Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 1 / 2

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.