by ADE singularities. Definition 10.5. Let X be dimension n with p ∈ X an isolated singularity. N PTX,p ⊂ P We say p is an n-fold point singularity if deg PTX,p = n. When −Z2 = 1 the point is smooth while if −Z2 = 2 it is a double 0 point. You can work this out by looking at h (O(n)) = deg OZ(n) + C = 2 n deg OZ(1) + C = −Z + C. This shows that the Hilbert polynomial has degree −Z2 which tells us that this is a rational n-fold point when −Z2 = n.

11. 2/5/20 Recall that a rational surface singularity is defined as follows: Suppose Y is a surface and y ∈ Y is a singular point and f : X → Y is a resolution of y with 1 special fiber E over y. Then, y ∈ Y is rational if R f∗OX = 0. Equivalently, χ(OX) = OY and for every effective Z with ∑i riEi with ri ≥ 0 and not all 0, then pa(Z) ≤ 0. We now have a surface X with some divisor E = ∪iEi with the intersection matrix |Ei · Ej| negative definite. and pa(Z) ≤ 0 for all Z > 0 supported on the Ei. Then we saw by Artin’s theorem, we can contract E. We saw last time that there is a fundamental cycle Z = ∑i riEi with ri > 0 and Z · Ei ≤ 0 for 0 0 all i, which is minimal in the sense that there is no Z with all ri ≤ ri such 0 that Z · Ei ≤ 0 other than Z itself. 22 AARON LANDESMAN

Theorem 11.1 (Artin). Suppose f : X → Y is a contraction, contracting E to y, as in Artin’s contraction theorem Theorem 9.2. Let Z be a fundamental cycle. Then,

(1)p a(Z) ≥ 0 (2)p a(Z) = 0 if and only if the contraction has a rational singularity at y ∈ Y.

This is nice because the second part shows that to check for rational singularities, there is just one thing we need to compute.

0 Proof. Note that pa(D1 + D2) = pa(D1) + pa(D2) + D1 · D2 − 1. If Z > 0 0 0 0 0 then Z 6≥ Z, there is some Ei with Z · Ei > 0. Then, pa(Z + Ei) = pa(Z ) + 0 pa(Ei) + Z · Ei − 1, using the above fact. Note that pa(Ei) ≥ 0 (since Ei is 0 integral) and Z · Ei − 1 ≥ 0 by assumption. Suppose pa(Z) = 0. Our goal is 0 0 to show that for any Z > 0 we have pa(Z ) ≤ 0. The idea will be to start with a Z1 and add suitable Eis and try to eventually get to Z, or a multiple of Z. The issue is to show that there will be some Ei which will “move Z1 toward Z” which satisﬁes Z1 · Ei > 0. 0 0 0 Suppose Z = ∑i riEi. Take Z = ∑i ri Ei with ri ≤ ri and not all ri are equal. 0 Then, we want to show there is some Ei < ri and Z · Ei > 0. 0 0 0 But, if ri = ri, then Z · Ei = ∑i rjEj · Ei ≤ rjEj · Ei = Z · Ei = 0. Therefore, 0 0 0 we’re “not allowed to increase Z by Ei” since Z · Ei < 0 if ri = ri. Therefore, 0 since there must be some i with Z · Ei > 0, it must occur at an i for which 0 ri ≤ ri. Remark 11.2. The proof gives a nice way of producing Z, since you can start with 0 and do this process, and just add Ei until you ﬁnish.

This ﬁnishes the ﬁrst part, so pa(Z) ≥ 0. We now want to show pa(Z) = 0 if and only y is a rational singularity. We have seen before that rational singularities satisfy pa(Z) = 0. So, we just want to see that if pa(Z) = 0 then y is a rational singularity. In the way we have built up Z, we have seen the genus of each Ei is 0, since we constructed Z as a sequence Z1 ⊂ Z2 ⊂ · · · ⊂ Z where pa(Zi + Ej) = pa(Zi) + pa(Ej) + Zi · Ej − 1, and the last two terms are at least 0. Since pa(Z) = 0, it must be that pa(Ej) = 0 for all j, since we start with 0 and add nonnegative values until we eventually get to 0. So the nonnegative values (which include all pa(Ej) must all be 0. For the same reason, we know that at each step, Zj · Ei − 1 = 0. 0 0 Now, we start with Z = 0 and add some Ei for which Z · Ei > 0. We 0 were following the strategy that if there exists some Ei with Z · Ei > 0, we add Ei, thereby increasing the genus (or at least keeping it constant). If 0 0 there is no Ei with Z · Ei > 0, we can consider Z − Z, which is effective by SURFACES NOTES 23

construction/deﬁnition of the fundamental cycle Z. We know 0 0 0 pa(Z − Z) + pa(Z) + Z · (Z − Z) − 1 = pa(Z ).

We are assuming pa(Z) = 0. Since Z is a fundamental cycle, we ﬁnd Z · 0 0 0 0 (Z − Z) ≤ 0. It follows that pa(Z − Z) ≥ pa(Z ). So, pa(Z − Z) is strictly 0 bigger than pa(Z ). 0 So, to summarize the process, either some ri < ri, and then we add Ei 0 to Z . If all ri ≥ ri, subtract Z. At each stage, the genus does not increase, and so the genus of Z0 is bigger than that of Z. This shows the singularity is rational. 12. 2/10/20 Suppose we are in Artin’s setup as in Theorem 9.2 and Z is a fundamental cycle. From last time, we owed the following lemma: Lemma 12.1. If D and E are effective divisors on a surface meeting in dimension 0 then pa(D + E) = pa(D) + pa(E) + D · E − 1.

