Consider an elliptic curve $E/ \mathbb{Q}$, with a regular model $\mathcal{E} / \mathbb{Z}$. We have (Beilinson regulator) maps $$ K_1(\mathcal{E})^{(2)} \to K_1(E)^{(2)} \to H_D^3(E_{/ \mathbb{R}} , \mathbb{R}(2) )$$ from (an Adams eigenspace of) K-theory (with rational coefficients) to Deligne cohomology of $E$. Call the first map $\iota$ and the second map $r$. Note that this map does NOT lie in the index range where the Beilinson conjectures predicts that $r$ is an isomorphism on the image of $\iota$ after tensoring with $\mathbb{R}$. Now, is anything known at all about $r$ or $r \circ \iota$, for elliptic curves in general or for some specific curve/class of curves? Unless I am mistaken, the Deligne cohomology group in question is always a one-dimensional real vector space. My main question is the following:

- After tensoring everything with $\mathbb{R}$, is the the map $r \circ \iota$ zero or surjective???

I would also be interested in the following questions:

Is anything known about the two K-groups here? Finite generation? Rank? Can you write down a nonzero element?

Is the map $\iota$ injective? (This could be asked in much more generality for K-groups of regular models.)

I'd be grateful for any hints, even those based on unproven conjectures.

EDIT: Maybe one can approach this question from another point of view. I am quite sure that the following is true (have to check though). The cokernel of $r \circ \iota$ can be identified with the Gillet-Soulé arithmetic Chow group $\widehat{CH}^2(\mathcal{E}) \otimes \mathbb{R}$. Furthermore, this group is generated by arithmetic cycles of the form $(Z,g) = (0,\alpha)$, where $\alpha$ is a real harmonic $(1,1)$-form on the complex torus $E(\mathbb{C})$. So the question becomes: Do all arithmetic cycles of this form lie in the group generated by arithmetic cycles of the forms $(div(f), - \log \| f \|^2)$ and $(0, \partial u + \bar{\partial} v)$?

Zagier's conjecture on $L(E,2)$which has results on $K_1$ of elliptic curves over algebraically closed fields. Maybe it adapts to the case where the base field is $\mathbf{Q}$ but I haven't thought about it. $\endgroup$ – François Brunault Apr 20 '11 at 15:31