Let $X$ be a smooth projective complex analytic space. We can cook up a complex analytic version of Bloch's cycle complex by declaring

$z^n(X^{\rm an}, m)$

is the free abelian group on all codimension $m$ analytic cycles on $X\times\Delta^n$ ($\Delta^n$ being the usual standard $n$ simplex in complex analytic spaces, ie. the spectrum of $\mathbf{C}\{u_0,\ldots, u_n\}/(u_0+\ldots+u_n -1)$) in good position (ie. intersecting every face in the appropriate codimension, as in Bloch's paper). The differential $d_m$ is the same as in Bloch's original definition, turning $(z^n(X^{\rm an}, m), d_m)$ into a complex of abelian groups.

Call $$\mathbf{Z}(n)_{\mathcal{M}} := (z^n(X^{\rm an}, m), d_m)[2m]$$ and its hypercohomology "motivic cohomology of $X$".

Here's the question. Is motivic cohomology of $X$ at all related to the Deligne cohomology of $X$? More optimistically, does there exist a quasi-isomorphism

$$\mathbf{Z}(n)_{\mathcal{M}}\to\mathbf{Z}(n)_{\mathcal{D}} ?$$

How should one think about Deligne cohomology, in other words? (if not as "the motivic cohomology of complex analytic spaces?)

**Remarks**

I can imagine a regulator map $\text{reg} : \mathbf{Z}(n)_{\mathcal{M}}\to\mathbf{Z}(n)_{\mathcal{D}}$ can be defined using currents, as done for the classical regulator.

This is for sure going to be a (rather uninteresting) quasi isomorphism, since $X$ is smooth, when $n = 0$.

For $n = 1$ this is likely going to be a quasi-isomorphism too (if one doesn't screw the definition of $\text{reg}$): both sides are just $\mathbf{G}_m[-1]$.

seriouslyinterested in the outcome, you should first check it for $n=0$. It's not as obvious (to me) as you seem to be suggesting. $\endgroup$ – Donu Arapura Jan 7 '18 at 1:43