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 ofcontents with the date.

Of course, these notes are not a faithful representation of the course, eitherin 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 thenotes are to be credited to the lecturer and not the scribe.1 Please emailsuggestions to aaronlandesman@gmail.com

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’llroughly linearly move through the book. One thing Ravi is interested in isunderstanding why proper surfaces are projective. Ravi is also interested inunderstanding the classification of algebraic surfaces in light of the minimalmodel 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 fieldk, and consider separated finite type schemes over k.

We’d like to have as few black boxes as possible. The first few weeks willfocus on

(1) Snappers theorem (with intersection theory in the proper, not neces-sarily 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 torational curves on algebraic varieties and Kleiman’s “Towards a numericalcriterion 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, theEuler characteristic

χ(F ⊗L n11 ⊗ · · · ⊗L nt

t )

is a polynomial in n1, . . . , nt from Zt → Z of total degree at most dim Supp F .

Remark 2.2. The key ideas of the proof is Devissage (a kind of Noetherianinduction).

Suppose some of the coherent sheaves on X are “good,” meaning theysatisfy the theorem statement. Here, “good” sheaves play well in exactsequences (so if two out of three of them are good, then the third is) and forall irreducible subvarieties V then OV is good. We want to show all sheavesare good.

The proof is now by induction on the dimension and length of the supportof the coherent sheaf. One can reduce to the case that the scheme V isirreducible. One can then further reduce to the case V again using inductionon 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 genericpoint. We want to reduce to a rank r− 1 sheaf. This can be accomplished inthe projective case by sufficiently twisting up until we get a section. For thereduction to the proper case, we can use Chow’s lemma.

Proof. We say a sheaf F is good if χ(F ⊗L n11 ⊗ · · · ⊗L nt

t ) is a polynomialof degree at most dim Supp F .

Suppose we are given F , L1, . . . , Lt on X. Let χ(F ⊗L n11 ⊗ · · · ⊗L n2

t )is a polynomial of degree ≤ dim Supp F .

When the dimension of the support of F is 0, we are certainly good. Also,OV is good for V irreducible. On V, χ

(L n1

1 ⊗ · · · ⊗L ntt)

is a polynomial.This can be seen in the projective case because every L is a difference oftwo very ample line bundles, and we can reduce to the very ample case. Weassume V is integral.

We can next reduce to the case V is normal by replacing V by its normal-ization. (This is not working strictly in the context of subvarieties.) Thereason is that χ(G ) for G a sheaf on the normalization ν : V → V thenχ(ν∗G ) = χ(G ).

So, now we may assume V is normal. We may take a rational section ofL1. Given a rational section of L1 we get a Weil divisor D+ − D− on V. SoL1 = 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 beCartier). If we take a cone and the difference of two lines through the conepoint, each line individually is a Weil divisor that is not Cartier, but thedifference 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)

0 ID+ ⊗L n11 ⊗ · · · ⊗L nt

t L n11 ⊗ · · · ⊗L nt

t ⊗L n11 ⊗ · · · ⊗L nt

t 0

(2.2)

0 ID− ⊗L n1−11 ⊗ · · · ⊗L nt

t L n1−11 ⊗ · · · ⊗L nt

t OD− ⊗L n1−11 ⊗ · · · ⊗L nt

t 0

We now see the first two terms of the above exact sequences agree. Thethird 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 firstsequence is a polynomial of the desired degree. �

Remark 2.4. What happened in Snapper’s theorem that was different in theproper vs. projective case? We had to deal with subvarieties V that wereproper, but not necessarily projective. Here we needed Chow’s lemma tomake Devissage work. We also need to use this trick that line bundles aredifferences of divisors, and then induct on the support dimension using theabove exact sequences.

Definition 2.5. We now define (L1 · · ·Lt ·F ) to be the coefficient of n1 · · · ntin Snapper’s polynomial χ(F ⊗L n1

1 ⊗ · · · ⊗L ntt ).

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 bewritten as D+ − D−, a difference of two effective divisors. Here are somefacts 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 kis ample if and only if L restricted to each irreducible component of Xredis ample. Hint: Use Devissage. Call a sheaf “good” if it satisfies the abovecohomological criterion on all components. Then filter the sheaf by sheavessupported 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 isample.

We’ll also need Snapper’s theorem from last time, which says that χ(F ⊗L ⊗n1

1 ⊗· · ·⊗L ntt ) is a polynomial in n1, . . . , nt of degree at most dim Supp F .

We also defined an intersection product (L1 · · ·Lt ·F ) as the coefficientof n1 · · · nt in the above polynomial defined by the Euler characteristic ofχ(F ⊗L ⊗n1

1 ⊗ · · · ⊗L ntt ).

Lemma 3.4. We have

(L1 · · ·Lt ·F ) = χ(F )− χ(F ⊗L ∨1 ) · · · − χ(F ⊗L ∨

t ) + χ(F ⊗L ∨1 ⊗L ∨

2 )) · · ·and this is symmetric and multilinear.

SURFACES NOTES 5

Proof. The key trick for multilinearity is that (L ·L ′ ·F ) = (L ·F ) + (L ′ ·F )− ((L ⊗L ′) ·F ), but we can assume the dimension of the support ofF 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 genericallyfinite. 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 bundleon X. Then, L is ample if and only if for all integral subvarieties V ⊂ X, we haveL · · ·L︸ ︷︷ ︸

dim V

V

=

L · · ·L︸ ︷︷ ︸dim V

·OV

X

> 0.

Proof. One implication is easy, so we will just focus on showing that if therestriction of L to each subvariety has positive self intersection, then L isample.

We can first reduce to the case X is integral, using Devissage, as we showedin 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 integralsubvarieties V ⊂ X.

Our goal is now to show the following. Suppose X is integral, L a linebundle, (L dim X) > 0 and L ample on all proper subvarieties, and F iscoherent on X. We want hi(X, L ⊗n) = 0 for i > 0 and n� 0. We can replaceX 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

Hi(OX(mD)) ' Hi(OX((m− 1)D)).

We also know χ(O(mD)) is a polynomial in m with leading term Ddim X

(dim X)! mdim X.

6 AARON LANDESMAN

One can see this using that if we consider χ(L n1 · · ·L nt) as a polynomialof degree at most t, with coefficient of n! · · · nt given by Ddim X. We also havethat 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(n1 + · · ·+ nk)

t.Actually, we’ll revise the proof, and won’t demonstrate the cohomological

criterion, but just produce the map to projective space. We know that form � 0 we have h0(mD) � 0 (since hi for i ≥ 2 vanish as we have shownabove). Hence, we may assume D is effective. We then have an exactsequence

(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 meanswe have a surjection

(3.5) H0(O(mD) H0(OD(mD)) 0 0

We basically know some multiple of D is generated by global sections, whichwill give us a map to projective space. This proof will be continued nexttime. �

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 theabove 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 propersurfaces. The normality hypotheses are so that we can work with Q-Cartierdivisors, and so that we can work with valuations along codimension 1points.

