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In the paper Iwasawa Theory and F-analytic Lubin-tate $(\phi,\Gamma)$-modules

Prop 3.4.2 says that for any $x\in{S}$, there exists (not uniuqely) $f(T)\in{B_{rig,F}^+}$ such that $f(u_n)=\log_{LT}(x_n)$ and $\psi_{q}(f(T))=\pi^{-1}{f(T)}$. Thm 3.4.5 and Thm 3.5.3 says that for any $n\geq1$, $V=K(\chi_{\pi})$, take $y=f(T)\otimes{t_\pi^{-1}u}$, we have $h_{F_n,V}^1(\partial{f(T)}u)=(q/\pi)^{-n}\delta(x_n)$ and any $j\leq{-1}$, $$\exp^*_{F_n，V^*(1-j)}(h_{F_n,V(\chi_{\pi}^j)}(\partial {f(T)}u\otimes{e_j}))=\frac{1}{(-1-j)!}(\pi/q)^n\partial_{V(\chi_\pi^j)}(\partial^{-j}f(T)\otimes{t_\pi^{-j-1}e_{1+j}}).$$ It seems that the LHS is independent of the choice of $f(T)$ (it is the twist (a la soule) of the system $(q/\pi)^{-n}\delta(x_n)$ by $\chi_{\pi}^j$), while the RHS determines $\varphi^{-q}(f(T))\in{K_n[[t_{\pi}]]}$ up to constants, what's wrong? Thanks. All my bests.

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    $\begingroup$ If you are not willing to explain more what the notation means you have to include at the very least a link to the paper. $\endgroup$
    – Chris Wuthrich
    Commented Apr 8, 2018 at 8:06
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    $\begingroup$ Thm 3.5.3 does not quite say what you wrote ($f$ should be $y$ on the RHS, and there are missing parts). In particular there is a $h^1_{F_n,V(\chi_\pi^j)}$ whose value, on the LHS, won't depend just on the values of $f$ at the $u_n$ but, as the thm says, on the values of its $-j$-th derivative. $\endgroup$
    – Laurent Berger
    Commented Apr 8, 2018 at 12:07
  • $\begingroup$ Thanks for pointing out the typos. In the special case of $V=K(\chi_{\pi})$, as in Thm 3.4.5, $y=f(T)\otimes{t_{\pi}^{-1}u}$ and $\nabla(y)=\partial{f(T)}u$. Doesn't the fact $h_{F_n,V}^1(\partial {f(T)}u)=(q/\pi)^{-n}\delta(x_n)$ and the general construction of ${h_{F_n,-}}$ implies that $h_{F_n, V(\chi_\pi^j)}(\partial{f(T)}u\otimes{e_j})$ is the twist (a la Soule) of $(q/\pi)^{-n}\delta(x_n)$ by $\chi^j_{\pi}$? $\endgroup$
    – GRH
    Commented Apr 8, 2018 at 16:11

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Let $K$ be a finite extension of $Q_p$ and let $(u_n)_n$ be an infinite subset of the $p$-adic open unit disk. A power series with bounded coefficients (in $K$) is determined by its values at the $u_n$ (that is : if $f(T)$ has bounded coeffts and $f(u_n)=0$ for all $n$, then $f(T)=0$).

This explains why this "twist à la Soulé" works in the classical case : if you know only the values $f(u_n)$ then you know $f(T)$ and hence the values of all the derivatives of $f(T)$ at the $u_n$.

However in our paper we look at possibly unbounded power series, so this argument is no longer valid. There is no simple "twist à la Soulé". If you wanted something like that, you'd need to bound the order of growth of $f(T)$, adjust accordingly the interpolation set, etc. You can see some of this appear in Colmez' papers on Perrin-Riou's map and (for the LT case) in Fourquaux' Phd.

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  • $\begingroup$ If we take $z=(z_n) $ with $Tr_{n+1/n}^{LT}(z_{n+1})=z_n$ and $x=(x_n)$ with $x_n=[q/\pi^n](z_n)$, then we have $x\in{S}$. Take $f(T)\in{B_{rig,F}}^+$ such that $f(u_n)=\log_{LT}(x_n)$ and $f(T)\in{B_{rig,F}^{+,\psi_q=1/\pi}}$, $V=F(\chi_{\pi})$, then we have $h_{F_n,V}^1(\partial{f(T)}u)=\delta(z_n)$ and the sequence $\{\delta(z_n)\}$ can be viewed as an element in $H^1_{Iw}(F,V)$. In this situation, do we have that $\{h_{F_n,V(\chi_{\pi}^j)}^1(\partial{f(T)}u\otimes e_{j})\}_{n\geq1}$ is the twist a la soule of $\{\delta(z_n)\}_{n\geq1}$ by $\chi_{\pi}^j$? $\endgroup$
    – GRH
    Commented Apr 9, 2018 at 10:46
  • $\begingroup$ There is no nontrivial sequence $\{z_n\}$ satisfying your condition if $F$ is not $Q_p$ (there are no "universal norms"). $\endgroup$
    – Laurent Berger
    Commented Apr 9, 2018 at 13:17

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