Intro.I would be deeply grateful if someone could please clarify the following to me.

**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 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 ofalgebraiccycles modulo rational equivalence, presumably compatible with cycle maps, so the rest of the question remains "quite puzzling".(2) Second: how to prove this?!

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 $H^n(X,\mathcal{H}^n(\mathbf{Z}(n)_{\mathcal{D}})\to H^n(X, \mathcal{H}^n(\mathbf{Z}(n)))$

agrees with the $n$-th cycle map?: $$\text{CH}^n(X)\to H^{2n}(X,\mathbf{Z}(n)).$$

This would answer this question 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

mustbe "no" as it cannot be this easy, butI would like to understand why.

**Another background reference** is this paper by Luca Barbieri-Viale.