# LECTURES ON SINGULARITIES AND ADJOINT LINEAR SYSTEMS

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1 LECTURES ON SINGULARITIES AND ADJOINT LINEAR SYSTEMS LAWRENCE EIN Abstract. 1. Singularities of Surfaces Let (X, o) be an isolated normal surfaces singularity. The basic philosophy is to replace the singularity by a manifold. This procedure is the resolution of singularities. Definition 1.1. Let (X, o) be a normal surface singularity. A resolution of the singularity (X, o) is a proper birational morphism f : Y X, where Y is smooth. The set f 1 (o) = E 1 E n is called the exceptional divisor. Remark 1.2. Since X is normal, then f 1 (o) is connected, hence has no isolated points. This implies that E i are distinct irreducible curves. ( ) Theorem 1.3. The intersection matrix (E i E j ) is negative definite, where E i are the exaction curves. Proof. We only need to show that for any non-trivial divisor D = ai E i, D 2 < 0. Assume for contradiction that D 2 0. We will reduce to that D is effective. Write D = A B such that A and B are effective and have no common components. Then D 2 = A 2 2A B + B 2 0. Since A B 0, hence A 2 0 or B 2 0. Assume that A 2 0. Moreover, we may assume that X and Y are projective. Take an ample divisor H on X. Then f H 2 = H 2 > 0 and A f H = 0. By Hodge index theorem, the intersection matrix over H is negative definite. This is a contradiction. 1

2 2 LAWRENCE EIN Definition 1.4. A divisor D on a projective variety X is called nef (numerically effective) if D C 0 for any curve C on X. We say X is minimal if the canonical divisor K X is nef. In higher dimension, to run the MMP, we will encounter singularities. But the singularities are not too bad. Theorem 1.5. Let ϕ : Y X be a proper birational map. If X is minimal, then ϕ is a morphism. Definition 1.6. Let f : Y X be a proper birational morphism. We say that a divisor D is f-nef if D C 0 for any f-exceptional curves C. Lemma 1.7 (Negativity Lemma). Let f : Y X be a proper birational morphism with exceptional divisor E i. If the divisor D = a i E i is a f-nef divisor, then D is effective. Proof. We will first proof the case that X and Y are surface. Write D = A B where A and B are effective divisors and have no common components. Since D is f-nef, hence D A = A 2 A B 0. On the other hand, A 2 0 and A B 0. Therefore, A 2 A B 0. This implies that A = 0. Therefore D = B is effective. For higher dimensional case, we may cut X by n 3 general hypersurface to reduce to the surface case. Proof of the Theorem. Take a resolution of the birational map ϕ. We get the a variety Z and two birational morphisms f : Z Y and g : Z X. We will use the nefness of K X to show that no such resolution is needed. Write K Z = f K Y + a i E i + f j F i = g K X + a ie i + g k G k, where E i are f- and g-exceptional divisors, F j are f-exceptional but not g-exceptional divisors and G k are g-exceptional but not f-exceptional

3 LECTURES ON SINGULARITIES AND ADJOINT LINEAR SYSTEMS 3 divisor. Then a i, a i, f j and g k are nonnegative. Consider the difference (g K X + a ie i + g k G k ) (f K Y + a i E i + f j F i ) = 0. We get g K X f K Y + g k G k = f j F i + (a i a i)e i. Note that the left hand side is f-nef since K X is nef and G k are not f-exceptional. Then by Negativity Lemma, f j F i + (a i a i )E i must be nef. Hence f j must be zero and a i a i. In other words, there is no f-exceptional but not g-exceptional divisors. Hence, there is no blowing up of ϕ needed. Therefore, ϕ : Y X is a morphism. Let X be a smooth projective variety. Denote by N 1 (X) R the space of 1-cycles of X modulo numerical equivalence. It is know that N 1 (X) is a finite dimensional space. Denote by NE 1 (X) the subspace of effective curves. Theorem 1.8 (Cone Theorem). NE 1 (X) = NE 1 (X) KX 0 + R + C i, where C i are rational curves such that K X C i dim X+1. Moreover, C i s form a countable collection. Those C i s a Assume that X is a surface. Then (1) if K X C i = 1 then C 2 i = 1 which implies that C i is a smooth (-1)-curve. (2) if K X C i = 2 then C 2 i = 0 which implies that X is a ruled surface. (3) if K X C i = 3 then X = P 2. Definition 1.9. Let f : Y 2 (X 2, o) be a resolution of an isolated normal surface singularity and f 1 (o) = E 1 E n. A effective

