# Psi operator on Phi-Gamma modules

This is a question about the base-rings appearing in the the theory of $(\varphi, \Gamma)$-modules in $p$-adic Hodge theory.

Let $p$ be prime, $n \ge 1$, and let $$\mathbf{A}_{\mathbf{Q}_p}^{\dagger, n} = \left\{ \sum_{k \in \mathbf{Z}} a_k T^k : a_k \in \mathbf{Z}_p, v_p(a_k) + \frac{k}{(p-1)p^{n-1}} \ge 0\, \forall{k} \text{ and \to \infty as k \to -\infty}\right\}.$$

(This ring also goes by the name of $\mathscr{O}_{\mathscr{E}}^{\dagger, n}$ in some works). There is a Frobenius map $\varphi: \mathbf{A}_{\mathbf{Q}_p}^{\dagger, n} \to \mathbf{A}_{\mathbf{Q}_p}^{\dagger, n+1}$ given by $T \mapsto (1 + T)^p - 1$.

There is also a map $\psi: \mathbf{A}_{\mathbf{Q}_p}^{\dagger, n+1} \to \mathbf{A}_{\mathbf{Q}_p}^{\dagger, n}[1/T]$, satisfying $\psi \circ \varphi = 1$ and $$(\varphi \circ \psi)(f) = \frac{1}{p} \sum_{\zeta: \zeta^p = 1} f( \zeta(1 + T) - 1).$$

Does the map $\psi$ send $\mathbf{A}_{\mathbf{Q}_p}^{\dagger, n+1}$ to $\mathbf{A}_{\mathbf{Q}_p}^{\dagger, n}$?

There is a sentence in the proof of Lemma V.1.4 of the paper "Theorie d'Iwasawa des representations p-adiques d'un corps local" by Cherbonnier and Colmez (JAMS 12(1), 1999) implying that this is the case, but the proof is only very briefly sketched, and I can't see how to make it work. On the other hand, Colmez's massive paper "Représentations de $\operatorname{GL}_2(\mathbf{Q}_p)$ et $(\varphi, \Gamma)$-modules" (in Asterisque 330, 2010) seems to avoid using this fact at points where it would seem (to me!) natural to do so, e.g. Lemma V.1.1, leading me to suspect that maybe it's not actually true. What's the correct statement here?

It seems to me that the statement follows from some formulas appearing in other papers of Colmez. Lemma I.9 (page 224) of "La série principale unitaire de $GL_2(Q_p)$" has some very precise estimates for the coefficients of $\psi(T^k)$ with $k<0$. It says that if $k<0$, then $\psi(T^k)= \sum_{i=k}^{\lfloor k/p \rfloor} b_{k,i} T^i$ with: $$v_p(b_{k,i}) \geq \frac{k-pi}{p-1}-1.$$ Note also that $v_p(b_{k,i}) \geq 0$. If you use this, then it seems to me that you can prove that $a_k T^k \in A^{\dagger,n+1}_{Q_p}$ implies $\psi(a_k T^k) \in A^{\dagger,n}_{Q_p}$. Don't forget to use the fact that all the $v_p(\cdot)$'s are integers: in particular I'm not sure that the claim would be true if you were to look at rings "with coefficients in $E/Q_p$" if $E$ is ramified - which is maybe why Colmez avoids it?
EDIT: even if the above works, I'd personally try a more direct proof along the following line: $$A^{\dagger,n+1}_{Q_p} = Z_p[[T]] \{ \frac{p}{T^{p^n(p-1)}} \} = Z_p[[T]] \{ \frac{p}{\phi(T)^{p^{n-1}(p-1)}} \},$$ and your claim then becomes clear.
• Thanks! The bound Colmez gives deals with all the coefficients other than the boundary case $i = \lfloor k/p \rfloor$, and that case you can bash out by hand. As you suggest, nailing this case does indeed need the fact that $v_p$ takes values in $\mathbf{Z}$ (otherwise $c/T$ is a counterexample for suitably chosen $c$). – David Loeffler Dec 18 '13 at 11:30