Math 203A, Solution Set 8.

Transcription

1 Math 20A, Solutio Set 8 Problem 1 Give four geeral lies i P, show that there are exactly 2 lies which itersect all four of them Aswer: Recall that the space of lies i P is parametrized by the Grassmaia G = G(1, ) which ca be realized as a quadric i P 5 via the Plucker embeddig For each lie L i defie We claim that X i = {M lie i P : M L i } G(1, ) P 5 X i = H i G for a hyperplae H i i P 5 Ideed, workig i Plucker coordiates, assume that L = L i has coordiates l ij ad M has Plucker coordiates m kl If these are calculated with respect to poits A, B over L ad poits C, D o M the we have a b = ij l ij e i e j c d = kl m kl e k e l The requiremet that L ad M meet is equivalet to a b c d = 0 sice the vector space spaed by a, b, c, d is dimesioal i this case This gives ( ) l ij e i e j m kl e k e l = 0 which gives ij l 12 m 4 l 1 m 24 + l 14 m 1 + l 2 m 14 l 24 m 4 + l 4 m 12 = 0 This is clearly a liear equatio i the variables m kl for each fixed l ij Now, the lies M that itersect L 1, L 2, L, L 4 are foud as the itersectio poits I other words, these poits correspod to kl X 1 X 2 X X 4 G(1, ) H 1 H 2 H H 4 G(1, ) P 5 We claim that this itersectio cosists of 2 poits i geeral We claim first that the itersectio H 1 H 2 H H 4 is a lie l i P 5 i geeral I ay case, the itersectio is give as the ull space of the 4 6 matrix of coefficiets describig the hyperplaes H i I geeral, this ull space is 1 dimesioal, so the itersectio is a lie, but it ca also be that the ull space has dimesio 2 or higher This coditio is 1

2 2 described as the rak of the matrix beig 4 or less i tur this is give by the vaishig of the 4 4 miors, so it is a closed subset Z i the space G G G G We assume (L 1, L 2, L, L 4 ) are chose away from Z Next, if l is the itersectio lie, we claim it itersects the quadric G i P 5 i 2 poits Ideed, we may assume that after a chage of coordiates, this lie is give by x 2 = x = x 4 = x 5 = 0 The quadric G will be give by a ij x i x j ad the itersectio of the lie l is obtaied by solvig a 00 x a 11 x a 01 x 0 x 1 = 0 which has exactly two solutios The oly exceptios correspod to a 2 01 = 4a 00 a 11 which correspods to oe solutio, or the case a 00 = a 01 = a 11 = 0 which correspods to ifiitely may solutios These are closed coditios determiig a closed set W as oe ca check Settig U = G \ (Z W ) we obtai that for (L 1, L 2, L, L 4 ) i U, there are exactly 2 lies itersectig L i The U is dese i G G G G if oempty To show oemptyess, we ca pick 4 lies L 1 = {x 0 = x 1 = 0}, L 2 = {x 0 = x 2 = 0}, L = {x 0 + x 1 = x 2 + x = 0} L 4 = {x 0 + 2x 1 = x 2 + 2x = 0} We claim this quadruple lies i U Ideed, oe ca easily ru the argumet above to fid the equatios of the hyperplaes H 1, H 2, H, H 4 above i terms of the Plucker coordiates We obtai m 01 = 0, m 02 = 0, m 1 + m 0 + m 12 + m 02 = 0, 4m 1 + 2m 0 + 2m 12 + m 02 = 0 We also have m 01 m 2 m 02 m 1 + m 0 m 12 = 0 for the equatio of the quadric These equatios oly have 2 commo solutios as oe checks immediately Problem 2 Show that the Segre embeddig has degree ( ) +m P P m P (+1)(m+1) 1 Aswer: Let Σ m, be the image of the Segre embeddig Degree l homogeeous polyomials o P (+1)(m+1) 1 restrict to Σ m, as polyomials i the variables x i o P ad y j o P m, bihomogeeous of degree l The dimesio of S(Σ m, ) (l) the equals ( )( ) l + l + m m

3 Expadig, we have ( )( ) l + l + m m = 1!m! l+m + lot which shows that the degree of Σ m, equals ( ) ( + m)! + m =! m! Problem Let X P be a projective scheme with Hilbert polyomial χ X Defie the arithmetic geus of X to be p a (X) = ( 1) dim X (χ X (0) 1) (i) Show that the geus of P is zero (ii) If X is a hypersurface of degree d i P, show that p a (X) = ( ) d 1 I particular, a cubic i P 2 has geus 1 (iii) If X is a complete itersectio of two surfaces of degree a ad b i P the p a (X) = 1 ab(a + b 4) I particular, itersectio of two quadrics i P has geus 1 Aswer: (i) We calculated the Hilbert polyomial of P to be χ(l) = ( ) l+ This yields immediately that p a (P ) = 0 (ii) We calculated the Hilbert polyomial of a degree d hypersurface to be ( ) ( ) + l + l d χ(l) = This yields ( ) ) d p a (X) = ( 1) (1 1 1 ( ) d = ( 1) = ( d 1 (iii) We claim that the Hilbert polyomial of the complete itersectio equals ( ) ( ) ( ) ( ) l + l + a l + b l + a b χ(l) = + b The, we fid ( ) ( ) ( ) a b a b χ(0) = 1 + a b which yields the aswer The claim about the Hilbert polyomial is justified as follows Let f ad g be the equatios of the two surfaces of degree a ad b i P whose itersectio is X There is a exact sequece 0 S(P ) (l a b) S(P ) (l a) S(P ) (l b) S(P ) (l) S(X) (l) 0 )

4 4 where the first two maps are give by P (gp, fp ) ad (P, Q) fp gq ad the last map is the restrictio We coclude by cosiderig dimesios Problem 4 Let X be a o-degeerate (ie, ot cotaied i ay hyperplaes) projective variety of degree d ad codimesio c i P (i) (Itersectig X with hyperplaes to cut dow the dimesio), show iductively that d c + 1 (iii) Show that equality holds for ratioal ormal curves i P, ad for the image v(p 2 ) of the Veroese embeddig v : P 2 P 5 (iii) Ca you classify the varieties of degree 2? Aswer: (i) Cosider a hyperplae H ad cosider the scheme X H By Bezout this has the same degree d as X, ad it has codimesio c i H, hece the iequality to prove for X is equivalet to the iequality to prove for X H (which is still odegeerate) By iductio, we reduce to the case whe X cosists of d poits i P I this case, we eed to show d + 1 which is clear sice if d, the X would be degeerate, as ay poits are cotaied i a hyperplae H This last statemet ca be see by arragig that the poits be p i = [0 : 0 : 1 : 0 : : 0] for 0 i 1, after a liear chage of coordiates, ad settig H = {x = 0} (ii) Both cases are particular examples of the Veroese embeddig whose degree we calculate below Cosider Veroese embeddig v d : P P N costructed from degree d moomials We claim that the image V d has degree d Ideed, degree l polyomials i N + 1 variables become, after restrictig to V d, polyomials of degree dl o P Hece the Hilbert fuctio of V d equals ( ) dl + χ(l) = = d l! + lot, cofirmig the claim

5 Now, it is easy to see that ( ) d + d = codim V d + 1 = holds for = 1 or for d = = 2 (iii) After passig to a smaller projective space, we may assume that X is odegeerate quadric Degree d = 2 forces c = 1 hece X is isomorphic to a projective 5