4 4 LAWRENCE EIN cycle Z nef = a i E i is called a minimal nef cycle, if Z nef is nef and Z nef D for any effective cycle D = d i E i such that D is nef. Proposition Z nef is well-defined and unique. Proof. Let D = d i E i and D = d ie i be two effective cycles such that D and D are nef. Define D = Min(D, D ) = min(d i, d i)e i. We claim that D is nef. Write D = D + R 1 and D = D + R 2. Then R 1 and R 2 have no common components. Therefore any exceptional curve E i can appear in at most one of the two cycles R 1 and R 2. Without lose of generality, we assume that E i does not appear in R 1. Then D E i = D E i R 1 E i 0. Therefore D is nef Rational singularities. Definition 1.11 (Rational Singularity). A morphism f : Y X is said to be a rational resolution if Y is smooth and f is a proper and birational morphism such that R i f O Y = 0 for i > 0. Proposition Let f : Y X be a rational resolution and f : Y X be another resolution. Then f : Y X is also a rational resolution. Proof. We have a birational map ϕ : Y Y. Successively Blowing up the undefined locus of ϕ, we get a variety Z and two proper birational morphisms g : Z Y and g : Z Y such that h := f g = f g. Since g is the composition of blowing-ups. Then R q g (O Z ) = 0 for q > 0. Apply the Leray spectral sequence E p,q 2 = R p f (R q g (F )) R p+q (f g) (F ). It follows that R i h O Z = 0 for i > 0. Apply the Leray spectral sequence to f g. It is easy to see that R 1 f O Y = 0. In fact, it fits in the following exact sequence 0 R 1 f O Y R 1 h O Z f R 1 g O Z. Since Y is smooth hence Y has a rational resolution. Now Z is another resolution of Y. By the above argument, we can conclude that

5 LECTURES ON SINGULARITIES AND ADJOINT LINEAR SYSTEMS 5 R 1 g O Z = 0. Apply the Leray spectral sequence to p + q = 2. We see that R 2 f O Y = 0. Hence R 2 g O Z = 0. By induction, we conclude that R p f O Y = 0 for p > 0. This shows that rational resolution is well-defined. Proposition Let f : Y (X, o) be a resolution of a rational surface singularity. Then χ(o D ) 1. Proof. Since (X, o) is rational and normal, then H 1 (O D ) = H 1 (O Y ) = 0. Therefore, χ(o D ) 1. What if R i f O Y 0? Let f : Y (X, o) be a resolution. Then R 1 f O Y is a finite length module supported at the origin. It is also an invariant of the singularity, called the geometric genus of o (See Kollár). Since smooth varieties has rational resolution, then R 1 g O Z = 0. Therefore, the Leray spectral sequence tells us that R 1 f O Y = R 1 f O Y. Reference: Miled Reid, Park city Lecture notes. Since R 1 f O Y is supported at the origin, by Serre-Grothendieck spectral sequence, we know that H 1 (O Y ) = R 1 f O Y. To compute higher direct image sheaves, besides the definition, we have the formal function theorem. Theorem Let f : Y X be a proper morphism and S be a subvariety of X. Then for any coherent sheaf F on Y, we have R p f F = lim k R p f (F /I k F ), where I is the defining ideal of S in Y and the left hand side is the completion along IO Y. Remark The left hand side in fact is isomorphic to R p f F since R p f F is coherent. A completion of a finitely generated module M