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 tothe 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 directionproving that L is ample.

Using induction, we saw we could assume that L is ample on everyscheme 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. Thisproves 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 issmaller dimensional, we know H1(D, O(mD)|D⊗) = 0 for m � 0. Sothis implies H1(X, L m) stabilizes, because there are surjections H1(X, (m−1)D)→ H1(X, mD).

We have

χ(L m) =

(L · · ·L︸ ︷︷ ︸dim X

)Xmdim 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 vanishingexactly on D.

We know H0(X, O(mD)) → H0(D, O(mD)|D) is surjective for m � 0because H1 stabilizes. Since O(mD)|D is very ample on D, it is base pointfree, so we can then lift this to a section of O(mD) on X, and hence this isbase point free. �

Hence, we get a map X → PN by O(mD). We claim this is finite onto theimage. We need to show the map has finite fibers (since it is automaticallyproper). 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 Cwere contained in a fiber, by the projection formula. This contradicts ourassumption. Therefore, there can be no curves contained in fibers. �

Ampleness is now a numerical property in terms of how a line bundleintersects all subvarieties. Assuming X is normal, we can define amplenessfor 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 subvarietiesY ⊂ X if

(L · · ·L︸ ︷︷ ︸dim Y

·Y) ≥ 0.

Remark 4.4. If L is ample, it is certainly nef. Further, L is nef on X if andonly if L is nef on each component.

Lemma 4.5. If L is nef and H is ample, then L ⊗H is ample. Said anotherway, “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 toargue the cross terms are positive. Above, we are using raising to the dim Ypower to indicate the self intersection dim Y times (and not the dim Ythtensor power). �

It turns out that a sum of nef divisors is again nef, but we haven’t yetshowed that. Kleiman showed that the ample cone is the interior of the nefcone.

Definition 4.6. Let L be a line bundle on X. We define nef≥r if for all integralY ⊂ 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 ispositive.

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 isample, then

(L s ·H d−s)Y ≥ 0.

This holds again because we can reduce to the case H is very ample, andthen slice by members in the divisor class of H and use the hypothesis thatL is nef.

SURFACES NOTES 9

Proof of Lemma 4.7. We can easily reduce to the following special case: Sup-pose X has dimension r and is integral, and L ∈ nef≤r−1. We want L dim X ≥0. If H is ample (which exists since we are assuming X is projective) con-sider D + tH with H = O(H) and t ∈ Q. Consider P(t) = (D + tH)r. IfDr < 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 weexpand (D + tH)r−1, we see that each monomial in the expansion is positive.Further, trHr is positive because t0 > 0. Therefore, this second term is strictlypositive, 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 useN1(X) as curves up to numerical equivalence and N1(X) denote divisors upto numerical equivalence. We consider these as vector spaces over Q. We letNE1 denote the classes of effective curves, and NE1 denotes its closure in theassociated real vector space.

(1) Then, a Q Cartier Q-divisor D is ample if and only if D · Z > 0 forevery 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 thenef cone whose interior is the ample cone.

5.1. Finishing the Proof of Kleiman’s criterion. From last time, we provethe 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 forN1 consisting of ample divisors. Let D1, . . . , Dn denote such a basis. Then,

10 AARON LANDESMAN

there is some fixed m so that H± 1m 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 abox as above, since we have bounded each of the coordinates. This shows itis 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 otherhand, by adjunction it is (KC×C + ∆) · ∆ = 2(2g− 2) + ∆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 thegraph. This has self intersection −(2g− 2). Since KC×C is ample. Hence,there are only finitely many numerical equivalence classes of curves Z withK · Z = 2g− 2.

Note that two different automorphisms correspond to different numericalequivalence classes because they intersect each other positively, while theyintersect themselves negatively (with degree 2g− 2. Further, the self inter-section of the graph of an automorphism agrees with the self intersectionof ∆ because there is an automorphism of the surface given by id×σ takingone 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 truethat X has finitely many automorphisms.

5.2. Behavior of cones under maps. For X an arbitrary irreducible projec-tive variety, consider N1(X) which is dual to N1(X) and N1(X)R which isdual to N1(X)R. We also have the ample cone and effective cone insideof these. We would like to understand how this interacts with maps be-tween varieties. Given X → Z we can factor this through a surjective mapπ : X → Y.

For surfaces, most X have a unique smooth minimal model. The sort ofthings 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 takethe canonical map it not ample, and quite negative, and maps the surfaceto a curve of lower dimension. In higher dimensions, we might hope forthe statement that when the canonical bundle ample, there will be a uniquesmooth minimal model.

SURFACES NOTES 11

Theorem 5.4. Let X be an arbitrary projective irreducible variety and a mapπ : X → Y, with Y normal. Let H be an ample divisor on Y. Then, π∗H restrictsto 0 on any curve which is contracted. There is a cone in NE1(X) ⊂ N1(X). Thepullback π∗H defines a functional on NE1(X), and defines a face (possibly of lowerdimension) of NE1(X) determines π.

We’ll come back to proving this theorem in several classes. What getscontracted 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 con-struct an ample divisor. Thanks to Nakai-Moishezon, we need to check it isnonnegative on every curve and has positive self intersection.

Start with an affine open U. The complement of U has pure codimension1, so it is a union of curves. The ample divisor we’ll come up with is somepositive combination of these curves. We need to show it meets every curvepositively, and itself non-negatively. This is true if the curve we are meetingmeet the affine open. So, we just need to cook up the multiples of the curvesin the complement of U so that the intersection of this with each such curveis positive.

5.4. Hodge Index Theorem. Here is a nice linear algebra fact Ravi reallylikes, but didn’t know before. This is in the lead up to the Hodge indextheorem.

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 isanother 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). Thenfor 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, thedimension of the radical is the number of connected components of the graph formedby joining ei to ej if Bq(ei, ej) > 0.

It seems this statement may have a typo because Ravi wants the radical tobe 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 arecontracted. For example, it could be a blow up, or it could be a sort offib-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 dueto 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 bothh0(nD) → ∞ and h0(−nD) → ∞. And similarly as n → ∞, we can’t haveboth h0(K + nD)→ ∞ and h0(K− nD)→ ∞.

