Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 1 (Smooth Representation Theory of P-adic Groups)

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.

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.

The algebraic dual of a direct sum is a product, rather than another direct sum. The mapis injective, since the summands span $V$. Given any family $({\ell }_i{)}_i$, define $\ell ((v_i{)}_i)={\sum }_i{\ell }_i(v_i)$. This sum is finite for each vector, regardless of whether the family of functionals has finite support. It defines a linear functional and is inverse to $R$. Hence this is a natural vector-space isomorphism.

The algebraic contragredient representation is $({\pi }^{\ast }(g)\ell )(v)=\ell (\pi (g^{-1})v)$. On the $i$th coordinate,Thus $R$ intertwines the actions and givesNo smoothness restriction has yet been imposed on that product.

Let $\ell =({\ell }_i{)}_i$ be a smooth vector in the product description of the algebraic contragredient representation. Its open stabilizer contains $U_m$ for some $m$. For every $i\ge m$, choose $u_i\in U_i\subseteq U_m$ with ${\chi }_i(u_i)\ne 1$. Invariance of $\ell$ at its $i$th coordinate givesforcing ${\ell }_i=0$. Hence a smooth functional has finite support, contained among $i<m$.

Conversely a finitely supported family has indices bounded by some $N$. Each inverse linear character has the same kernel as its original linear character, and all those kernels contain $U_{N+1}$. Thus this subgroup fixes the family. The smooth dual is thereforeThis is the principle that conductors force finite support in a smooth dual. For instance a product vector with every coordinate nonzero is a perfectly valid algebraic functional on the original direct sum, but is not smooth.

Write $d=\mathrm{diag}({\varpi }^{-(n-1)},\dots ,1)$ for the diagonal matrix called $\delta$ in this question. If $v\in V^K$ and $k_i=d^{i}k{d}^{-i}\in K_i$, thenThus $\rho (d^i)$ maps the fixed-vector space $V^K$ into $V^{K_i}$. Its inverse $\rho (d^{-i})$ gives the reverse map, so

For an upper-unipotent matrix $u$, inverse conjugation changes each entry above the diagonal byEvery such factor tends to zero as $i\to \infty$. Because $U_0$ is compact, its finitely many coordinate functions are uniformly bounded; because it is open, it contains a neighbourhood of the identity in $U$. Therefore this contraction is uniform on $U_0$ and gives $d^{-i}{U}_0{d}^i\subseteq U_0$ for every sufficiently large $i$. The diagonal group commutes with $d$, soConjugating this inclusion gives $K\subseteq K_i$ for all sufficiently large $i$. This is an eventual inclusion; arbitrary $U_0$ need not give a monotone sequence at every earlier step.

Consequently $V^{K_i}\subseteq V^K$ eventually. Admissibility makes these spaces finite-dimensional, while the isomorphism above makes their dimensions equal. HenceThis proves contraction and admissible fixed spaces by both inclusion and dimension, not by an assumption that conjugate fixed spaces are literally equal from the outset.

Take any $v\in V$. A smooth representation has an open stabilizer, so some sufficiently small congruence subgroup of $B$ fixes $v$. It can be written $K=T_m{U}_m$, where the diagonal entries in $T_m$ belong to $1+{\mathfrak{p}}^m$ and all entries above the diagonal in $U_m$ belong to ${\mathfrak{p}}^m$, with $m\ge 1$. These are compact-open in their respective groups, $T_m$ normalizes $U_m$, and their product is compact-open in $B$.

By part (a), this vector lies in $V^{K_i}$ for every sufficiently large $i$. Given any $u\in U$, the coordinate contraction $d^{-i}u{d}^i\to 1$ eventually puts it in $U_m$. Choose $i$ large enough for both properties. Then $u\in d^i{U}_m{d}^{-i}\subseteq K_i$, so $\rho (u)v=v$. Since $u$ and $v$ were arbitrary,Thus every admissible representation of this upper-triangular group factors through its diagonal quotient. The argument uses finite-dimensional fixed spaces essentially; smoothness alone does not force the unipotent action to be trivial.

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.

Apply part (a) with $\chi =\xi =1$. It gives $\dim {\mathrm{End}}_G(V(1))=1$, so every endomorphism is a scalar multiple of the identity. The local Steinberg representation of GL2 in the supplied sequence is nonzero: $V(1)$ is the space of locally constant functions on $B\backslash G\cong {\mathbb{P}}^1(F)$, and the characteristic function of the compact-open affine set $\mathfrak{o}$ is not constant. Its class in the quotient by constants is therefore nonzero.

If the sequence split, $V(1)$ would be $1\oplus \mathrm{St}$ with both summands nonzero. Projection onto either summand would be a $G$-endomorphism. These two projections are linearly independent: restrict a proposed linear dependence first to one summand and then to the other. This contradicts the one-dimensional endomorphism space. Equivalently the projection onto the trivial summand would be a nonzero idempotent different from the identity, whereas a scalar idempotent can only be zero or the identity. The principle that scalar endomorphisms obstruct a split extension givesNo assertion of irreducibility of $V(1)$ is needed; this representation has a proper invariant line of constants.