show/hide this revision's text 2 Proposed candidate for class with non-trivial regulator

Additional Comments.

(1) When you write "surjective after tensoring with $R$" I guess you really mean does the image generate the real vector space $H_D^3(E_{/ \mathbb{R}} , \mathbb{R}(2) )$?

(2) I think one can construct elements of $K_1^{(2)}$ with non trivial regulator as follows. Take a curve $E$ with rank at least one, so that there is a rational point $P$ which is not torsion. Take a conjugate pair of points $Q_1$, $Q_2$ in a real quadratic extension $F$ of $\mathbb Q$ such that $Q_1+Q_2+P=0$ in the elliptic curve (such points exist by taking a line with rational slope through $P$, when we embed $E$ in the projective plane). Now take a non trivial unit $\alpha$ in the ring of integers $\mathcal{O}_F$ of $F$. The pair $(Q_1,Q_2)$ determines a point $q:Spec(\mathcal{O}_F)\to \mathcal{E}(\mathcal{O}_F)$. Push $\alpha$ forward by $q$. Then I think (but have not double checked) that the regulator of this class will be non-zero, and essentially equal to the regulator of $\alpha$.

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This is not a full answer, more a lenghty comment, since I think the key part of your question is whether there are elements in $K_1(\mathcal{E})^{(2)}$ with non-trivial regulator.

Conjecturally $K_1(\mathcal{E})^{(2)}$ is finitely generated -- this a particular case of what is known as Bass's conjecture which is that the K-theory of a regular f.g. $\mathbb Z$-algebra is finitely generated in each degree. For rings of integers in global fields this is a theorem of Quillen, but as I wrote in my comment above, I don't believe that this is known for any non-rational arithmetic surfaces. The map $K_1(\mathcal{E})^{(2)} \to K_1(E)^{(2)}$ is far from being bijective, since by the localization sequence its cokernel is the kernel of the map $$\bigoplus_p K'_0(\mathcal{E}_p)^{(1)}\to K_0(\mathcal{E})^{(2)}$$ where $\mathcal{E}_p$ is the curve mod $p$. By Bloch-Kato-Saito, the target of this map is finite, while the source is an infinite sum of non-trivial groups. This is analogous to the situation for number fields.

There is a completely trivial reason that $r\circ \iota$ is not onto: the source is countable while the target is not. As I wrote above what is of more interest is to exhibit elements of $K_1(\mathcal{E})^{(2)}$ with non-trivial regulator. My impression is that the standard method of constructing elements that I described in the comment should give such elements, but I do know where this might have been done.