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Jul 23, 2011 at 4:37 answer added jkl timeline score: 1
Jul 18, 2011 at 11:51 comment added David Loeffler @Jupiter Jones: let $F$ be such a function. Then $F$ has to satisfy $F(x + y + xy) = F(x)F(y)$ (because this is true when $x$ and $y$ are powers of $1 + p$ and these are Zariski-dense). Hence its restriction to the set of points $S = \{\zeta - 1 : \zeta \in \mu_{p^\infty}\}$ would have to be a homomorphism $\mu_{p^\infty} \to \mathbb{C}_p^\times$, all of which are given by elements of $\mathbb{Z}_p$.
Jul 17, 2011 at 13:55 comment added Jupiter Jones @David: why is there no Iwasawa function interpolating powers of $\Omega_p$?
Jun 15, 2011 at 7:49 comment added jvo This is just a comment, because I have not read the through Katz's construction in detail. My impression is that the problem could be a very hard one, because there would be an implicit need to compare the p-adic period(s) in Katz with the period used in Hida/Perrin-Riou (the Petersson inner product of the underlying eigenform). Also, note that the two-variable measure of Hida/Perrin-Riou can be seen to take integral values in the field of definition, as a consequence of the fact that Hida's bounded linear form sends integral-valued forms to integral-valued forms (which is easy to check).
Jun 14, 2011 at 20:45 comment added David Loeffler @Francois, Olivier: there is a character $\chi$ (the Galois character corresponding to the Groessencharacter of E) such that the value of Katz's L-function at $\chi^k$ is $\Omega_p^k$ times an element of $\overline{Q}$, so just "dividing out by $\Omega_p$" doesn't work -- moreover, there is no element of the Iwasawa algebra taking the value $\Omega_p^k$ at $\chi^k$ for all $k$. @Jeremy: Thanks for the suggestion, I will look at Tsuji's paper.
Jun 14, 2011 at 19:49 comment added Jeremy Teitelbaum Work of Tsuji on Explicit Reciprocity Laws and Bloch-Kato is relevant: Tsuji, Takeshi(J-TOKYOGM) Explicit reciprocity law and formal moduli for Lubin-Tate formal groups. (English summary) J. Reine Angew. Math. 569 (2004), 103–173.
Jun 14, 2011 at 17:58 comment added Olivier Again, I really don't know but, generally speaking, p-adic periods arise as determinants of comparison isomorphisms. If your motive is ordinary to begin with, these comparison isomorphisms and the determinants in question live in your original ring of coefficients. At any rate, this is certainly what happens on the algebraic side, which is the only one I really know anything about. But perhaps I am talking non-sense here, and I really should strop writing before I know more about this $\Omega_{p}$.
Jun 14, 2011 at 17:44 comment added François Brunault @David: maybe one just needs to divide the Katz L-function by $\Omega_p$ ? Indeed there is no "p-adic period" in the definition of the classical p-adic L-function.
Jun 14, 2011 at 17:34 comment added David Loeffler @Olivier: I'm sorry, that's not true. (The values of Katz's L-function at algebraic characters involve a period $\Omega_p$ which is transcendental over $\mathbb{Q}_p$.)
Jun 14, 2011 at 17:34 comment added François Brunault +1. I would be also interested to know the answer! I remember being told that the Katz two-variable p-adic L-function specializes to the classical (one-variable) p-adic L-function of E, but I don't know about your more general question.
Jun 14, 2011 at 17:30 comment added Olivier My understanding is that if your E is CM and ordinary, then Katz's measure can be shown to live in in some finite extension of Qp as well, but this is based more on general expectations that on any direct knowledge I have of the construction of Katz.
Jun 14, 2011 at 16:54 history edited David Loeffler CC BY-SA 3.0
added a class number condition
Jun 14, 2011 at 15:58 history asked David Loeffler CC BY-SA 3.0