# Multiplication in Deligne cohomology: explicit formula for $p=q=1$

[This is a double of my question of math.stackexchange https://math.stackexchange.com/questions/3214962/multiplication-in-deligne-cohomology-explicit-formula-for-p-q-1]

In the very beginning of [1] the geometric meaning of Deligne cohomology $$H^q(X, \mathbb{Z}(p))_D$$ and multiplicative structure on it is being discussed. In particular, it is not hard to see that $$H^q(X, \mathbb{Z}(1))$$ can be canonically identified with $$H^{q-1}(X, \mathcal{O}^{\times}_X)$$.

The group $$H^2(X, \mathbb{Z}(2))_D$$ is identified with the group of holomorphic rank $$1$$ bundles with holomorphic connection (group structure is given by tensor product)

The $$\cup$$-multiplication gives us a map $$H^1(X, \mathbb{Z}(1))_D \otimes H^1(X, \mathbb{Z}(1))_D = H^0(X, \mathcal{O}^{\times}_X) \otimes H^0(X, \mathcal{O}^{\times}_X) \to H^2(X, \mathbb{Z}(2))_D$$ In other words, given two nowhere vanishing holomorphic functions $$f$$ and $$g$$ on $$X$$ we obtain a holomorphic line bundle with holomorphic connection on $$X$$.

Though in [1] the explicit formula for this in terms of Čhech cocycles is given, I am looking for another description of the same operation.

First of all, observe that each pair of functions $$f, g \in H^0(X, \mathcal{O}^{\times}_X)$$ define a holomorphic map $$F_{f,g} \colon X \to (\mathbb{C}^{\times})^2$$. Following Esnault and Viehweg, denote the resulting line bundle with holomorphic connection $$f \cup g$$ by $$r(f, g)$$. Then it seems clear from functoriality reasons that $$r(f, g) = F_{f,g}^*r(z,w),$$ where $$z$$ and $$w$$ are coordinate functions on $$\mathbb{C}^{\times}\times \mathbb{C}^{\times}$$. Thus, I'd be happy to understand, what $$r(z, w)$$ is.

Since $$(\mathbb{C}^{\times})^2$$ is a product of two Stein manifolds, there are no non-trivial holomorphic line bundles. Therefore, the only ''interesting'' part of $$r(z,w)$$ is the holomorphic connection. Any holomorphic connection on trivial bundle is given by $$\nabla = d + \eta$$, where $$\eta$$ is a holomorphic $$1$$-form. So my questions are:

• What is this $$1$$-form $$\eta$$ on $$\mathbb{C}^{\times} \times \mathbb{C}^{\times}$$? It seems to me, that $$\frac{dz}{z} - \frac{dw}{w}$$ would be nice (at least, if this is the case, it satisfies the properties of $$r(f, g)$$ given in [1]), however I'm not able do deduce this explicitly form Esnault-Viehweg formulae.
• From my speculations it follows that the underlying line bundle for any $$r(f,g)$$ is trivial. Is this at least true? If not, than where is my mistake?

[1] -- H. Esnault, E. Viehweg. Deligne-Beilinson cohomology. in: Beilinson's Conjectures on Special Values of L-Functions ( Ed.: Rapoport, Schappacher, Schneider ). Perspectives in Math. 4, Academic Press (1988) 43 - 91 (http://page.mi.fu-berlin.de/esnault/preprints/ec/deligne_beilinson.pdf)

There is indeed a "universal" holomorphic bundle with connection on $$\mathbb{C}^\times \times \mathbb{C}^\times$$ which induces the bundles $$r(f,g)$$ defined in Esnault-Viehweg. This universal bundle has been constructed by D. Ramakrishnan using the Heisenberg group (Bulletin AMS vol. 5 n. 2, 1981, https://doi.org/10.1090/S0273-0979-1981-14942-9 ).

