version corrected on 21/9/12

Transcription

1 ( ) version corrected on 21/9/12

2 Degrees of Noether-Lefschetz loci Israel Vainsencher joint with Fernando Cukierman, Angelo Lopez, José Alberto Maia & Adriana Rodrigues partially supported by

3 The Noether-Lefschetz theorem tells us that very general surfaces of degree 4 are not supposed to contain curves besides complete intersections with other surfaces.

4 Asking the surface to contain say, a bunch of lines, a conic, a twisted cubic, etc, defines subvarieties of the appropriate P N = O P 3(d). There are polynomial formulas for their degrees.

5 For example, while a general surface f d of degree d 4 contains no line, the locus consisting of f d s that do contain some (varying) line is a subvariety of P N = O P 3(d) with codimension d 3 ( ) d and degree (3d 4 + 6d d d + 24)/4! subs(d = 3,%); ( )/24 27.

6 We show how to obtain similar formulas for the locus NL(W, d) of all f d s containing some member of a suitable family, W, of curves. deg NL(W, d) is a polynomial in d.

7 In all cases we ve handled so far, namely 1, 2 or 3 lines plane curves twisted cubics elliptic quartics, the polynomial formula obtained is of degree twice the dimension of the family,w, of curves, though the potential degree bounded by GRR is 3 dimw.

8 there is a polynomial formula By Castelnuovo-Mumford regularity, W carries an exact sequence of vector bundles D d > F d E d where the fiber of D d over C W is H 0 (P 3, I C (d)) H 0 (P 3, O P 3(d)) =: F d for all d d 0 = regularity.

9 universal curve: C W P 3 q 2 P 3 q 1 W D d > F d E d ÑL(W, d) = P(D d ) W P N. E d = q 1 (O C(d)) w := dim W deg NL(W, d) = c w (E d ).

10 GRR express the Chern character of E d = q 1 (O C(d)) as the image of ch(o C)ch(O P 3(d))todd(P 3 ), a polynomial in d of degree 3. c w (E d ) is a polynomial of degree w on the coefficients of the Chern character c w (E d ) is a polynomial in d of degree 3w.

11 2 lines in a surface in P 3 deg (2 lines) (d) = 1 3 5( ) d (45d d d d d d d d d d d d d ). d = Compatible with : : the combinatorics of the 27 lines (José Alberto Duarte Maia, 2010) (Adriana Rodrigues, 2011)

12 double point formula Consider the morphism X := {(l, f) G P (F d ) l f} p 2 P N. Direct application of the double point formula (Laksov, Fulton) yields an answer which differs from the one gotten above.

13 The class of the double point locus of ϕ := p 2 in X is presently given by D = ϕ ϕ [X] c d 3 (ϕ T P N T X )) [X], cycle supported on the set of all x X such that ϕ 1 ϕ(x) is a subscheme strictly larger than x, due to existence of x x with ϕ(x ) = ϕ(x), or ramification at x.

14 D d = image of double point locus in P N. deg D d = ( d+1 4 ) (45d d d d d d d d d d d d )/2. The difference to our formula deg (2 lines) (d) = 1 3 5( ) d (45d d is accounted for by

15 deg \/ 2304( (d) = 1 ) ( d 3d 11 9d d d d d d d d d d ) which you ll sureley recognize as the degree of the locus of f d s containing an unordered pair of incident lines! (singular conic)

16 parameter spaces At least morally, the calculation for the case of more lines should be done by means of the Hilbert scheme, Hilb n G, of points of G = Grass[1, 3].

17 Hilb 2 G is simple to describe: the Grassmannian is isomorphic to a smooth quadric G P 5. One shows Hilb 2 G is the blowup of the Grassmannian G[1, 5] of lines in P 5 along the locus of lines contained in the Plücker quadric. this is a line in G:

18 Easier to bring the situation to a more naïve set up. Start with the blowup of the diagonal, G (2) G G. G (2) is a natural parameter space for a flat family of subschemes of P 3 with generic member a union of two lines.

