Are Kato's zeta elements integral? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T11:07:10Z http://mathoverflow.net/feeds/question/47135 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47135/are-katos-zeta-elements-integral Are Kato's zeta elements integral? David Loeffler 2010-11-23T19:00:06Z 2013-05-07T09:28:38Z <p>Let $E$ be an elliptic curve over $\mathbb{Q}$ and $T$ the $p$-adic Tate module of $E$. Kato's Euler system, constructed in the paper "P-adic Hodge theory and values of zeta functions of modular forms" (Asterisque 295, 2004), gives rise to an element <code>$\mathbf{z}_{\rm Kato}$</code> lying in the Iwasawa cohomology <code>$H^1_{\mathrm{Iw}}(\mathbb{Q}_p, T)[\frac{1}{p}]$</code>. In Theorem 12.5(4) of the paper, Kato shows that if the image of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ in $\mathrm{Aut}(T)$ contains $\mathrm{SL}_2(\mathbb{Z}_p)$, then in fact <code>$\mathbf{z}_{\rm Kato} \in H^1_{\mathrm{Iw}}(\mathbb{Q}_p, T)$</code>.</p> <p>Is it known if there are weaker conditions that are sufficient to ensure that <code>$\mathbf{z}_{\rm Kato}$</code> has this integrality property? Are there examples where it is genuinely non-integral, or is it conjectured that it should always be so?</p> <p>I would be the last person to claim I understand Kato's argument, but it looks to me as if he only actually uses the weaker statement that the mod $p$ Galois representation $T/pT$ is irreducible. I'd be interested to know if this weaker condition is indeed sufficient, and whether anything is known in this direction if the weaker condition doesn't hold (i.e. if $E$ admits a $p$-isogeny). </p> <p>(EDIT: Added more detail and references.)</p> http://mathoverflow.net/questions/47135/are-katos-zeta-elements-integral/47225#47225 Answer by Olivier for Are Kato's zeta elements integral? Olivier 2010-11-24T12:54:14Z 2010-11-24T22:40:45Z <p>First, let us assume that $p$ is odd for safety.</p> <p>Then I think your impression is correct: the first statement of 12.5 (4), i.e the integrality of the module generated by the $\mathbf{z}^{(p)}_{\gamma}$, is true under the hypothesis that $T/pT$ is irreducible. Hypothesis (12.5.2) is used only to prove the second (and much more spectacular) statement, i.e the Iwasawa-theoretic divisibility.</p> <p>When $E$ admits a $p$-isogeny, then C.Wuthrich has a result towards integrality. However, the published version is flawed and I admit I can't quite understand the correction on his webpage. Because C.Wuthrich is a frequent poster here, we can count on his explanation.</p> <p>In general, proving that Euler systems are integral should be quite a difficult task because it is generally intimately linked with the question of whether the $\mu$-invariant vanishes.</p> <p>UPDATE: Let me answer the question of where Kato requires the full force of (12.5.2). First of all, there is something tricky in the sense that the answer is "nowhere directly in the article we are discussing". However, this hypothesis is crucial to the method of Euler system, so it is used it in the reference to his prior article Euler systems, Iwasawa theory and Selmer groups (there, it is hypothesis $ii_{str}$).</p> <p>Finally, though it is true that integrality of Euler systems elements does not in itself imply that $\mu$ vanishes by the usual divisibility, it is nevertheless true that by playing around with reciprocities laws, you can extract the $\mu$-invariant from integrality properties of compatible systems of norms (and in particular Euler systems). This is work of T.Fukaya (unpublished, I think) in the case of $\textrm{GL}_{1}$. Now that the most general form of the reciprocity law is known by the work of Colmez, it might be within reach to link integrality of Kato's Euler system to the $\mu$-invariant of modular forms. Whence my remark to this effect at the end of my original answer. Contrary to the case of the cyclotomic extension, the full-force of this idea can be exploited in the case of the anticyclotomic extension of an imaginary quadratic field (and here again the vanishing of the $\mu$-invariant is almost equivalent to the integrality of the Euler system).