Let $u$ be a $d \times d$ matrix with coefficients in $\mathbb{Q}_\ell$, the $\ell$-adic numbers. Let $n \in \widehat{\mathbb{Z}}$. Something I'm reading claims (and says it's easy and leaves the proof to the reader) that the limit in $M_{d}(\mathbb{Q}_\ell)$ (topological space of $d\times d$ matrices with product topology of the topology on $\mathbb{Q}_\ell$) of
$$\lim_{m \mapsto n, \ m \in \mathbb{Z}} u^m$$
exists (for all $n \in \widehat{\mathbb{Z}}$) if and only if the eigenvalues of $u$ in an algebraic closure of $\mathbb{Q}_\ell$ have norm $1$. Why?
I can see what this is a necessary condition when $u$ is diagonalizable, as then $u^m$ is diagonal with eigevalues the $m$th powers of the eigenvalues of $u$, and if these eigenvalues don't have norm 1, then by choosing $m$ to go to $\infty$ or $-\infty$ (while also converging to $n \in \widehat{\mathbb{Z}}$) we can make the norm go to $\infty$.
Here is the excerpt from Fontaine-Ouyang p.51, my question is basically prove prop. 1.10 (the equivalence to the characteristic polynomial is clear)
Best Answer
$\newcommand{\Z}{{\mathbb Z}}\newcommand{\Q}{{\mathbb Q}} \newcommand{\N}{{\mathbb N}}$ Since $\hat\Z$ is metrizable, it is sufficient to discuss the existence of the limit $\lim_{k\to\infty}u^{m_k}$ for a given invertible matrix $u\in M_d(\Q_\ell)$ and any sequence $m_k$, $k\in\N$ of integers converging to some element $\hat n$ in $\hat\Z\setminus \Z$.
Consider a finite extension $L$ of $\Q_\ell$ containing all eigenvalues of $u$. It is well known that there exists a unique extension of the $\ell$-adic valuation to $L$ which we call $v$. We will also use the integral domain $A\subset L$ of all elements of non-negative valuation, its maximal ideal $M$ of all elements of positive valuation and the residue class field $F=A/M$. $F$ is a finite extension of the residue class field $\Z/\ell\Z$ of $\Z_\ell$. Hence $F$ is a finite field of $\ell^f$ elements with some $f\in\N$. Therefore $x^{\ell^f-1}=1$ for all $x\in F^*$. Hence $\mu^{\ell^f-1}\equiv 1\mod M$ for all $\mu\in A$ of valuation 0.
There is an invertible matrix $T\in M_d(L)$ such that $J=T^{-1}uT$ is in block diagonal form $J=\mbox{bl.diag.}(J_1,\dots,J_r)$, where $J_i=\mu_i(I_{d_i}+Z_i)$ and $\mu_i$ is an eigenvalue of $u$ (hence non-zero), $I_{d_i}$ is the identity matrix of size $d_i$ and $Z_i$ is a nilpotent matrix, $Z_i^{d_i}=0$. It is sufficient to study the convergence of $J_i^{m_k}$, $k\to\infty$, $i=1,\dots,r$ in $M_{d_i}(L)$.
We write $\hat n=(n_i+i\Z)_{i\in\N}$ where $n_i\equiv n_j\mod j$ if $j$ divides $i$. $m_k\to\hat n$ means that for any $i\in\N$ there is a $K\in\N$ such that $m_k\equiv n_i\mod i$ for all $k\geq K$. In particular, $m_k$ converges in $\Z_l$ to $\check n$ determined by $\check n\equiv n_{\ell^r}\mod \ell^r$ for all $r$. As a consequence, all binomial coefficients $\binom{m_k}{s}$ converge in $L$ to $\binom{\check n}s$ as $k\to\infty$.
A first consequence of this for nilpotent $Z$ with $Z^d=1$ is that $(I+Z)^{m_k}=I+\sum_{s=1}^{d-1}\binom{m_k}sZ^s$ converges to $I+\sum_{s=1}^{d-1}\binom{\check n}sZ^s=:(I+Z)^{\check n}$.
It remains to study the convergence of $\mu^{m_k}$ for $\mu\in L$.
Case 1: $v(\mu)\neq0$: Then clearly $\mu^{m_k}$ cannot converge as $m_k$ and hence $v(\mu^{m_k})$ does not remain bounded.
Case 2: $v(\mu)=0$: Since $\mu^Q\equiv 1\mod M$ for $Q=\ell^f-1$, it is convient to write $m_k=n_Q+Q\tilde m_k$ for all sufficiently large $k$. This is possible because $m_k\equiv n_Q\mod Q$ for large $k$. Clearly, $\tilde m_k$ converges to some limit $\hat{\tilde n}$ in $\hat Z$ and hence also to $\check{\tilde n}$, say, in $Z_\ell$. We have $\mu^{m_k}=\mu^{n_Q}(1+d)^{\tilde m_k}$ where $d=\mu^{Q}-1$ has positive valuation. The convergence of $(1+d)^{\tilde m_k}$ follows from the expansions $$(1+d)^{\tilde m_k}=1+\sum_{s=1}^\infty \binom{\tilde m_k}s\,d^s\mbox{ and } \binom{\tilde m_k}s\to\binom{{\check{\tilde n}}}s\mbox{ for all }s.$$
Combining the above statements shows that $u^{m_k}$ converges in $M_d(L)$ and hence in $M_d(\Q_\ell)$ as $k\to\infty$ if and only if all eigenvalues of $u$ have valuation 0.