6 6 LAWRENCE EIN over a Noetherian ring R can be obtained by extension of scalars: ˆM = M R ˆR. By the formal function theorem, we see that there is an isomorphism H 1 (O Y ) = lim Supp(D) f 1 (o) H 1 (O D ). On the other hand, we have morphism H 1 (O Y ) H 1 (O D ). Therefore, there is a D such that H 1 (O Y ) = H 1 (O D ) for all D D. Definition 1.16 (Cohomology cycle). Let f : Y (X, 0) be a resolution of an isolates normal surface singularity (X, o). A divisor D supported on the exceptional locus is called a cohomology cycle if H 1 (O Y ) = H 1 (O D ). Theorem There exists a unique minimal cohomology cycle. The proof is similar to the proof of existence of minimal nef cycle. Proof. Let D = d i E i and D = d ie i be two cohomology cycles. We claim that D = min(d, D ) = min d i, d ie i is also a cohomology cycle. Write D = D + R 1 and D = D + R 2. Then R 1 and R 2 have no common components. Therefore, I D D = O X ( D ) I Σ, where Σ = R 1 R 2. We have the following exact sequences. which induces an exact sequence 0 I D D I D I D I D D 0, 0 O D D O D I D O D D 0. Therefore, we have an exact sequence H 1 (O D D ) H 1 (O D ) H 1 (O D ) H 1 (O D D ) 0. It is also easy to check that 0 O Σ ( D ) O D D O D 0

7 LECTURES ON SINGULARITIES AND ADJOINT LINEAR SYSTEMS 7 is exact by snake lemma. Therefore, h 1 (O D ) h 1 (O D ) + h 1 (O D ) h 1 (O D D ) = h 1 (O Y ). We conclude that H 1 (O Y ) H 1 (O D ) is an isomorphism. Let f : Y (X, o) be a resolution of a rational surface singularity. Assume that X is affine. Then H 1 (O Y ) = 0. Moreover, H 2 (O Y ) = 0. From the exponential sequce we see that 0 Z O Y O Y 0 Pic(Y ) = H 1 (O Y ) = H 2 (Y, Z) = n Z L i, where L i are divisors such that L i E j = δ ij. A divisor L on Y is then determined by the intersection numbers a i = L E i. Slightly move L i to L i, we get the same intersection number. Hence O Y (L i ) is globally generated. A divisor L on Y is nef if and only if a i 0. Line bundles associated to nef divisors are globally generated. The nef cone Nef(Y ) is Z + L i. Let f : Y (X, o) be a resolution of an isolated surface singularity. Denote the maximal ideal associated to o by m. Then mo Y = O Y ( Z max ) I Σ where dim Σ = 0. Morevore, Z max is nef and Z max Z nef. Definition 1.18 (Fundamental cycle). The cycle Z max is called the fundamental cyle. Remark In Miles Reid s book, Z max is called the fiber cycle. Theorem 1.20 (Artin). Assume that (X, o) is an rational surface singularity. Then (1) mo Y = O Y ( Z max ), i.e. I Σ = O Y. Moreover, Z max = Z nef. (2) m/m 2 = H 0 (O Zmax ( Z max ) and the embedding dimension of o is dim m/m 2 = Z 2 max + 1. i=1 (3) The multiplicity of o is Z 2 max.

8 8 LAWRENCE EIN Proof. We may assume that X is affine. Since (X, o) is rational and Z max is nef, then O Y ( Z max ) is globally generated, equivalently, H 0 (Y, O Y ( Z max )) O Y O Y ( Z max ) is surjective. Notice that m = H 0 (Y, O Y ( Z max )) and mo Y O Y ( Z max ). Therefore, mo Y = O Y ( Z max ). Since Z max Z nef, hence m = H 0 (O Y ( Z max )) H 0 (O Y ( Z nef )). Therefore H 0 (O Y ( Z nef )) = m. Since O Y ( Z nef ) is also globally generated, then mo Y = O Y ( Z nef ) = O Y ( Z max ). Take the cohomology of the following short exact sequence 0 O Y ( 2Z max ) O Y ( Z max ) O Zmax ( Z max ) 0. Note that H 1 (O Y ( 2Z max ) = 0 and H 1 (O Zmax ( Z max ) = 0 because Z max is nef and big. We then have the following short exact sequence 0 m 2 = H 0 (O Y ( 2Z max ) m = H 0 (O Y ( Z max ) H 0 (O Zmax ( Z max )) 0. The embedding dimension is given by h 0 (O Zmax ( Z max )) = χ(o Zmax ( Z max )) = Z 2 max+1 p a (O Zmax ) = Z 2 max+1. Apply the same trick to m k /m k+1, we see that dim m k /m k+1 = h 0 (O Zmax ( kz max )) = kz 2 max + 1. Hence mult o X = Z 2 max. By Artin s theorem, we know that a rational surface singularity is a hypersurface singularity if Z 2 max = Du Val singularities. Definition Let f : Y (X, o) be a minimal resolution of an isolated surface singularity. Denote the fundamental cycle by Z. We say (X, o) is a Du Val singularity if K Y E = 0 for any exceptional curve E. Theorem An isolated surface singularity (X, o) is Du Val if and only if (X, o) is a rational double point, equivalently a rational singularity such that Z 2 = 2.