19 Whenever the support of the subscheme is just one line, it carries a double structure with arithmetic genus 1, so the Hilbert polynomial remains 2t + 2. When the support is the union of two incident lines, the point of intersection becomes an embedded point.

20 x 2 0, x 0 x 1, x 0 x 2, x 1 x 2

21 Well, any surface containing such a thick subscheme is singular at the (embedded) point of incidence. This tells us how to sort out the good configurations (i.e., flat limits of two skew lines), from the bad ones; we get in this way a correction for the faulty answer given by the double point formula.

22 3 lines Similarly, construct a parameter space G (3) G (2) G G (2) via a composition of three additional blowups along explicit, smooth, geometrically meaningful centers.

23 The closure in P N of set of surfaces of degree d containing 3 lines in general position is of codimension 3d ( d 1 2 and degree ) ( 10395d d d d d d d d d d d d d d d d d d d d d ) d

24 The degree in d is 24 = 2 dim W; d = 3 720, again in agreement with the combinatorics of triplets of skew lines contained in a smooth cubic surface.

25 surfaces containing a plane curve of degree k 2 OˇP3( 1) F k 1 > ˇP 3 F k F k F k = vector bundle with fiber the space of equations of plane curves of degree k; kernel = multiples of the equation of a plane. Set W = P(F k ); W parameterizes the family of plane curves of degree k in P 3.

26 Form the diagram of vector bundles over W = P(F k ), OˇP3( 1) F d 1 OˇP3( 1) F d 1 F d,k > W F d O Fk ( 1) F k > F d (take an equation of a plane curve of degree k and multiply by all equations of degree k = d k)

27 F d,k X F d vector subbundle of the trivial bundle; rank F ( ) ( ) d,k = d+2 + d k P( F d,k ) = {(p, C, f) plane, curve, surface...} dim P( F d,k ) = dim X + rank F d,k 1 ( ) ( ) ( ) = 1 + k+2 + d+2 + d k+2 ; d >> k P( F d,k ) P N gen. inj.

28 script with(schubert); #from Katz & Stromme with luv i:=2; DIM:=3; fkb:=unapply(symm(k,4-o(-h)),k); dimk:=k->subs(t=0,fkb(k))-1; DIM:=3+dimk(i); ftili:=mtaylor(o(-h),h,4)*symm(d-1,4)+ o(-hk)*fkb(d-i): u:=mtaylor(chern(dim,symm(d,4)-ftili),h,4): seg:=mtaylor(chern(-fkb(i)),h,4): u:=collect(u,hk): lo:=binomial(i+2,2)-1; u:=subs(hk^lo=1,u): for j from lo+1 to DIM do u:=subs(hk^j=coeff(seg,t^(j-lo)),u): od: u:=factor(subs(h=1,u));

29 conics For all d 5 the locus of surfaces in P 3 of degree d containing a conic is a subvariety of O(d) of codimension 2d 7 and degree F 2 (d) := ( d 4 ) (d 2 d + 6)(d 2 d + 8) ( 207d 8 288d d d d d d 2 ) d / Compare degree in d with dim W : = 2 8. subs d = 4 in F 2 (d) 5016 but...

30 (i) For d = 4, the correct degree is 5016/2 due to Bézout. (ii) If we restrict the conics say to move on a given plane, we get the following formula for the degree of NL(W, d) O P 3(d) (codim 2d 4): 1 d 960( ) ( d 2 d + 2i ).

31 1 d 960( ) ( d 2 d + 2i ). For d = 2 the formula gives 1 as expected. For d = 3, we get 21, which coincides with the degree of the locus of cubic surfaces bitangent to a fixed plane, same as the number of binodal plane cubics through 7 general points...

32 F 3 (d) := plane cubics ( d 1 4 ) 5 (d 2 3d i) (225d d d d d d d d 63360) subs(d = 5, [F 2 (d), F 3 (d)]) [282880, ]. 0

33 plane quartics ( d 2 4 ) 10 (d 2 5d i) 0 (5301d d d d d d d d ) degree in d = 34 = 2 (3 + 14).

