Let $X$ be a smooth projective complex analytic space, $i,p\ge 0$ integers, $\mathbf{Z}(p)_{\mathcal{D}}$ the Deligne complex of $X$, $H^i_{\mathcal{D}}(X,\mathbf{Z}(p))$ its hypercohomology.

What properties does the subgroup of torsion elements of $H^i_{\mathcal{D}}(X,\mathbf{Z}(p))$ have?

  • For instance, is it finite? Does it contain divisible elements?

Since $H^i_{\mathcal{D}}(X,\mathbf{Z}(p))$ is an extension of a finitely generated $\mathbf{Z}$-module by a quotient of a graded in the Hodge filtration on de Rham cohomology of $X$, one expects the answer depends on $i,p$.

More precisely, the question might as well be asked about the quotient group

$$J^{i,p}(X/\mathbf{C}) := \frac{H^i_{\rm dR}(X/\mathbf{C})}{F^pH^i_{\rm dR}(X/\mathbf{C}) + H^i(X,\mathbf{Z}(p))}$$

  • What if one restricts attention to the set $J^{i,p}(X/k)$ of those classes in $J^{i,p}(X/\mathbf{C})$ that come from algebraic cycles on the algebraization of $X$, and are defined on a fixed subfield $k\subset\mathbf{C}$?

More precisely, if $\mathcal{X}$ is a smooth projective algebraic $k$-variety such that $(\mathcal{X}\otimes_k\mathbf{C})^{\rm an}\simeq X$, and $c : H^i_{\mathcal{M}}(\mathcal{X},\mathbf{Z}(p))\to H^i_{\mathcal{D}}(X,\mathbf{Z}(p))$ is the cycle map, define $J^{i,p}(X/k)$ to be

$$J^{i,p}(X/k) := J^{i,p}(X/\mathbf{C})\times_{H^i_{\mathcal{D}}(X,\mathbf{Z}(p))}H^i_{\mathcal{M}}(\mathcal{X},\mathbf{Z}(p))$$

What can be said about the torsion subgroup of $J^{i,p}(X/k)$?

If $k$ is algebraically closed, do we have $$J^{i,p}(X/k)_{\rm tor} = J^{i,p}(X/\mathbf{C})_{\rm tor}\ ?$$

Example: the case $i=2,p=1.$

As suggested in the comment, one can think about the case $i = 2$, $j=1$ first. Here $\mathbf{Z}(1)_{\mathcal{D}} \simeq\mathbf{G}_{\rm m}[-1]$ and $H^2_{\mathcal{D}}(X,\mathbf{Z}(1)) = H^1(X,\mathbf{G}_{\rm m}) = \text{Pic}(X)$. In this case, we have an exact sequence:

$$H^1(X,\mathbf{G}_{\rm a}) \to \text{Pic}(X)\xrightarrow{c_1} H^2(X,\mathbf{Z}(1))$$

that identifies the extension

$$0\to J^{i,p}(X/\mathbf{C})\to H^i_{\mathcal{D}}(X,\mathbf{Z}(p))\to\text{Hdg}^{i,p}(X/\mathbf{C})\to 0$$


$$0\to \text{Pic}^0(X)\to\text{Pic}(X)\to\text{NS}(X)\to 0$$

whence $J^{2,1}(X/\mathbf{C}) = \text{Pic}^0(X)$. By GAGA we have $\text{Pic}^0(X)\simeq\text{Pic}^0(\mathcal{X}_{\mathbf{C}})$ and since $\mathbf{C}$ is separably closed $$\text{Pic}^0(\mathcal{X}_{\mathbf{C}}) = \underline{\text{Pic}}^0_{\mathcal{X}_{\mathbf{C}}/\mathbf{C}}(\mathbf{C}) = (\underline{\text{Pic}}^0_{\mathcal{X}/k}\times_k\mathbf{C})(\mathbf{C}).$$

If $k$ is separably closed too, then indeed the torsion subgroup of $\text{Pic}^0(X)$ agrees with that of $\underline{\text{Pic}}^0_{\mathcal{X}/k}(k)$, since the kernel of multiplication by $n$ on $\underline{\text{Pic}}_{\mathcal{X}/k}^0$ is an étale group scheme for every $n$ (since $k$ of characteristic zero).