Altogether, Riemann Roch as above shows as n→ ±∞, we know one ofh0(nD)→ ∞ or h0(K− nD)→ ∞, but the same one cannot go to infinity bythe 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) → ∞ asn→ ∞ 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, thenD2 ≤ 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 Hample 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 andD2 = 0. Whet if D 6≡ 0? Then, there is some E with which D · E 6= 0, becausethe pairing on N1 and N1 is nondegenerate.

SURFACES NOTES 13

Set F = E + αH and note that D · F = D · E = 0 because D · H = 0. Wehave

H · F = H · E + αH · HChoose 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.Now, replace D by Dn := nD + F. Then, D2

n = n2D2 + 2nDF + F2 =2nDF + F2 and note that D · F 6= 0. But Dn · H = 0. By choosing n verypositive or negative, we know D2

n > 0

Exercise 6.5. To make the above proof complete, check that N1(X) = N1(X).

�

Corollary 6.6. The signature of the quadratic form on N1(X)R given by intersect-ing 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 spaceto the ample class H is negative definite. The previous hodge index theoremshowed it was semidefinite. �

Exercise 6.7. Let X be a projective smooth surface D1, D2 with D21 > 0. Show

D21D2

2 ≤ (D1D2)2.

6.1. A Proposition from linear algebra.

Proposition 6.8. Let B be a symmetric bilinear form on a Q vector space M. Letei 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 ofconnected components of the adjacency graph of the ei (where ei is connected to ej ifei · ej 6= 0).

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

Proof. We can begin by replacing ei by aiei to assume z = ∑i ei. Write x =∑i ciei. Then,

x · x = ∑i

c2i ei · ei + 2 ∑

i<jcicj(ei · ej) ≤∑

ic2

i ei · ej + ∑i<j

(c2i + c2

j )(eiej)

= ∑i

c2i (ei ∑ ei)

= 0.

For the statement about null spaces, we now assume the adjacency graphis connected. We then conclude from AM-GM that in order for equality tohold, we must have all ci = cj (using connectivity of the adjacency matrix).So, only ∑i ei = 0 can lie in the null space.

14 AARON LANDESMAN

Remark 6.9. In fact, the proof shows that the only possible dependencerelation 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 bignessmore. 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) =

mn Dn

n! + O(mn−1).

Proof. We’ll start with the first part. Write D = E1 − E2. We have exactsequences

(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

h0(OX(mD)) ≤ hi(OX((m− 1)D)) + O(mn−1).

Therefore, hi(OX(m0D)) = ∑m0m=0 hi(OX(mD))− hi(OX(m− 1)D) = m0O(mn−1

0 ) =O(mn

0).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

h1(OX(mD)) ≤ h1(OX((m−m0)D)) + O(mn−2)

and by induction this means h1(OX(mD)) = O(mn−1).

SURFACES NOTES 15

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

lim supm→∞

h0(X, mD)

mn > 0.

Remark 7.3. A line bundle is nef and big then it is in particular nef, soDn > 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. Wehave already seen it is nef. We need to check the sum of two big divisors isbig, assuming they are nef. Just expand (D1 + D2)

n and we know each termin the expansion is non-negative because they are nef and then Dn

1 is strictlypositive, so (D1 + D2)

n is strictly positive.

Lemma 7.5. Suppose X is projective. If D is nef and big, then there is an effectiveE 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

For m� 0, we know OH(mD) is O(mn−1) while h0(OX(mD)) = O(mn), soh0(OX(mD−H)) is effective. This shows there is some E with E′ = mD−H.Let mE = E′ which is an effective Q-divisor. Then, D− E = 1

m H, which isample.

We will now show D − tE is ample for 0 < t ≤ 1. Indeed, D − tE =tm H + (1− t)D which is the sum of an ample and a nef. Here, (Q-)amplemeans that some multiple is very ample. �

Remark 7.6. Something we might want to say that if D is nef and big, thenthere exists m > 0 so that O(mD) gives a rational map X → Pr with imageof dimension n. We take R := ⊕H0(X, mD), and take proj. Suppose thisis finitely generated (which is hard). This gives a map to projective spacecorresponding to the surjection H0(O(mD) ⊗Ok OX → O(mD) given byevaluation of sections and using bigness tells us the dimension, because ithas 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 withgeometrically connected fibers, any fiber has intersection form which is non-positive definite, and has only a single 0 eigenvalue (and all other eigenvaluesstrictly 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 smoothY 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 · Ejis negative definite.

Remark 8.2. The vague idea is that we can add in an additional divisorwhich intersects every component quite positively, and then the resultingmatrix satisfies the linear algebra lemma we saw last time.

The flavor of the general statement is that given anything negative youshould 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 grouplaw, if p corresponds to the identity on the elliptic curve. There is no curvein P2 which meets E only in the point x0, since x0 is not torsion.

Now, blow up P2 at x0. Then, blow up the point over X0 in the propertransform of E in the blow up. The self intersection is now 8. Then repeatthis 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 imageof the contraction. Choose a curve in Y missing y. The preimage in thesuccessive 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 nottorsion, which means there is no D meeting E only at x0.

Indeed, because O(d)|E = NO(x) we obtain Nx = 3dp, which meansx− p is 3d torsion.

Remark 8.4. Ravi said he would bet up to $10 that if x0 were torsion in theabove 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 geometricallyconnected 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 isomorphismexcept in the fiber over a single point y whose preimage is E. Additionally,χ(OX) = χ(OY).

SURFACES NOTES 17

(2) The intersection matrix (EicEj) is negative definite and pa(Z) ≤ 0 for eachpositive 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 takea linear combination of the Ei and compute the genus using adjunction.

Example 8.7. Suppose p ∈ X is a rational double point (meaning the singu-larity is rational, and is a double point, which is something like the tangentcone has degree 2)

It is also true that if you have a −1 curve, you can contract it, and theresulting contracted surface is smooth. The smoothness is the hard part, andyou can compute the nth formal neighborhood of a point in the contractedsurface and its preimage corresponds to powers of the ideal sheaf of the −1curve.

It is also true in the case of rational double points that the preimage of thenth power of the ideal sheaf of the double point is the nth power of the idealsheaf of the contracted curve.

8.1. Facts about curves which analytically locally embedable in P2. LetE1 ∪ · · · ∪ En be a geometrically connected scheme in a surface X and letZ = ∑ri≥0 riEi with ri ∈ Z.

Lemma 8.8. The following are equivalent:(1) H1(OZ) = 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/Z′ OZ OZ′ 0.

Taking the long exact sequence in cohomology we get

H1(Z, OZ)→ H1(OZ′)→ 0

which shows H1(OZ′) = 0. This implies pa(Z′) = 1− h0(OZ′) + h1(OZ′) =1− h0(OZ′) ≤ 0.

