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Ruled Quartic Surfaces (I) Ruled quartic surfaces (I) Pedro A. Castillejo 08.01.2014 Abstract This is a talk of the seminar on Algebraic Geometry at FU Berlin, orga- nized by Joana Cirici, on the Winter Semester 2013-14. The aim of the talk is to partially describe the results of [1]. 1 Some properties of the Grassmannian In this section we will mention some properties of the Grassmannian Gr(2; 4) that will be useful in the following sections. We begin with a vector space V of dimension 4 over an algebraically closed field K. As usual, the points of Gr := Gr(2; 4) will be the lines of P3 := P(V ). We fix a basis e1; e2; e3; e4 of V , and this allows us to identify K and the exterior V4 product of V , so to say we have the isomorphism V ! K : e1 ^ e2 ^ e3 ^ e4 7! 1. V2 Note that given a plane W = hv1; v2i ⊂ V we can define a line in V by considering hv1 ^ v2i. Recall that by definition, V ^ V = (V ⊗ V )=(v ⊗ v). Hence, different planes of V are mapped to different lines of V2 V . Note that w 2 V2 V different from zero can be identified with a plane if and only if1 w is decomposable (ie V2 w = v1 ^ v2), so this identification can't be extended to the whole V . However, we can write the following identifications, where we will understand that we are identifying just what we can identify: P(V ) ! V ! we line W = hv1; v2i plane V2 V2 ∼ 5 ! V ! P( V ) = P V2 W = hv1 ^ v2i = hwi line w = Kw point With these identifications, we have that Gr consists of all the points w coming from a line we of P(V ), and by the previous paragraph, we have that a point w 2 V2 P( V ) belongs to Gr if and only if w is decomposable. But w is decomposable if and only if w ^ w = 0. Hence, we have a criterion to decide wether a point of V2 P( V ) belongs to Gr or not. Let's see what happens if we introduce coordinates. 1One direction is clear. For the other, note that the identification is injective, so it will be bijective with its image. 1 V2 Denote eij := ei ^ ej for 1 ≤ i < j ≤ 4. They form a basis for V with the lexicographic ordering. The coordinates of w 2 V2 V with respect to this basis are P the Pl¨ucker coordinates: w = i<j pijeij. What happens to the coordinates if we want w to be in Gr? The condition is equivalent to say w ^ w = 0. We know that eij ^ ekl 6= 0 if and only if i; j; k; l are different, and if this is the case, the (wedge) product is equal to 1 or −1, depending of the order. Hence, 0 = w ^ w P P = i<j pijeij ^ i<j pijeij = p12p34 − p13p24 + p14p23 + p23p14 − p24p13 + p34p12 = 2p12p34 − 2p13p24 + 2p14p23 So, as we already knew from Vincent's talk, w = [p12 : ::: : p34] 2 Gr if and only if w lies on Kleins quadric p12p34 − p13p24 + p14p23 = 0. From now on, we will abuse the notation and we will use w with w ^ w = 0 for the point of Gr and for the line we defined in P(V ). Now, we write some remarks that will be useful in the following sections. Some of them were mentioned in Gabriel's talk: (i) Two lines w1; w2 of P(V ) intersect if and only if dim(hv11; v12; v21; v22i) = 3, and this is equivalent to saying that w1 ^ w2 = v11 ^ v12 ^ v21 ^ v22 = 0. V2 (ii) Every hyperplane of P( V ) has the form fzj w ^ z = 0g, where w is a fixed point. It is clear that this is an hyperplane because this set is the V2 projectivization of the kernel of the linear map γw : V ! K : z 7! w ^ z. We will denote this hyperplane as Hw. (iii) If w is indecomposable, then the intersection of Hw and Gr is a non degenerate quadric, because Kleins quadric is non degenerate and w2 = Gr, so Hw is not a tangent plane2. We are intersecting Gr with an hyperplane in general position. If w is decomposable, then Hw is the tangent space of Gr at the point w, TGr;w, so the intersection will have a singularity. In particular, TGr;w \ Gr can be identified3 with the cone in P4 over a non singular quadric in P3 with vertex w. Note that if K is algebraically closed, all the non singular quadrics are equivalent, so the cone is always the same. This intersection can be identified with σ1(w) := the collection of all lines l ⊂ P(V ) with l \ w 6= ;, because of (i). (iv) Let p0 be a point of P(V ). Then, σ2(p0) := the collection of all lines through 2 p0. If we consider the lines as points, we obtain a P , because we can consider a sphere with center p0 and then the lines identify the antipodal points. When 2Recall that a non degenerate quadric give us a canonical way of going from a vector space to its dual space. In this particular case, the isomorphism is w 7! Hw, where Kleins quadric defines the wedge product. We are doing the inverse process: we start with a quadric and we obtain a bilinear form (uniquely determined up to scalar product). In this way, we can define the wedge product geometrically starting from Kleins quadric 3The reader can think in a lower dimensional example: if we start with an hyperboloid H, then 2 TH;p \ H is a cone in P over a non singular quadric in P, ie, a pair of lines. 2 we go to the grassmannian, we obtain again a 2-plane. For example, if p0 = e1, P then σ2(e1) = f 1<j≤4 p1je1 ^ ejg is a 2-plane in Gr. (v) Let h0 be a plane of P(V ). Then, σ1;1(h0) := the collection of all lines in the 2 plane h0. Again, it has the structure of a P (just consider, in h0, the dual space. Then, you obtain the lines in h0), and when you look at Gr, you have again a plane: indeed, if you choose a basis such that h0 = he1; e2; e3i, then P σ1;1(h0) = f 1≤i<j≤3 pijei ^ ejg. (vi) σ2;1(p0; h0) := the collection of all lines in h0 through p0. Note that σ2;1(p0; h0) = σ2(p0) \ σ1;1(h0), so this is a line on Gr. (vii) Every plane in Gr has the form σ2(p0) or σ1;1(h0), and every line in Gr has the form σ2;1(p0; h0). (viii) We can classify three dimensional projective subspaces P of P(^2V ) with re- spect to their relation with Gr: (a) Gr \ P is a non degenerate quadric surface. Then, the matrix associated to this quadric surface has rank 4, so we can choose the basis in such a way that the equations of P are p12 = p34 = 0. But the hyperplane p12 = 0 is 4 precisely He34 , and this is the tangent space TGr;e34 . Analoguely, p34 = 0 defines the hyperplane TGr;e12 . Hence, P is the intersection of two tangent spaces. (b) Gr \ P is an irreducible degenerate quadric surface. In this case, the equations of P are p34 = 0 and p13 + p24 = 0 for a suitable basis of V . In this case, P lies on only one tangent space, TGr;e12 and Gr \ P is the cone 2 p13 + p14p23 = 0, p34 = 0 and p13 + p24 = 0. The cone is over the quadric 2 curve p13 + p14p23 = 0. (c) Gr \P is reducible. The equations for P in this case are p12 = p13 = 0 for a basis of V , and Gr \ P is the union of the planes p14 = 0 and p23 = 0. 2 Ruled surfaces and curves on Gr Given a ruled surface S in P(V ), we can consider the set of lines of the surface. This set, in general, can be viewed as family of lines, and therefore as family of points Ce in Gr. In this section we will study the relation between ruled surfaces and curves on Gr, and will answer the following question: given the ruled surface S, what do we obtain in Gr? We start with a naive example that we have already mentioned. If we consider a plane h0 ⊂ P(V ), then we have a lot of lines. Indeed, as seen in (v), this set is precisely σ1;1(h0), and this is a 2-plane on Gr. Note that in general, one starts with a ruled surface that consists on all the lines passing through a curve, but in this configuration there are a lot of lines that appear in the surface, and therefore we obtain a two-parameter family. We expect that this is not the general case. 4 If z = [p12 : ::: : p34], then z 2 He34 iff 0 = z ^ e34 = p12. 3 We are also interested in the inverse problem, so to say: given a curve C ⊂ Gr, do we obtain a ruled surface? We will start with this second question, and afterwards we will study the first one. 2.1 From a curve in Gr to a ruled surface in P(V ) Again, we can also look at an easy example. If we start with a line C on Gr, we know from (vi) that the only lines on Gr have the form σ2;1(p0; h0), and we obtain a plane in P(V ) (all the lines on h0 that pass through p0 define the plane h0).
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