证明
Lemma 1. Let Φ=T_ _X (-⊗_X(1)) . There is a natural isomorphism of exact…Theorem 1.1. Every smooth complex quintic threefold admits a numerical…观点
The Weak Lefschetz Theorem (in the integral form recalled in ) and Poincar\'e duality give H^ 2j (X,)= H^j for 0≤j≤3 ; the integral version also gives H^2(X,)= H . Thus the even Chern character is precisely (E)=r+cH+dH^2+eH^3. The tangent sequence gives c(T_X)=(1+H)^5/(1+5H) , so that c_1(T_X)=0 , c_2(T_X)=10H^2 , and td(X)=1+56H^2. Hirzebruch--Riemann--Roch, obtained by taking the proper map to a point in , now gives, and in particular χ(E)=5(e(E)+56c(E)). The matrix of the bilinear form is J= 0, 25/6, 0, 5; -25/6, 0, -5, 0; 0, 5, 0, 0; -5, 0, 0, 0 , J=625. Moreover the four vectors v(_X(k))=(1,k,k^2/2,k^3/6), 0≤k≤3, span ^4 : their row matrix has determinant one. It follows that the kernel of v on K_0(X) is exactly the radical of the Euler pairing. Indeed one implication follows from Riemann--Roch, and the other follows by testing against these four spanning vectors and using the nondegeneracy of J . Hence v identifies the numerical group with its image, which spans ^4 . For any class of K_0(X) , integrality of Chern classes and _2=(c_1^2-2c_2)/(2), _3=(c_1^3-3c_1c_2+3c_3)/(6) give d∈1 10 and e∈1 30 ; integral Weak Lefschetz gives c∈ . The image is therefore a subgroup of the discrete lattice in. Since it contains the four spanning line-bundle classes, it is itself a full lattice. In particular its real span is the full space ^4 . Tensoring by _X(1) sends (r,c,d,e) (r,c+r,d+c+ r2,e+d+ c2+ r6). The twist T_ _X then subtracts χ(E(1)) from the rank and leaves the other three coordinates unchanged. Formula therefore gives M as displayed. Expanding (zI-M) gives ; in particular its four eigenvalues are the nonidentity fifth roots of unity, each with multiplicity one.