
> 
> **Intro.** *I would be deeply grateful if someone could please clarify the following to me.*
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**The question.** (the main point is (4))


Let $X$ be a smooth projective variety over $\mathbf{C}$, and $\mathbf{Z}(n)\_{\mathcal{D}}$ the Deligne complex on $X$.


We have a triangle:
$$\to\Omega^{<n}\_X \to\mathbf{Z}(n)\_{\mathcal{D}}\to\mathbf{Z}(n)\to \Omega^{<n}\_X[1]$$


of bounded complexes of abelian groups. We take hypercohomology, and sheafify with respect to the analytic topology on $X$ the cohomology groups obtained.


We get: $\mathcal{H}^\*(\Omega^{<n}\_X)$, $\mathcal{H}^\*(\mathbf{Z}(n)\_{\mathcal{D}})$, $\mathcal{H}^\*(\mathbf{Z}(n))$, and a spectral sequence


$$H^p(X, \mathcal{H}^q(\mathbf{Z}(n)\_{\mathcal{D}}))\Rightarrow H^{p+q}(X,\mathbf{Z}(n)\_{\mathcal{D}})$$


In this [paper](https://pdfs.semanticscholar.org/bfe4/03501f5a05ef464564a9132d2722381b3cdf.pdf) by Gillet (Thm. 2(ii)) it is claimed without proof that
$$H^n(X,\mathcal{H}^n(\mathbf{Z}(n)\_{\mathcal{D}})) \simeq\text{CH}^n(X).$$



> 
> (1) First off, it is not clear if by $\text{CH}^n(X)$ one means the Chow group of **analytic cycles**. In any event, by GAGA, $\text{CH}^n(X^{\rm an})$ at least receives a surjection from the Chow group of **algebraic** cycles modulo rational equivalence, presumably compatible with cycle maps, so the rest of the question remains "quite puzzling".
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> (2) Second: how to prove this?!
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Most importantly.


We get exact sequences of analytic sheaves on $X$:


$$\to \mathcal{H}^\*(\Omega\_X^{<n})\to\mathcal{H}^\*(\mathbf{Z}(n)\_{\mathcal{D}})\to \mathcal{H}^\*(\mathbf{Z}(n))\to$$



> 
> (3) Is there an isomorphism $H^n(X,\mathcal{H}^n(\mathbf{Z}(n)))\simeq H^{2n}(X,\mathbf{Z}(n))$ compatible with the one for $\mathcal{H}^\*(\mathbf{Z}(n)\_{\mathcal{D}})$ and such that the induced map $H^n(X,\mathcal{H}^n(\mathbf{Z}(n)\_{\mathcal{D}})\to H^n(X, \mathcal{H}^n(\mathbf{Z}(n)))$
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> agrees with the $n$-th cycle map?:
> $$\text{CH}^n(X)\to H^{2n}(X,\mathbf{Z}(n)).$$
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This would answer this [question](https://mathoverflow.net/questions/294856/cycle-maps-as-edge-maps) of mine.


I am very puzzled by this, however.


Essentially by design, the map $H^{2n}(X,\mathbf{Z}\_{\mathcal{D}}(n))\to H^{2n}(X,\mathbf{Z}(n))$ surjects onto the subgroup of Hodge classes $\text{Hdg}^n(X)$.


**The main point.**



> 
> (4) Doesn't this mean that, if (3) is true and in light of (1) and by choosing $p = q = n$ in the above spectral sequence, the cycle map $\text{CH}^n(X)\to H^{2n}(X,\mathbf{Z}(n))$ surjects onto Hodge classes? This would be the Hodge Conjecture, so the answer **must** be "no" as it cannot be this easy, but **I would like to understand why**.
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**Another background reference** is this [paper](https://msp.org/akt/2016/1-1/akt-v1-n1-p02-p.pdf) by Luca Barbieri-Viale.



A quite puzzling question on Deligne cohomology sheaves and cycle maps
1) presumably it's algebraic cycles. The linked announcement works with schemes of finite type over $\operatorname{spec} \mathbb{C}$. For a smooth projective variety, which you're using, Chow's theorem should force analytic cycles to agree with algebraic cycles. Then equivalence relation comes from function fields of subschemes, and these will all be algebraic.


2) If you believe Theorem 2.i then this should just be plugging and chugging the spectral sequence. The claim is $E^{p,p}\_2(X,p) \simeq CH^p(X)$. But $E^{p,p}\_1(X,p)$ is $\oplus\_{x \in X^{(p)}}H^0(X, \mathbb{Z}\_{\mathscr{D}}(n))$ which up to twists is exactly $Z^p(X)$ the group of formal $p$-cycles. So you just need to identify the maps on page 1 of the spectral sequence with the maps from function fields, which I don't expect is too bad.


3) There can't be such an iso in general. As far as your question regarding mapping $CH^p(X)$ into $H^{2p}(X, \mathbb{Z}(p))$, this should be given by the current of integration along the submanifolds in a cycle. Unfortunately I don't know a great citation for this. It might be in some of James King's papers from the early 70s or it might just be in de Rham's book. 


4) The integral Hodge conjecture is known to be false, so it can't be that!


