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:


1. 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:


1. Is anything known about the two K-groups here? Finite generation? Rank? Can you write down a nonzero element?
2. 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 $(0, \partial u + \bar{\partial} v)$?



A question on K_1 of an elliptic curve
Let me explain why Beilinson's conjecture implies that $\iota$ is the zero map (thus your first question has conditionally a negative answer).


Let $\mathcal{E}$ be a proper regular model of $E$ over $\mathbf{Z}$. The morphism $E \to \mathcal{E}$ induces a $\mathbf{Q}$-linear map $\iota : K\_1(\mathcal{E})^{(2)} \to K\_1(E)^{(2)}$. The image of $\iota$ is the *integral subspace* $K\_1(E)^{(2)}\_{\mathbf{Z}}$, which is also written $H^3\_{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2))$ in cohomological notations.


Now, what does Beilinson's conjecture predict for this group? We are concerned here with the motive $h^2(E)$, whose $L$-function is $L(H^2(E),s)=\zeta(s-1)$, and we are looking at the point $s=2$. Thus we are in the case of the "near central point" (see for example 
Schneider, *Introduction to the Beilinson conjectures*, Section 5, Conjecture 
II, or the articles by Beilinson and Nekovar mentioned in my comment).


Since the $L$-function has a pole, we have to introduce the group $N^1(E)=(\operatorname{Pic}(E)/\operatorname{Pic}^0(E)) \otimes 
\mathbf{Q}$ which is isomorphic to $\mathbf{Q}$ (more generally, the dimension should be equal to the order of the pole). There is a natural injective map $\psi : N^1(E) \to H^3\_{\mathcal{D}}(E\_{\mathbf{R}},\mathbf{R}(2))$. Then Beilinson's conjecture asserts that $r \oplus \psi$ induces an isomorphism


$$(H^3\_{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2)) \otimes\_{\mathbf{Q}} \mathbf{R}) \oplus (N^1(E) \otimes\_{\mathbf{Q}} \mathbf{R}) \xrightarrow{\cong} H^3\_{\mathcal{D}}(E\_{\mathbf{R}},\mathbf{R}(2)).$$


Since the target space is $1$-dimensional and $N^1(E) \cong \mathbf{Q}$, this predicts in particular that $H^3\_{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2))=0$.


Moreover, it can be shown that the map $r$ is nonzero. As pointed out by profilesdroxford, the group $H^3\_{\mathcal{M}}(E,\mathbf{Q}(2))$ is generated by symbols of the form $(P,\lambda)$ where $P$ is a closed point of $E$ and $\lambda \in \mathbf{Q}(P)^\*$. I found a reference for this in Beilinson, *Notes on absolute Hodge cohomology* (Beilinson attributes this construction to Bloch and Quillen). Furthermore, in the same article the regulator of such elements is computed (in a more general setting). After some computations it turns out that $r([P,\lambda])$ is proportional to $\log | \operatorname{Nm}\_{\mathbf{Q}(P)/\mathbf{Q}}(\lambda) |$. Thus $r$ is nonzero. Another useful reference is Dinakar Ramakrishnan's article on regulators.


It would be also interesting to compare the above construction with the construction proposed by profilesdroxford in his answer.


