1
A Proof of the Hodge Conjecture
Kazuhisa MAEHARA∗ Nov.14, 2006
1 Introduction
In this paper we show that the Hodge conjecture and a part of the Tate conjec- ture hold. Since it is difficult to find algebraic cycles in general, the strategy is to proceed by induction argument to vanish a certain subspace of a cohomology of an open affine subvariety of an affine variety which is obtained excluding a general hyperplane from a given variety.
2 In Case of Non Sigular Varieties
Let k beafieldwithanalgebraicclosurek and X a smooth geometrically irreducible variety over k. There exists the canonical cycle map for =char k
r r 2r cl :CH(X) −→ Het (Xk, Q (r))
This image is included in the fixed part 2r Gk Het (Xk, Q (r)) where Gk = Gal(k/k).Tatesconjecturesaysthatifk is finitely generated r 2r Gk as a field, the image of cl generates Het (Xk, Q (r)) . Fix an isomorphism ι : Q → C. Let k = C. One has the canonical cycle map
clr :CHr(X) −→ H2r(X(C), Q(2π i)r) This image is included into
H2r(X(C), Q(2π i)r) ∩ Hr,r(X(C))
∗Associate Professor, General Education and Research Center, Tokyo Polytechnic University Received Sept. 28, 2006 2 ACADEMIC REPORTS Fac. Eng. Tokyo Polytech. Univ. Vol. 29 No.1 (2006)
Hodge conjecture says that the image of clr generates H2r(X(C), Q(2π i)r) ∩ Hr,r(X(C)). Let U be a smooth quasiprojective variety over k. The images of the canonical cycle maps are 2r Gk Het (Uk, Q (r)) for finitely generated k r 2r 2r F H (U, C) ∩ W2rH (U, Q(r)) for k = C
Let U be a smooth quasi-projective variety over k and X a smooth projective compactification of U. One denotes by cl∗ the following cycle maps clDR, cl , clH ; 2r 2r 0 2r (a) ΓDR(HDR(U)(r)) = W0(HDR(U)(r)) ∩ F (HDR(U)(r))
2r 2r Gk 2r (b) Γ (H (U)(r)) = H (U)(r) ∩ W0(H (U)(r))) 2r 2r 0 2r (c) ΓH (Hσ (U)(r)) = W0(Hσ (U)(r) ∩ F (Hσ (U)(r)) ⊗ C)
Lemma 1 It suffices to prove it for a smooth affine variety over k.
Proof. It is well known that it is enough to treat it in the case of dim X =2d. Choose a smooth irreducible hyperplane Y on X.ThusX −Y is a smooth affine variety. One obtains the following commutative diagram:
d−1 2d CH (Y) −→ Γ∗H∗,Y(X)(d) ↓↓ d 2d CH (X) −→ Γ∗H∗ (X)(d) ↓↓ d 2d CH (X − Y) −→ Γ∗H∗ (X − Y)(d) ↓↓ 00
2d ∼ Here the vertical sequences are exact. By duality, one has H∗,Y(X)(d) = 2d−2 H∗ (Y)(d − 1). Assume conjectures hold for X − Y . Induction hypothesis for d 2d−2 CH (Y) −→ Γ∗H∗ (Y)(d − 1) implies conjectures. Take general strict normalcrossing divisors D1, ··· ,D2d on X excluding Y . Since X − Y is affine, let A denote Γ(X − Y,O), which is geometrically regular commutative kalgebra of finite type. Thus X − Y =SpecA.LetI1, ··· ,I2d be ideals of A such that V (I1)=D1, ···, V (I2d)=D1 ∩···∩D2d.ByN¨other’s nor- malization lemma, one obtains algebraically independent elements x1, ··· ,x2d of A such that
(a) A is integral over k[x1, ··· ,x2d].
(b) Ii ∩ k[x1, ··· ,x2d]=(x1, ··· ,xi).
Let K = k(x1, ··· ,x2d)andL = Q(A) the total fraction field of A.LetK ⊂ L be a finite extension and M the smallest quasi-Galois extension of L/K . Take the integral closure B of A over k[X]inM .LetG =Gal(M/K). Lemma 2 The action of G on M keeps B to be invariant. A Proof of the Hodge Conjecture 3
Proof. Since A is regular; hence normal, A coincides with the integral closure of k[x1, ··· ,x2d]inL .Leta ∈ A with the minimal polynomial hand let σ ∈ G. Then hσ = h. Take any b ∈ B , which satisfies the minimal polynomial g(b)=0 with coefficients in A .Thusgσ has its coefficients in A. Hence gσ(bσ)=0, which implies b is an integral element of M. Therefore bσ ∈ B .
Let K bethefieldoftheinvariantsofM by G and C the integral closure of k[x1, ··· ,x2d]. . Note that K is a radical extension of K of finite degree.
Lemma 3 (i) BG = C
(ii) Spec C → Spec k[x1, ··· , x2d] is a finite, surjective and radical morphism, which is a universal homeomorphism.