For the other direction, once you’ve chosen Z, one can induct on Z byadding an appropriate Ei. But we omit the proof. �

Fact 8.9. This is a fairly difficult fact.There is a map d : PicZ/k(Spec k)→ Zn 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

Moreover there is a map Zn → PicZ/k(Spec k) is given by sending (ti) toOX(∑i tiEi)|Z.

If the above four equivalent conditions (1), (2), (3) and (3′) are all true,then H1(OZ) = 0 then every line bundle on Z has degEi

(L) ≥ 0 globallygenerated and h1(L) = 0 Riemann Roch implies h0(L) = ∑i ridi + h0(OZ).

A surface which has resolution satisfying any of the above properties, issaid to have rational singularities.

9. 1/31/20

Suppose we start with a map f : X → Y which contracts some config-uration of curves Ei on X with E := ∪iriEi connected. Suppose f−1(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.(d) Let Z = ∑i riEi with ri ≥ 0 (not all 0). Then H1(OZ) = 0.(e) Let Z = ∑i riEi with ri > 0. Then there is a canonical map

PicZ/k(Spec k)→ Z⊕n

L 7→(

degE1L1, . . . , degEn

L)

which is injective and has image of finite index Equivalently there is amap Zn → Zn sending L := O(∑i riEi)|E to

(degEi

L)

, and that mapshould factor through PicZ/k(Spec k) with finite kernel.

(a) R1 f∗OX = 0(b) χ(OX) = χ(OY).

Proof. The idea is that c and d imply all components are rational and form atree, which will imply e. To go back from e to d, we can see the curve mustbe a tree of genus 0 curves, and H1(OZ) is the tangent space to Pic, whichwill 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-ates, so Hi(Y, OY) ' Hi(Y, f∗OX) ' Hi(Y, OY). The equivalence of (a) and(d) follows because R1 f∗OX Let E = ∑i riEi is supported on the point y and

is equivalent to H1(E, OE) = 0. (R1 f∗OX)y = limri→∞ H(OZ). �

SURFACES NOTES 19

Theorem 9.2 (Artin). Suppose we have a map f : X → Y as in the previoustheorem. The following are equivalent

(1) The map f has rational singularities(2) the matrix

∣∣(Ei · Ej)∣∣

i,j is negative definite and pa(Z) ≤ 0 for all Z =

∑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 isunique, in the sense that any other Y′ which realizes such a contraction is isomorphicto Y.

Proof. We start with a very ample line bundle H on X. We want to tweak Hby 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→(

∑i

riEi

)Ej

with image contained in k(Z⊕n) for some integer k. This is contained inim ψ.

So if we replace H by kH, we will have the restriction map lying in theimage. Hence, we can arrange H + ∑i riEi = Z with ri > 0. We haveZ · Ei = 0 for all i.

We will contract this divisor Z. There will also be a section not vanishingat any given point which gives basepoint freeness; we’ll see this by realizinga certain cohomology vanishes. We’ll similarly see it separates points andtangent vectors which gives an isomorphism away from these curves and allthe curves get mapped to a point. �

10. 2/3/20

Let X be a smooth projective surface over k and let E = E1 ∪ · · · ∪ En. Wehad Artin’s theorem on contractions:

Theorem 10.1 (Artin contraction). Let X be a smooth surface over a field k andE = E1 ∪ · · · ∪ En a connected divisor. Then, the following are equivalent:

(1) There exists f : X → Y contracting only E and χ(OX) = χ(OY).(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

PicX/k(Spec k)→ (Zn)∨ .

20 AARON LANDESMAN

There is a composition map

Zn → (Zn)∨

(r1, . . . , rn) 7→((

∑i

riEi

)· E1, . . . ,

(∑

iriEi

)· En

).

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 haveZ · Ei < 0 since H is very ample.

Lemma 10.2. Suppose Z = ∑i riEi. Suppose the intersection matrix (Ei · Ej) isnegative 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 thatthe Ei’s in the support of Z1 are disjoint from those in the support of Z2. Thisimplies 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.

Taking the long exact sequence on cohomology gives h1(OX(H)) = 0. So weget an exact sequence on H0. h0(OE) = 1. So, the sections of OX(H + Z) arebasepoint free away from E and also basepoint free away from E, using thatH is ample. They separate points and tangent vectors away from E, whilethe map contracts E. We can produce a section which vanishes on E anddoes not vanish at a specified other point using the exact sequence. This tellsus 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.Consider the map Zn → (Zn)∨ given by sending a line bundle L = ∑i riEito its pairing with each of the curves E1, . . . , En.

Lemma 10.4. There is a smallest effective class Z = ∑i riEi meeting each Ejnegatively. This Z is called the fundamental cycle.

Proof. Suppose Z′ = ∑i r′iEi and Z′′ = ∑i r′′i Ei both meet each of the Ei’snegatively. I claim that Z = ∑ min(r′i, r′′i )Ei works too.

Suppose r′1 ≤ r′′1 . Then, we can compute E1 · Z as

r′1E1 · E1 + ∑i>1

min(r′i, r′′i

)E1 · Ei ≤ r′1E1 · E1 + ∑

i≥1r′iE1 · Ei = Z′ · E1 < 0.

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 abovelemma. We’ll see for f : X → Y a contraction, O(−nZ) = f−1(mn

Y). Youcan then use this information to work out the degree of the singularity, theembedding deformation, and tangent information, etc. You can work out thedegree of the singularity by looking at sections of O(−nZ) as n grows, andso you get a formula in terms of the geometry of Z. You can find when thesingularity is a double point, which are classified by ADE singularities.

Definition 10.5. Let X be dimension n with p ∈ X an isolated singularity.PTX,p ⊂ PN 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 doublepoint. You can work this out by looking at h0(O(n)) = deg OZ(n) + C =n deg OZ(1) + C = −Z2 + C. This shows that the Hilbert polynomial hasdegree −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 isa surface and y ∈ Y is a singular point and f : X → Y is a resolution of y withspecial fiber E over y. Then, y ∈ Y is rational if R1 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 intersectionmatrix |Ei · Ej| negative definite. and pa(Z) ≤ 0 for all Z > 0 supported onthe Ei. Then we saw by Artin’s theorem, we can contract E. We saw last timethat there is a fundamental cycle Z = ∑i riEi with ri > 0 and Z · Ei ≤ 0 forall i, which is minimal in the sense that there is no Z′ with all r′i ≤ ri suchthat 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) pa(Z) ≥ 0(2) pa(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 rationalsingularities, there is just one thing we need to compute.

