TITLE: Can we glue characteristic 0 and characteristic p representations of a finite group given equality of (Brauer) characters?
QUESTION [11 upvotes]: Suppose I have a prime $p$ and a finite group $G$ together with representations $\sigma: G \to GL_n(\mathbb{Q}_p)$ and $\pi: G \to GL_n(\mathbb{F}_p)$.  My question is:

When does there exist a representation $\rho: G \to GL_n(\mathbb{Z}_p)$ such that $\sigma$ is isomorphic to the base change of $\rho$ to $\mathbb{Q}_p$, and $\pi$ is isomorphic to the base change of $\rho$ to $\mathbb{F}_p$?

One necessary condition is that for any $p$-regular element $g \in G_{reg}$, the trace of $\sigma(g)$ is equal to the Brauer character of $\pi(g)$.  However, I can't tell if this is sufficient or not - the results in Serre's Linear Representations book only give me existence theorems up to semisimplification.  I tried to cook up a suitable $G$-stable $\mathbb{Z}_p$-lattice using affine buildings, but I got hopelessly lost.

REPLY [13 votes]: The condition on Brauer characters is not sufficient.
Let $G$ be a $p$-group, $\pi$ any nontrivial representation over $\mathbb{F}_p$, and $\sigma$ the trivial representation over $\mathbb{Q}_p$ of the same degree as $\pi$. Then a representation over $\mathbb{Z}_p$ whose base change to $\mathbb{Q}_p$ is isomorphic to $\sigma$ must also be trivial, so its base change to $\mathbb{F}_p$ can’t be isomorphic to $\pi$.
But the condition on Brauer characters is trivially satisfied.