Técnicas e instrumentos para la recolección de datos

3.3.1 Projective generation

Suppose we havemdifferentr-dimensional linear systems|Li|of hypersur- faces of degreesdiinPn. Choose projective isomorphismsφi:Pr→ |Li|and consider the variety

Z={(λ, x)∈Pr×Pn :x∈φ1(λ)∩. . .∩φm(λ)}. (3.32)

The expected dimension of a general fiber of the first projection pr1:Z→Pr is equal ton−m. Assume

• Zis irreducible of dimensionr+n−m;

• the second projection pr₂:Z→Pnis of finite degreekon its imageX.

Under these assumptions,Xis an irreducible subvariety of dimensionr+n−

m.

Proposition 3.3.1

degX=sr(d1, . . . , dm)/k,

wheresris ther-th elementary symmetric function inmvariables.

Proof It is immediate that Z is a complete intersection inPr×Pn of m

divisors of type(1, di). LetΠbe a general linear subspace inPnof codimen-

sionn−m+r. We use the intersection theory from [232]. Let¯h1and¯h2be the natural generators ofH2₍

Pr×Pn,Z)equal to the preimages of the co-

homology classesh1, h2of a hyperplane inPrandPn, respectively. We have

(pr₂)∗([Z]) =k[X].By the projection formula,

(pr₂)∗([Z]) = (pr2)∗( m Y j=1 (¯h1+dj¯h2)) = (pr2)∗( m X j=1 sj(d1, . . . , dm)¯h j 1¯h m−j 2 ) = m X j=1 sj(d1, . . . , dm)hm2−j(pr2)∗(¯hj1) =sr(d1, . . . , dm)hm2−r.

Since through a general point inPn passes a unique member of a pencil,

k= 1ifr= 1.

The following example isSteiner’s constructionof rational normal curves of degreenin_Pn_{. We have already used it in the case of conics, referring the} reader for the details to [268].

Example3.3.2 Letr = 1, m =nandd1 =. . . =dn = 1. Letp1, . . . , pn be linearly independent points inPn and let Pi be the pencil of hyperplanes passing through the codimension 2 subspace spanned by all points exceptpi. Choose a linear isomorphismφi:P1→ Pisuch that the common hyperplane

H spanned by all the points corresponds to different parametersλ∈P1.

Let Hi(λ) = φi(λ). A line contained in the intersection H1(λ)∩. . .∩

Hn(λ)meetsH, and henceHmeets eachHi(λ). IfHis different from each

Hi(λ), this implies that the base loci of the pencilsPi meet. However this contradicts the assumption that the pointspiare linearly independent. IfH =

Hi(λ)for somei, thenH∩Hj(λ)is equal to the base locus ofPj. Thus the intersectionH1(λ)∩. . .∩Hn(λ)consists of the pointpi. This shows that, under the first projection pr₁:Z→P1, the incidence variety (3.32) is isomorphic to P1. In particular, all the assumptions on the pencilsPiare satisfied withk= 1. Thus the image ofZinPn is a rational curveRnof degreen. Ifφi(λ) =H, then the previous argument shows thatpi ∈Rn. Thus all pointsp1, . . . , pnlie onRn. Since all rational curves of degreeninPnare projectively equivalent,

we obtain that any such curve can be projectively generated bynpencils of hyperplanes.

More generally, letP1, . . . ,Pnbenpencils of hyperplanes. Since a projec- tive isomorphismφi:P1→ Piis uniquely determined by the images of three different points, we may assume thatφi(λ) =V(λ0li+λ1mi)for some lin- ear formsli, mi. Then the intersection of the hyperplanesφ1(λ)∩. . .∩φn(λ) consists of one point if and only if the system ofnlinear equations withn+ 1 unknowns

λ0l1+λ1m1=. . .=λ0ln+λ1mn= 0

has a 1-dimensional space of solutions. Under some genericity assumption on the choice of the pencils, we may always assume it. This shows that the ra- tional curveRnis projectively generated by the pencils, and its equations are expressed by the condition that

rank l0 l1 . . . ln m0 m1 . . . mn ≤1.

Observe that the maximal minors of the matrix define quadrics inPn of rank

Example3.3.3 Take two pencilsPi of planes inP3 through skew lines`i.

Choose a linear isomorphismφ:_P1_{→ P}

i. Then the union of the linesφ1(λ)∩

φ2(λ)is equal to a quadric surface inP3containing the lines`1, `2.

3.3.2 Projective generation of a plane cubic

We consider a special case of the previous construction wheren = 2, r = 1 andm= 2. By Proposition3.3.1,Xis a curve of degreed1+d2. Assume that the base locus of the pencilPiconsists ofd2i distinct points and the two base loci have no points in common. It is clear that the union of the base loci is the set ofd2

1+d22points onX.

Take a pencil of linesP1and a pencil of conicsP2. We obtain a cubic curve

Ccontaining the base point of the pencil of lines and four base points of the pencil of conics. The pencilP2cuts out onCag12. We will use the following. Lemma 3.3.4 For anyg1

2on an irreducible reduced plane cubic curve, the lines spanned by the divisor fromg12intersect at one point on the curve. Proof The standard exact sequence

0→ O_P2(−2)→ O_P2(1)→ OC(1)→0

gives an isomorphismH0(P2,OP2(1)) ∼= H

0_{(C,OC(1)). It shows that the} pencilg1

2is cut out by a pencil of lines. Its base point is the point whose exis- tence is asserted in the Lemma.

The point of intersection of lines spanned by the divisors from ag1 2 was called by Sylvester thecoresidual pointofC(see [493], p. 134).

LetCbe a nonsingular plane cubic. Pick up four points onC, no three of them lying on a line. Consider the pencil of conics through these points. Letq

be the coresidual point of theg1₂onCdefined by the pencil. Then the pencil of lines throughqand the pencil of conics projectively generateC.

Note that the first projection pr₁ :Z → P1is a degree 2 cover defined by

theg1

2cut out by the pencil of conics. It has four branch points corresponding to linesφ1(λ)which touch the conicφ2(λ).

There is another way to projectively generate a cubic curve. This time we take three nets of lines with fixed isomorphismsφi toP2. Explicitly, ifλ =

[λ0, λ1, λ2] ∈ P2 andφi(λ) = V(a(i)0 t0+a (i) 1 t1 +a (i) 2 t2), wherea (i) j are linear forms inλ0, λ1, λ2, thenCis given by the equation

det    a(1)₀ a(1)₁ a(1)₂ a(2)₀ a(2)₁ a(2)₂ a(3)₀ a(3)₁ a(3)₂   = 0.

This is an example of a determinantal equation of a plane curve which we will study in detail in the next Chapter.