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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 $\mathbf{G}_m[-1]$.

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    $\begingroup$ I saw your question when you first posted it. I'm fairly skeptical, but I don't have counterexample. If you are seriously interested 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
  • $\begingroup$ Done. Yes I agree in the end it is not so obvious even in the case $n = 0$, but upon mimicking the Kerr-Lewis-MuellerStach construction to obtain $\text{reg}$, if the source is the "analytic motivic cohomology" above then one can construct an inverse up to homotopy, at least for $n = 0,1$. In other words, the question asks if anybody ever thought about this and if one already has a negative answer to share. $\endgroup$ – user92332 Jan 7 '18 at 2:02
  • $\begingroup$ The answer by @SUSY points out an obstruction to a positive answer beyond the smooth projective case, for instance. I'm unable to prove $\text{reg}$ is a quasi-isomorphism for $n>1$, for now, and it seems hard, in fact. The point of the question is that if this is correct, then one can rewrite the usual regulator map as "analytification of cycles" followed by $\text{reg}$ $\endgroup$ – user92332 Jan 7 '18 at 2:05
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The answer is no.

So far, you have checked only very special cases, such as weight $n$ and degree $2n$, for $n = 0,1$.

Consider the analytic hypercohomology of your $\mathbf{Z}(n)_{\mathcal{M}}$ in degree $2n$, denoted $H^{2n}_{\rm an}(X,\mathbf{Z}(n))$. The same (elementary) argument as in several references (eg. Bloch's "Algebraic Cycles and Higher $K$-Theory", 1986) shows

$$H^{2n}_{\rm an}(X,\mathbf{Z}(n)) \simeq\text{CH}^n(X^{\rm an})$$ the Chow group of analytic cycles on $X^{\rm an}$. By GAGA, there is a canonical isomorphism of abelian groups $\text{CH}^n(X)\simeq\text{CH}^n(X^{\rm an})$.

Upon running the same construction of the cycle map to Deligne cohomology as in Kerr-Lewis, as you suggest, you must end up with a cycle map

$$\text{CH}^n(X^{\rm an})\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$$ that, under the above isomorphism from GAGA, should better agree with the usual cycle map $\text{CH}^n(X)\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$.

Rem. If your conjecture were true, then, in particular, the cycle map $\text{CH}^n(X)\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$ should be surjective, this latter group being surjecting onto the finitely generated abelian group of Hodge cycles $\text{Hdg}^{n,n}(X^{\rm an})$, the composition being the cycle map to Betti cohomology. As a consequence, the Hodge conjecture is implied by your contention.

There are two problems. I am not aware of any expectation along the lines of surjectivity of the cycle map onto Deligne(-Beilinson) cohomology. Rather, the only claim being usually made is surjectivity onto the abelian group of Hodge classes.

On the other hand, if your contention were correct, then the cycle map $\text{CH}^n(X)\to H^{2n}_{\mathcal{D}}(X^{\rm an},\mathbf{Z}(n))$ should be injective too, which rarely happens, as this puts strong constraints on the Hodge numbers of $X^{\rm an}$.

A paper of Esnault and Levine proves that if such cycle map is injective, then it is also surjective, and implies strong conditions on the shape of the Hodge diamond of $X^{\rm an}$. There are many examples of an $X$ that does not meet such conditions.

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  • $\begingroup$ OK. Thanks! I'll wait a bit before awarding the bounty. $\endgroup$ – user92332 Feb 12 '18 at 6:19
  • $\begingroup$ @Merlin: Could you please elaborate on why GAGA implies $$CH^n(X) \cong CH^n(X^{an})$$? Thanks $\endgroup$ – guest Feb 12 '18 at 16:19
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    $\begingroup$ @guest: I think the author means Chow's theorem that a closed analytic subspace of a projective analytic space is algebraic. Although proven before GAGA, it also follows from it, cf. Proposition 13 of Serre's GAGA paper. $\endgroup$ – R. van Dobben de Bruyn Apr 21 '18 at 4:15
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Wouldn't your motivic cohomology group in degree $(2m, m)$ be the Chow group of codimension $m$ cycles? But the corresponding Deligne cohomology group surjects onto the Hodge classes in cohomological degree $2m$. For them to be quasi-isomorphic you would (at least) need the Hodge conjecture, which is false for a general Kahler manifold.

As you said any explicit formula for the usual regulator should extend to this setting. The paper of Kerr, Lewis, Mueller-Stach https://arxiv.org/abs/math/0409116 does this at least for nullhomologous classes.

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    $\begingroup$ By Chow group you must mean the Chow group of analytic cycles. The case the question is asking about is mostly the one when $X$ is algebraizable. You might as well assume, at the very beginning of the question, $X$ is smooth projective. $\endgroup$ – user92332 Jan 7 '18 at 0:14
  • $\begingroup$ sorry I didn't see "projective." $\endgroup$ – SUSY student Jan 7 '18 at 1:41

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