Download Algebraic Geometry Bucharest 1982: Proceedings of the by Lucian BĂdescu (auth.), Lucian Bădescu, Dorin Popescu (eds.) PDF

By Lucian BĂdescu (auth.), Lucian Bădescu, Dorin Popescu (eds.)

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Extra resources for Algebraic Geometry Bucharest 1982: Proceedings of the International Conference held in Bucharest, Romania, August 2–7, 1982

Example text

For every p ~ o let U ~ (Xo,Ou/mP+IOu) denote the dimk(mP+l P ) and set N(p) = p th infinitesimal neighbourhood of X o in U (U ° = Xo_ / p+2 m ). In both cases, for every p ~ o consider the standard exponential sequence: ~ mP+IOu /m P+20 U ~ 0 ~( P) -- o > 0~U > 0 ~. ~OxN(P), o ~ ~ PiG(Up+ l) £~ ~ PiG(up)-- ~ H2" N(p)) ~0Xo i Since we assumed that H (0X ) = o for i = 1,2, we get that the maps m ~# are isomorphisms for every p ~ o ° (in the analytic case we implicitly used a result of GAGA-type).

Proof. I, ~5) to obtain E (see [3] for more details). By restricting the exact sequence that f~Y=@, we get that the splitting (d,-d); a simple argument given extension is i). We choose a 2-vector bundle given by such an exten- and c2(E)=0. above to a fibre f, provided such that ty- shows that the integer r from the is the second numerical invariant of E, hence E is of (d,r). (3) Let us denote Consider the spectral Ll=0x(-dCo-rfo)8~' (LI) and i2=Ox(dCo+rfo)~;r~*(L2 ). sequence of term E~,q=HP(x,E~xt~ (Iy@il,i2)),which X 38 converge to EP+q=ExtP+q(Iy~Ll,i2 ).

We deduce that there n e c e s s a r i l y exist m u l t i p l e fibres. But this implies in its turn that r~4. Indeed if we w o u l d have r=3, taking a m u l t i p l e fibre mE and i n t e r s e c t i n g it w i t h H we would get 3=m(HE). Since m~2 we w o u l d get that E is a straight line. 565, if an elliptic fibration has a fibre of the form mE w i t h E integral then Pa(E)=l. We deduce that r~4 hence there is exactly one m u l t i p l e fibre. We get K~(m-I)E. 3=(m-I)HE and since HE#l, Taking degrees we get as we already have seen, we get HE=3 and m=2 hence r=6.

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