It is nicely explained in R. Hain, Classical polylogarithms. Put $$\begin{equation*} H_{\mathbb{C}} = \begin{pmatrix} 1 & \mathbb{C} & \mathbb{C} \\ 0 & 1 & \mathbb{C} \\ 0 & 0 & 1 \end{pmatrix} \end{equation*}$$ and $$\begin{equation*} H_{\mathbb{Z}} = \begin{pmatrix} 1 & \mathbb{Z}(1) & \mathbb{Z}(2) \\ 0 & 1 & \mathbb{Z}(1) \\ 0 & 0 & 1 \end{pmatrix}. \end{equation*}$$ The exponential map gives a canonical projection $$H_{\mathbb{Z}}\backslash H_{\mathbb{C}} \to \mathbb{C}^\times \times \mathbb{C}^\times$$ with fiber $$\mathbb{C}/\mathbb{Z}(2) \cong \mathbb{C}^\times$$, which is the bundle you want. This bundle is not trivial but becomes trivial after pulling-back to $$\mathbb{C} \times \mathbb{C}$$ using this exponential map.

Denoting by $$u,v$$ the coordinates on $$\mathbb{C} \times \mathbb{C}$$ (they are just the matrix coefficients using the Heisenberg description), the pull-back of the connection is given by $$\begin{equation*} \nabla s = ds - s \cdot u dv/2\pi i \end{equation*}$$ This defines a connection on $$\mathbb{Z}(2)\backslash H_{\mathbb{C}}$$ which descends to $$H_{\mathbb{Z}} \backslash H_{\mathbb{C}}$$.

• Thank you, this is precisely what I was looking for. May 9, 2019 at 16:54

Since $$(\mathbb{C}^\times)^2$$ is Stein it means that topological and holomorphic classification of line bundles coincides, not that there are no nontrivial ones. $$(\mathbb{C}^\times)^2$$ is homotopy equivalent to the real 2d torus $$(S^1)^2$$, which has the volume form $$d\theta_1\wedge d\theta_2$$, which is the curvature (up to proportionality constant) of a connection on the line bundle corresponding to the generator of $$H^2((S^1)^2,\mathbb{Z}) \simeq \mathbb{Z}$$. We can take this connection to be $$\theta_1 \mathrm{d}\theta_2$$.
Since $$H^2((S^1)^2,\mathbb{Z}) \simeq H^2((\mathbb{C}^\times)^2,\mathbb{Z}) \simeq H^1((\mathbb{C}^\times)^2,\mathcal{O}^\times)$$, we get a unique holomorphic line bundle on $$(\mathbb{C}^\times)^2$$ whose underlying topological line bundle is the one pulled back along the retraction $$(\mathbb{C}^\times)^2 \to (S^1)^2$$.
• Note that the connection on the holomorphic line bundle on $(\mathbb{C}^\times)^2$ should locally be $\log(x) \frac{dy}{y}$. May 6, 2019 at 7:29
• Yes, your are absolutely right! However, in my situation is still needs to be trivial, since a line bundle, which admits a global holomorphic connection is flat, isn't it? (there is a natural characteristic class in $H^1(X, \Omega^1)$ which vanishes iff a line bundle admits holomorphic connection, but I'm a bit confused if it coincides with the Chern class in non-proper case) May 9, 2019 at 7:11
• well, in one hand , our line bundle lies in the kernel of homomorphism $Pic(X) \to H^1(X, \Omega^1)$ and it's first Chern class lies in the kernel of the homomorphism $H^2(X, \mathbb{C}) \to H^2(X, \mathcal{O})$. On the other hand, if you look at Hain's paper quoted by Francçois Brunault, you will see, that the curvature form for this connections is $\frac{1}{2\pi i} \frac{dx}{x} \wedge \frac{dy}{y}$. This gives a non-trivial class in cohomology, and what is more interesting, this $2$-form is holomorphic (which shows us that Lefschetz (1,1)-theorem fails in non-compact case). May 9, 2019 at 16:51
• Finally, I would like to notice that if the image of our manifold inside $(\mathbb{C}^{\times})^2$ is a curve (which is often the case), then the pull-back of this ''universal'' bundle is flat, since it's curvature is a holomorphic $2$-form. Interesting! May 9, 2019 at 16:54