Proof. By adjunction (K + C) · C = 2pa(C) − 2. We want pa(D + E) = (pa(D) − 1) + (pa(E) − 1) + D · E. We can see this because (k + D + E)(D + E (K + D)D (K + E)E = + + D · E. 2 2 2 12.1. Computing the fundamental cycle. We want a fundamental cycle Z with Z ≥ E1 + ··· + En with Z · Ei ≤ 0 which is minimal. Let’s see some examples.

Example 12.2. Let’s look at an An singularity, where we have a chain of −2 curves meeting in nodes. The intersection matrix looks like   −2 1 0  1 −2 1  0 1 −2

We can see that E1 + ··· + En works. The dual graph is a line with n marked points on it.

Example 12.3. Let’s consider a Dn singularity, which has dual graph given by the Dn Dynkin diagram. I.e., there are two rational curves meeting a third, and then we have a chain joining the third to the fourth and so on. We start with ∑i Ei. We ﬁnd we need to make E3 larger because the current 0 intersection is positive −2 + 1 + 1. So we then have Z := E3 + ∑i Ei We 24 AARON LANDESMAN

0 0 next find E4 does has negative intersection with Z so far. And we get Z := n−1 n E3 + E4 + ∑i Ei. Continuing in this way, we find Z := ∑i=3 Ei + ∑i=1 Ei is the fundamental cycle. We now want to check the above algorithm we were implicitly following works. The crux is the following, which we basically saw last class. 0 0 Lemma 12.4. Suppose Z = ∑i siEi < Z and Z · Ei > 0 then (Z + Ei) ≤ Z. 0 Proof. Let Z = ∑i siEi and Z = ∑i riEi. We know ∑j sjEj · Ei > 0 and ∑j rjEj · Ej ≤ 0. We can find, rj ≥ sj. Theorem 12.5 (Artin, 3.28 in Badescu). Suppose we are set up as in Theorem 9.2, so we have f : X → Y. If (Y, y) is a rational singularity on the target corresponding ∗ n n to an ideal sheaf m ⊂ OY (so m = Iy/Y. Then, f m = IZ/X. Then, 0 n 0 n H (OY/m ) ' H (OX/IZ ). Further, h0(mn/mn+1) = 1 − n(Z · Z), and h1(mn/mn+1) = 0, Proof. We’ll come back to this proof later. Corollary 12.6. If Z · Z = −1 then Y is smooth at y.

Proof. From the above theorem, we find dim TY,y = 1 − Z · Z so y is smooth because it has 2 dimensional tangent space. Definition 12.7. If y ∈ Y is an isolated singularity, we define the multiplicity of the singularity at y ∈ Y as the degree of PTCY,y ⊂ P, the projectivized • 0 2 tangent cone. Here P is Sym H (m/m ) and PTCY,y is the graded ring ⊕H0(mn/mn+1). Corollary 12.8. The multiplicity of a rational singularity at y as above is −Z · Z. Proof. To find the degree of a variety, we can look at the dimension of O(n) restricted to the subvariety. In particular, Z2 = −2 if and only if the singularity is a rational double point and Z2 = −1 if and only if the singularity is smooth. Theorem 12.9. Suppose we are set up as in Theorem 9.2. The following are equivalent. 2 (1)p a(Z) = 0 and Z = −2, i.e., Y has a rational double point singularity. (2)Z 2 = −2 and Z · K = 0. 2 If there are no Ei with both (Ei = −1 and pa(Ei) = 0) then the above are equivalent to 2 (c)p a(Ei) = 0 and Ei = −2. SURFACES NOTES 25

2 Proof. First, let’s assume There are no Ei with both (Ei = −1 and pa(Ei) = 0). Let’s show that (b) implies (c). 2 We have K · Ei = 2pa(Ei) − 2 − Ei ≥ 0 If Z = ∑i riEi then Z · K = 0. Then 2 0 = Z · K = ∑ riK · Ei = ∑ 2pa(Ei) − 2 − Ei i

Then, by our assumption that the intersection matrix of the Ei is negative 2 definite but we do not have both Ei 6= −1 and pa(Ei) = 0, we find each term 2 in the sum is nonnegative. This implies pa(Ei) = 0 and so Ei = −2. Then, Z · K = 0. We’ll now show (c) implies (a). We know pa(Z) ≥ 0 because Y is a contraction of X, and so any effective nonzero divisor has positive genus. 2 2 We have 2pa(Z) − 2 = (K + Z) · Z = Z . IT follows that Z ≥ −2. Further, 2 2 Z ≤ 0 by negative definiteness, and we see Z is even since it is 2pa(Z) − 2. Therefore, Z2 ≤ −2, and so Z2 = −2. This shows (a). Observe that (a) and (b) are equivalent using the formula 2pa(Z) − 2 = (K + Z)Z, from adjunction.

