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HODGE THEORY LEFSCHETZ PENCILS 1. More on Lefschetz

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HODGE THEORY LEFSCHETZ PENCILS 1. More on Lefschetz HODGE THEORY LEFSCHETZ PENCILS JACOB KELLER 1. More on Lefschetz Pencils When studying the vanishing homology of a hyperplane section, we want to view the hyperplane as one element of a Lefschetz pencil. This is because associated to each singular divisor in the pencil, there is an element of the homology of our hyperplane that is not detected by the Lefschetz hyperplane theorem. In order to do this we should know more about Lefschetz pencils, and that’s how we will start this lecture. We want to be able to produce Lefschetz pencils containing a given hyperplane section. So we should get a better characterization of a Lefschetz pencil that is easier to compute. Let X be a projective variety embedded in PN that is not contained in a hyperplane. We consider the variety N∗ Z = {(x, H)|x ∈ X ∩ H is a singular point} ⊆ X × P To better understand the geometry of Z, we consider it’s projection π1 : Z → X. For any x ∈ X and N hyperplane H ⊆ P , the tangent plane to H ∩ X at x is the intersection of H and TxX. Thus X ∩ H will −1 be singular at x if and only if H contains TxX. Thus for x ∈ X, π1 (Z) is the set of hyperplanes that contain TxX. These hyperplanes form an N − n − 1 dimensional projective space, and thus π1 exhibits Z as a PN − n − 1-bundle over X, and we deduce that Z is smooth. N∗ Now we consider the projection π2 : Z → P . The image of this map is denoted DX and is called the discriminant of X. To better understand it’s local structure, and connect it to Lefschetz pencils, we prove the following lemma N Lemma 1.1. For a point z = (x, H) ∈ Z, Tzπ2 : TzZ → TH P is injective if and only if x is an ordinary double point of H ∩ X. N N Proof. Let x1, . , xN be local coordinates on a chart C ⊆ P . In these local charts, projective hyperplanes are represented by elements of the space K of affine hyperplanes N X k = aixi + bi = 0. i=1 Then Z is defined by {(x, k)|k(x) = 0, dk(x) = 0} ⊆ X × K where the condition dk(x) means that the differential of k vanishes at x as a differential on X (not as N a differential on C ). Because X is smooth, we can choose local coordinates z1, . , zn on X. In these coordinates the equations of Z are ∂k k(y) = (y) = 0 ∂zi for i = 1, . , n. We can split the tangent space of X × K, at the point (x, H), into TxX × TH K. The ∂H derivatives of our equations with respect to vectors u ∈ TxX are denoted by duH and du( ). And by ∂zi ∂k linearity, the derivatives with respect to vectors k ∈ TH K = K are just k(x) and (x). Now to find the ∂zi tangent space to Z in the product space TxX × TH K we simply add those derivatives together and set them equal to 0. Thus we find ∂H ∂k T(x,H)Z = (u, k) ∈ TxX × TkK | duH + k(y) = du + (x) = 0 . ∂zi ∂zi 1 In these coordinates, the derivative of π2 is simply the projection to TkK. The kernel of that linear map is ∂H then seen to be the elements (u, k) such that k = 0 and duH = du = 0. This is exactly the kernel of ∂zi the Hessian of H, considered as a quadratic form on TxX. Thus Tzπ2 is injective if and only if the Hessian is, which is equivalent to the point x being an ordinary double point. Corollary 1.2. The image of Tzπ2 is the set of hyperplanes passing through x. Through this we see that if a hyperplane section of X has an ordinary double point then the general hyperplane section in DX (general makes sense because DX is the image of the irreducible variety Z) has an ordinary double point. This happens if and only if the dimension of DX is N − 1. In fact the corollary shows that the dimension of the locus of singular hyperplane sections having more than one ordinary double point has dimension at most N − 2. This is because if two different points x and y in the same hyperplane section were both ordinary double points, then the tangent space to DX at that hyperplane would contain all hyperplanes through x and all hyperplanes through y, and thus would be the whole space. These hyperplanes are thus singularities of DX and the general element of DX only has one ordinary double point (as long as there exists one hyperplane with an ordinary double point). 