9 LECTURES ON SINGULARITIES AND ADJOINT LINEAR SYSTEMS 9 Proof. Assume that (X, o) is a rational double point. Then K Y Z = 2p a (Z) 2 Z 2 = 2χ(O Z ) Z 2 = 0. Because Y is minimal. We know that K Y is nef. As the fundamental cycle, Z supported on all exceptional curves. Therefore, K Y E i = 0. Assume that (X, o) is a Du Val singularity. We need to show that R 1 f O Y = 0. By formal function theorem, it suffices to show that H 1 (O D ) = 0 for all effective D supported on exceptional curves. We do this by induction. Write O Y = O Y (K Y + A). Then A E = 0 for any exceptional curves E. Assume that D = E is irreducible. Then O D = (K Y + A) D = K D + (A E) )D. Note that deg((a E) )D) = D 2 0. Therefore, H 1 (O D ) = H 1 (K E + (A E) E ) = H 0 ( (A E) E ) = 0. Now write D = D + E. The classification of Du Val singularities are well known. following are all the possible Du Val singularities. A n : x 2 + y 2 + z n+1 = 0, D n : x 2 + y 2 z + z n 1 = 0, E 6 : x 2 + y 3 + z 4 = 0, E 7 : x 2 + y 3 + yz 3 = 0, E 8 : x 2 + y 3 + z 5 = 0. The 1.3. Canonical model. Let X be a minimal surface of general type. Then K X is nef and KX 2 > 0. By Reider s theorem, which will be discussed later, X can = Proj H 0 (mk X ) is a surface. There is a canonical birational morphism X X can. The curves being contracted are the curves C such that C K X = 0. Therefore, X can has only Du Val singularities Gorenstein Singularities. Let X be a normal surface. We can a canonical sheaf ω X by extending the canonical bundle O U (K U ), where

10 10 LAWRENCE EIN U is the smooth locus of X. A normal variety X is Gorenstein if ω X is a line bundle. Example Let S be the surface x 2 + y 2 + z 2 = 0. Then ω S = O S (K A 3 S ) N S. Assume that f : Y (X, o) is a minimal resolution. Since the intersection matrix is negative definite, there is a unique Q-cycle Z K = ai E i such that K Y E i = Z K E i for any exceptional curve E i. By negativity Lemma, Z K is effective. Proposition Assume that (X, o) is a normal Gorenstein surface singularity. Let f : Y (X, o) be the minimal resolution. (1) K Y is linear equivalent to an integral cycle Z K = a i E i where a i 0. Moreover, if O KY is nontrivial, then a i > 0. (2) Z K = Z coh. Proof. (1) Choose an effective cycle G = D + b i E i K Y, where the components of D are not f-exceptional. Then f G = D K X 0. Now note that f D 0. So KY G f f G = a i E i. We see that K Y is linearly equivalent to a ie i. We know that K Y is also numerical equivalent to ) a i E i. Since (E i E j is negative definite. So a i = a i and K Y Z K. (2) Now O Y ( a i E i ) = O Y ( Z K ) = O Y (K Y ). Consider the short exact sequence 0 O Y ( Z K ) O Y O ZK 0. By the Grauert-Riemannschneider Theorem, R i f O Y (K Y ) = 0 for all i > 0. Therefore, H 1 (O Y ) = R 1 f O Y R 1 f O Z = H 1 (O Z ) is an isomorphism. So Z K is a cohomology cycle. Since Z con is the unique minimal cohomology cycle, then Z coh Z K. If Z K Z con is strictly effective, then we may write Z K = E +D such that E is an irreducible component and Z coh D. So

