Galois representations with semisimple residue representation

Question

$\DeclareMathOperator\GL{GL}$Let $\mathbb{Z}_p$ be the ring of integers of $p$-adic numbers $\mathbb{Q}_p$, $G$ a profinite group (e.g. Galois group of local field or global field) and $\rho:G\to \GL_n(\mathbb{Z}_p)$ a continuous homomorphism. If $\rho$ is semisimple as a representation of $G$, then it's not necessarily true that the reduction $\overline{\rho}:G\to \GL_n(\mathbb{F}_p)$ of $\rho$ is also semisimple. It leads to the following question:

Let $\rho:G\to \GL_n(\mathbb{Z}_p)$ be a continuous homomorphism, $\overline{\rho}$ its reduction and $\smash{\overline{\rho}}^\text{ss}$ the semisimplification of $\overline{\rho}$. Is there a continuous homomorphism $\rho':G\to \GL_n(\mathbb{Z}_p)$ such that it has the same semisimplification as $\rho$ and its reduction is $\smash{\overline{\rho}}^\text{ss}$?

---

As pointed out in the comments, the answer to the above question is no. However, we may ask the following modified question:

Let $\rho:G\to \GL_n(\mathbb{Z}_p)$ be a continuous homomorphism, $m$ a positive integer and $\overline{\rho}_m:G\to \GL_n(\mathbb{Z}_p)\to \GL_n(\mathbb{Z}_p/p^m \mathbb{Z}_p)$ its mod $p^m$ reduction, i.e., the composite of $\rho$ and the natural surjective morphism $\GL_n(\mathbb{Z}_p)\to \GL_n(\mathbb{Z}_p/p^m \mathbb{Z}_p)$. Is there a continuous homomorphism $\rho':G\to \GL_n(\mathcal{O}_L)$ where $\mathcal{O}_L$ is the ring of integers for some finite extension $L/\mathbb{Q}_p$ such that -1 it has the same semisimplification as $\rho$; -2 its mod $\pi_L^m$ reduction is semisimple (as a $(\mathcal{O}_L/\pi_{L}^{m}\mathcal{O}_L)[G]$-module $M$ where $(\mathcal{O}_L/\pi_{L}^{m}\mathcal{O}_L)[G]$ is the group ring of $ G $ over $ \mathcal{O}_L/\pi_{L}^{m}\mathcal{O}_L $ and $M$ corresponds to $\bar{\rho'}_m$) where $\pi_L$ is a uniformizer of $\mathcal{O}_L$ -3 the ramification of $L/\mathbb{Q}_p$ is less than $\phi(n)$ where $\phi(n)$ is some function of $n$?

Answer 1

As pointed out in the comments, the answer to the modified question for the case of $m=1$ is Yes, and I try to give my own proof as follows. But I do not know if there is also an affirmative answer for $m>1$.

We may assume that $ \bar{\rho}_{1} $ is non-semisimple and the matrix representation of $ \bar{\rho}_{1} $ has the form $$ \begin{pmatrix} A_1 & \ast & \cdots & \ast \\ 0 & A_2 & \cdots & \ast \\ \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & A_r \end{pmatrix} $$ with only zeros below the boxed diagonal where each $ A_{i} $ is a block matrices corresponding to the irreducible components of $ \bar{\rho}_{1} $ for some integer $ r\geq 2 $. Suppose that $ A_{i} $ is a $ n_{i}\times n_{i} $ matrix. Then the semisimplification of $ \bar{\rho}_{1} $ turns this matrix into $$ \begin{pmatrix} A_1 & 0& \cdots & 0 \\ 0 & A_2 & \cdots & 0 \\ \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & \cdots & A_r \end{pmatrix}. $$ Moreover, the matrix representation of $ \rho $ must have the form $$ \widetilde{A}=\begin{pmatrix} \widetilde{A_{1}} & \ast & \cdots & \ast \\ pB_{21} & \widetilde{A_{2}} & \cdots & \ast \\ \cdots & \cdots & \cdots & \cdots \\ pB_{r1} & pB_{r2} & \cdots & \widetilde{A_{r}} \end{pmatrix} $$ where each $ \widetilde{A_{i}} $ is a block matrix such that its mod $ p $ reduction is $ A_{i} $ for all $ i $ and $ B_{21},\cdots $ are some block matrices. Let $ L:=\mathbb{Q}_p(\sqrt[r]{p}) $ which is a totally ramified extension of $ \mathbb{Q}_p $ with degree $ r $. Then $ \pi_{L}=\sqrt[r]{p} $. We put $$ Q:=\begin{pmatrix} (\sqrt[r]{p})^{r}I_{n_{1}} & 0 & \cdots & 0 \\ 0 & (\sqrt[r]{p})^{r-1}I_{n_{2}} & \cdots &0 \\ \cdots & \cdots & \cdots &\cdots \\ 0 & 0 & 0 & (\sqrt[r]{p})I_{n_{r}} \end{pmatrix} $$ where $ I_{n_{i}} $ is the identity matrix of of size $ n_{i} $ for each $ i $. A simple calculation shows that $$ Q\widetilde{A}Q^{-1}=\begin{pmatrix} \widetilde{A_{1}} & (\sqrt[r]{p})\ast & \cdots & (\sqrt[r]{p})\ast \\ (\sqrt[r]{p})\ast & \widetilde{A_{2}} & \cdots & (\sqrt[r]{p})\ast \\ \cdots & \cdots & \cdots & \cdots \\ (\sqrt[r]{p})\ast &(\sqrt[r]{p})\ast & \cdots & \widetilde{A_{r}} \end{pmatrix}. $$ Then we define $ \rho' $ to be the composite of $ \rho $ and the conjugate by $ Q $. We are done.

We remark that the above argument is no longer valid for $m>2$.