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An admissible representation is a smooth representation for which every fixed-vector space $V^K$ for a compact-open subgroup has finite dimension. Admissibility can hold for an infinite-dimensional representation; it bounds each local fixed level, not their union.

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.

A compact-open subgroup is both a compact subset and an open subgroup. Its coset space is discrete, and compact subsets meet only finitely many cosets. Finite intersections of compact-open subgroups are again compact-open, making them useful common stabilizers for finitely many vectors.

For the upper-triangular subgroup of ${\mathrm{GL}}_n(F)$, conjugation by $\mathrm{diag}({\varpi }^{-(n-1)},\dots ,1)$ expands the upper-unipotent coordinates. For a decomposed compact-open subgroup $K=T_0{U}_0$, its conjugates eventually contain $K$, while their fixed-vector spaces are isomorphic. Finite dimension therefore forces those spaces to be equal. Inverse conjugation contracts any fixed unipotent element into $U_0$, proving that the unipotent radical acts trivially in every admissible representation of this triangular group.

Evaluation of an intertwiner at the identity gives a $B$-equivariant functional. Conversely a functional $\ell$ gives $A(v)(g)=\ell (\pi (g)v)$, which has the required covariance and is fixed on the right by any compact-open subgroup fixing $v$. These inverse constructions prove the displayed adjunction. If $\xi$ is trivial on the unipotent radical, it is equivalent to a character-valued functional on the Jacquet module.

A locally profinite group is a Hausdorff, locally compact, totally disconnected topological group. It has a neighbourhood basis at the identity consisting of compact-open subgroups. In smooth representation theory this is also called an $\ell$-group; that terminology is unrelated to a lattice-ordered group.

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.

Write $U_m=1+{\mathfrak{p}}^m$, $m\ge 1$, for the principal-unit filtration of a local field. Each linear character ${\chi }_i$ is a smooth character, since its kernel contains $U_{i+1}$. A vector in the algebraic direct sum has only finitely many nonzero coordinates. Taking $m$ larger than all their indices gives a common subgroup $U_m$ fixing the vector, so the direct sum is smooth.

If $i<m$, then $U_m\subseteq U_{i+1}$, so $V_i$ is fixed by $U_m$. If $i\ge m$, the subgroup $U_i$ lies inside $U_m$, and ${\chi }_i$ is nontrivial on $U_i$; hence $V_i^{U_m}=0$. It follows thatEvery compact-open subgroup $K$ of $F^{\times }$ contains some $U_m$, because it is an identity neighbourhood. Therefore $V^K\subseteq V^{U_m}$ is finite-dimensional. The direct sum is an admissible representation, despite its infinite total dimension.

Use unnormalized parabolic induction, the convention fixed by the supplied sequence and by the occurrence of constants in $V(1)$. Its functions satisfy $f(bg)=\chi (b)f(g)$, and $G$ acts by right translation. Characters in this smooth category are smooth characters. Put $\lambda ={\chi }^w{\delta }^{-1}$, retaining the question's modulus $\delta (\mathrm{diag}(a,d))=|d/a|$.

First derive the evaluation adjunction for unnormalized induction. A $G$-intertwiner $A:V(\chi )\to V(\xi )$ gives a functional $\ell (v)=A(v)(1)$. Since $A$ intertwines and induced functions have left covariance, $\ell (\pi (b)v)=\xi (b)\ell (v)$. Conversely such a functional definesThis function has the required left covariance. Any compact-open subgroup fixing $v$ fixes it on the right, so it is a smooth induced function; compactness of $B\backslash G$ removes any additional support issue. The construction is $G$-equivariant, and the two operations are inverse. Since $\xi$ is trivial on $U$, these functionals factor precisely through the Jacquet module, givingNow use the given exact sequence $0\to \lambda \to J\to \chi \to 0$, where $J=V(\chi {)}_U$ has dimension two.

