# Existence of regular semisimple elements in linear group over local field

Let $$L$$ be a finite extension of $$p$$-adic numbers $$\mathbb{Q}_p$$. Let $$\text{GL}_{n}(L)$$ denote the general linear group $$\text{GL}_{n}(L)$$ over $$L$$ equipped with the topology induced from the $$p$$-adic field $$L$$. An element $$g\in \text{GL}_{n}(L)$$ is called regular semisimple if it has pairwise distinct eigenvalues in the algebraic closure $$\overline{\mathbb{Q}}_{p}$$ of $$\mathbb{Q}_p$$. A linear group over $$L$$ is a subgroup of $$\text{GL}_{n}(L)$$.

My question is the following: When does a linear group $$\Gamma$$ have a regular semisimple element?

As pointed out in the comments, it is not the case only under the assumption that $$\Gamma$$ has no open solvable subgroup.

• "has distinct" means "has at least two distinct" or "has pairwise distinct"? (given the title, you probably mean the latter)
– YCor
Commented Mar 27, 2022 at 18:33
• A trivial counterexample would then be the diagonal copy of $\mathrm{GL}_2$ in $\mathrm{GL}_4$.
– YCor
Commented Mar 27, 2022 at 18:36
• here "has distinct" refer to "has pairwise distinct". Commented Mar 27, 2022 at 19:07
• If the subgroup is irreducible, that should be true and should follow from the proof of the Tits alternative. Commented Mar 27, 2022 at 23:05
• "subgroup is irreducible" means " if the representation $\Gamma \to \text{GL}_n(L)$ is irreducible" or "it is an irreducible topological space"? Is it possible to point out references or a more detailed explanation, thanks! Commented Mar 28, 2022 at 17:32

The group $$\mathrm{GL}_N(\mathbf{Q}_p)$$ acts by conjugation on trace zero matrices, and the corresponding subgroup $$\Gamma \subset \mathrm{GL}_{N^2 - 1}(\mathbf{Q}_p)$$ (which is isomorphic to $$\mathrm{PGL}_N(\mathbf{Q}_p)$$) is both absolutely irreducible and has the property that every element $$\gamma \in \Gamma$$ has $$1$$ as an eigenvalue with multiplicity at least $$N-1$$.
• What exactly is group $\Gamma$? Commented Apr 1, 2022 at 21:07
• @YufanLuo If a group $G$ acts on a vector space $V$, there is a map $$G \rightarrow \mathrm{GL}(V)$$ and it has some image. In this case, $G = \mathrm{GL}_N(\mathbf{Q}_p)$, $V = M^0(\mathbf{Q}_p)$ is the space of trace zero matrices which is a vector space of dimension $N^2 - 1$, so there is a map $$\mathrm{GL}_N(\mathbf{Q}_p) \rightarrow \mathrm{GL}_{N^2-1}(\mathbf{Q}_p),$$ the image is then some group $\Gamma$. The kernel of the map above is the center (scalar matrices) and so $\Gamma$ is identified with $\mathrm{PGL}_N(\mathbf{Q}_p)$. I'm not sure I can say it any simpler. Commented Apr 2, 2022 at 2:28
• But why your $\Gamma$ is contained in $\text{GL}_{N^2-1}(\mathbb{Z}_p)$? Commented Apr 2, 2022 at 13:51