A question on K_1 of an elliptic curve - MathOverflow

A question on K_1 of an elliptic curve

2011-04-15T18:32:21Z

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 $(0, \partial u + \bar{\partial} v)$?

Answer by SGP for A question on K_1 of an elliptic curve

2011-04-15T22:23:03Z

not an answer, but

in general, the image of K-theory depends on the choice of the regular model, see the work of Rob de Jeu (on his webpage) on further counterexamples to Beilinson's conjecture (this is incorrect: see Cisinski's comment below)
there is some work on the function field analogue, by Kondo and Yasuda, available here

Answer by profilesdroxford54 for A question on K_1 of an elliptic curve

2011-04-18T13:19:35Z

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.

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 $\alpha$.

Answer by François Brunault for A question on K_1 of an elliptic curve

2011-04-23T06:55:35Z

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.