Proof. Note that pa(D1 + D2) = pa(D1) + pa(D2) + D1 · D2 − 1. If Z′ > 0then Z′ 6≥ Z, there is some Ei with Z′ · Ei > 0. Then, pa(Z′ + Ei) = pa(Z′) +pa(Ei) + Z′ · Ei − 1, using the above fact. Note that pa(Ei) ≥ 0 (since Ei isintegral) and Z′ · Ei − 1 ≥ 0 by assumption. Suppose pa(Z) = 0. Our goal isto 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 eventuallyget to Z, or a multiple of Z. The issue is to show that there will be some Eiwhich will “move Z1 toward Z” which satisfies Z1 · Ei > 0.

Suppose Z = ∑i riEi. Take Z′ = ∑i r′iEi with r′i ≤ ri and not all ri are equal.Then, we want to show there is some Ei < ri and Z′ · Ei > 0.

But, if r′i = ri, then Z′ · Ei = ∑i r′jEj · Ei ≤ rjEj · Ei = Z · Ei = 0. Therefore,we’re “not allowed to increase Z′ by Ei” since Z′ · Ei < 0 if r′i = ri. Therefore,since there must be some i with Z′ · Ei > 0, it must occur at an i for whichr′i ≤ ri.

Remark 11.2. The proof gives a nice way of producing Z, since you can startwith 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 justwant 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, sincewe 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. Sincepa(Z) = 0, it must be that pa(Ej) = 0 for all j, since we start with 0 and addnonnegative 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 ateach step, Zj · Ei − 1 = 0.

Now, we start with Z′ = 0 and add some Ei for which Z′ · Ei > 0. Wewere following the strategy that if there exists some Ei with Z′ · Ei > 0, weadd Ei, thereby increasing the genus (or at least keeping it constant). Ifthere 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

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 ·(Z′ − Z) ≤ 0. It follows that pa(Z′ − Z) ≥ pa(Z′). So, pa(Z′ − Z) is strictlybigger than pa(Z′).

So, to summarize the process, either some r′i < ri, and then we add Eito Z′. If all ri ≥ ri, subtract Z. At each stage, the genus does not increase,and so the genus of Z′ is bigger than that of Z. This shows the singularity isrational. �

12. 2/10/20

Suppose we are in Artin’s setup as in Theorem 9.2 and Z is a fundamentalcycle.

From last time, we owed the following lemma:

Lemma 12.1. If D and E are effective divisors on a surface meeting in dimension 0then 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 + E2

=(K + D)D

2+

(K + E)E2

+ D · E.

�

12.1. Computing the fundamental cycle. We want a fundamental cycle Zwith Z ≥ E1 + · · ·+ En with Z · Ei ≤ 0 which is minimal. Let’s see someexamples.

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

1 −2 10 1 −2

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

Example 12.3. Let’s consider a Dn singularity, which has dual graph givenby 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 currentintersection is positive −2 + 1 + 1. So we then have Z′ := E3 + ∑i Ei We

24 AARON LANDESMAN

next find E4 does has negative intersection with Z′ so far. And we get Z′ :=E3 + E4 + ∑i Ei. Continuing in this way, we find Z := ∑n−1

i=3 Ei + ∑ni=1 Ei is

the fundamental cycle.

We now want to check the above algorithm we were implicitly followingworks. The crux is the following, which we basically saw last class.

Lemma 12.4. Suppose Z′ = ∑i siEi < Z and Z′ · Ei > 0 then (Z + Ei) ≤ Z.

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 correspondingto an ideal sheaf m ⊂ OY (so m = Iy/Y. Then, f ∗mn = I n

Z/X. Then,

H0(OY/mn) ' H0(OX/I nZ ).

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 smoothbecause it has 2 dimensional tangent space. �

Definition 12.7. If y ∈ Y is an isolated singularity, we define the multiplicityof the singularity at y ∈ Y as the degree of PTCY,y ⊂ P, the projectivizedtangent cone. Here P is Sym• H0(m/m2) 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 singu-larity is a rational double point and Z2 = −1 if and only if the singularity issmooth. �

Theorem 12.9. Suppose we are set up as in Theorem 9.2. The following are equiva-lent.

(1) pa(Z) = 0 and Z2 = −2, i.e., Y has a rational double point singularity.(2) Z2 = −2 and Z · K = 0.

If there are no Ei with both (E2i = −1 and pa(Ei) = 0) then the above are equivalent

to(c) pa(Ei) = 0 and E2

i = −2.

SURFACES NOTES 25

Proof. First, let’s assume There are no Ei with both (E2i = −1 and pa(Ei) = 0).

Let’s show that (b) implies (c).We have K · Ei = 2pa(Ei)− 2− E2

i ≥ 0 If Z = ∑i riEi then Z · K = 0. Then

0 = Z · K = ∑i

riK · Ei = ∑(

2pa(Ei)− 2− E2i

)Then, by our assumption that the intersection matrix of the Ei is negativedefinite but we do not have both E2

i 6= −1 and pa(Ei) = 0, we find each termin the sum is nonnegative. This implies pa(Ei) = 0 and so E2

i = −2. Then,Z · K = 0.

We’ll now show (c) implies (a). We know pa(Z) ≥ 0 because Y is acontraction of X, and so any effective nonzero divisor has positive genus.

We have 2pa(Z)− 2 = (K + Z) · Z = Z2. IT follows that Z2 ≥ −2. Further,Z2 ≤ 0 by negative definiteness, and we see Z2 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, thesingularities are isolated. We can resolve the singularities (even to a minimalproper regular surface in a canonical way) so that we can understand thesingularities exactly in terms of the configuration of curves over them. Atthe very least, if you take some resolution, your singularity is determined byits fiber.

As usual, we retain the setup as in Theorem 9.2. If no Ei is a −1 curve, i.e.,Ei has genus 0 and E2

i = −1 recall last time we showed the equivalence of

(1) pa(Z) = 0, Z2 = −2(2) pa(Z) = 0, Z · K = 0(3) pa(Ei) = 0, E2

i = −2.These are rational double point singularities.

A map is crepant if the pullback of the canonical divisor is the canonicaldivisor. 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.

Example 13.2. If we consider x2 + y2 + zn+1 = 0 is an An singularity. Wecan similarly get An singularities in higher dimension xn+1

0 + ∑i x2i = 0.

26 AARON LANDESMAN

Example 13.3. There are also Dn singularities x2 + y2z + zn−1 = 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 onlyrational 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 doublepoint singularities.

(1) Suppose we have a configuration of curves giving a rational doublepoints.

(2) We know each divisor has to have genus 0. Hence, the only singulari-ties 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 intersectionmatrix. Indeed, (2E0 + E1 + E2 + E3 + E4)

2 = −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 compo-nents) we get an An singularity.

(7) If there is a triple point, we can’t have a triple point Ei where each ofthe 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, thedivisor (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 withvalence 2.