13. 2/12/20 We’d like to understand normal surface singularities. By normality, the singularities are isolated. We can resolve the singularities (even to a minimal proper regular surface in a canonical way) so that we can understand the singularities exactly in terms of the conﬁguration of curves over them. At the very least, if you take some resolution, your singularity is determined by its ﬁber. As usual, we retain the setup as in Theorem 9.2. If no Ei is a −1 curve, i.e., 2 Ei has genus 0 and Ei = −1 recall last time we showed the equivalence of 2 (1) pa(Z) = 0, Z = −2 (2) pa(Z) = 0, Z · K = 0 2 (3) pa(Ei) = 0, Ei = −2. These are rational double point singularities. A map is crepant if the pullback of the canonical divisor is the canonical divisor. This can be checked by verifying K · Ei = 0, or something like that. Today, we’d like to answer the following question: Question 13.1. What are the rational double point singularities? The answer will be the ADE singularities. 2 2 n+1 Example 13.2. If we consider x + y + z = 0 is an An singularity. We n+1 2 can similarly get An singularities in higher dimension x0 + ∑i xi = 0. 26 AARON LANDESMAN

2 2 n−1 Example 13.3. There are also Dn singularities x + y z + z = 0.

Example 13.4. There is the E6 singularity x2 + y3 + z4 = 0

the E7 singularity x2 + y3 + yz3 = 0

and the E8 singularity x2 + y3 + z5 = 0

It turns out that the above An, Dn, E6, E7, E8 singularities are the only rational double point singularities.

Remark 13.5. E6 turns out to be the most complicated singularity in nature, they appear in waves, apparently.

When you blow up the An singularities (and similarly Dn singularities) you find an inductive structure. Lemma 13.6. The only rational double point singularities are ADE singularities. Proof. We’ll not make some observations to characterize rational double point singularities. (1) Suppose we have a configuration of curves giving a rational double points. (2) We know each divisor has to have genus 0. Hence, the only singularities can be nodes. (3) Also, the configurations of curves has to be a tree. (4) We also know the self intersections of all curves have −2. (5) We cannot get a configuration with 4 curves meeting a single common curve Ei. The reason is that this has non-negative definite intersection 2 matrix. Indeed, (2E0 + E1 + E2 + E3 + E4) = −2(4 + 1 + 1 + 1 + 1) + 4 ∗ 4 = 0. So, the valence of any curve is always at most 3. (6) If there are no triple points (i.e., components meeting 3 other components) we get an An singularity. (7) If there is a triple point, we can’t have a triple point Ei where each of the three curves E1, E2, E3 meeting it are adjacent to a further curve, i.e, Ei is adjacent to E3+i for i ∈ {1, 2, 3}. Indeed, in this case, the divisor (3E0 + 2(E1 + E2 + E3) + E4 + E5 + E6) = −2(3 · 1 + 3 · 4 + 9) + 2(3 · 2 + 3 · 6) = 0. (8) So, at each triple point, we can have at most 2 adjacent points with valence 2. SURFACES NOTES 27

(9) We cant have E0 a triple point with branches of length 1, 3, 3 coming out of it and can’t have E0 a triple point with extensions of length 1, 2, 5. In the ﬁrst case, we put a 4 on the triple point, 2 on the vertex chain of length 1 and 3, 2, 1 on the two other chains. Taking self intersection we get −2(1 + 4 + 9 + 16 + 9 + 4 + 1 + 4) + 2(2 + 6 + 12 + 12 + 6 + 2 + 8) = 0 In the second case we have the triple point with multiplicity 6, the 1 branch of multiplicity 3, and length 2 branch with multiplicities 2, 4 and the length 5 chain with multiplicities 5, 4, 3, 2, 1. −2(4 + 16 + 36 + 25 + 16 + 9 + 4 + 1 + 9) + 2(8 + 24 + 30 + 20 + 12 + 6 + 2 + 18) = −120 + 120

(10) It only remains to show we can’t have two triple points. Say E0 and En both have triple points with F, G adjacent to E0 and H, K adjacent to En. Then put 1s on F, G, H, K and 2s on all An. Remark 13.7. There is no moduli in the rational double point singularities. In other words if we have a given diagram, is it etale´ locally given by the speciﬁc singularity equation we have written. For example if we want an E8 singularity, they are etale´ locally isomorphic to x3 + y3 + z5. Or, relatedly, the to understand the An singularity, you can contract one curve at a time, and relate the sections of the neighborhoods as one builds the tower by resolving.

Exercise 13.8. Check this! Also, it’s nice that all of these are quotient singularities. Lemma 13.9. Suppose we have f : X → Y = Spec A a projective morphism with Y affine. Let m be a singularity in the target Y and a projective morphism with fiber 1 1 over m connected. Assume R f∗OX = H (OX) = 0. Then, for m ∈ Spec A, we have f ∗(m) is an effective Cartier divisor, (of pure dimension 1 with no associated points) or equivalently, the ideal sheaf of the pullback of m is O(−D) for D an effective Cartier divisor. The strategy is to show the pullback is reflexive. We’ll do this next time. 14. 2/14/20 Today, we’ll again defer the proof of Artin’s contractibility criterion and jump forward in time! We’ll go back to the following question: Example 14.1. Suppose we have some variety X and we want to understand contractions in the hopes of making a minimal model or otherwise understanding its structure. For example, if we have a fibration over a curve, we 28 AARON LANDESMAN

