Prove
Lemma 4 ([Rational descent of a cycle span]). Let E be a finite-dimensional…Lemma 3 ([Degree projectors]). Every cohomological K\"unneth projector on a…Proposition
Suppose first that f:Y→S satisfies. Let D=Alb(Y) , let a:Y→D be an Albanese map, and put b= D . Albanese pullback identifies holomorphic one-forms . Hence there is β∈H^ 2,0 (D)=^2H^ 1,0 (D), y=f_*a^*β≠0. In particular b≥2 and y∈H^ 2,0 (S)⊂V_ ℂ . The surface correspondence (a,f)_*[Y] has codimension b on D×S . Its transpose acts by a_*f^*:H^2(S)⟶H^ 2b-2 (D)(b-2). This action is nonzero on V : the projection formula gives ∫_Dβ∪a_*f^*y =∫_Ya^*β∪f^*y =∫_Sy∪y≠0. We give an algebraic degree adjustment. For an ample line bundle L on D , written additively, set E=m^*L-p_1^*L-p_2^*L^1(D×D), where m is addition. The class of E is purely mixed. Put U=H^1(D,ℚ) and write this class as ∑_i u_i⊗v_i , choosing a basis (u_i) of U . Its matrix M:U^∨⟶U, M(φ)=∑_iφ(u_i)v_i is invertible: it is the alternating tensor associated with c_1(L) , whose top exterior power is nonzero since L is ample. Intrinsically it is a Hodge map M:U^∨→U(1) ; in the matrix calculation we use the underlying rational spaces. With the Koszul convention for K\"unneth products, cl(E)^2 =-2∑_ i<j (u_i∧u_j)⊗(v_i∧v_j). Poincar\'e duality identifies H^ 2b-2 (D) with (^2U)^∨ on underlying rational spaces, by integration against degree-two classes. Consequently (E^2)_* is -2^2M after this identification, and is an isomorphism (E^2)_*:H^ 2b-2 (D)(b-2) ∼H^2(D). The twists can also be checked from H^ 2b-2 (D)(b-2)≃H^2(D)^∨(-2) and ^2M:H^2(D)^∨→H^2(D)(2) . Composing with (f,a)_*[Y] yields a cycle of codimension b+2-b=2 on S×D , nonzero on V by. This uses a square of an explicit divisor; it assumes no algebraic inverse Lefschetz operator for a general variety. Conversely suppose that α_*:V→H^2(B) is nonzero and Hg(V)=SO(V,q) . The standard representation is irreducible, so α_*|_V is injective. For 0≠σ∈H^ 2,0 (S) , put η=α_*σ≠0 in H^ 2,0 (B) . Write c= B≥2 and let ℓ be an ample divisor class on B , replaced by a very ample multiple if necessary. The rational surface cycle C=α·p_B^*ℓ^ c-2 _2(S×B)_ ℚ satisfies ∫_ S×B cl(C)∪p_S^*σ∪p_B^*η =∫_Bη∪η∪ℓ^ c-2 ≠0. Here rationality gives α_*σ=η , and the last pairing is definite up to a fixed nonzero sign on H^ 2,0 (B) by the Hodge--Riemann bilinear relations. An actual representative of C is a finite rational combination of integral surfaces; for example use general hyperplanes to represent the indicated intersections. At least one component has nonzero pairing in. Resolve that component projectively and call the resulting smooth connected surface Y . Its projections give morphisms f:Y→S and a':Y→B with ∫_Y f^*σ∪a'^*η≠0. Thus f^*σ≠0 , which forces f to have two-dimensional image. It is therefore dominant and generically finite. The projection formula also gives f_*a'^*η≠0 . Since η is a sum of wedges of holomorphic one-forms on B , its pullback lies in the wedge image in. This proves the converse.