A question on K_1 of an elliptic curve - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T03:29:13Z http://mathoverflow.net/feeds/question/61852 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61852/a-question-on-k-1-of-an-elliptic-curve A question on K_1 of an elliptic curve Andreas Holmstrom 2011-04-15T18:32:21Z 2011-04-26T15:26:02Z <p>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:</p> <ol> <li>After tensoring everything with $\mathbb{R}$, is the the map $r \circ \iota$ zero or surjective??? </li> </ol> <p>I would also be interested in the following questions:</p> <ol> <li><p>Is anything known about the two K-groups here? Finite generation? Rank? Can you write down a nonzero element?</p></li> <li><p>Is the map $\iota$ injective? (This could be asked in much more generality for K-groups of regular models.)</p></li> </ol> <p>I'd be grateful for any hints, even those based on unproven conjectures.</p> <p>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 $(0, \partial u + \bar{\partial} v)$?</p> http://mathoverflow.net/questions/61852/a-question-on-k-1-of-an-elliptic-curve/61871#61871 Answer by SGP for A question on K_1 of an elliptic curve SGP 2011-04-15T22:23:03Z 2011-04-16T00:31:59Z <p>not an answer, but</p> <ul> <li><p>in general, the image of K-theory depends on the choice of the regular model, see the work of Rob de Jeu <a href="http://www.few.vu.nl/~jeu/" rel="nofollow">(on his webpage)</a> on further counterexamples to Beilinson's conjecture (this is incorrect: see Cisinski's comment below)</p></li> <li><p>there is some work on the function field analogue, <a href="http://www.kurims.kyoto-u.ac.jp/preprint/file/RIMS1564.pdf" rel="nofollow">by Kondo and Yasuda, available here</a> </p></li> </ul> http://mathoverflow.net/questions/61852/a-question-on-k-1-of-an-elliptic-curve/62120#62120 Answer by profilesdroxford54 for A question on K_1 of an elliptic curve profilesdroxford54 2011-04-18T13:19:35Z 2011-04-18T20:11:36Z <p>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.</p> <p>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.</p> <p>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. </p> <p>Additional Comments. </p> <p>(1) When you write "surjective after tensoring with $R$" I guess you really mean does the image generate the real vector space $H_D^3(E_{/ \mathbb{R}} , \mathbb{R}(2) )$? </p> <p>(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 $\alpha$. </p> http://mathoverflow.net/questions/61852/a-question-on-k-1-of-an-elliptic-curve/62712#62712 Answer by François Brunault for A question on K_1 of an elliptic curve François Brunault 2011-04-23T06:55:35Z 2011-04-26T15:26:02Z <p>Let me explain why Beilinson's conjecture implies that $\iota$ is the zero map (thus your first question has conditionally a negative answer).</p> <p>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 <em>integral subspace</em> <code>$K_1(E)^{(2)}_{\mathbf{Z}}$</code>, which is also written <code>$H^3_{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2))$</code> in cohomological notations.</p> <p>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, <em>Introduction to the Beilinson conjectures</em>, Section 5, Conjecture II, or the articles by Beilinson and Nekovar mentioned in my comment).</p> <p>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</p> <p>$$(H^3_{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2)) \otimes_{\mathbf{Q}} \mathbf{R}) \oplus (N^1(E) \otimes_{\mathbf{Q}} \mathbf{R}) \xrightarrow{\cong} H^3_{\mathcal{D}}(E_{\mathbf{R}},\mathbf{R}(2)).$$</p> <p>Since the target space is $1$-dimensional and $N^1(E) \cong \mathbf{Q}$, this predicts in particular that $H^3_{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2))=0$.</p> <p>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, <em>Notes on absolute Hodge cohomology</em> (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.</p> <p>It would be also interesting to compare the above construction with the construction proposed by profilesdroxford in his answer.</p>