SURFACES NOTES 27

(9) We cant have E0 a triple point with branches of length 1, 3, 3 comingout of it and can’t have E0 a triple point with extensions of length1, 2, 5. In the first case, we put a 4 on the triple point, 2 on the vertexchain of length 1 and 3, 2, 1 on the two other chains. Taking selfintersection 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 1branch of multiplicity 3, and length 2 branch with multiplicities 2, 4and 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 andEn both have triple points with F, G adjacent to E0 and H, K adjacentto 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 thespecific singularity equation we have written. For example if we want an E8singularity, they are etale locally isomorphic to x3 + y3 + z5. Or, relatedly, theto understand the An singularity, you can contract one curve at a time, andrelate 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 withY affine. Let m be a singularity in the target Y and a projective morphism with fiberover m connected. Assume R1 f∗OX = H1(OX) = 0. Then, for m ∈ Spec A, wehave f ∗(m) is an effective Cartier divisor, (of pure dimension 1 with no associatedpoints) or equivalently, the ideal sheaf of the pullback of m is O(−D) for D aneffective 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 andjump forward in time! We’ll go back to the following question:

Example 14.1. Suppose we have some variety X and we want to understandcontractions in the hopes of making a minimal model or otherwise under-standing 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 lookfor whether something should secretly be fibered over a curve by lookingfor a configuration of curves where the intersection number is nonpositivedefinite, with one 0 eigenvector.

We’d now like to understand contractions in terms of the intersectiontheory of the variety.

Example 14.2. Consider X = Blpt P2. Here, N1 := N1(X) = Z2. This isgenerated by E and L, where E is the exceptional divisor and L is the class ofa 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 obtaina map π∗ : N1(X) → N1(Y) and π∗ : N1(Y) → N1(X). For β ∈ N1(X), α ∈N1(Y) then (π∗α) · β = α · (π∗β). If π is surjective, then π∗ is surjective andπ∗ is injective. Then N1(π) := ker π∗ and N1(π) := coker π∗.

Lemma 14.3. For π : X → Y, Y is integrally closed in K(X) over K(Y) if andonly if OY → π∗OX is an isomorphism (i.e., π is O-connected).

Proposition 14.4. Given X, Y, Y′ projective integral and normal, π : X → Y andπ′ : X → Y′ projective with π and π′ O-connected. Suppose further N1(π) =N1(π

′).(1) Then there exists a unique isomorphism Y′ → Y.(2) If further N1(π) ⊂ N1(π

′) there exists a unique map ρ : Y → Y′ making

(14.1)X Y

Y′

π

π′

ρ

commute

Proof. Note the second point implies the first. The plan is to find a linebundle L on X and L basepoint free with L · C if and only if C ∈ N1(π).We may come back to this later. �

Example 14.5. Now we’ll return to the example of LpP2. We have N1 andits effective cone is spanned by L− E and E. The ample cone in N1(X) is theinterior of the cone spanned by L− E and L. Note that L− E contracts L− Ewhich 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 eitherfinite or contracts to a point. This is related to saying that maps from theHirzebruch surface are either finite or contract the P1 fibers of the fibrationto a point.

SURFACES NOTES 29

Example 14.6. Suppose we have a flat family X → Y where the fibers aresmooth genus 0 curves. We’d like to ask whether it is a P1 bundle? We cancanonically 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 linebundle on X over Y.

Note that if we have this relative degree 1 line bundle, we can pushforwardthe line bundle and get an embedding of X into a P1 bundle over Y which isan 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. LetX → Y be the universal family of conics. There is a map X → P2. X sits asa dense open inside the relative universal conic (including singular conics)and U is a P4 bundle over P2.

(14.2)X U P2

Y = P5 − ∆ P5

We know Pic U = Z⊕2. We’d like to check that the intersection of the class ofa 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 weare looking at conics meeting a line. The desired intersection with this classis 2 because a conic meets a line in 2 points. One can check that the otherline bundle restricts to a degree 0 line bundle on a conic (essentially one cancheck the appropriate restriction is the trivial bundle), and so both pullbackshave even degree, and there is no degree 1 divisor.

15. 2/19/20

Last time we were discussing Tsen’s theorem. The main statement wasthat if you have an etale P1 bundle over a smooth curve, it is a Zariski P1

bundle.

Lemma 15.1. An etale P1 bundle is the same as a conic bundle (where conic bundlemeans the fibers are smooth).

Proof. Note that being an etale P1 bundle is the same as being a conic bundle,because you can embed X → B into P2

B by using the anticanonical embed-ding, and the fibers are conics. Conversely, any conic bundle etale locally hasa 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 meetsthe conics transversely. �

Last time we had a universal family of conics

(15.1)U

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 argumentshows that the pullback of U to Pk is a Pk−1 bundle over P2, so long as wechoose 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 everypencil of conics has basepoints. So there will be some fibers where thedimension jumps from 0 to 1, and the pullback of U will actually be a blowup of P2 at the base points.

Start with a conic bundle (with some singular fibers)

(15.2)X P2

B

B

for B a projective curve. Then, X is given as

V

(∑i,j

aijxixj

)a nondegenerate quadratic form. We’d like to show X → B actually hasa rational section. It suffices to show there are sections of OB(Np) for p apoint on B. We know the number of sections of OB(Np). So, as N � 0 therewill exist sections x1, x2 of OB(Np) such that ∑i,j aijx2

0) = 0. Something iswrong with the above argument because we didn’t use the base field wasalgebraically closed.

16. 2/21/20

Last time, we started with a projective smooth curve B over an alge-braically closed field k.

Proposition 16.1. Over K(B) for B a smooth projective curve over an algebraicallyclosed field. any etale conic bundle is trivial. In other words, H1(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 corre-sponds to a map from K(B) to P5, the space of conics, which gives a mapB→ P5 by the valuative criterion for properness. Suppose this map is givenby L a line bundle on B. The conic bundle is defined by ∑ij aijxixj withaij ∈ H0(B, L ) for deg L > 0.

Now, let xi ∈ H0(B, L ⊗m). Suppose d = deg L > 0. The dimension ofthe space of choices of 3 sections is 3(md + 1− g) where g is the genus ofB. 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 spaceto another. We want a nonzero point mapping to the origin. We know 0 inthe source maps to the origin. We want to to show there is some nonzero kpoint mapping to the origin, since this then corresponds to the three desiredsections. But indeed, this follows by upper semicontinuity of dimensionon the source. Therefore, the fiber over 0 contains a positive dimensionalscheme, 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 theeffective cone Eff ⊂ N1(X) which is the closure of effective Q-curves. This isa closed cone. A face of the cone is a supported hyperplane An extremal rayof the cone is a 1-dimensional face of the cone.