11 LECTURES ON SINGULARITIES AND ADJOINT LINEAR SYSTEMS 11 H 1 (O D ) = H 1 (O Y ). We will show that this can not be the case. Consider the exact sequence 0 O E ( D) O ZK O D 0. We want to show that H 1 (ω ZK ) H 1 (O D ) has non-trivial kernel. By Serre duality, H 1 (O ZK ) = H 0 (ω ZK ) and H 1 (O D ) = H 0 (ω D ). Note that ω ZK = (K Y + Z K ) ZK = O ZK and ω D = (K Y + D) D = O D ( E). Now consider the exact sequence, 0 O D ( E) O ZK O E 0. We see easily that H 0 (ω D ) H 0 (ω ZK ) is a nonzero map. This completes the argument. Theorem Assume that f : Y (X, o) be a minimal resolution of a normal surface singularity. If K Y is numerically equivalent to Z K = a i E i where a i Z + and Z K = Z con, then (X, o) is a Gorenstein singularity. Lemma Pic(Z K ) = Pic(Y ) Proof. Note that Supp(Z K ) is a deformation retract of Y. Apply fivelemma to cohomolgy sequences of the exponential sequences of Y and Z K, we get the isomorphism. Lemma K Y = OY ( Z K ). Proof. By our assumption, L = K Y + Z K is numerically trivial. We only need to show that L Z = ω Z is trivial. Then by the previous lemma L is trivial. We will show that L Z is globally generated. If so, we will have a morphism O Z L Z. Since L Z is numerically trivial, then O Z L Z must be an isomorphism. Write Z K = E + D, where E is an irreducible component of Z K. By applying snake lemma, we get the following exact sequence 0 O Y (K Y + D) D L Z ω Z E 0.

12 12 LAWRENCE EIN Since Z K is the minimal cohomology cycle, then H 1 (O E ( D)) H 1 (O ZK ) is a non-zero map. By Serre duality, H 0 (O ZK ) H 0 (ω ZK E ) have a non-zero map. Since ω Z is numerically trivial. So ω ZK E = O E and the section 1 of H 0 (O E ) can be lifted to ω ZK. This shows that ω ZK is generated by the section. We conclude that ω ZK = O ZK. 2. Flip constructed from linear algebra Let V = C a+1 and W = C b+1 with b a 0. Set H = Hom(V, W ) = C (a+1)(b+1). Consider the subvariety X = {ϕ H rankϕ 1} of H. Outside the origin, actions of GL(V ) and GL(W ) will move ϕ X transitively. Therefore, X has an isolated singularity at the origin. We will see that (X, o) has two natural resolutions. One way to say that ϕ X is rank one is a 1-dimensional quotient space Im(ϕ) = V/U of some a-dimensional subspace U W. Another way is that Im(ϕ) is a 1-dimensional subspace of W. Denote by P = P a = P(V ). 1-dimensional quotients of V are classified by O P (1). Let Y = H om(o P (1), W O P ) = W O P ( 1). As the total space of a vector bundle Y is smooth. We have a natural morphism f : Y X H, which maps a morphism O P (1) W O P to the composition V O P O P (1) W O P. It is easy to see that the pre-image of a point outside the zero point of X is a single point. Hence Y X is a resolution. In fact, Y X collapse the zero section of Y. Denote the zero section by Z. Then the normal bundle N Z is isomorphic to W O P ( 1). Therefore, Ω Y/P = N Z = W O P and K Y Z = OP ( a 1). Hence K Y = π K P K Y/P = π O P (b a) which is nef by our assumption. Denote by Q = P b = P(W ) and Y = H om(v O Q, O Q ( 1)) = V O Q ( 1). Then Y is also a resolution which collapse the zero section of Y. The morphism Y X sent a morphism V O Q O Q ( 1) to the compostions V O Q O Q ( 1) W O P. Similarly, we get K Y = π O Q (a b) which is not nef.

13 LECTURES ON SINGULARITIES AND ADJOINT LINEAR SYSTEMS 13 This is an example of flips. Moreover, X has a rational singularity at the origin. Denote T = H P(W ). T can be viewed as a trivial vector bundle over P(W ) or a projective bundle of the trivial vector bundle W O H over H. Let p : T H and q : T P(W ) be the two projections. Y is the vector bundle V O P(W ) ( 1). Consider the universal maps p V O T p W O T on T. We also have the following exact sequence on T p W O T O T (1) 0. Therefore, V can be identified as the zero locus of a section of p V O T ( 1). This will give a resolution of O Y. By chasing the resolution, we can prove that R i p O Y = 0 for i > Singularities of theta divisors Let C be a smooth projective curve of genus g. We denote the d-th symmetric product of C by C (d). Let Pic d (C) be the degree d component of the Picard group Pic(C). Set Wd r = {[L] Picd (C) h 0 (L) > r}. There is a morphism D = ϕ : C (d) W 0 d Pic d (C) d p i C (d) [O C (D)]. i=1 Assume that L is a line bundle on C such that r = h 0 (L) > 0. The fiber ϕ 1 ([L]) = {D D 0 and O X (D) L} = P(H 0 (L) ) = P r 1. When deg L = g 1, χ(l) = h 0 (L) h 1 (L) = deg L + 1 g = 0. So h 0 (L) = h 0 (K C L). Denote by Θ = ϕ(c (g 1) ) which is called the theta divisor. For general L, h 0 (L) = h 1 (L) = 1. (??? Why?). This says that ϕ : C (g 1) Θ is birational. Theorem 3.1 (Mumford). dim W r g 1 (g 1) 2r 1.

