I have the following problem:
Let $(\pi,V)$ be a smooth complex representation of a locally profinite group $G$. Let $U$ be a $G$-subspace of $V$. Show that there is a natural representation of $G$ on the quotient $V/U$ which is smooth. The definition of smooth that I'm working with is that for all $v \in V$, there exists a compact open subgroup $K$ of $G$ such that $\pi_k(v)=v$ for all $k \in K$.
My attempt:
Given an element $\pi_g \in \text{Aut}_{\mathbb{C}}(V)$, we define an element $\overline{\pi_g} \in \text{Aut}_{\mathbb{C}}(V/U)$ via $\overline{\pi_g}(v+U)=\pi_g(v)+U$. It's clear that $\overline{\pi_g}$ is a linear map, with inverse $\overline{\pi_{g^{-1}}}$. Furthermore, for any $g,h \in G$ it holds \begin{equation} \overline{\pi_{gh}}(v+U)=\pi_{gh}(v)+U=\pi_g\pi_h(v)+U=\overline{\pi_g}(\pi_h(v)+U)=\overline{\pi_g}\overline{\pi_h}(v+U) \end{equation} thus $(\overline{\pi},V/U)$ is a representation of $G$. Given a representative element $v$ of the element $v+U \in V/U$, there exists a compact open subgroup $K$ of $G$ such that $\pi_k(v)=v$ for all $k \in K$. As \begin{equation} \overline{\pi_k}(v+U)=\pi_k(v)+U=v+U, \end{equation} we conclude $(\overline{\pi},V/U)$ is also smooth.
I'm trying to understand why we need to assume that $U$ is a $G$-subspace. It seems that it's required to ensure $\overline{\pi_g}$ is linear for all $g$. For if there was some $g$ such that $\pi_g(U) \not\subset U$, then there exists $u \in U$ such that $\pi_g(u) \not\in U$ and so \begin{equation} \overline{\pi_g}(U)=\overline{\pi_g}(u+U)=\pi_g(u)+U \neq U \end{equation} contradicting that $\overline{\pi_g}$ is linear. Is this correct, or is there another reason?
Lastly, in what sense is the representation $(\overline{\pi},V/U)$ "natural"?
I'm also a new learner on these stuff, so I am so sorry if I got it wrong.
Question: Where the $G$-stable property of $U$ is used?
Answer: To make the quotient representation, i.e. the representation $\overline{\pi}: G \rightarrow \mathrm{Aut}(V/U)$ well-defined. Recall that in your definition of $\overline{\pi}_g$, $$ \overline{\pi}_g (v+U) := \pi_g(v) +U. $$ But is this definition independent of the choice of the representative of the coset $v+U$? In other words, if we choose another representative $v^{\prime} \in V$ such that $v+U = v^{\prime} + U$, does $\pi_g(v) +U$ and $\pi_g(v^{\prime}) +U$ really equal as two cosets in $V/U$. To check this, we shall prove that $$\pi_g(v^{\prime}) - \pi_g(v) \in U.$$ Indeed, since $\pi_g$ is linear, $\pi_g(v^{\prime}) - \pi_g(v) = \pi_g(v^{\prime}-v)$ . Now as $v+U = v^{\prime} + U$, we see $v^{\prime} - v \in U$. Hence by the $G$-stable property of $U$, we have $\pi_g(v^{\prime} - v) \in U$. This is exactly what we want!
The discussion above has nothing to do with the smoothness of the representation and is a general procedure when one hopes to define something like "quotient action". Only after checking the things above can we say that a representation of $G$ on $V/U$ is defined.
Question: What does the word natural mean?
Answer: From my own naiive point of view, it just means that it is the canonical way to define such a representation, i.e. anyone in the world that is familiar with algebra will first come up with the same definition like yours. I'm not sure if there is a categorical meaning of "natural" here.
Answer': Packaging the smooth representations of $G$ as an abelian category, the quotient defined above is indeed the quotient object of an object $V$ with respect to the subobject $U$.