have some sort of negative definiteness in the fibers. We can try to look for whether something should secretly be fibered over a curve by looking for a configuration of curves where the intersection number is nonpositive definite, with one 0 eigenvector. We’d now like to understand contractions in terms of the intersection theory of the variety. 2 2 Example 14.2. Consider X = Blpt P . Here, N1 := N1(X) = Z . This is generated by E and L, where E is the exceptional divisor and L is the class of a line. Note that L − E is also effective. In fact, L − E and L generate N1 Suppose we have a morphism π : X → Y of projective varieties. We obtain ∗ 1 1 a map π∗ : N1(X) → N1(Y) and π : N (Y) → N (X). For β ∈ N1(X), α ∈ 1 ∗ N (Y) then (π α) · β = α · (π∗β). If π is surjective, then π∗ is surjective and ∗ 1 ∗ π is injective. Then N1(π) := ker π∗ and N (π) := coker π . Lemma 14.3. For π : X → Y, Y is integrally closed in K(X) over K(Y) if and only if OY → π∗OX is an isomorphism (i.e., π is O-connected). Proposition 14.4. Given X, Y, Y0 projective integral and normal, π : X → Y and 0 0 0 π : X → Y projective with π and π O-connected. Suppose further N1(π) = 0 N1(π ). (1) Then there exists a unique isomorphism Y0 → Y. 0 0 (2) If further N1(π) ⊂ N1(π ) there exists a unique map ρ : Y → Y making X π Y (14.1) π0 ρ Y0 commute Proof. Note the second point implies the first. The plan is to find a line bundle L on X and L basepoint free with L · C if and only if C ∈ N1(π). We may come back to this later. 2 Example 14.5. Now we’ll return to the example of LpP . We have N1 and its effective cone is spanned by L − E and E. The ample cone in N1(X) is the interior of the cone spanned by L − E and L. Note that L − E contracts L − E which induces a map to P1 and shows it is a P1 bundle. This is related to the fact that a map from Pn to somewhere is either finite or contracts to a point. This is related to saying that maps from the Hirzebruch surface are either finite or contract the P1 fibers of the fibration to a point. SURFACES NOTES 29

Example 14.6. Suppose we have a ﬂat family X → Y where the ﬁbers are smooth genus 0 curves. We’d like to ask whether it is a P1 bundle? We can canonically embed X → Y in a P2 bundle by the anticanonical embedding. This is a Zariski P1 bundle if and only if there is a relative degree 1 line bundle on X over Y. Note that if we have this relative degree 1 line bundle, we can pushforward the line bundle and get an embedding of X into a P1 bundle over Y which is an isomorphism. There is always such a line bundle when Y is a curve, by Tsen’s theorem. However, this is not the case in general. We can see this in the universal example. Take Y ,→ P5, the Hilbert scheme of smooth conics in P2. Let X → Y be the universal family of conics. There is a map X → P2. X sits as a dense open inside the relative universal conic (including singular conics) and U is a P4 bundle over P2. X U P2 (14.2)

Y = P5 − ∆ P5 We know Pic U = Z⊕2. We’d like to check that the intersection of the class of a ﬁber of X → Y intersects to even values with the two generators of Pic U. One of the line bundles is the pullback from P2 of points on a line. So we are looking at conics meeting a line. The desired intersection with this class is 2 because a conic meets a line in 2 points. One can check that the other line bundle restricts to a degree 0 line bundle on a conic (essentially one can check the appropriate restriction is the trivial bundle), and so both pullbacks have even degree, and there is no degree 1 divisor.

15. 2/19/20 Last time we were discussing Tsen’s theorem. The main statement was that if you have an etale´ P1 bundle over a smooth curve, it is a Zariski P1 bundle. Lemma 15.1. An étale P1 bundle is the same as a conic bundle (where conic bundle means the fibers are smooth). Proof. Note that being an etale´ P1 bundle is the same as being a conic bundle, 2 because you can embed X → B into PB by using the anticanonical embedding, and the fibers are conics. Conversely, any conic bundle etale´ locally has a section. You can see this by choosing a line meeting the conic transversely. Near the given point, we get an etale´ double cover, and this locally gives a 30 AARON LANDESMAN

section. One gets an etale´ cover away from the locus where the line meets the conics transversely. Last time we had a universal family of conics U (15.1)

P5 P2 where U is a P4 bundle over P2. This is a conic bundle which is not a P1 bundle. We can choose a map Pk → P5. When k ≥ 2 the same argument shows that the pullback of U to Pk is a Pk−1 bundle over P2, so long as we choose the map from Pk so that the conic bundle does not have basepoints. However, when k = 1, it is no longer a P0 bundle over P2 because every pencil of conics has basepoints. So there will be some ﬁbers where the dimension jumps from 0 to 1, and the pullback of U will actually be a blow up of P2 at the base points. Start with a conic bundle (with some singular ﬁbers)

2 X PB (15.2)

B for B a projective curve. Then, X is given as ! V ∑ aijxixj i,j a nondegenerate quadratic form. We’d like to show X → B actually has a rational section. It sufﬁces to show there are sections of OB(Np) for p a point on B. We know the number of sections of OB(Np). So, as N 0 there 2 will exist sections x1, x2 of OB(Np) such that ∑i,j aijx0) = 0. Something is wrong with the above argument because we didn’t use the base ﬁeld was algebraically closed.

