Suppose $K$ is an imaginary quadratic field (with class number 1, for simplicity), $p \ne 2$ a prime split in $K$, and $K_\infty$ the $\mathbb{Z}_p^2$-extension of $K$.

If $E / \mathbb{Q}$ is an elliptic curve with CM by $K$, then there is a construction (due to Katz) for a "two-variable $p$-adic $L$-function" attached to $E$, which is a $p$-adic measure on the Galois group $K_\infty / K$, interpolating $L$-values of the twists of the Groessencharacter of $K$ attached to $E$ by finite-order characters of p-power conductor. See e.g. de Shalit's book "Iwasawa theory of elliptic curves with complex multiplication" (Academic Press, 1987)

If $E / \mathbb{Q}$ is any elliptic curve with good ordinary reduction at $p$ (or more generally any ordinary modular form of weight $\ge 2$), but not necessarily with CM by $K$, there is also a construction of a two-variable $L$-function attached to $E$, written down by Perrin-Riou (J. London Math. Soc 38 (1988), 1-32) based on earlier work by Hida and others. This interpolates $L$-values of the twists of $E$ by certain 2-dimensional Artin representations of $\mathbb{Q}$, obtained by inducing up finite-order characters of $\operatorname{Gal}(K_\infty / K)$.

My question is this: if we apply Perrin-Riou's method to an $E$ which happens to have CM by $K$, then what is the relation between the $L$-functions coming from the two constructions?

(My impression is that Perrin-Riou's construction should give the product of Katz's $L$-function with its conjugate, corresponding to the decomposition of the Tate module of $E$ as a $\operatorname{Gal}(\overline{K} / K)$-representation into the direct sum of two conjugate characters. But I'm puzzled by the discrepancy of coefficient fields: Perrin-Riou's measure takes values in some finite extension of $\mathbb{Q}_p$, while Katz's lives in the completion of the unramified $\mathbb{Z}_p$-extension of $\mathbb{Q}_p$, which is far larger.)

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    $\begingroup$ 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. $\endgroup$
    – Olivier
    Jun 14, 2011 at 17:30
  • $\begingroup$ +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. $\endgroup$ Jun 14, 2011 at 17:34
  • $\begingroup$ @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 $\mathbb{Q}_p$.) $\endgroup$ Jun 14, 2011 at 17:34
  • $\begingroup$ @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. $\endgroup$ Jun 14, 2011 at 17:44
  • $\begingroup$ 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 $\Omega_{p}$. $\endgroup$
    – Olivier
    Jun 14, 2011 at 17:58

1 Answer 1


The conditions on K in Perrin - Riou's paper (Heegner hypothesis) probably exclude the case that we can take K to be the field of complex multiplication.

  • $\begingroup$ The Heegner hypothesis, as I understand it, is that all primes $\ell \mid N$ are split in $K$. P-R doesn't assume this, but she does assume that all primes dividing $N$ are unramified in K. It is true that this does actually exclude the case I'm looking at here, but I'm pretty sure that this is more for simplification than because it's essential to the method. $\endgroup$ Jul 23, 2011 at 7:14

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