14 14 LAWRENCE EIN Let Σ = {x C (g 1) dim ϕ 1 (ϕ(x)) 1}. By Mumford s theorem, we know that that dim Σ g 3. Hence ϕ is a small resolution. The theta divisor is not smooth in general. Theorem 3.2 (Riemann). mult [L] Θ = dim H 0 (L). Riemann singularity theorem tells us Θ has the singular locus W 1 (g 1). However, singularities of Θ are not to bad. The following theorem tells us the Θ has rational singularities. Theorem 3.3 (Kempf). The theta divisor Θ has rational singularities. The proof is not very difficult. We will use the fact that Θ is a hypersurface. Proof. We want to show that R i ϕ O C (g 1) = 0 for i > 0. Since Θ is a hypersurface, then K Θ is a line bundle. Because ϕ is a small resolution, we see that K C (g 1) = ϕ K Θ. By Grauert-Riemannshneider theorem and projective formula, we see that R i ϕ O C (g 1) = R i ϕ (K C (g 1) ϕ (K 1 Θ )) = Ri ϕ K C (g 1) K 1 Θ = 0. Using the fact that Θ is a hypersurface, in fact a determinantal variety, at a point [L], Θ can be defined by a polynomial f = f m + f m+1 +, where f m is a degree m homogenious polynomial. Blowing up [L], the tangent cone is define by f m in P g 1. Let Λ = ϕ 1 ([L]) = P r C (g 1) and E be the exceptional divisor of the blowing up ψ : Bl Λ C (g 1) C (g 1). We know that ϕ is a rational resolution, so is the composition ϕ ψ. Therefore, dim k m t+1 /m t = h 0 (O E ( te)), where f is the maximal ideal of [L]. So we get mult [L] Θ = c 1 (O E ( E)) g 2. Note

15 LECTURES ON SINGULARITIES AND ADJOINT LINEAR SYSTEMS 15 that the exceptional divisor E is the projectivization of the normal bundle N = N Λ/C (g 1). So O E ( E) = O P(N) (1). On P(N), we have the universal sequence 0 Ω 1 N/Λ(1) ψ N O P (1) 0. So c 1 (O P (1)) = ψ (c 1 (N )) and c 1 (O E ( E)) g 2 = ψ (c 1 (N )) (g 2). Now we compute the normal bundle N. Consider the product C Λ, where Λ = P(H 0 (L) ). Let p : C Λ C and q : C Λ Λ be the two projections. Then H 0 (p L q O Λ (1)) = H 0 (L) H 0 (O Λ (1)) = H 0 (L) H 0 (L) = End(H 0 (L)). Let s H 0 (p L q O Λ (1)) be the section corresponding to the identity element in EndH 0 (L) and D = div(s) be the universal divisor. We then have the following exact sequence 0 O C Λ p L q O Λ (1) O D (D) 0. Apply q, we have the exact sequence 0 O Λ H 0 (L) O Λ (1) q O D (D) H 1 (O C ) O Λ H 1 (L) O Λ (1) 0, because D Λ is a finite morphism and R 1 q O D (D) = 0. We observe that the cockerel of O Λ H 0 (L) O Λ (1) = T Λ and q O D (D) = T C (g 1) Λ. (Why the second equality?????) So we obtain the following exact sequence 0 N H 1 (O C ) O Λ H 1 (L) O Λ (1) 0. Hence we get c 1 (N ) = c 1 ((O Λ (r + 1))). Therefore, mult [L] Θ = c 1 (O E ( E)) g 2 = r + 1 = h 0 (L). (Why (ψ c 1 ((O Λ (r + 1)))) g 2 = r + 1????) 4. Singularities in higher dimension Recall that a normal surface singularity (X, o) is Du Val, if there is a minimal resolution f : Y X such that K Y E i = 0, equivalently, K Y = f K X.

