Let $E$ be an elliptic curve and $\rho$ an Artin representation of $\operatorname{Gal}(\overline{\mathbb{Q}} / \mathbb{Q})$. Then there is a "twisted L-function" $L(E, \rho, s)$, corresponding to the compatible system of $\ell$-adic representations $H^1_\ell(E) \otimes \rho$.

Let $p$ be a prime where $E$ has good supersingular reduction. Is it known (maybe just in some specific examples) whether there exists a $p$-adic $L$-function interpolating the values $L(E, \rho \otimes \chi, 1)$, where $\rho$ is fixed but $\chi$ varies over all Dirichlet characters of $p$-power conductor? I gather this is known in some cases when $E$ has good ordinary reduction at $p$, but I'm specifically interested in the supersingular case.

EDIT: Since this question seems to have come alive again after over a year's inactivity, I will add a little clarification. What I'm hoping for are distributions of finite order on the cyclotomic Galois group $\Gamma \cong \mathbb{Z}_p^\times$ satisfying some reasonable interpolation formula linking their values at finite-order characters to the $L$-values I mentioned above; and I expect that one will have to make a choice of eigenvalue of crystalline Frobenius of E at p satisfying some small slope condition, as one does in the Amice-Velu-Vishik construction for $\rho = 1$. I fully expect that there will be some story involving decomposing these potentially unbounded distributions in terms of auxiliary bounded distributions as in Pollack's $\pm$-construction, but I'm not asking about that here.

  • $\begingroup$ Very interesting question. My guess is that there are no such examples, yet. $\endgroup$ Jun 8, 2010 at 14:10
  • $\begingroup$ David, what are the results you know in the case of an $E$ which has good ordinary reduction. $\endgroup$
    – Joël
    Oct 2, 2011 at 20:21
  • $\begingroup$ Trying to clarify my mind about this question, I realized I didn't know the answer to a much more basic question: For which Artin representation $\rho$ can we define a $p$-adic $L$-function interpolating the values $L(\rho \otimes \chi,1)$? The case where $\rho(c)$ is a scalar would follow from the Brauer method and Deligne-Ribet, but is the general case known? $\endgroup$
    – Joël
    Oct 2, 2011 at 21:21
  • $\begingroup$ In lots of cases these will all be zero for trivial reasons (e.g. if $\rho$ is the regular representation of $\operatorname{Gal}(K / \mathbb{Q})$ for $K$ a number field that isn't totally real or CM). So one really wants to interpolate (suitably normalized) leading terms at 1, not just values. I asked about this case at: mathoverflow.net/questions/18884/… $\endgroup$ Oct 3, 2011 at 7:21

1 Answer 1


Hi David,

I don't think you can really hoped for such a p-adic L function. Already if \rho=1 "the" p-adic L function does not exist - Robert Pollack has this \pm construction. In terms of Coates-Fukaya-Kato-Sujatha-Venjakob Non-commutative Iwasawa Theory, I would conjecture that supersingular curves don't have S-torsion Selmer groups, preventing the main Conjecture to be stated. I hope this makes some sense.

  • $\begingroup$ The $p$-adic functions exist when $\rho = 1$. This is an old story. Perrin-Riou has some papers where she formulates the main conjecture and the $p$-adic BSD conjecture despite the fact that the Selmer group is not $\Lambda$-torsion. The $\pm$ construction came later and is prettier. So I am certain that there is a statement of the main conjecture in the non-commutative setting, too. $\endgroup$ Oct 2, 2011 at 17:20
  • $\begingroup$ @Filippo: You are selectively (mis)interpreting the question in such a way that it has a trivial answer. $\endgroup$ Oct 2, 2011 at 20:10
  • $\begingroup$ @David: Sorry, I read now your EDIT and see your point better (and agree that my answer was unfair). For Chris, you should be right but for the moment I really don't see how in the localization sequence of Non-Comm. Iwasawa th. one can get around the trouble of Selmer not being torsion. But this should definitely come from too narrow a point of view we (rather say: "I") have on the non-comm. MC. $\endgroup$ Oct 5, 2011 at 9:40

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