I was reading the following paper: "Beilinson’s Hodge Conjecture For Smooth Varieties". They study the cycle class map $cl_{m,r}: H^{2r-m}_{\mathcal{M}}(U, \mathbb{Q}(r))\rightarrow \text{hom}_{MHS}(\mathbb{Q}(0), H^{2r-m}(U, \mathbb{Q}(r)))=\Gamma(H^{2r-m}(U,\mathbb{Q}(r)))$. The left side is the motivic cohomology and the right side is the hom in the category of mixed Hodge structures. On the second page the state the following three conjectures:

  1. (S1) $cl_{r,m}:H^{2r-m}_{\mathcal{M}}(X, \mathbb{Q}(r))\rightarrow \Gamma(H^{2r-m}(X, \mathbb{Q}(r)))$ is surjective for smooth complex projective varieties $X$.

  2. (S2) $cl_{r,m}:H^{2r-m}_{\mathcal{M}}(U, \mathbb{Q}(r))\rightarrow \Gamma(H^{2r-m}(U, \mathbb{Q}(r)))$ is surjective for smooth complex quasi-projective varieties $U$.

  3. (S3) $cl_{r,m}:H^{2r-m}_{\mathcal{M}}(\mathbb{C}(X), \mathbb{Q}(r))\rightarrow \Gamma(H^{2r-m}(\mathbb{C}(X), \mathbb{Q}(r)))$ is surjective for function fields $\mathbb{C}(X)$ of complex varieties $X$.

Now for $m=0$ they claim that all three are equivalent for all $r\geq 0$. The direction going from (S1) to (S2), follows from localization sequence and using the fact that $\Gamma(H^{2r}(X, \mathbb{Q}(r)))\rightarrow \Gamma(H^{2r}(U, \mathbb{Q}(r)))$ is surjective. Going from (S2) to (S3) is just a limiting argument. I was not able to figure out how to go from (S1) to (S3). I'd appreciate if anyone can give a brief overview or point to the right source.

Another similar question one can ask is whether something similar holds for Tate conjecture? (the reduction to the generic point)

Couple of points: Note that $H_{\mathcal{M}}^{2r}(\mathcal{C}(X), \mathbb{Q}(r))=0$, so (S3) (when $m=0$) is just equivalent to saying $\Gamma(H^{2r}(\mathbb{C}(X), \mathbb{Q}(r)))=0$. It is possible by using Gersten type resolutions to show that $\Gamma(H^{2r}(\mathbb{C}(X), \mathbb{Q}(r)))=0$ implies that $\Gamma(H^{2r}(R, \mathbb{Q}(r)))=0$ where $R$ is the spectrum of a local ring (or semi-local) of a variety. But I am not sure how one can go from local rings to arbitrary quasi-projective varieties.


1 Answer 1


This turned out to be much simpler than what I expected and it just follows from the localization. I will sketch the proof from (S3) to (S2). We have the following sequence from the localization for the motivic cohomology: $$H^{2n-2c}_{\mathcal{M}}(Z, \mathbb{Q}(n-c))\rightarrow H^{2n}_{\mathcal{M}}(X, \mathbb{Q}(n))\rightarrow H^{2n}_{\mathcal{M}}(U, \mathbb{Q}(n))$$ Where $Z$ is a smooth closed subscheme of the smooth scheme $X$ with codimension $c$ and $U$ is the open complement. The idea is that we can take the limit of the above sequence and get to the following sequence: $$H_{\mathcal{M}, \mathcal{Z}}^{2n}(X, \mathbb{Q}(n))\rightarrow H^{2n}_{\mathcal{M}}(X, \mathbb{Q}(n))\rightarrow H^{2n}_{\mathcal{M}}(\mathbb{C}(X), \mathbb{Q}(n)) \text{ }\text{ }(1)$$ where here $\mathcal{Z}$ is the family of smooth supports in $X$ with positive codimension.

A similar exact sequence holds for the Hodge cohomology side i.e. we have: $$\Gamma(H_{\mathcal{Z}}^{2n}(X, \mathbb{Q}(n)))\rightarrow \Gamma(H^{2n}(X, \mathbb{Q}(n)))\rightarrow \Gamma(H^{2n}(\mathbb{C}(X), \mathbb{Q}(n)))\text{ }\text{ }(2)$$ Note that the right most group in the sequence $(1)$ is zero and also the right most group in $(2)$ is zero if we assume (S3). Now we will prove (S2) by induction on the dimension of the variety. There is a vertical map from sequence $(1)$ to the sequence $(2)$ by the induction hypothesis the vertical map on the left group is surjective so we can see this implies that the vertical map on the middle groups is also surjective finishing the induction.

So the Hodge conjecture is simply equivalent to the vanishing of the groups $\Gamma(H^{2n}(\mathbb{C}(X), \mathbb{Q}(n)))$. The similar equivalence also holds for the Tate conjecture the same exact argument works in the $l$-adic setting too.


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