16. 2/21/20 Last time, we started with a projective smooth curve B over an algebraically closed field k. Proposition 16.1. Over K(B) for B a smooth projective curve over an algebraically 1 closed field. any étaleconic bundle is trivial. In other words, H (K(B), PGL2) = 0 SURFACES NOTES 31

Proof. We want to show that if we have a conic bundle over K(B) it is a P1 bundle. In P2 × B, over an open set, we have some quadratic polynomial, determined by the conic bundle. We have a family of conics, which corresponds to a map from K(B) to P5, the space of conics, which gives a map B → P5 by the valuative criterion for properness. Suppose this map is given by L a line bundle on B. The conic bundle is defined by ∑ij aijxixj with 0 aij ∈ H (B, L ) for deg L > 0. 0 ⊗m Now, let xi ∈ H (B, L ). Suppose d = deg L > 0. The dimension of the space of choices of 3 sections is 3(md + 1 − g) where g is the genus of B. When we plug into the equation, we get a section of L ⊗2m+1. Therefore, we have an affine space of dimension 3(md + 1 − g) to a space of dimension (2m + 1)d + 1 − g. We therefore have a quadratic map from one affine space to another. We want a nonzero point mapping to the origin. We know 0 in the source maps to the origin. We want to to show there is some nonzero k point mapping to the origin, since this then corresponds to the three desired sections. But indeed, this follows by upper semicontinuity of dimension on the source. Therefore, the fiber over 0 contains a positive dimensional scheme, and hence a nonzero k point because k is algebraically closed. Now, given a projective surface X, we are looking for contractions X → Y. We are looking at the curve classes inside N1(X). We have the closure of the effective cone Eff ⊂ N1(X) which is the closure of effective Q-curves. This is a closed cone. A face of the cone is a supported hyperplane An extremal ray of the cone is a 1-dimensional face of the cone. We’d next like to describe a correspondence between certain irreducible curves in X and extremal rays in the effective cone. Proposition 16.2. Suppose X is a smooth projective surface and C is an irreducible curve in X with C2 = 0. Then, C lies in ∂Eff(X). Proof. Recall that if for any effective D with C · D ≥ 0, we know the class of [C] lies in the closure of the effective locus Eff(X). If C does not lie in ∂Eff(X) then C + tH is effective for all ample H and small t. Then, C(C + tH) ≥ 0 which implies tCH ≥ 0. So, if C · C = 0, not only is C on the boundary, but we can even see that when we move in direction −H, we will leave the boundary of the cone. Proposition 16.3. If C ⊂ X is irreducible and C2 < 0, then C spans an extremal ray of ∂Eff(X).

Proof. Say C = Z1 + Z2 with Z1, Z2 lying in the effective cone. Then, we want to show Zi are actually multiples of C. Say Zi = limn→∞ Di,n for Di,n effective Q-divisors. where limits are taking in the real vector space N1(X). 32 AARON LANDESMAN

0 Now, we’ll write Di,n = ai,nC + Di,n (here there are no copies of C in the 0 0 support of Di,n). Note C · Di,n ≥ 0. Then, H · C = H · Z1 + H · Z2. Therefore, a1,n is bounded by the above, because both are nonnegative. We can therefore replace the ai,n by a convergent 0 subsequence. Then, lim Di,n = Zi − aiC exists where ai = limn ai,n. Then, 0 0 (1 − a1,n − a2,n) C = D1,n + D2,n.

Intersecting with C yields 1 − a1,n − a2,n ≤ 0, but intersecting with H yields yields something nonnegative and gives 1 − a1,n − a2,n ≥ 0. Therefore, we 0 0 must have a1,n + a2,n = 1. This implies limn→∞ D1,n · H + D2,n · H = 0 which 0 0 means D1,n, D2,n must both tend to 0. This is what we wanted since it means Z1 and Z2 both tend to multiples of C. Proposition 16.4. Suppose C is an irreducible curve on X over an algebraically closed field with C2 = 0 and K · C < 0. Then, C spans an extremal ray and X is a P1 bundle over a curve. In particular X has Picard number 2. Proof. Adjunction shows that K · C = 2g − 2 which implies g = 0 so C ' P1 since we are over an algebraically closed field. Remark 16.5. Suppose we have a divisor D ⊂ X with D · H > 0 and (K − mD) · H < 0 when m 0. In particular, K − mD is not effective. Therefore, H0(K − mD) = 0 and H2(mD) = 0 by Serre duality. By the above remark, we find H2(mC) = 0 for m 0. Therefore, 0 1 h (mC) − h (mC) = χ(X, OX) + m. We’ll finish this next time. 17. 2/24/20 We have a smooth projective surface X. We’d like to understand surjective maps X → Y. Said another way, we want to find all contractions of X. The place to look for these is Eff1(X) ⊂ Eff1(X) ⊂ N1(X)R. Last time, we saw certain irreducible curves on X yield extremal rays, and visa versa. We saw: Theorem 17.1. Let X be a smooth projective surface. Let C be an irreducible curve in X with C2 = 0 then C lies in ∂Eff(X). If C is irreducible and C2 < 0 then C spans an extremal ray of Eff1(X). Remark 17.2. The proof idea from last time was to show that in the latter 2 case when C < 0, we supposed C = Z1 + Z2 for Zi ⊂ Eff1(X). Then, we said each Zi was a limit of effectives. We could then write each of these effectives as a multiple of C and something not meeting C. We saw we could take a subsequence where the coefficients of C stabilize and hence the class of the remainder stabilized. We were then able to use this to deduce the SURFACES NOTES 33