34 twisted cubics Use Norwegian expertise, kindly provided by Ellingsrud & Strømme. The Hilbert scheme component is well described: Altogether, there are 130 fixed points. Everything is sufficiently explicit to feed into Bott s formula.

35 deg P (c(e. )) = x F x P T (c(e. )) c T 12 (τ xw) sum over all fixpoints of a suitable C -action.

36 We find the formula for the degree of the subvariety of P N = O P 3(d) consisting of surfaces which contain a twisted cubic curve:

37 ( d 1 )( d d d d d d d d d d d d d d d d d d d d d ) /( ). Next, a little latino expertise

38 (Avritzer &,1992): W =irreducible component of Hilb of elliptic quartic curves in P 3. Starting point: X = G(2, F 2 ), Grassmannian of pencils of quadrics in P 3.

39 G(19, F 4 ) W = X Ê G(8, F 3 ) X X Ẽ Ỹ G(2, F 2 ) =: X Z Y where Z ˇP 3 G(2, F 1 ) pencils with a fixed plane; Y {(p, l) p l} = closed orbit of Z; Ỹ Y = P 2 bundle of degree 2 divisors on the varying lines l p; X = the blow-up of X along Z and X = the blow-up of X along Ỹ.

40 Ỹ = { } Z = { } Y = { }

41 ÑL(W, d) = P(D d ) NL(W, d) P N is gen. injective for d 5 by NLL. Setting m = dim NL(W, d) deg NL(W, d) = H m ÑL(W, d) = X s 16 (D d ) = X c 16 (E d ).

42 Fixed points in W are easily listed as well as the corresponding fibers of E d ; run Bott s formula as we learned from Meurer, get the formula for the degree of Noether-Lefschetz locus of

43 ( d 2 3 elliptic quartics, d 5: ) ( d d d d d d d d d d d d d d d d d d d d d d d d d d d d ) d / ( ).

44 extra twist for the case d = 4 Presently p 2 : ÑL(W, 4) NL(W, 4) is no longer generically injective. It shrinks dimension by one: a general fiber is a disjoint union of two P 1 s, as we learn from Noether-Lefschetz-Lopez.

45 Explicitly, say F 4 = A 1 Q 1 + A 2 Q 2, deg A i = deg Q i = 2, everything in sight as general as needed. Then our quartic surface F 4 = (A 1 tq 2 )Q 1 + (A 2 + tq 1 )Q 2, contains a whole pencil of elliptic quartics A 1 tq 2, A 2 + tq 1, t P 1 ; setting t =, we find Q 1, Q 2.

46 Similarly, get Q 1 ta 2, Q 2 + ta 1 ; this is one and the same pencil of elliptic 4ics in F 4. But there is also the pencil A 1 ta 2, Q 2 + tq 1. In general, these 2 pencils are disjoint. As curves in G(2, F 2 ), we actually get a P 1 - linearly Plücker embedded as a conic (A 1 ta 2 ) (Q 2 + tq 1 ) = A 1 Q 2 + t(a 1 Q 1 A 2 Q 2 ) t 2 A 2 Q 1 anddisjoint from the blowup center Y ; capping against the Plücker hyperplane class Π, we find 1, oops, 2 (correction thanks to K. Ranestad, 9/18/2012:-).

47 deg NL(W, d) = H 33 NL(W, d). The cycle p 2H 33 NL(W, d) can be represented by a sum of deg NL(W, d) disjoint unions of pairs of P 1 s. Each P 1 cuts the Plücker hypln. class Π twice; 4 deg NL(W, d) = 1 Π H 33 X Π p 1 (H = 1 4 ÑL(W, d) ÑL(W, d)) X Π c 15(E 4 ). The latter integral can be computed via Bott and we find 38475, presently dedicated to Steve (now with Kristian s seasoning:-)

48 H A P PY B I RT H D AY