16 16 LAWRENCE EIN The analogue in higher dimension is canonical singularities. Let f : Y X be a log resolution, i.e. Y is smooth, f is proper, birational and the exceptional divisors are simple normal crossing. Write K Y f K X = a i E i, where a i is called the log discrepancy of X along E i. We say that X has canonical singularities, if all a is are non-negative. Theorem 4.1. Assume that X is an algebraic variety with only canonical singularities and f : Y X is a log resolution of X. Then + m=0 H 0 (O X (mk X )) = + m=0 H 0 (O Y (mk Y )). Proof. Let E be the exceptional divisor. Notice that f O E (E) = 0. Apply f to the exact sequence 0 O Y O Y (E) O E (E) 0, we see that f O Y (E) = O Y. Apply the projective formula, we obtain the equality. Definition 4.2 (Analytic version). Let X be a smooth affine variety and I Z =< f 1, f 2,..., f r > be the ideal of a subvariety Z. For an positive number λ, the multiplier ideal of Z of weight λ is defined as I (I λ Z) = {g O X g 2 ( f i 2 ) λ L1 loc}. Example 4.3. Let A =< z a > be an ideal in k[z]. The integral z b dxdy = r b rdrdθ exists if and only if b + 1 > 1. Therefore, I ( λ ) =< z [λa] >. More general, I ((z a 1 i zn an ) λ ) = ((z λa 1 i zn λan )). Let f : Y X be a log resolution of the pair (X, Z), where Z is a subvariety of X. The support Supp(Exc(f) + f 1 (Z)) = E 1 E 2 E r is simple normal crossing. Let z 1,, z n be local coordinates of X and w 1,, w n be local coordinates of Y. Then dw 1 dw n d w 1 d w n = Jac(f) 2 dz 1 dz n d z 1 d z n.

17 LECTURES ON SINGULARITIES AND ADJOINT LINEAR SYSTEMS 17 Write relative canonical divisor K Y/X = div(det(jac(f))) = k i E i and I z O Y = O Y ( b i E i ). Then g I (IZ λ) if and only if ord E i g [λb i ] k i. So we can define multiplier ideal sheaves algebraically. Definition 4.4 (Algebraic version). Let f : Y X be a log resolution of the pair (X, Z). Then the multiplier ideal sheaf of Z of weight λ is I (I λ Z) = f O Y (K Y/X [λe]), where E is the exceptional locus, i.e. O Y ( E) = I Z O Y. Definition 4.5. Let (X, D) be a pair where D is an effective Q-divisor. The pair (X, D) is said to have Kawamata log terminal (kit) singularity if and only if the multipler ideal sheaf I (X, D) = O X, equivalently, k i b i > 1; terminal singularity if and only if k i b i > 0; canonical singularity if and only if k i b i 0; log canonical singularity if and only if k i b i 0. Theorem 4.6 (Kawamata-Viehweg vanishing theorem). Let (X, ) be a pair, where = δ i E i is simple normal crossing and X is smooth. Let A be a nef and big Q-divisor such that K X + A + is numerically equivalent to a line bundle L. Then H i (L) = 0 for i > 0. Theorem 4.7 (Nadel vanishing theorem). Let X be a smooth variety and D be a Q-divisor on X. Assume that L is an integral divisor such that L D is nef and big. Then H i (O X (K X + L) I (D)) = 0, for all i > 0, where I (D) is the multiplier ideal sheaf of D. Let A be an abelian variety of dimension g and Θ be a line bundle on A. Then the following are equivalent (1) c i (Θ) g = g!,

