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The space $C_c^{\infty }(X)$ consists of locally constant complex functions with compact support. On a locally profinite group, finite compact-open coset covers of the support show that every such function is fixed by one common compact-open subgroup under right translation. The symbol infinity describes local constancy in this context, not real differentiability.

Interpret the map as the action on ${\mathbb{C}}^n$. A smooth representation requires every vector to have an open stabilizer in the locally profinite group ${\mathrm{GL}}_n(F)$. Take the first standard basis vector $e_1$. Every identity neighbourhood contains matrices $d_a=\mathrm{diag}(1+a,1,\dots ,1)$ with $a\ne 0$ sufficiently small in the local field. We may require $|a|<1$, so $1+a\ne 0$. Injectivity of the field embedding givesThus no identity neighbourhood fixes $e_1$, and its stabilizer is not open. Consequently the representation is not smooth for any $n\ge 1$. This includes $n=1$, when a nonzero vector's stabilizer is just the identity. The field embedding need not be continuous, but no continuity assumption would repair this stabilizer obstruction. The existence of the stated embedding itself forces characteristic zero; in positive characteristic no field embedding into $\mathbb{C}$ exists.

An $\ell$-group here means a locally profinite group, so identity neighbourhoods contain compact-open subgroups. The right action is a group representation because $\rho (g)\rho (h)f(x)=f(xgh)=\rho (gh)f(x)$.

Let $f$ be a compactly supported locally constant function with support $S$. For every $x\in S$, local constancy supplies a compact-open subgroup $K_x$ with $f$ constant on $x{K}_x$, and this coset lies in $S$. Compactness gives a finite cover $S={\bigcup }_{j=1}^r{x}_j{K}_j$. Put $K={\bigcap }_j{K}_j$. Each coset in this cover is right $K$-invariant, so $S$ is right $K$-invariant. Within the support, $xk$ and $x$ lie in the same chosen $x_j{K}_j$ and have the same function value; outside the support, right $K$-invariance of $S$ ensures that both values remain zero. Hence $\rho (k)f=f$ for all $k\in K$. The zero function is fixed by every subgroup. This proves that the smooth right regular representation is smooth.

For a fixed compact-open subgroup $K$, right-invariant functions are functions on the discrete coset space $G/K$. Compact support becomes finite support there, since the image of a compact set in a discrete space is finite. Conversely every finitely supported function on $G/K$ lifts to a compactly supported locally constant function on $G$. Thuswhere the superscript indicates finite support. Its basis is the set of characteristic functions $1_{gK}$, and its dimension is finite exactly when $[G:K]$ is finite.

If $G$ is compact, the discrete quotient $G/K$ is compact and therefore finite for every such $K$; hence the representation is admissible. Conversely, admissibility gives finite index for any one compact-open subgroup. Then $G$ is a finite union of compact cosets, so is compact. Therefore the representation is admissible if and only if $G$ is compact.

A smooth vector in any abstract group representation of a locally profinite group has an open stabilizer, equivalently is fixed by some compact-open subgroup. Smooth vectors form an invariant linear subspace: use intersections for addition and conjugation for translation. A smooth representation is one in which every vector has this property.