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

Question

[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?

Thank you for any comments!

[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)

Answer 1

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 $H_{\mathbb{Z}} \backslash H_{\mathbb{C}}$.

Answer 2

Not a complete answer, but addressing the second point.

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 $(\mathbb{C}^\times)^2 \to (S^1)^2$.