We’d next like to describe a correspondence between certain irreduciblecurves in X and extremal rays in the effective cone.

Proposition 16.2. Suppose X is a smooth projective surface and C is an irreduciblecurve 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 andsmall 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 indirection −H, we will leave the boundary of the cone. �

Proposition 16.3. If C ⊂ X is irreducible and C2 < 0, then C spans an extremalray of ∂Eff(X).

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

32 AARON LANDESMAN

Now, we’ll write Di,n = ai,nC + D′i,n (here there are no copies of C in thesupport of D′i,n). Note C · D′i,n ≥ 0.

Then, H · C = H · Z1 + H · Z2. Therefore, a1,n is bounded by the above, be-cause both are nonnegative. We can therefore replace the ai,n by a convergentsubsequence. Then, lim D′i,n = Zi − aiC exists where ai = limn ai,n. Then,

(1− a1,n − a2,n)C = D′1,n + D′2,n.

Intersecting with C yields 1− a1,n − a2,n ≤ 0, but intersecting with H yieldsyields something nonnegative and gives 1− a1,n − a2,n ≥ 0. Therefore, wemust have a1,n + a2,n = 1. This implies limn→∞ D′1,n ·H + D′2,n ·H = 0 whichmeans D′1,n, D′2,n must both tend to 0. This is what we wanted since it meansZ1 and Z2 both tend to multiples of C. �

Proposition 16.4. Suppose C is an irreducible curve on X over an algebraicallyclosed field with C2 = 0 and K · C < 0. Then, C spans an extremal ray and X is aP1 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,h0(mC)− h1(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 surjectivemaps X → Y. Said another way, we want to find all contractions of X. Theplace to look for these is Eff1(X) ⊂ Eff1(X) ⊂ N1(X)R. Last time, we sawcertain 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 curvein X with C2 = 0 then C lies in ∂Eff(X). If C is irreducible and C2 < 0 then Cspans an extremal ray of Eff1(X).

Remark 17.2. The proof idea from last time was to show that in the lattercase when C2 < 0, we supposed C = Z1 + Z2 for Zi ⊂ Eff1(X). Then, wesaid each Zi was a limit of effectives. We could then write each of theseeffectives as a multiple of C and something not meeting C. We saw we couldtake a subsequence where the coefficients of C stabilize and hence the classof 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 ofthe closure of the effective cone as limits of effective divisors again severaltimes.

Lemma 17.3 (Lemma A). For any D with D · H > 0 then H2(X, mD) = 0 form� 0.

Proof. The proof is essentially Serre duality and which implies that this isH0(X, K−mD) which vanishes because it has negative intersection with Hfor m large. �

Theorem 17.4. Let C is an irreducible curve on a smooth projective surface X withC2 = 0 and K · C < 0 spanning an extremal ray of Eff1(X). Then there is a mapX → B for B a smooth projective curve over a field and X is a P1 bundle (reallymeaning 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 nowto contract C. We will use O(mC) for m � 0. This will contract C becauseC · C = C ·mC = 0.

We saw in the above lemma that H2(X, O(mC)) = 0. We’d like to checkthat 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 bundleon a genus 0 curve. Taking cohomology, this gives us

(17.3)

0 H0(OX((m− 1)C)) H0(OX(mC)) k

H1(OX((m− 1)C)) H1(OX(mC)) 0

H2(OX((m− 1)C)) H2(OX(mC)) 0.

Remark 17.5. As an aside this shows H2(X, mC) = 0 for all m.

For some m we must have H0(OX((m − 1)C)) → H0(OX(mC)) is notsurjective, since H1 is finite, and so can only drop finitely many times. Infact, we can even bound m by h1(OX) since there must be some time in

34 AARON LANDESMAN

the first m steps where the map H1(OX((m− 1)C)) → H1(OX(mC)) is anisomorphism, which implies the map on H0 is not an isomorphism. Thisthen means the map to k from H0(OX(mC)) is surjective and so the linearsystem 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 = 0and K · F = −2a. From adjunction we find 2gF − 2 = −2a. If we are incharacteristic 0, generic smoothness tells us the fiber is reduced. Probablythe 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 fibermust be supported only on C because if it has other curves, the fact thatC · C is 0 and the fiber is connected (because this is the stein factorizationX → B→ PN) forces the fiber to be supported exactly on C. �

Lemma 17.6 (Lemma B). We have{Z ∈ N1(X)R : Z2 > 0, HZ > 0

}⊂ Eff(X)R

Proof. We first reduce to showing{Z ∈ N1(X)Q : Z2 > 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 sufficesto prove the statement for all these Q-divisors.

By Lemma A, we know H2(X, mZ) = 0. Therefore,

χ(mZ) = h0(mZ)− h1(mZ) =(mZ)2

2− mZK

2+ χ(OX).

This is eventually positive because the m2 term dominates. This forcesh0(mZ) to eventually be positive so mZ is effective and Z is effective. �

Proposition 17.7. If r spans an extremal ray of Eff1(X) then r2 ≤ 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 inthe closure of the effective cone. If rH = 0, then r2 < 0 by the Hodge indextheorem (if it is perpendicular to the positive H then it lies in the negativedefinite part of the intersection form).

Therefore, we may assume rH > 0. If r2 > 0, by Lemma B above, we knowr lies in the effective cone r ∈ Eff(X)0

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, wealso obtain that r is ample. �

SURFACES NOTES 35

Theorem 17.8 (Mori’s cone theorem for surfaces). If X is a smooth projectivesurface 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 < 0is 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 ofelliptic curves. The cone lies in 3 space for general such E with no complexmultiplication.

18. 2/26/20

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

Proof. We can write r = lim Dm for Dm ∈ Eff1(X)Q. Therefore, r · Dm < 0for m� 0 because r2 < 0. This implies there is a component C of Dm withr · C < 0. This implies Dm = am[C] + D′m with am ∈ Q. Now, for H ample,

HDm = amC · H + DmH

gives an upper bound on am. Therefore, we can replace this by a subsequencesuch that am converges to a with a ≥ 0. We conclude D′m converges tor− a [C].

D′m · C converges to r · C− aC2. The left hand side is positive because D′mis 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.

As r is extremal, we find r = aC + lim D′m, and since both C and lim D′mare effective divisors, and r is extremal, we find C and lim D′m must bedependent and so r is a multiple of C. Hence r is spanned by the class of anirreducible curve. �

We now restate the cone theorem for surfaces.

