Lecture 19: Smoothness, Canonical Bundles, the Adjunction Formula

Lecture  18.725 Algebraic Geometry I Lecture 19: Smoothness, Canonical Bundles, the Adjunction Formula Last time we defined ΩX , Tx X and smoothness. We proved that any X contains an open dense smooth subset, that X is smooth at x if and only if ΩX is locally free around x, and X is smooth if and only if ΩX is locally free. Here’s a trivial observation: suppose we have a surjection f : A → B and mB ∈ B an maximal ideal, let mA = f −1 mB , then mB /m2B = mA /m2A + I where I = ker(f ). If Y = Spec B contains X = Spec A, y ∈ X ⊆ Y , then Ty∗ Y = Ty∗ X/(dfi ), where fi are generators of I. Corollary 1. We have the following: 1. If X ⊆ An is a hypersurface given by the equation IX = (P ), then x ∈ X is smooth if and only if ∂P 6= 0 for some i. dP |x ∈ Tx∗ An 6= 0, i.e. ∂xi x 2. Suppose X ⊆ An has dimension n − m where IX = (f1 , . . . , fm ) (this is not true for all X), then X is ∗ n n smo  oth at  a point x if and only if dfi |x ∈ Tx A = k are linearly independent, i.e. the m-by-m matrix ∂fi has rank m. ∂xj x Proof. The first claim is a particular case of the second. dim(X) ≥ dimx X ≥ n − m =⇒ dimx X = n − m. Now apply the definition of smoothness, and that Tx∗ X = T ∗ An /(dfi |x ). If X ⊆ Pm has dimension n − = (F1 , . . . , Fm ) for homogeneous polynomials, then x = (x0 , . . . , xn ) m, IX  ∂fi is a smooth point if and only if has rank m. To see this, note that X is smooth at x if and only if ∂xj x CX (the cone) is smooth at x̃ because CX is locally isomorphic to X × A1 , and note that T(∗x,y) (X × Y ) = Tx∗ X ⊕ Ty∗ Y . Proposition 1. Suppose X ⊆ An , x ∈ X is a smooth point if and only if ∃f1 , . . . fm ∈ IX ⊆ k[x1 , . . . , xn ] which locally generate IX and dfi |x are linearly independent. Proof. If f1 , . . . , fn as above exists, then dim(Tx∗ X) = n − m while dimx X ≥ n − m also dimx X ≤ (dim Tx∗ X) =⇒ dimx X = dim Tx∗ X i.e. x is a smooth point. Conversely, suppose X is smooth at x, pick f1 , . . . , fm ∈ IX such that fi |x form a basis in ker(Tx∗ An → Tx∗ X). Then by the first part of the proof, Z = (f1 , . . . , fm ) is smooth at x with dimx Z = n − m = dimx X, where Z ⊇ X. So we are done if we know that Z is locally irreducible, which follows from the next lemma: n ∼ [ Lemma 1. O Z,x = lim k[Z]/mx = k[[t1 , . . . , tn−m ]] (i.e. is a free ring). ← − n Why does this imply Z locally irreducible? Z locally irreducible means OZ,x has no zero divisors, which [ would follo \w from the fact that OZ,x ⊆ OZ,x which follows from Nakayama. In particular, ker(OZ,x → [ O mnx which is a finitely generated ideal OZ,x is Noetherian, and we have mx I = I =⇒ I = 0. Z,x ) = n Remark 1. This lemma is equivalent to that M mnx /mn+1 (the associated graded ring) is isomorphic to x n k[t1 , . . . , tn ]. The general case is given in the next lemma. Lemma maximal ideal, a ∈ A. Suppose a ∈ mp , write a ∈ Ap = mp /mp+1 and M 2. Let A be a ring, m a M n n+1 A= m /m . Then A/(a) = (A/a)n /(A/a)n+1 = A/(a) if a is not a zero divisor. n Proof. (A/(a))n = mn /(mn+1 +(aA∩mn )), A/(a) = mn /mn+1 +amn−p . For any x ∈ mk , we have ax ∈ mk+p ; if a is not a zero divisor, then x ∈ / mk+1 , then ax ∈ / mk+p+1 . 1 Lecture  18.725 Algebraic Geometry I Now we return to the first lemma. f1 , . . . , fn have linearly independent differential at x, by induction ∼ check k[x1 , . . . , x\ n ]/(f1 , . . . , fi ) = k[[t1 , . . . , tn−1 ]], i.e. gr(k[x1 , . . . , xn ]/(f1 , . . . , fi )) = k[t1 , . . . , tn−1 ], and if so, wlog can assume fi+1 = t1 . ∼ [ Proposition 2. X is smooth at x iff O X,x = k[[t1 , . . . , td ]] where d = dimx X. Proof. The forward direction follows from the proof of the previous proposition where we deduced this from the fact that X is locally given by equations with independent differentials. For the other direction, assume bX,x ∼ O = k[[t1 , . . . , td ]] then we want to conclude d = dim Tx∗ X. It suffices to check that dimx X ≥ d. Pick f1 , . . . , fd ∈ mx with linearly independent differentials, and we claim that (f1 , . . . , fd ) is a regular sequence, i.e. fi+1 is not a zero divisor in OX,x /(f1 , . . . , fi ). Then fi+1 = 6 0 on each component of Zf1 ,...,fi passing through x, so we get X ) Z1 ) Z2 . . . ) Zd 3 x, where Zi is a component in Zf1 ,...,fi . Why is it a regular sequence? because OX,x /\ (f1 , . . . , fi ) ∼ = k[[t1 , . . . , tm−i ]] ⊇ OX,x /(f1 , . . . , fi ) (check by induction). This concludes the proof of the proposition. Proposition 3. Suppose Z ⊆ X is a closed subvariety. 1. We have an exact sequence IZ /IZ2 = IZ |Z → ΩX |Z → ΩZ → 0 (recall F |Z = OZ ⊗OX F = F /IZ F allows us to identify a sheaf F on Z with i∗ F ). 2. If for all x ∈ Z, IZ is locally (around x) generated by f1 , . . . , fm such that dfi |x are linearly independent at x, then the sequence is short exact, and IZ /IZ2 is a locally free sheaf of rank m where m is the codimension. In the situation of (2), IZ /IZ2 is called the conormal bundle. Example 1. X, Z are smooth irreducible, dim(Z) = dim(X) − 1, Z = D is a divisor, IZ = O(−D) is an invertible sheaf. A local section of it is a function vanishing on Z. We can send f to a 1-form df vanishing on D, and it defines a section of the conormal bundle. Definition 1. If X is a smooth irreducible variety of dimension d, then Ω(X) is a locally free sheaf of rank ^d d. Then the top exterior power ω(X) = Ω(X) is a locally free sheaf of rank 1. We call it the canonical line bundle or the canonical sheaf (“canonical” because any smooth variety gets it for free). If 0 → A → B → C → 0 is a short exact sequence of locally free sheaves, then we have ^top (B) = ^top (C) ⊗ ^top (A). Corollary 2 (Adjunction Formula). ωD = ωX (−D)|D . One last comment: the graded algebra has a nice geometric property as follows: Definition 2. Spec(gr(OX,x )) is called the tangent cone to X at x. Proposition 4. The tangent cone is the cone over the exceptional locus in the blowup at x. 2