</p> http://mathoverflow.net/questions/47135/are-katos-zeta-elements-integral/47243#47243 Answer by Chris Wuthrich for Are Kato's zeta elements integral? Chris Wuthrich 2010-11-24T15:50:44Z 2013-05-07T09:28:38Z <p>There are two issues. Let $H=H^1_{\mathrm{Iw}}(\mathbb{Q},T)$ where $T=V_{\mathbb{Z}_p}(f)(1)$ and $f$ is the modular form associated to the isogeny class of $E$.</p> <p>(1) What is $T$ ?</p> <p>$T$ will correspond to the lattice $\Lambda$ in $\mathbb{C}$ generated by all modular symbols. This is because $V_{2,\mathbb{Z}}(f)$ is the image of $V_{2,\mathbb{Z}} (Y_1(N)) = H_1 \bigl( X_1(N)(\mathbb{C}),{\text{cusps}},\mathbb{Z}\bigr)$ inside $H_1(E(\mathbb{C}),\mathbb{Z})$. The lattice $\Lambda$ will contain the lattice $\Lambda_0$ of the $X_0$-optimal (strong Weil) curve $E_0$, but it may be strictly larger if $E_0$ has rational torsion points, e.g. for 11a. Note also that $\Lambda$ need not be the lattice of an elliptic curve in the isogeny class, but rather $\tfrac{1}{2}\Lambda_E$ for some elliptic curve $E$. in case $E_0[2] \subset E_0(\mathbb{Q})$, e.g. for 17a. But can only happen for $p=2$ and for $p$ of additive reduction. So if $p$ is a prime of odd semi-stable reduction, then $T$ is the Tate-module of an elliptic curve $E_*$ which is an étale quotient of $E_0$. So let $E$ be this curve for the rest of the answer.</p> <p>(2) Are Kato's elements integral in $H$ ?</p> <p>There are two kinds of them. The $z_{\gamma}$ and the ${c,d}$ $z_{m}$. (sorry I don't seem to be able to produce indices before the symbol in MathJax) The latter are in $H$, see 8.1 of Kato, but they depend on the choices of $c$ and $d$. They are useful for bounding the Selmer group as, for a fixed $c$ and $d$ they form an Euler system.</p> <p>The $z_{\gamma}$ instead is linked to the $p$-adic $L$-function and they are independent of the choices. They are obtained by dividing by $\mu(c,d)$, page 229 of Kato. So they need not be integral anymore. The appendix A in Delbourgo's book "Elliptic curves and big Galois representations" discusses this in detail. Kato shows that they are in $H\otimes \mathbb{Q}_p$</p> <p>Kato shows that $z_{\gamma}$ is integral if $H$ is a free $\Lambda$-module of rank 1, e.g. as shown in 12.4.(3) if $T/pT$ is irreducible. In fact it is not hard to show that $H$ is free also if $E(\mathbb{Q})[p]$ is trivial. </p> <p>Now if the curve admits an isogeny of degree $p$, one can show that for all curves $A$ in the isogeny class $H^1_{\text{Iw}}(\mathbb{Q},T_p A)$ is a free $\Lambda$-module of rank $1$, except for at most a single one of them (I mean up to non-$p$-isogenies of course). This exception - if present - will always be the minimal curve $E_{\text{min}}$ in the class. Moreover if $E_{\text{min}}$ is does not have a free $H^1_{\text{Iw}}$, then there is an embedding of it into $\Lambda$ with image equal to the maximal ideal of $\Lambda$. One can now conclude from the interpolation property of the $p$-adic $L$-function that even if $E_*=E_{\text{min}}$ then $z_{\gamma}$ will be in $H$.</p> <p>Using this one can prove that the $p$-adic $L$-function is integral and that $\mu\geq 0$, but it would not say anything about Greenberg's $\mu=0$ conjecture. Furthermore one gets a proof of the divisibility in the main conjecture as in 12.5.(3), if the $T/pT$ is reducible. However this conclusion can not be extended at present to all odd semi-stable primes, because there may be primes for which the Galois representation is not surjective, yet $T/pT$ is irreducible; because the Euler system method requires and element $\binom{1\ 1}{0\ 1}$ in the image of the Galois representation, see Hyp($K_{\infty},T)$ in Rubin's book. So the integrality won't help, yet. </p> <p>In summary, all zeta-elements of Kato are integral with respect to $T$ in the case of an elliptic curve. For Kato's divisibility on the other hand, the surjectivity of the representation to $\mathrm{GL}_(\mathbb{Z}_p)$ or its reducibility is still needed.</p> <p>(edits: quite a few in the whole answer above, now that I konw the full answer to the question)</p>