Theorem 18.2. Let X be a smooth projective surface. Then there exists a countablefamily C1, C2, . . . , of irreducible rational curves such that −3 ≤ K · C < 0 and

Eff1(X) = {r + ∑i

aiCi ∈ Eff1(X) : K · r ≥ 0, ai ∈ R≥0, almost all ai = 0}

Further, Eff1(X) only accumulates at K · r = 0, meaning that when we are boundedaway 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 minimalmodel.

36 AARON LANDESMAN

Proof. Previously we saw if C2i > 0 then −K is ample and so ρ(X) = 1. We

also saw that if C2i = 0 then ρ(X) = 2 and the surface is a projective bundle

over a curve B. In these cases, we know the statement holds.Hence, we assume C2

i < 0. Then, K · C + C2i = 2g − 2. This means

K · Ci = −1 and C2i = −1. Therefore, by Castelnuovo’s theorem, we may

blow down Ci and repeat. This must terminate because the Picard numberto start was finite. �

Example 18.4. Suppose we start with a del Pezzo surface which is Bl8 pts P2.This is Fano and ρ = 8 + 1 = 9. This has 240 extremal rays.

When you blow down one of the curves corresponding to these extremalrays, you get P2 blown up at 7 points. This has 56 extremal rays, and theremaining roots are those perpendicular to the root you blew down.

They are perpendicular because when you blow down a curve, the −1curves which remain are those not intersecting the −1 curve you blew down,because if they intersect it their self intersection number in the blow downwill increase.

Example 18.5. Consider a complete intersection of two transverse cubics inP2. This has Picard number 10. There are in fact infinitely many −1 curveson 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 usethat the normal bundle of the section in the surface is related to the tangentbundle of the fibers.

We then have E1 as our identity section and we have 8 other sections whichdo not meet each other. We claim that we have found all the −1 curves asinteger linear combinations of (E1 − Ei) for 2 ≤ i ≤ 9, though we won’tcheck 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 theclasses of two fibers and the diagonal. Any curve has C2 ≥ 0.

By the Hodge index theorem, H · z = 0. We have a cone with canonicalbundle equal to 0. There is therefore the full effective cone lies in the K · r ≥ 0part using Mori’s theorem.

When one works out the intersection matrix, one finds three classes on theboundary 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 algebraicallyclosed field. Then any etale 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 → Bwhich becomes X′ ' B′ ×Pn−1 → B′ for B′ → B etale. To trivialize it, wewant to construct a line bundle L on X which is a relative O(1) which meansit defines an isomorphism X → Pπ∗L for π : X → B. We then have somesubschemes of X in B that wants to be the Hilbert scheme of planes in X overB. We want to show this has a B-point.

Then, we want to find a section of the dual bundle X∨ which correspondsto finding a B-point of X. So we just need to know that any etale Pn−1 bundleover B has a section.

The key idea is to produce something on B′. We can write X′ = PVon B′. This PV may not descend, but it turns out that End(V, V) is rankn2 which does descend. This is because the ambiguity of PV descendingwas a unit, which disappears when you look at End(V, V). So we obtain avector E = End(V, V)B on B which base changes to End(V, V) on B′. Thenon invertible endomorphisms of V is a discriminant locus, so when youdescend this it appears as a map O → Symn E ∨, which is usually called thereduced norm.

So, we have E a locally free sheaf on B of rank n2 and the reduced normis a section of Symn(E ∨). We then have this section defining a divisor ∆in PE . Then, ∆ is given by ∑m am ∏j x

njj of homogeneous degree n and

am ∈ H0(B, L ). This gives us a map from B to a projective bundle. Wethen choose xi ∈ Γ(B, L ⊗m). When we plug these into the formula for ∆ weobtain a map

H0(B, L ⊗m)n2 → H0(B, L ⊗1+nm).

This is an algebraic map given by the polynomial defining ∆ and this is a mapof 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, sothere 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 B′ which has a map which is nonzeroand not invertible. The kernel of this map defines a quotient of V whichdoes 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 soour original variety has a point as well. �

Example 19.2. We’d like to understand the effective cone of an abelian sur-face. Start with Eff1(E × E) for E an elliptic curve with no complex mul-tiplication and let X := E × E. We claim, without proof that N1(X)Q isisomorphic to Q⊕3 which is generated by F1, F2, ∆ with Fi the two fibers ofthe projections and ∆ the diagonal. Note that N1(X)Q has dimension at least3, since these divisors are independent.

The hodge diamond has middle row 1, 4, 1 and the Lefschetz 1, 1 theoremtells us N1(X)Q has dimension at most 4, and we’re claiming (though haven’tunderstood why) this has rank 4 exactly when E has complex multiplication(meaning a bigger endomorphisms group than Z).

The intersection matrix of F1, F2, ∆ is0 1 11 0 11 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.

Eff1(X) ={

z ∈ N1(X) : z2 ≥ 0, H · z ≥ 0}

.

Proof. Given a curve, we can translate it to have no components in commonwith 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 severalclasses 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 ofthe 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, butwe’ll probably see this next time.

20. 3/2/20

20.1. Abelian surfaces. Let A be an abelian surface over an algebraicallyclosed 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 againobtains C because otherwise C would intersect this translate at p and havepositive 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 shouldimply ρ(E× E) = 3, though Ravi is still not sure why. Consider the curve

C := {(x1, x2) ∈ E× E : ax1 + bx2 = 0}for a, b ∈ Z. We can write C = αF1 + βF2 + γ∆. We find CF1 = a2 =β + γ, CF2 = b2 = α + γ, C∆ = (a + b)2 = 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, Raviclaims one can do a computation to check that every solution to this equationis 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 projectivevariety. Suppose the map is defined away from finitely many points. Weobtain

(20.1)Γπ ⊂ X×Y

X Yp

Zariski’s main theorem implies that p has connected fibers. If p−1(x) is asingle point then p is finite over a neighborhood of x. This implies p is anisomorphism above U, where U. Then, at the points of X where the map πis 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 atpoints with smooth centers, after which X → Y is a morphism.

40 AARON LANDESMAN

Proof. Suppose we have a rational map X → Pn. Take D1, D2 as two divisorsin O(1) on Pn. Choose a reduced point s ∈ D1 ∩ D2 We have Bls X → X. LetD1 and D2 denote the strict transforms of D1 and D2 in Bls(X). The stricttransforms satisfy D1 · D2 < D1 · D2. This invariant decreases, and hencethis process terminates. �

Corollary 20.2. Suppose we have a rational map π : X → Y and Y has no rationalcurves. 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 amorphism.

Proof. We can blow up X to resolve it to obtain a morphism X → Y. Therewill 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 projec-tive 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 surfaceand KY is nef. Then π is a morphism.

At this point, the course abruptly ended. More lectures were planned, butwe stopped meeting as Arizona winter school happened the next week, andthe remaining lectures in the quarter were cancelled.