result. We’re going to have to do this analytic process as writing elements of the closure of the effective cone as limits of effective divisors again several times. Lemma 17.3 (Lemma A). For any D with D · H > 0 then H2(X, mD) = 0 for m 0. Proof. The proof is essentially Serre duality and which implies that this is H0(X, K − mD) which vanishes because it has negative intersection with H for m large. Theorem 17.4. Let C is an irreducible curve on a smooth projective surface X with 2 C = 0 and K · C < 0 spanning an extremal ray of Eff1(X). Then there is a map X → B for B a smooth projective curve over a ﬁeld and X is a P1 bundle (really meaning a conic bundle over non-algebraically closed ﬁelds) and ρ(X) = 2. Proof. Adjunction implies K · C = −2 and C has genus 0. The plan is now to contract C. We will use O(mC) for m 0. This will contract C because C · C = C · mC = 0. We saw in the above lemma that H2(X, O(mC)) = 0. We’d like to check that O(mC) is basepoint free. To see this, consider

(17.1) 0 OX(−C) OX OC 0 and twisting by mC we get

(17.2) 0 OX((m − 1)C) OX(mC) OC(mC) 0

The last bundle is trivial because OC(C) is O since it is a degree 0 line bundle on a genus 0 curve. Taking cohomology, this gives us

0 0 0 H (OX((m − 1)C)) H (OX(mC)) k

(17.3) 1 1 H (OX((m − 1)C)) H (OX(mC)) 0

2 2 H (OX((m − 1)C)) H (OX(mC)) 0.

Remark 17.5. As an aside this shows H2(X, mC) = 0 for all m. 0 0 For some m we must have H (OX((m − 1)C)) → H (OX(mC)) is not surjective, since H1 is ﬁnite, and so can only drop ﬁnitely many times. In 1 fact, we can even bound m by h (OX) since there must be some time in 34 AARON LANDESMAN

1 1 the first m steps where the map H (OX((m − 1)C)) → H (OX(mC)) is an isomorphism, which implies the map on H0 is not an isomorphism. This 0 then means the map to k from H (OX(mC)) is surjective and so the linear system mC is basepoint free. Now, let F be the class of a fiber associated to the linear system O(mC), with m 0 as above. We can write F = aC for a ∈ Q. Since F2 = 0 and K · F = −2a. From adjunction we find 2gF − 2 = −2a. If we are in characteristic 0, generic smoothness tells us the fiber is reduced. Probably the same argument works in characteristic p, but Ravi wasn’t sure. This implies a = 1 because a is non-negative. We can argue that the fiber must be supported only on C because if it has other curves, the fact that C · C is 0 and the fiber is connected (because this is the stein factorization N X → B → P ) forces the fiber to be supported exactly on C. Lemma 17.6 (Lemma B). We have

n 2 o Z ∈ N1(X)R : Z > 0, HZ > 0 ⊂ Eff(X)R

Proof. We ﬁrst reduce to showing

n 2 o Z ∈ N1(X)Q : Z > 0, HZ > 0 ⊂ Eff(X)Q The point is that any R divisor satisfying the above statement has many Q divisors surrounding it, satisfying the same relations. Therefore, it suffices to prove the statement for all these Q-divisors. By Lemma A, we know H2(X, mZ) = 0. Therefore, (mZ)2 mZK χ(mZ) = h0(mZ) − h1(mZ) = − + χ(O ). 2 2 X This is eventually positive because the m2 term dominates. This forces h0(mZ) to eventually be positive so mZ is effective and Z is effective. 2 Proposition 17.7. If r spans an extremal ray of Eff1(X) then r ≤ 0 or both (Pic X = 1 and r is ample). Proof. We first reduce to the case rH > 0. We know rH ≥ 0 because it lies in the closure of the effective cone. If rH = 0, then r2 < 0 by the Hodge index theorem (if it is perpendicular to the positive H then it lies in the negative definite part of the intersection form). Therefore, we may assume rH > 0. If r2 > 0, by Lemma B above, we know 0 r lies in the effective cone r ∈ Eff(X)Q. The only way r can be an extremal ray and in the interior of the effective cone is if ρ(X) = 1. In this case, we also obtain that r is ample. SURFACES NOTES 35

Theorem 17.8 (Mori’s cone theorem for surfaces). If X is a smooth projective surface there is a countable family of irreducible rational curves C1 on X with −3 ≤ KX · C1 < 0 which are all extremal ray classes in Eff(X) where K · C1 < 0 is polyhedral, with only ﬁnitely many faces satisfying K · r < ε. Example 17.9. Consider P2 → P1 given by two general cubics. This is P2 blown up at 9 points. There are countably many −1 curves on this blow up. Example 17.10. Consider an abelian surface E × E which is a product of elliptic curves. The cone lies in 3 space for general such E with no complex multiplication. 18. 2/26/20

Proposition 18.1. If r ∈ N1(X)R spans an extremal ray of Eff1(X) for X a smooth projective surface and r2 < 0 then the ray is spanned by the class of an irreducible curve.

