Algebraic contragredient representation ($({\pi }^{\ast }(g)\ell )(v)=\ell (\pi (g^{-1})v)$)

The algebraic contragredient acts on the entire algebraic dual space. It need not be a smooth representation, even if the original representation is smooth. Taking the inverse in this formula is necessary to obtain a group action, and inverts a one-dimensional linear character.

For a finite-dimensional complex group representation, combining inverse and conjugate transpose gives another representation: both operations reverse multiplication order. This is the complex conjugate of the algebraic dual representation in matrices. If $W$ is invariant under $D^{\#}$, its orthogonal complement is invariant under $D$, because $⟨w,D(g)v⟩=⟨D^{\#}(g^{-1})w,v⟩=0$ for $w\in W$, $v\in W^{\perp }$. Thus irreducibility is preserved. Applying the construction twice returns $D$, so the equivalence of irreducibility works in both directions for arbitrary groups in finite dimension.

The smooth dual consists of the smooth vectors of the algebraic contragredient representation. Their union is a vector subspace because finite intersections of open stabilizers remain open; it is invariant because translation conjugates stabilizers.

If one-dimensional smooth linear characters of a local multiplicative group have conductors tending to infinity, their direct sum can be admissible. Its algebraic dual is the product of inverse linear characters, but a smooth vector in this product has only finitely many nonzero coordinates. A common open stabilizer contains one fixed principal-unit subgroup, on which every sufficiently ramified linear character is nontrivial. Thus the smooth dual is again a direct sum.