18 18 LAWRENCE EIN (2) h 0 (Θ) = 1, (3) The morphism ϕ Θ : A Pic 0 A = A given by ϕ Θ (x) = T x Θ Θ 1, where T x is the translation by x, is an isomorphism. Definition 4.8. An abelian variety A together with a line bundle Θ satisfying one of the equivalent conditions is called a principal polarized abelian variety (p.p.a.v. for short), denoted by (A, Θ). Theorem 4.9. A p.p.a.v. (A, Θ) is log canonical if and only if the multiplier ideal sheaf I ((1 ε)θ) = O A for any ε > 0. We need the following lemma. Lemma Let (A, Θ) be a p.p.a.v. and Z be a closed subscheme of A. If H 0 (I Z Θ P ) 0 for all P Pic 0 (A), then Z =. Proof. Since (A, Θ) is p.p.a.v., then Θ P = T x Θ for some x A. By the assumption, H 0 (I Z Θ P ) 0, for all P. Therefore, Z T x Θ for all x A. However, T x Θ = which forces Z to be empty. Proof of Kollar s theorem. Assume for the contradiction that I ((1 ε)θ) O A for some ε. Let Z be the subvariety such that the ideal sheaf I Z = I ((1 ε). It is clear that H 0 (I Z Θ) 0. Since Θ is nef and big, then H i (I Z Θ P ) = 0 for all i > 0 and P Pic 0 (A) by Nadel vanishing theorem. Therefore, χ(i Z Θ) > 0. Since P Pic 0 (A), then χ(i Z Θ P ) > 0 which implies that H 0 (I Z Θ P ) 0. So we see that Z =. Theorem 4.11 (Ein-Lazarsfeld). Assume that (A, Θ) is a p.p.a.v. and Θ is irreducible. If Θ has canonical singularities then Θ has rational singularities. 5. Adjoint linear systems on surfaces Conjecture 5.1 (Fujita). Let X be a smooth projective variety of dimension n and A be an ample divisor on X. Then

19 LECTURES ON SINGULARITIES AND ADJOINT LINEAR SYSTEMS 19 (1) K X + (n + 1)A is base-point-free. (2) K x + (n + 2)A is very ample. For surfaces, Reider proved the conjecture. The base-point-freeness of K X + (n + 1)A in 3 and 4 dimensional was proved by Ein-Lazarsfeld and Kawamata respectively. The conjecture is open for higher dimensional varieties. Reider s proof uses Bogomolov unstability theorem Theorem 5.2 (Bogomolov). Let E be a rank 2 vector bundle on a smooth projective surface X. The the following are equivalent (1) c 2 1(E) 4c 2 (E) > 0. (2) Let L = det(e ). There exists a divisor B, a 0-dimensional subscheme W X and an exact sequence 0 O X (L B) E I W O X (B) 0 such that (L 2B) 2 > 4 deg W and (L 2B) H > 0 for any ample divisor H. For higher dimensional variety, so far we don t have any analogue of Bogomolov s theorem. The proofs of Ein-Lazarsfeld and Kawamata use multiplier ideal sheaves and Kawamata-Viehweg vanishing theorem. Theorem 5.3 (Reider). Let X be a smooth projective surface and A be a nef and big divisor on X. Assume that A 2 > 4, then the linear system K X + A is base point free at a point p X, unless there is a curve B passing through p such that (1) B 2 = 1 and (K X + A) B = 0, or (2) B 2 = 0 and (K X + A) B = 1.

20 20 LAWRENCE EIN Proof. Write L = K X + A. Consider the exact sequence 0 O X (L) I p O X (L) O X (L) p 0, where I p is the ideal sheaf of p in O X. Since A is nef and big, then H 1 (O X (L)) = 0. The obstruction of L being base point free at p is in H 1 (O X (L) I p ). By Serre duality, we have (H 1 (O X (L) I p )) = Ext 1 (O X (L) I p, O)X(K X )) = Ext 1 (O X (A) I p, O X ). If L is not base point free at p, then there is an nonzero element η Ext 1 (O X (A) I p, O X ). So we have an extension 0 O X E O X (A) I p 0. It is easy to check that c 1 (E) 2 4c 2 (E) = A 2 4 > 0. By Bogomolov theorem, we have an exact sequence 0 O X (A B) E O X (B) I W 0, such that (A 2B) 2 > 4 and (A 2B) H > 0 for any ample divisor H. Observe that the composition of morphism O X (A B) E O X (A) I p is nontrivial. So we see that H 0 (O X (B) I p ) 0. Then there is an effective divisor D linearly equivalent to B and passing through p. Since c 2 (E )) = 1. Then (A D) D + deg W = 1. To prove the theorem, it suffices to show that (A D) D = 1 and D 2 0.

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