I am reading the lecture notes. Let $G$ be a reductive group and $(\pi, V)$ a representation of $G$. For $v \in V$, define $\operatorname{Stab}(v)=\{g\in G \mid \pi(g)v=v\}$ and $V^{\infty}=\{v\in V \mid \operatorname{Stab}(v) \text{ is open} \}$.
I have some questions about the proof of the lemma on Page 13. Suppose that $v_1, v_2 \in V^{\infty}$. Could we conclude that there is a compact open subgroup $K$ such that $K \subset \operatorname{Stab}(v_1) \cap \operatorname{Stab}(v_2)$? Is it possible that $\operatorname{Stab}(v_1) \cap \operatorname{Stab}(v_2) = \emptyset$? Thank you very much.
Two subgroups of a group, such as the stabilizers of $v_1$ and $v_2$, can never have empty intersection, because the identity element is always in both of them. If $v_1$ and $v_2$ are smooth, i.e., their stabilizers are open, then the intersection of the stabilizers will also be open, as well as a subgroup. The compact open subgroups of the group of $\mathbf{Q}_p$-points of a linear algebraic group over $\mathbf{Q}_p$ form a base of opens around the identity, so any open subgroup will contain one. In particular, the intersection of the stabilizers of $v_1$ and $v_2$ will contain a compact open subgroup.