\section{Irreducible constituents and nonzero Whittaker functionals}
\label{fl:sec:unitary-extraction}

The constructions of \cite{HuIII} give spaces of
cusp forms stable under right translation, together with a
Whittaker functional that does not vanish on them. To state results
about generic representations we must pass to an irreducible
subrepresentation on which the functional is still nonzero. This
short section provides the tool: such spaces are semisimple, because
they are admissible and carry an invariant inner product after the
coefficients are transported to $\mathbb C$.

Fix an isomorphism of fields $\iota:k\xrightarrow{\sim}\mathbb C$
which maps $q^{1/2}$ to the positive square root of $q$.
We use it to transport functions and representations; algebraic
identities are preserved. Let $\Xi$ be a central lattice. A
cusp form of a given level which is invariant under $\Xi$ has
compact support modulo $\Xi$: its support is compact modulo the
centre \cite[Chapter~I, Proposition~5.3]{Morris1982}, and the
quotient of the adelic points of the centre by its rational points
and by $\Xi$ is compact. So the cusp forms invariant under $\Xi$
are square integrable on $G(F)\backslash G(\mathbb A)/\Xi$, and
right translation, which commutes with $\Xi$, is unitary on them.
At a fixed level they form a space of finite dimension
\cite[Theorem~0.1]{VLa18}. So every space of cusp forms invariant
under $\Xi$ which is stable under right translation is an
admissible subspace of the smooth vectors of a unitary
representation. No character of the centre is fixed here: when the
centre of $G$ is not connected, a space of this kind can contain
several of them.

The lemma below is a statement about smooth representations. After
it has been applied, an irreducible subrepresentation and a
functional which does not vanish on it are transported back to
$k$. The excursion operators commute with this transport. A given
vector lies in a finite-dimensional space of functions of fixed
level, so it is defined over a finite extension of the field of
coefficients already in use.

\begin{lemma}[Admissible subspaces of unitary representations]
\label{fl:lem:unitary-admissible-semisimplicity}
Let $\mathcal H$ be a unitary representation of a totally
disconnected locally compact group on a Hilbert space, and $V$ a
subspace of smooth vectors which is stable under the group and
admissible. Then $V$ is a semisimple representation. Every vector
of $V$ lies in a finite sum of irreducible subrepresentations,
and a linear form which does not vanish on the vector does not
vanish on one of them.
\end{lemma}

\begin{proof}
Let $U\subset V$ be a subrepresentation and $\overline U$ its
closure in $\mathcal H$. For a compact open subgroup $K$, the
average over $K$ is a continuous projection of $\mathcal H$ onto
the vectors fixed by $K$. It maps a sequence in $U$ converging to
a vector of $(\overline U)^K$ to a sequence in $U^K$ with the
same limit. Hence
\[
 (\overline U)^K=\overline{U^K}=U^K ,
\]
the last equality because $U^K$ has finite dimension, by
admissibility, and is therefore closed.

Since the representation is unitary, $\overline U$ and its
orthogonal complement are stable under the group, and the
orthogonal projection onto $\overline U$ commutes with the group.
It maps $V^K$ into $(\overline U)^K=U^K$. Every vector of $V$ is
fixed by some $K$, so the projection maps $V$ onto $U$. Thus
every subrepresentation of $V$ is a direct summand, and $V$ is
semisimple. The subrepresentation generated by one vector is
semisimple and finitely generated, hence a finite sum of
irreducible subrepresentations. A linear form which vanishes on
all of them vanishes on the vector.
\end{proof}

\begin{remark}
Admissibility is used in the equality $(\overline U)^K=U^K$. Without
it, orthogonal projection would only map $V$ to the closure of $U$,
and the algebraic statement would not follow.
\end{remark}

\begin{proposition}[Images of restricted tensor products]
\label{fl:prop:restricted-tensor-unitary-image}
Let $G/F$ be split connected reductive. For every place $v$ let
$V_v$ be a smooth admissible complex representation of $G(F_v)$
of finite length whose largest semisimple quotient $W_v$ is
irreducible, and let $\mathcal T=\bigotimes'_vV_v$ be the
restricted tensor product with respect to vectors which, for
almost all $v$, are fixed by $G(\mathcal O_v)$ and have nonzero
image in $W_v$. Let
\[
 T:\mathcal T\longrightarrow\mathcal H^\infty
\]
be a nonzero $G(\mathbb A)$-equivariant map into the smooth
vectors of a unitary representation $\mathcal H$ of
$G(\mathbb A)$. Then the image of $T$ is irreducible and
isomorphic to $\bigotimes'_vW_v$.
\end{proposition}

\begin{proof}
Fix a place $v$ and a pure tensor $w$ in the restricted tensor
product of the $V_{v'}$ with $v'\ne v$. The map
$x\mapsto T(x\otimes w)$ from $V_v$ to $\mathcal H^\infty$
commutes with $G(F_v)$. Its image is admissible: taking the
vectors fixed by a compact open subgroup is an exact functor on
smooth complex representations, by averaging over the subgroup.
The image is a subrepresentation of the restriction of
$\mathcal H$ to $G(F_v)$. By
Lemma~\ref{fl:lem:unitary-admissible-semisimplicity}
the image is semisimple, so the map factors through $W_v$. This
holds for all $v$ and $w$. Therefore $T$ vanishes on the sum of
the subspaces obtained by replacing one factor $V_v$ by the
kernel of $V_v\to W_v$. The quotient of $\mathcal T$ by this sum
is $\bigotimes'_vW_v$: this is clear for finite tensor products,
and the restricted tensor product is their union.

The representation $\bigotimes'_vW_v$ is irreducible, as a
restricted tensor product of irreducible admissible
representations. So the nonzero map which $T$ induces on it is
injective.
\end{proof}