# Poincaré duality for Deligne (co)homology

My question is about the papers

(both from the Beı̆linson's conjectures volume).

Let $X$ be a smooth complex variety. We may define Deligne-Beı̆linson cohomology groups $$H^\bullet_\mathcal{D} (X, A (n)),$$ where $\mathbb{Z} \subseteq A \subseteq \mathbb{R}$, and $n \in \mathbb{Z}$ is some twist. (See the paper by Esnault and Viehweg.) There is also Deligne homology defined by Jannsen: $$H_\bullet^\mathcal{D} (X, A (n)).$$ Jannsen establishes the "twisted Poincaré duality" $$\tag{*} H^i_\mathcal{D} (X, A (n)) \cong H^\mathcal{D}_{2d-i} (X, A (d-n)),$$ where $d = \dim_\mathbb{C} X$.

Does it give some classic duality result as a particular case?

My guess is that it's supposed to be some generalization of the duality between singular cohomology and Borel-Moore homology. According to Lemma 1.11 from the same paper of Jannsen, $$H^\mathcal{D}_\bullet (X, \mathbb{Z} (0)) \cong H^\mathrm{BM}_\bullet (X, \mathbb{Z}).$$ So that the above duality gives us as a particular case $$H^i_\mathcal{D} (X, \mathbb{Z} (d)) \cong H^\mathrm{BM}_{2d-i} (X, \mathbb{Z}).$$ But what is the left hand side?

Edit: The above is false, I misread the statement of Lemma 1.11.

I hope someone can explain me the motivation behind (*).

Thank you.

No, lemma 1.11 of Jannsen doesn't say what you wrote. I checked because the isomorphism you gave is usually false. The right side is a finitely generated abelian group but the other side usually isn't. For a smooth projective variety, Deligne cohomology is an extension of a subspace of singular cohomology (the Hodge cycles) by the intermediate Jacobian, which is a complex torus.

Here's one thing an algebraic geometer might use Poincaré duality for. There is a fairly easy to construct cycle map from the Chow group to singular homology. Using duality, you can flip it into cohomology, which is better for some purposes. You can play the same sort of game with Deligne (co)homology because one also has duality for it. The cycle map extends to a map to Deligne (co)homology, so it detects more than the previous map. This is often why one cares about it.

• Oh, I see. I totally misread the statement: I thought it was ${}' C^\bullet_\mathcal{D} (X,Y,\mathbb{Z})$, and what is written there is ${}' C^\bullet (X,Y,\mathbb{Z})$... – AAA Oct 6 '17 at 18:30
• I guessed as much. – Donu Arapura Oct 6 '17 at 18:33
• Thank you very much! Still, if we take $n = 0$ in the duality isomorphism, then the formula reads $H^i (X, \mathbb{Z}) \cong H^\mathcal{D}_{2d-i} (X, \mathbb{Z} (d))$. What is the right hand side? – AAA Oct 6 '17 at 19:01
• @DonuArapura Could you give a reference for the fact about Deligne cohomology being an extension of a finitely generated abelian group by a complex torus? For which degrees and weight is this true? Thanks – user97068 Nov 29 '17 at 10:39

Deligne complex $A(p)$ is the complex

$$A((2\pi i)^p) \rightarrow \mathcal O \rightarrow \Omega^1 \rightarrow {}...{} \rightarrow \Omega^{p-1}.$$

You can work with similar bigraded complexes, taking into account not only holomorphic, but also antiholomorphic forms. Let us define $A(p,q)$ as the complex of sheaves

$$A((2\pi i)^{p+q}) \rightarrow \mathcal O \oplus \overline{\mathcal O} \rightarrow \Omega^1\oplus \overline{\Omega^1} \rightarrow {}...{} \rightarrow \Omega^{p-1} \oplus \overline{\Omega^{p-1}} \rightarrow \overline{\Omega^p} \rightarrow ... \rightarrow \overline{\Omega^{q-1}}.$$

(Here it's supposed that $q>p$). In this situation, we may take $A$ to be a subring of $\Bbb C$, not necessarily of $\Bbb R$. Suppose that $A = \Bbb C$. Then we can write a resolution of this complex of sheaves by differential forms, and it's possible to see that $A(p,q)$ is quasiisomorphic to the complex

$$... \stackrel{d}{\rightarrow} \mathcal{A}^{p-2, q-1} \oplus \mathcal{A}^{p-1, q-2} \stackrel{d}{\rightarrow} \mathcal{A}^{p-1, q-1} \stackrel{\partial\bar\partial}{\rightarrow} \mathcal{A}^{p, q} \stackrel{d}{\rightarrow} \mathcal{A}^{p, q-1} \oplus \mathcal{A}^{p-1, q} \stackrel{d}{\rightarrow}{} ...$$

where the term $\mathcal{A}^{p-1, q-1}$ has the grading $p+q-2$. (The $\partial\bar\partial$ operator arises because of local $\partial\bar\partial$-lemma. You can find the proof that these complexes are quasiisomorphic in M. Schweitzer's text https://arxiv.org/abs/0709.3528).

In particular, $\Bbb H^{p+q-2}(X,{~}A(p,q))$ is known as Aeppli cohomology $H_{Ae}^{p-1,q-1}$, and $\Bbb H^{p+q-1}(X,{~}A(p,q))$ is Bott-Chern cohomology $H_{BC}^{p,q}$. If your $X$ satisfies global $\partial\bar\partial$-lemma (for example, if it is compact Kaehler), then these are just isomorphic to the corresponding Dolbeaut cohomology; on a general complex manifold, they are different.

The Deligne complex with real coefficients $\Bbb R(p)$ is isomorphic to the fixed points of complex conjugation on $\Bbb R(p,p)$, so your can take real forms in the Aeppli-Bott-Chern resolution and obtain the resolution for the real Deligne cohomology.

Now, from the looks of the resolution and a bit of harmonic theory one can conclude that there's a non-degenerate pairing between $\Bbb H^{k}(X,{~}\Bbb C(p,q))$ and $\Bbb H^{2n+1-k}(X,{~}\Bbb C(n+1-p,n+1-q))$ (at least when $X$ is compact, otherwise you need to take cohomology with compact support on one side). This is complex version of the duality on Deligne cohomology, and when $k=p+q$ it specializes to the well-known duality between Bott-Chern and Aeppli cohomology (and to Poincare duality when $X$ satisfies $\partial\bar\partial$-lemma).

I don't know what happens if you work with Deligne-Beilinson complex instead of Deligne complex (with logarithmic forms instead of regular), but I'd suppose that everything works the same and gives you the duality between Bott-Chern cohomology and Aeppli cohomology with compact support on an open part of your variety as a particular case.

I don't know whether the duality you mention for Deligne cohomology gives back Poincaré duality; in my mind it is more another instance of the same principle, which is Poincaré duality for motives. The punchline is that the dual of the motive (say, in Voevodsky's category of motives) M(X) of a smooth proper variety of dimension d is again M(X), but Tate-twisted by -d and shifted by -2d. This result leads to, in a unified way, duality results for various cohomologies. See the paper of Cisinski and Déglise "Mixed Weil cohomologies", §2.6, for an explanation of this for mixed Weil cohomologies. Deligne cohomology isn't a mixed Weil cohomology because of the presence of the twist in the definition of Deligne cohomology, but the ideas in the above-mentioned paper work for Deligne cohomology as well, since Deligne cohomology is also representable by a motivic ring spectrum. This is discussed in a paper by Holmstrom and Scholbach "Arakelov motivic cohomology I".