$\sigma$-conjugate iff $p$-adically close

2012-12-05T14:08:37Z

First some notations. Let $p$ be a prime, $k$ a perfect field of characteristic $p$, $W=W(k)$ the ring of Witt vectors over $k$, $\sigma : W \rightarrow W$ the Frobenius, $R$ a commutative $\mathbb{Z}_p$-algebra, $R_W = R \otimes_Z W $, $\phi$ the $R$-linear extension of $\sigma$ to $R_W$. Also denote by $\phi$ the endomorphism of $R_W((u))$ given by $\phi(\sum_i a_i u^i)=\sum_i \phi(a_i) u^{pi}$.

$LG(R) = GL_d(R_W((u)))$
$LG^+(R) = GL_d(R_W[[u]])$
$LG^{\leq m}(R) = { A \in GL_d(R_W((u))) \mid A, A^{-1} \in u^{-m}M_d(R_W[[u]]) }$

Proposition 2.2 in Pappas, Rapoport - $\phi$-modules and coefficient spaces, available here states the following

Suppose $n > \frac{2m}{p-1}$

For each $g\in U_n(R)$, $A\in LG^{\leq m}(R)$ there is a unique $H\in U_n(R)$ such that $g^{-1} A \phi(g) = H^{-1}A$.
Conversely for each $A\in LG^{\leq m}(R)$, $H\in U_n(R)$ there is a unique $g\in U_n(R)$ such that $g^{-1} A \phi(g) = H^{-1}A$.

In the footnote the authors state that the analogous fact in classical Dieudonné theory is also true. However it seems that the proof given there does not work in that case, the problem being that the Frobenius on the Witt vectors leaves the uniformizer $p$ fixed whereas in the situation of Proposition 2.2 above the uniformizer $u$ is raised to the $p$-th power by $\phi$. In the first statement this forces me to assume $m$ to be $0$, but the comparison of the coefficients in uniqueness part of 2. doesn't seem to be working at all.

How can one prove this result in the classical case?

EDIT: Scholze uses a similar thing in 'The Langlands-Kottwitz Method and Deformation Spaces of p-Divisible Groups'. This is Lemma 4.4 there. But he does not get explicit bounds.

$\sigma$-conjugate iff $p$-adically close - MathOverflow

$\sigma$-conjugate iff $p$-adically close