G G Proof. Since M ⊃ B ⊃ A ⊃ C ⊃ k[x1, ··· ,x2d] , one has K = M ⊃ B ⊃ G G A ⊃ C = C.SinceB is a finite k[x1, ··· ,x2d]module and k[x1, ··· ,x2d]isa N¨otherian ring, every submodule of B is a finite k[x1, ··· ,x2d] module. Every element of BG is integral over C. Hence BG = C since C is integrally closed. φ :SpecC→ Spec k[x1, ··· , x2d] For every point x of Spec C one has κ(x)isa radical extension of κ(φ(x)); thus φ :SpecC→ Spec k[x1, ··· , x2d]isaradical morphism. It is clear that the mophism is finite and surjective; hence a universal homeomorphism. Let F be a suitable smooth sheaf on X. One has a trace map : trL/K : L → K , which naturally extends to a map trA/C :A→ C. If f ∈ C, one further has 1 1 a map trA[ 1 ]/C[ 1 ] : A[ ] → C[ ] and a cohomological map f f f f j 1 j 1 trA[ 1 ]/C[ 1 ] :H Spec A[ ], F → H Spec C[ ], F . f f f f
1 1 If f ∈ k[x1, ··· ,x2d], φ :SpecC[f ] → Spec k[x1, ··· , x2d, f ] is a finite, surjective and radical morphism; hence a universal homeomorphism. Thus one has an isomorphism j 1 j 1 H Spec C[ ], F → H Spec k[x1, ··· , x2d, ], F . f f
Let H =Gal(M/L).
Lemma 4 Assume that dimSpecA=n=2d≥ 4.Letf1, ··· ,fn be k-linear combinations of x1, ··· ,xn in k[x1, ··· ,xn]. Assume that the intersection locus V (f1, ··· ,fn) is void in Spec k[x1, ··· , xn], i.e., the hyperplanes intersect one n 1 point in infinity. One obtains Γ∗H Spec A[ ], F =0. f1···fn Proof. If n>2, by the affine vanishing theorem, one has
H2n−1 (Spec A, F)=0, H2n−2 (Spec A, F)=0. 4 ACADEMIC REPORTS Fac. Eng. Tokyo Polytech. Univ. Vol. 29 No.1 (2006)
One has a Cechˇ Spectral sequence: p,q q 1 p+q E1 = ⊕|I|=p+1H ∩i∈ISpec A[ ], F ⇒ H (Spec A − V(f1, ··· , fn), F) . fi
pq Note that E1 =0forq>nby affine vanishing theorem and that p>n− 1by Cechˇ cohomology theory. One obtains
n−1,n 2n−1 E∞ =H (Spec A − V(f1, ··· , fn), F)=0,
n−2,n n−2,n n−2 2n−2 E2 = E∞ = Gr H ((Spec A, F)=0,
n−4,n+1 n,n−1 since E2 = E2 = 0. Note that there exists the following exact se- quence:
Γ 2d−2 F − −→ ∗H∗ (V(fik ), )(d 1)
2d 1 2d 1 Γ∗H (Spec A[ ], F)(d) −→ Γ∗H (Spec A[ ], F)(d) −→ 0. ∗ ··· ∗ ··· fi1 fik−1 fi1 fik Thus, n 1 n 1 ⊕|I|=n−2Γ∗H ( Spec A[ ], F →⊕|I|=n−1Γ∗H ( Spec A[ ], F i∈I fi i∈I fi
and n 1 n 1 ⊕|I|=n−1Γ∗H ( Spec A[ ], F →⊕|I|=nΓ∗H ( Spec A[ ], F i∈I fi i∈I fi
n−3,n+1 n+1,n−1 n,n are surjections, respectively. Using E2 = E2 = E1 = 0, one has n−1,n n−3,n+1 n−1,n n+1,n−1 E3 =H E2 → E2 → E2 =
n−1,n n−2,n n−1,n n,n n−1,n E2 =H E1 → E1 → E1 =E∞ =0.
Hence, the homomorphism
n−2,n n−1,n E1 → E1
is a surjection. n−4,n+1 n,n−1 Applying E2 = E2 = 0, one has n−2,n n−4,n+1 n−2,n n,n−1 E3 =H E2 → E2 → E2 = A Proof of the Hodge Conjecture 5
n−2,n n−3,n n−2,n n−1,n n−2,n E2 =H E1 → E1 → E1 =E∞ =0.