0 Let DX be the subset of DX consisting of singular hyperplane sections containing exactly one ordinary 0 double point. We have shown that if DX is N −1 dimensional, then DX is a dense open set of smooth points of DX . 1 I would like to remind everyone now of the definition of a Lefschetz pencil. Let Xt t ∈ P be a pencil such that X0 is defined by σ0 = 0, X∞ by σ∞, and then Xt is defined by σ0 + tσ∞ for t ∈ C, and lastly the base locus B is defined by σ0 = σ∞ = 0 Definition 1.3. The pencil Xt is called a Lefschetz pencil if • The base locus is the smooth complete intersection of σ0 and σ∞. This means that each Xt is smooth along B • Each Xt is smooth except for at most one singular double point. With this in mind we will now prove a useful characterization of Lefschetz pencils. Proposition 1.4. A pencil Xt is a Lefschetz pencil if and only if one of the following two conditions is satisfied 1 • The discriminant locus has codimension 1 and the P parametrizing the pencil intersects DX transver- 0 sally in the open dense set DX . • The discriminant locus has codimension higher than 1 and the P1 does not meet it. Proof. It is clear that if the second condition occurs then it is a Lefschetz pencil, with each member smooth. If the first condition occurs then what we need to check is that the base locus is smooth, because we know 0 from the definition of DX that each Xt has at most one ordinary double point. If x ∈ H is the ordinary double point of a singular hypersurface section H, then the P1 is transverse to DX at H if and only if it is not contained in the tangent plane to DX at H. By the corollary, this tangent plane is the set of hyperplanes containing x. Therefore the intersection is not transverse if and only if x is in the base locus. Lastly, the base locus is smooth if and only if the singular points of the pencil do not lie in the base locus. This ends the proof. Here https://www.desmos.com/calculator/qac5q9gi88 is an example of a Lefschetz pencil whose general member is an elliptic curve. It has one singular member which is a nodal cubic. The discriminant locus in this case is the zero locus of the familiar discriminant which is a function of the coefficients. The zero locus is exactly when the roots of the cubic are repeated. Here https://www.desmos.com/calculator/cfx2alwgdn is a non-example of a Lefschetz pencil. The origin is contained in the base locus, and is the singular point of the general member of the pencil. Corollary 1.5. If X ∈ PN is a smooth projective variety then the general pencil of hyperplane sections of X is Lefschetz. 2 2. Cones over vanishing cycles Now we turn back to the topology of a smooth member of a Lefschetz pencil. In particular, we want to describe the relative homology H∗(X, Y ), where Xt is a Lefschetz pencil and Y = X0 is a smooth member. 1 For each singular member of the pencil at points pi ∈ P , we chose a small disk, ∆i 3 pi, along with points ∗ n ti ∈ ∆i whose corresponding fibers contained vanishing spheres Si ⊆ Xti . We also constructed the cones n n n over the vanishing spheres which are balls Bi contained in X whose boundaries are Si . One reason the Si are called vanishing spheres is because as ti tends to pi they contract to a point inside Xpi . Theorem 2.1. Let f : X → ∆ be a proper holomorphic map from a variety to the unit disk such that f is ∗ a submersion over ∆ . Further, assume that X0 has only one ordinary double point. Then X deformation retracts onto the union of a smooth fiber Xt and a ball with its boundary glued along the vanishing sphere. ˜ We draw paths γi ∈ C \{p1,...} such that γi starts at 0 and ends at ti. Denoting by X the blow up of ˜ X along the base locus of our pencil, then we know that the restricted family Xγi is trivial by Ehresmann’s ˜ ∼ theorem, and so Xγi = X0 × [0, 1]. Under this identification we consider the set 0n n n Bi = Bi ∪ (S × [0, 1]) n where Bi is now identified with a subset of X0. This set is then really a cone over the vanishing sphere 0n n Si = Si × 1 ⊆ X0. Because this is a cycle whose boundary lies in X0, we can orient it and this determine a class ˜ Γi ∈ Hn(X − X∞,X0, Z). ˜ 0n A theorem proved by Elham in lecture shows that X \ X∞ deformation retracts onto X ∪i Bi .
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