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Say $X$ is a smooth projective variety over $\mathbf{C}$, and $\mathcal{X} = X^{\rm an}$ its $\mathbf{C}$-analytic space.

For what integers $i,d$ is the Deligne cohomology $H^i_{\mathcal{D}}(\mathcal{X},\mathbf{Z}(d))$ of $\mathcal{X}$, an extension of a finitely generated abelian group by a compact group?

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    $\begingroup$ Did you mean to ask when it would be an extension of a finitely generated Abelian group by a compact group? In all of the interesting cases that I know of, the group is neither compact nor finitely generated, but sometimes it is an extension of a finitely generated group by a compact group. $\endgroup$ – Jason Starr Nov 29 '17 at 1:10
  • $\begingroup$ Yes, correct. This is what I actually had in mind. I will modify the question now. Thanks $\endgroup$ – user97068 Nov 29 '17 at 2:31
  • $\begingroup$ Is there an integer-valued function $i\mapsto d_i$, and a subset $I\subset\mathbf{Z}$ of the integers, such that $H^i(\mathcal{X}, \mathbf{Z}(d_i))$ is an extension of a finitely generated abelian group by a compact group, for all $i\in I$, and for all such $\mathcal{X}$? eg. $I = 2\mathbf{Z}$ and $d_i = i+1$, just to fix ideas. $\endgroup$ – user97068 Nov 29 '17 at 2:35
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    $\begingroup$ For $I = 2\mathbf{Z}$, $d_i := i/2$, $H^i_{\mathcal{D}}(\mathcal{X},\mathbf{Z}(d_i))$ is an extension of a finitely generated abelian group by a compact torus: we have a ses of abelian groups $0\to J^n(\mathcal{X})\to H^{2n}_{\mathcal{D}}(\mathcal{X},\mathbf{Z}(n))\to \text{Hdg}^n(\mathcal{X},\mathbf{Z})\to 0$ where $J^n(\mathcal{X})$ is the $n$th intermediate Jacobian of $\mathcal{X}$, and $\text{Hdg}^n(\mathcal{X},\mathbf{Z})$ the finitely generated abelian group of integral Hodge classes. $\endgroup$ – user87684 Nov 29 '17 at 10:44
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    $\begingroup$ See Esnault-Viheweg, $\S$7.9. $\endgroup$ – user87684 Nov 29 '17 at 10:48
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The following ought to answer your question.

Prop. Let $\mathcal{X}$ be a smooth proper complex analytic space. We have a long exact sequence of abelian groups:

$$\cdots \to H^i_{\mathcal{D}}(\mathcal{X},\mathbf{Z}(j))\to H^i(\mathcal{X},\mathbf{Z})\to H^i(\mathcal{X},\mathbf{C})/F^jH^i(\mathcal{X},\mathbf{C})\to H^{i+1}_{\mathcal{D}}(\mathcal{X},\mathbf{Z}(j))\to\cdots$$

where $F^{\bullet}H^i(\mathcal{X},\mathbf{C})$ is the Hodge filtration on $H^i(\mathcal{X},\mathbf{C})$.

Sketch of Proof. By design of the Deligne complex $\mathbf{Z}(j)_{\mathcal{D}}$, we have a triangle in $D(\text{Ab})$: $$\Omega_{\mathcal{X}/\mathbf{C}}^{\le j-1}[1]\to \mathbf{Z}(j)_{\mathcal{D}}\to\mathbf{Z}.$$ Form hypercohomology and prove $\mathbb{H}^i(\mathcal{X},\Omega_{\mathcal{X}/\mathbf{C}}^{\le j-1}) = H^i(\mathcal{X},\mathbf{C})/F^jH^i(\mathcal{X},\mathbf{C})$. To see this, use the triangle: $$\Omega_{\mathcal{X}/\mathbf{C}}^{\ge j}\to \Omega_{\mathcal{X}/\mathbf{C}}^{\bullet}\to \Omega_{\mathcal{X}/\mathbf{C}}^{\le j-1}$$ and reduce to check $\mathbb{H}^i(\mathcal{X},\Omega_{\mathcal{X}/\mathbf{C}}^{\ge j})=F^jH^i(\mathcal{X},\mathbf{C})$, which follows almost by definition of the Hodge filtration. QED

In particular, for $i=2j$, we get: $$0\to J^j(\mathcal{X})\to H^{2j}_{\mathcal{D}}(\mathcal{X},\mathbf{Z}(j))\to \text{Hdg}^{j}(\mathcal{X},\mathbf{Z})\to 0$$ upon being careful about grading conventions (eg. I have seen $J^n$ often denoted $J^{2n}$, etc.)

Rem. Note that $H^i(\mathcal{X},\mathbf{Z})$ is always finitely generated, hence your question really is about when the image of $H^i(\mathcal{X},\mathbf{Z})$ in $H^i(\mathcal{X},\mathbf{C})/F^jH^i(\mathcal{X},\mathbf{C})$ under the above edge map, is a full lattice. If $i$ is odd this is true.

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