Moreover, since n−2,n n 1 E2 =H ⊕|I|=n−2H (Spec A[ ], F) −→ i∈I fi
n 1 n 1 ⊕|I|=n−1H (Spec A[ ], F) →⊕|I|=nH (Spec A[ ], F) =0 i∈I fi i∈I fi and the homomorphism
n 1 n 1 ⊕|I|=n−2Γ∗H (Spec A[ ], F) →⊕|I|=n−1Γ∗H (Spec A[ ], F) i∈I fi i∈I fi is surjective and a functor Γ∗ is exact, it implies that
n 1 n 1 ⊕|I|=n−1Γ∗H (Spec A[ ], F) →⊕|I|=nΓ∗H (Spec A[ ], F) i∈I fi i∈I fi is a zero map. On the other hand,
n 1 n 1 ⊕|I|=n−1Γ∗H (Spec A[ ], F) →⊕|I|=nΓ∗H (Spec A[ ], F) i∈I fi i∈I fi is a surjection. Thus one concludes that
n 1 n 1 Γ∗H (Spec A[ ], F)=⊕|I|=nΓ∗H (Spec A[ ], F)=0. f1 ···fn i∈I fi
Theorem 5 For 0 ≤ k ≤ n the images of
d 1 2d 1 CH (Spec A[ ]) −→ Γ∗H (Spec A[ ])(d) f1 ···fk f1 ···fk generate the targets. In particular, the image of
d 2d CH (Spec A) −→ Γ∗H∗ (Spec A)(d) generates the target.
Proof. One continues to proceed by induction argument: Hn (Spec A, F) → V(f1) Hn (Spec A, F) → Hn Spec A[ 1 ], F f1 ··· 6 ACADEMIC REPORTS Fac. Eng. Tokyo Polytech. Univ. Vol. 29 No.1 (2006)
Hn Spec A[ 1 ], F → Hn Spec A[ 1 ], F → Hn Spec A[ 1 ], F V(fn) f1···fn−1 f1···fn−1 f1···fn One has the following commutative diagram, whose vertical sequences are exact:
d−1 2d−2 CH (V(fk)) −→ Γ∗H∗ (V(fk))(d − 1) ↓↓ d 1 2d 1 CH (Spec A[ ]) −→ Γ∗H (Spec A[ ])(d) f1···fk−1 ∗ f1···fk−1 ↓↓ d 1 2d 1 CH (Spec A[ ]) −→ Γ∗H (Spec A[ ])(d) f1···fk ∗ f1···fk ↓↓ 00
In Case of Singular Varieties
Uwe Jannsen proved that the Hodge conjecture and the Tate conjecture for singular varieties are deduced by the original conjectures. For the readers con- vienience we explain it. For a smooth variety X of dimension d one has the ∼ 2d−2i Poincar´e duality H2d(X, i) = H (X, d − i). There is no such duality in general for non smooth varieties. Fundamental classes induce a cycle map: cli : Zi(X) −→ H2i(X, i), which factors through the canonical cycle map that Fulton defines CHi(X) −→ H2i(X, i). The Hodge conjecture for singular varieties says that for all i ≥ 0themap
−i −i cli ⊗ Q : Zi(X) ⊗ Q −→ Γ∗H2i(X, Q)(i) = (2πi) W−2iH2i(X, Q) ∩ F H2i(X, C)
is surjective.
Theorem 6 The Hodge conjecture is true for singular varieties.
Proof. By Chow’s lemma and Hironaka’s resolution of singularities for a sin- gular non complete variety X there exist π : X → X a projective and sujective morphism with X quasi-projective and smooth and α : X → X an open immersion with X projective and smooth forming a diagram of varieties
α X −→ X π ↓ X
∗ ∗ α π Zi(X ) ⊗ F −→ Zi(X ) ⊗ F −→ Zi(X) ⊗ F
↓ cli ↓ cli ↓ cli
∗ ∗ Γα Γπ ΓH2i(X , i) −→ ΓW0H2i(X , i) −→ ΓW0H2i(X, i) A Proof of the Hodge Conjecture 7
Since H2i(X , i) is a semi-simple object, W0H2i(X, i) is a direct factor of ∗ ∗ H2i(X , i) via π∗ ◦ α and so Γπ∗ ◦ Γα is surjective. Thus Zi(X) ⊗ F → ΓW0H2i(X, i) is surjective.
Appendix
2 Theorem 7 Let f0 :: X0 → Y0 a projective morphism, ∈ H (X0, Q (1)) the first Chern class of an f0-ample invertible sheaf and F0 a perverse sheaf over X0 for i ≥ 0. The following map is an isomorphism
i p −i ∼ p i : H f∗F0 = H f∗F0(i)
Lemma 8 It suffices to prove the Hodge conjecture in case of i =2d =dimX.
Proof. By the strong Lefschetz theorem it reduces to the case i =2p>2d. Let Y be a general hyperplane section of X. By the weak Lefschetz one has an exact sequence Hi−2(Y, F) → Hi(X, F) → Hi(X − Y, F) = 0. The following commutative diagram completes the proof:
p−1 2p−2 CH (Y) −→ Γ∗H (Y)(p − 1) ↓↓ p 2p CH (X) −→ Γ∗H∗ (X)(p) ↓↓ p 2p CH (X − Y) −→ Γ∗H∗ (X − Y)(p) ↓↓ 00
Theorem 9 Let f : X → Y be an affine morphism. The functor
b b Rf∗ : Dc(X, Q ) → Dc(Y,Q ) is right t-exact. In particular, Let k be an algebraically closed field and F an etale sheaf on X. Hi(X, F)=0for i>dim X.
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