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Unramified cohomology and the entire Hodge conjecture. (Cohomologie non ramifiée et conjecture de Hodge entière.) (French. English summary) Zbl 1244.14010
Let $$X$$ be a complex projective manifold of dimension $$d$$. For $$i \geq 1$$ we set $$\mathbb Z(i):=\mathbb Z(1)^{\otimes i}$$ where $$\mathbb Z(1):=\mathbb Z(2 \pi i) \subset \mathbb C$$. We define the group of integral Hodge classes $Hdg^{2i}(X, \mathbb Z) \subset H^{2i}(X(\mathbb C), \mathbb Z(i))$ as the preimage of the usual Hodge classes $$Hdg^{2i}(X, \mathbb Q)$$ under the natural map from integral to rational cohomology. We define $$H^{2i}_{ alg}(X(\mathbb C), \mathbb Z(i))$$ as the image of the map from the Chow group $CH^i(X) \rightarrow H^{2i}(X(\mathbb C), \mathbb Z(i)).$ Then we have an inclusion $$H^{2i}_{ alg}(X(\mathbb C), \mathbb Z(i)) \subset Hdg^{2i}(X, \mathbb Z)$$ and we set $Z^{2i}(X) := Hdg^{2i}(X, \mathbb Z)/H^{2i}_{ alg}(X(\mathbb C), \mathbb Z(i)).$ If the Hodge conjecture holds in degree $$2i$$ the group $$Z^{2i}(X)$$ is torsion and equal to $H^{2i}(X(\mathbb C), \mathbb Z(i)) / H^{2i}_{ alg}(X(\mathbb C), \mathbb Z(i)),$ in particular it would be a finite group. If the group $$Z^{2i}(X)$$ is equal to zero this means that the Hodge conjecture holds even for integral coefficients, a statement that is true for $$i=1$$, but known to be false in general for $$i=2$$ by examples of M. F. Atiyah and F. Hirzebruch [Topology 1, 25–45 (1962; Zbl 0108.36401)] and J. Kollár [in: Classification of irregular varieties. Minimal models and Abelian varieties. Lecture Notes in Mathematics. 1515. Berlin etc.: Springer-Verlag (1992; Zbl 0744.00029), Lemma on p. 134].
In the paper under review the authors consider the group $$Z^4(X)$$ which is particularly interesting since it is a birational invariant of the manifold $$X$$. Their main theorem relates this group to unramified cohomology groups $$H^i_{ nr}(X, \mathbb Z/n)$$ introduced by J.-L. Colliot-Thélène and M. Ojanguren [Invent. Math. 97, No. 1, 141–158 (1989; Zbl 0686.14050)]. More precisely they prove the following statement: let $$X$$ be a complex projective manifold such that the Chow group of zero cycles $$CH_0(X)$$ has support on a surface. Then we have an isomorphism of finite groups $H^3_{ nr}(X, \mathbb Q/\mathbb Z(2)) \simeq Z^4(X)$ where the first group is the union of its subgroups $$H^3_{ nr}(X, \mu_n^{\otimes 2})$$. Based on this fundamental result it is possible to compare and translate results obtained by methods of algebraic geometry and algebraic $$K$$-theory. For example it implies that for any smooth projective threefold that is covered by rational curves, the group $$H^3_{ nr}(X, \mathbb Q/\mathbb Z(2))$$ is zero since $$Z^4(X)$$ is zero by a theorem of C. Voisin [Advanced Studies in Pure Mathematics 45, 43–73 (2006; Zbl 1118.14011)]. Vice versa an example of a non-rational manifold by Colliot-Thélène and Ojanguren [loc. cit.] gives an example of a projective manifold of dimension $$6$$ that is unirational, but $$Z^4(X)=Z^4(X)_{ tors} \neq 0$$. This gives a negative answer to a question of C. Voisin [Jpn. J. Math. (3) 2, No. 2, 261–296 (2007; Zbl 1159.14005)]. The authors also consider the case of a threefold admitting a fibration $$X \rightarrow \Gamma$$ onto a smooth curve, under certain conditions they establish a link between the group $$H^3_{ nr}(X, \mathbb Q/\mathbb Z(2))$$, the group $$Z^4(X)$$ and the index of the fibration, i.e., the greatest common divisor of the degrees of all the multisections.

##### MSC:
 14E05 Rational and birational maps 14F99 (Co)homology theory in algebraic geometry
##### Keywords:
Hodge conjecture; unramified cohomology
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