If $\lambda \ne \chi$, choose $t_0\in T$ on which their values differ. Its operator on $J$ is upper triangular with distinct eigenvalues, so has two one-dimensional eigenspaces. Since $T$ is abelian, all its operators commute with this operator and preserve the eigenspaces. They are the linear character lines $\lambda$ and $\chi$. This proves that distinct-character extensions of an abelian group split, so $J\cong \lambda \oplus \chi$. A map between one-dimensional linear character spaces is zero unless the linear characters agree, and otherwise is one-dimensional. Therefore the required dimension is one for $\xi =\lambda$ or $\xi =\chi$, and zero for any other linear character.

The case $\lambda =\chi$ needs an extra argument: the exact sequence alone would also permit a split self-extension with two quotient functionals. Write $\chi (\mathrm{diag}(a,d))={\chi }_1(a){\chi }_2(d)$. Equality with $\lambda$ forces ${\chi }_1/{\chi }_2=|\cdot |$, so $\chi =(\eta \circ \det ){\chi }_0$, where $\eta ={\chi }_2$ and ${\chi }_0(\mathrm{diag}(a,d))=|a|$. Multiplication of induced functions by $\eta (\det g)$ identifies this with a determinant-character twist of $V({\chi }_0)$. Twisting is invertible, so it is enough to prove nonsplitting for ${\chi }_0$.

Use the open Bruhat decomposition coordinateFor $x\ne 0$ the matrix factorizationshows that $g_f(x)=|x{|}^{-1}f(1)$ for all sufficiently large $|x|$: the lower-unipotent matrix approaches the identity, where $f$ is locally constant. The coordinate map is injective, since this open cell is dense in the projective-line quotient and induced functions are locally constant sections.

For $t=\mathrm{diag}(a,d)$, direct multiplication givesThere exists a section $f_0$ with $f_0(1)=1$ and coordinate functionIndeed the displayed factorization changes to the lower chart near infinity, where this tail corresponds to the constant function one. Thus it extends smoothly across the missing point. With $t=\mathrm{diag}(\varpi ,1)$ and $|\varpi |=q^{-1}$, the difference is the compactly supported functionNormalize additive Haar measure by $\mathrm{vol}(\mathfrak{o})=1$. ThenThis difference survives in the Jacquet module. To prove that assertion, the unipotent action translates coordinates: $g_{\pi (n(y))f}(x)=g_f(x+y)$. Any such translation difference is compactly supported because the tail is $f(1)|x{|}^{-1}$ and $|x+y|=|x|$ when $|x|$ is sufficiently large. Its integral is zero: integrate over a sufficiently large additive ball, invariant under translation by $y$, and change variables. A finite sum of unipotent translation differences also has integral zero. Hence $h$, whose integral is nonzero, cannot be a coinvariant relation. In particular, on $J$ the operator $\pi (t)-{\chi }_0(t)I$ is nonzero. This proves the equal-character Jacquet self-extension for GL2 is nonsplit.

Any $T$-equivariant functional $J\to {\chi }_0$ must kill the nonzero image of that operator. Since $J$ has dimension two, its space of such functionals has dimension at most one; the quotient in the supplied exact sequence already provides one. Twisting back gives the same conclusion for the common linear character $\chi =\lambda$. For any different $\xi$, a surjection $J\to \xi$ is impossible because both composition factors of $J$ are $\chi$. Combining the two cases givesWhen the two listed linear characters coincide, the dimension remains one; they are not two independent contributions.

A complex group representation $\pi :G\to \mathrm{GL}(V)$ is smooth when every vector is fixed by some compact-open subgroup. Equivalently each vector's stabilizer is open. Smoothness is local constancy of orbit maps with the vector space regarded discretely, not ordinary continuity into a finite-dimensional complex topology.

The right action on compactly supported locally constant functions is a smooth representation. For a compact-open subgroup $K$, its fixed-vector space has basis the characteristic functions of cosets in $G/K$, since compact support on that discrete quotient is finite support. Consequently this representation is admissible exactly when $G$ 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.