Since $\underline{\text{Pic}}^0_{\mathcal{X}/k}(k) = J^{2,1}(X/k)$, we indeed get

$$J^{2,1}(X/\mathbf{C})_{\rm tor} = J^{2,1}(X/k)_{\rm tor}.$$

  • $\begingroup$ The Deligne cohomology group with $i = 2$ and $j=1$ is isomorphic to the Picard group of $X$, so, in this case, the answer to your second question also depends on $X$. $\endgroup$ – Tony Feb 7 '18 at 18:18
  • $\begingroup$ After the edit, $j = p$ in my comment above. $\endgroup$ – Tony Feb 7 '18 at 19:40
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    $\begingroup$ As you note, $H^i_{\mathcal{D}}(X,\mathbf{Z}(p))$ contains a copy of a quotient of a $\mathbf{Q}$-vector space, hence lots of divisible elements. If you ask yourself the question whether $J^{i,p}(X/\mathbf{C})$ contains torsion divisible elements, then the answer is yes as soon as $H^i(X,\mathbf{Z}(p))$ is not contained in $F^pH^i_{\rm dR}(X/\mathbf{C})$. If so, then $J^{i,p}(X/\mathbf{C})$ is a $\mathbf{Q}$-vector space, hence torsion free. It typically never happens that $H^i(X,\mathbf{Z}(p))\subset F^pH^i_{\rm dR}(X/\mathbf{C})$, so usually $J^{i,p}(X/\mathbf{C})_{\rm tor}$ has div elements $\endgroup$ – user92332 Feb 8 '18 at 3:10

The following is not an answer, since the last comment to your question already answers most of it.

It is, rather, a "better question".

What you really want to ask is: let $f: X\to\text{Spec}(k)$ be a proper, geometrically connected, geometrically reduced scheme over a perfect field.

For integers $p,q$, $p\ge 0$, does there exist a $k$-group scheme $P^{p,q}_{X/k}$?, such that:

  • $P^{p,q}_{X/k}$ is locally of finite type over $k$ and quasi-separated.
  • The component of the identity $(P^{p,q}_{X/k})^0$ is connected, finite type and separated.
  • Formation of $P^{p,q}_{X/k}$ commutes with separable field extensions on $k$.
  • If $X$ is smooth over $k$, then $(P^{p,q}_{X/k})^0$ is smooth.
  • If $X$ is smooth and projective over $k$, then $\text{NS}^{p,q}(X) := P^{p,q}_{X/k}/(P^{p,q}_{X/k})^0$ is a constant étale $k$-group scheme whose value group is finitely generated.
  • If $X$ is smooth and projective over $k$, and $k$ is a subfield of $\mathbf{C}$, then $$(P^{p,q}_{X/k}\times_k\mathbf{C})(\mathbf{C}) = H^p_{\mathcal{D}}(X_{\mathbf{C}}^{\rm an},\mathbf{Z}(q)),\ \ ((P^{p,q}_{X/k})^0\times_k\mathbf{C})(\mathbf{C}) = J^{p,q}(X^{\rm an}_{\mathbf{C}}/\mathbf{C}),\ \ (\text{NS}^{p,q}(X)\times_k\mathbf{C})(\mathbf{C}) = \text{Hdg}^{p,q}(X_{\mathbf{C}}^{\rm an}/\mathbf{C}).$$

In other words, you're actually looking for "higher Picard functors", representable under the assumptions stated at the beginning of this post, such that their connected component of the identity is an extension of a connected smooth affine $k$-group scheme by an abelian $k$-variety, and such that an analog of the theorem of the base holds true.

For $q=1$, this problem is related to the question whether $R^{p-1}f_{\rm fppf, *}\mathbf{G}_m$ is representable, which, for $p=2$, we now to be the case by work of Artin. Even the case $q=1$, for arbitrary $p\ge 2$, is extremely hard. A seemingly easier question is whether $R^{p-1}f_{\rm fppf, *}\mu_n$ is representable, and even this question (posed by Artin when $n$ is prime) is very hard too, and answered positively only in very special cases (mostly by work of Lieblich).

If the above dream were true, then your question becomes indeed very interesting, if asked about $(P^{p,q}_{X/k})^0(k^{\rm sep})^{\text{Gal}(k^{\rm sep}/k)}$ instead of $J^{p,q}(X_{\mathbf{C}}/\mathbf{C})$, and when $k$ is a finitely generated field.

The functors $P^{p,q}_{X/k}$, which I'm not going to define here, occurred at some point in my work, where I was able to show their representability in some special cases I was interested in (by direct construction, ie. not via Artin's axioms). The general question, open for now, seems to be very hard (if true at all).

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    $\begingroup$ Trivial comment: since $P^{p,q}_{X/k}$ is being asked to be an lft group scheme rather than "just" an lft group algebraic space, it is automatically separated (and its identity component is automatically of finite type, even if "just" an algebraic space). Of course, perhaps a "better" dream is that it is just an lft algebraic space, though then under quasi-separatedness it is a scheme anyway. $\endgroup$ – nfdc23 Feb 8 '18 at 5:05
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    $\begingroup$ This is a very interesting way to think about it. Thanks. $\endgroup$ – user113393 Feb 8 '18 at 5:39

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