Proof. We can write r = lim Dm for Dm ∈ Eff1(X)Q. Therefore, r · Dm < 0 2 for m 0 because r < 0. This implies there is a component C of Dm with 0 r · C < 0. This implies Dm = am[C] + Dm with am ∈ Q. Now, for H ample, HDm = amC · H + Dm H

gives an upper bound on am. Therefore, we can replace this by a subsequence 0 such that am converges to a with a ≥ 0. We conclude Dm converges to r − a [C]. 0 2 0 Dm · C converges to r · C − aC . The left hand side is positive because Dm is effective and has no component in common with C. On the other hand, a ≥ 0 and r · C is negative by assumption, so altogether this implies C2 < 0. 0 0 As r is extremal, we ﬁnd r = aC + lim Dm, and since both C and lim Dm 0 are effective divisors, and r is extremal, we ﬁnd C and lim Dm must be dependent and so r is a multiple of C. Hence r is spanned by the class of an irreducible curve. We now restate the cone theorem for surfaces. Theorem 18.2. Let X be a smooth projective surface. Then there exists a countable family C1, C2,..., of irreducible rational curves such that −3 ≤ K · C < 0 and Eff1(X) = {r + ∑ aiCi ∈ Eff1(X) : K · r ≥ 0, ai ∈ R≥0, almost all ai = 0} i

Further, Eff1(X) only accumulates at K · r = 0, meaning that when we are bounded away from K · r = 0, there will only be ﬁnitely many such extremal rays. Lemma 18.3. Assuming the cone theorem, for every surface, we can ﬁnd a minimal model. 36 AARON LANDESMAN

2 Proof. Previously we saw if Ci > 0 then −K is ample and so ρ(X) = 1. We 2 also saw that if Ci = 0 then ρ(X) = 2 and the surface is a projective bundle over a curve B. In these cases, we know the statement holds. 2 2 Hence, we assume Ci < 0. Then, K · C + Ci = 2g − 2. This means 2 K · Ci = −1 and Ci = −1. Therefore, by Castelnuovo’s theorem, we may blow down Ci and repeat. This must terminate because the Picard number to start was ﬁnite.

2 Example 18.4. Suppose we start with a del Pezzo surface which is Bl8 pts P . This is Fano and ρ = 8 + 1 = 9. This has 240 extremal rays. When you blow down one of the curves corresponding to these extremal rays, you get P2 blown up at 7 points. This has 56 extremal rays, and the remaining roots are those perpendicular to the root you blew down. They are perpendicular because when you blow down a curve, the −1 curves which remain are those not intersecting the −1 curve you blew down, because if they intersect it their self intersection number in the blow down will increase.

Example 18.5. Consider a complete intersection of two transverse cubics in P2. This has Picard number 10. There are in fact inﬁnitely many −1 curves on it. Choose one of the −1 curves and call it the 0 section.

Exercise 18.6. Check that every rational section is a −1 curve. Possibly use that the normal bundle of the section in the surface is related to the tangent bundle of the ﬁbers.

We then have E1 as our identity section and we have 8 other sections which do not meet each other. We claim that we have found all the −1 curves as integer linear combinations of (E1 − Ei) for 2 ≤ i ≤ 9, though we won’t check this.

Example 18.7. Let E be an elliptic curve without complex multiplication. Then, E × E has Picard number ρ(E × E) = 3. These are generated by the classes of two ﬁbers and the diagonal. Any curve has C2 ≥ 0. By the Hodge index theorem, H · z = 0. We have a cone with canonical bundle equal to 0. There is therefore the full effective cone lies in the K · r ≥ 0 part using Mori’s theorem. When one works out the intersection matrix, one ﬁnds three classes on the boundary with lots of other stuff. We’ll continue this example next time. SURFACES NOTES 37

19. 2/28/20 The version of Tsen’s theorem we explained last time is that any etale´ P1 bundle over a curve over an algebraically closed field is a Zariski P1 bundle. In fact this holds for Pn bundles as well. Lemma 19.1. Let B be a smooth projective connected curve over an algebraically closed field. Then any étale Pn−1 bundle is a Zariski Pn−1 bundle. Sketch of proof. The key plan is to start with such an etale´ Pn bundle X → B which becomes X0 ' B0 × Pn−1 → B0 for B0 → B etale.´ To trivialize it, we want to construct a line bundle L on X which is a relative O(1) which means it defines an isomorphism X → Pπ∗L for π : X → B. We then have some subschemes of X in B that wants to be the Hilbert scheme of planes in X over B. We want to show this has a B-point. Then, we want to find a section of the dual bundle X∨ which corresponds to finding a B-point of X. So we just need to know that any etale´ Pn−1 bundle over B has a section. The key idea is to produce something on B0. We can write X0 = PV on B0. This PV may not descend, but it turns out that End(V, V) is rank n2 which does descend. This is because the ambiguity of PV descending was a unit, which disappears when you look at End(V, V). So we obtain a 0 vector E = End(V, V)B on B which base changes to End(V, V) on B . The non invertible endomorphisms of V is a discriminant locus, so when you descend this it appears as a map O → Symn E ∨, which is usually called the reduced norm. So, we have E a locally free sheaf on B of rank n2 and the reduced norm is a section of Symn(E ∨). We then have this section defining a divisor ∆ nj in PE . Then, ∆ is given by ∑m am ∏j xj of homogeneous degree n and 0 am ∈ H (B, L ). This gives us a map from B to a projective bundle. We ⊗m then choose xi ∈ Γ(B, L ). When we plug these into the formula for ∆ we obtain a map 2 H0(B, L ⊗m)n → H0(B, L ⊗1+nm). This is an algebraic map given by the polynomial defining ∆ and this is a map of affine spaces. If deg L = e then the source has dimension n2(em + 1 − g) for g the genus of the curve. The dimension of the target is (1 + nm)e + 1 − g. Since n2e dominates, we see the map has positive relative dimension, so there is a nonzero point in the fiber over 0, which gives us a section of ∆ → B. We then get a map from E → E over B0 which has a map which is nonzero and not invertible. The kernel of this map defines a quotient of V which does descend to the base B. This gives us a smaller subbundle our original 38 AARON LANDESMAN

Brauer-Severi variety X over B, which then has a point by induction, and so our original variety has a point as well. Example 19.2. We’d like to understand the effective cone of an abelian surface. Start with Eff1(E × E) for E an elliptic curve with no complex multiplication and let X := E × E. We claim, without proof that N1(X)Q is ⊕3 isomorphic to Q which is generated by F1, F2, ∆ with Fi the two ﬁbers of the projections and ∆ the diagonal. Note that N1(X)Q has dimension at least 3, since these divisors are independent. The hodge diamond has middle row 1, 4, 1 and the Lefschetz 1, 1 theorem tells us N1(X)Q has dimension at most 4, and we’re claiming (though haven’t understood why) this has rank 4 exactly when E has complex multiplication (meaning a bigger endomorphisms group than Z). The intersection matrix of F1, F2, ∆ is   0 1 1 1 0 1 1 1 0 with the ﬁrst column and row F1 and the second F2 and the third ∆. Now, we’ll assume the cone theorem. Proposition 19.3. n 2 o Eff1(X) = z ∈ N1(X) : z ≥ 0, H · z ≥ 0 . Proof. Given a curve, we can translate it to have no components in common with itself to make z2 ≥ 0 and H · z ≥ 0 is true for any effective divisor. We need to check the reverse containment, but we have proven this several classes ago.

Given a class in N1(X)Q, we can write H = F1 + F2. Write z = aF1 + bF2 + c∆. The quadratic form is z2 = ab + bc + ca. We are looking for the part of the quadratic form with z · H = a + b + 2c ≥ 0 and ab + bc + ca ≥ 0. We should see that this is an example with a non-polyhedral cone, but we’ll probably see this next time.

20. 3/2/20 20.1. Abelian surfaces. Let A be an abelian surface over an algebraically closed ﬁeld. Let C be an irreducible curve on A containing the origin 0 ∈ C ⊂ A. Suppose C2 = 0. If one translates C by a point p on C, one again obtains C because otherwise C would intersect this translate at p and have positive self intersection number. Therefore, C must be an abelian subvariety, and therefore an elliptic curve. SURFACES NOTES 39

Let A = E × E for E a curve with no complex multiplication. This should imply ρ(E × E) = 3, though Ravi is still not sure why. Consider the curve

C := {(x1, x2) ∈ E × E : ax1 + bx2 = 0} 2 for a, b ∈ Z. We can write C = αF1 + βF2 + γ∆. We find CF1 = a = 2 2 β + γ, CF2 = b = α + γ, C∆ = (a + b) = α + β. This implies a2 + ab + b2 = α + β + γ. This implies α = b(a + b) β = a(a + b) γ = −ab. The boundary of the effective cone is the set of z with H · z ≥ 0 and z2 = 0. This corresponds to the condition αβ + βγ + γα = 0. We can see α, β, γ satisfying the above relations satisfy this. And then, Ravi claims one can do a computation to check that every solution to this equation is of the form α = b(a + b) β = a(a + b) γ = −ab, though one would have to do a computation if one wanted to check this. 20.2. Indeterminacy of rational maps from a smooth projective surface. Suppose we have π : X → Y a map from a smooth surface to a projective variety. Suppose the map is defined away from finitely many points. We obtain

Γπ ⊂ X × Y (20.1) p X Y Zariski’s main theorem implies that p has connected fibers. If p−1(x) is a single point then p is finite over a neighborhood of x. This implies p is an isomorphism above U, where U. Then, at the points of X where the map π is not defined, it follows that the fiber of p is a curve. Lemma 20.1. In the same situation as above, there is a sequence of blowups of X at points with smooth centers, after which Xe → Y is a morphism. 40 AARON LANDESMAN

n Proof. Suppose we have a rational map X → P . Take D1, D2 as two divisors n in O(1) on P . Choose a reduced point s ∈ D1 ∩ D2 We have Bls X → X. Let Df1 and Df2 denote the strict transforms of D1 and D2 in Bls(X). The strict transforms satisfy Df1 · Df2 < D1 · D2. This invariant decreases, and hence this process terminates. Corollary 20.2. Suppose we have a rational map π : X → Y and Y has no rational curves. Suppose Y is projective and X is a smooth surface. Then, π is a morphism. Remark 20.3. In particular, any rational map from to an abelian variety is a morphism. Proof. We can blow up X to resolve it to obtain a morphism Xe → Y. There will be some P1 in the blow up that is not contracted when mapping to Y, which will force Y to have a rational curve. 20.3. The minimal model program for surfaces. Let X be a smooth projective surface. We’d like to understand when two such surfaces are birational. To start, we take X and blow down all −1-curves. By the cone theorem, either KX is nef, or there exists a rational curve. One result in the subject is the following. Proposition 20.4. Suppose π : X → Y is a birational map from a smooth surface and KY is nef. Then π is a morphism. At this point, the course abruptly ended. More lectures were planned, but we stopped meeting as Arizona winter school happened the next week, and the remaining lectures in the quarter were cancelled.