\section{Purity of elliptic parameters}
\label{fl:sec:common-arithmetic}
\label{fl:sec:arithmetic-purity}

The geometric construction needs to know that the eigenvalues of
Frobenius under the parameter, in every representation of the dual
group, are algebraic numbers of absolute value one. The main
theorems assume nothing of the kind. This section shows that for
an elliptic parameter with values in a semisimple group the
property is automatic
(Theorem~\ref{fl:thm:arithmetic-strong-purity}). By the theorem of
L.~Lafforgue, an irreducible representation of $\Gamma_F$ has
this property as soon as its determinant has finite order. A
determinant of infinite order would be an algebraic character with
infinite image of the Zariski closure of the image of the
parameter, and such characters are excluded because the centre of
the monodromy group is finite. The property is needed
after every finite extension of $F$, and after such an extension
the closure of the image becomes smaller. The proof therefore
passes through the identity components of the closures, which do
not change.

Throughout this section $k=\overline{\mathbb Q}_\ell$, with
$\ell\ne p$. All closures of subgroups are Zariski closures. We use
the degree map $\deg_F:\Gamma_F\to\widehat{\mathbb Z}$, for which
the arithmetic Frobenius of the field of constants has degree one;
its kernel is denoted $\Gamma_F^{\mathrm{geom}}$.

\subsection{Three lemmas}

\begin{lemma}[A coefficient field for a compact group]
\label{fl:lem:arithmetic-finite-coefficients}
A compact subgroup $\mathscr K\subset\mathrm{GL}_n(k)$ is contained in
$\mathrm{GL}_n(E)$ for some finite extension $E/\mathbb Q_\ell$.
It preserves an $\mathcal O_E$-lattice in $E^n$.
\end{lemma}

\begin{proof}
The field $k$ is the union of countably many finite extensions
$E_j$ of $\mathbb Q_\ell$, so $\mathscr K$ is the union of the
countably many closed subgroups
$\mathscr K\cap\mathrm{GL}_n(E_j)$. A compact group is a Baire
space, so one of these subgroups has nonempty interior and is
therefore open, hence of finite index. Adjoining to $E_j$ the
entries of representatives of the finitely many cosets gives $E$
(see also \cite[Theorem~1]{ConradCompactSubgroups}). For the second
assertion, the $\mathcal O_E$-module spanned by the translates of
a lattice under the compact group $\mathscr K$ is bounded, so it is
a lattice, and it is stable under $\mathscr K$.
\end{proof}

\begin{lemma}[Torus characters on geometric Galois]
\label{fl:lem:torus-geometric-image}
Let $\chi$ be a continuous homomorphism from $\Gamma_F$ to the
points of a torus over $k$, unramified at almost all places. Then
$\chi(\Gamma_F^{\mathrm{geom}})$ is finite.
\end{lemma}

\begin{proof}
In coordinates on the torus, $\chi$ is a finite family of
characters, and it suffices to treat one character
$\chi:\Gamma_F\to k^\times$. By
Lemma~\ref{fl:lem:arithmetic-finite-coefficients} its values lie
in $\mathcal O_E^\times$ for a finite extension $E$ of
$\mathbb Q_\ell$. By class field theory $\chi$ is a character of
the idele class group. At a place $v$ where it is ramified, the
group of principal units of $F_v$ is a pro-$p$ group, while
$\mathcal O_E^\times$ has an open subgroup which is pro-$\ell$; so
the image of the principal units is finite, and $\chi$ is trivial
on the units congruent to one modulo a power of the maximal ideal.
Hence $\chi$ is trivial on an open subgroup of the idele class
group.

The group of idele classes of degree zero is compact
\cite[Chapter~II, \S16]{CasselsFrohlich}, so its image under
$\chi$ is finite. The reciprocity map identifies the image of
$\Gamma_F^{\mathrm{geom}}$ in the abelianization of $\Gamma_F$
with the group of idele classes of degree zero
\cite[\S1.2]{Zywina2013}. So $\chi(\Gamma_F^{\mathrm{geom}})$ is
finite.
\end{proof}

\begin{lemma}[Finite centre and algebraic characters]
\label{fl:lem:finite-centralizer-algebra}
Let $H$ be an algebraic group over a field of characteristic zero
whose identity component is reductive. If the centre of $H$ is
finite, every algebraic character of $H$ has finite image.
\end{lemma}

\begin{proof}
Let $S=Z(H^\circ)^\circ$, a torus, on which the finite group
$H/H^\circ$ acts by conjugation. The fixed points of this action
form the group $S\cap Z(H)$, which is finite. So the lattice of
characters of $S$ invariant under $H/H^\circ$ is zero: its rank is
the dimension of the fixed points. The restriction to $S$ of a
character $\chi$ of $H$ is invariant, hence trivial. Since
$H^\circ$ is generated by $S$ and by its derived group, on which
every character is trivial, $\chi$ is trivial on $H^\circ$ and
factors through the finite group $H/H^\circ$.
\end{proof}

\subsection{Monodromy groups}

For a parameter $\sigma$ with values in a group $H$, the
\emph{arithmetic monodromy group} $H_{\mathrm{ar}}$ is the closure
of $\sigma(\Gamma_F)$, and the \emph{geometric monodromy group}
$H_{\mathrm{geom}}$ is the closure of
$\sigma(\Gamma_F^{\mathrm{geom}})$. The second is a normal subgroup
of the first.

\begin{lemma}[Arithmetic monodromy]
\label{fl:lem:arithmetic-monodromy-quotient}
Suppose that $H$ is semisimple and that
$\sigma:\Gamma_F\to H(k)$ is continuous, finitely ramified,
semisimple and elliptic. Then $H_{\mathrm{ar}}/H_{\mathrm{geom}}$
is a finite cyclic group.
\end{lemma}

\begin{proof}
Let $A=H_{\mathrm{ar}}/H_{\mathrm{geom}}$. The image of an element
of degree one generates a dense subgroup of $A$, because
$\Gamma_F/\Gamma_F^{\mathrm{geom}}=\widehat{\mathbb Z}$. So $A$ is
commutative, and it suffices to prove that it is finite; a finite
group with a dense cyclic subgroup is cyclic.

The identity component of $H_{\mathrm{ar}}$ is reductive because
$\sigma$ is semisimple, and the centre of $H_{\mathrm{ar}}$ is
contained in the centralizer of $\sigma$, which is finite because
$\sigma$ is elliptic and $H$ is semisimple. By
Lemma~\ref{fl:lem:finite-centralizer-algebra} every character of
$H_{\mathrm{ar}}$ has finite image. If $A$ were infinite, its
identity component would be a torus of positive dimension, since
$A$ is commutative with reductive identity component. A nontrivial
character of that torus has a power which extends to a character
of $A$, and this would be a character of $H_{\mathrm{ar}}$ with
infinite image.
\end{proof}

\begin{proposition}[Determinants after base extension]
\label{fl:prop:arithmetic-finite-determinants}
Under the hypotheses of
Lemma~\ref{fl:lem:arithmetic-monodromy-quotient}, let $F'/F$ be a
finite extension and $r:H\to\mathrm{GL}(V)$ an algebraic
representation. Every irreducible constituent of
$r\sigma|_{\Gamma_{F'}}$ has a determinant of finite order.
\end{proposition}

\begin{proof}
A purely inseparable extension does not change the Galois group,
so we may assume $F'/F$ separable. Then $\Gamma_{F'}$ has finite
index in $\Gamma_F$, and
$\Gamma_{F'}^{\mathrm{geom}}=\Gamma_{F'}\cap\Gamma_F^{\mathrm{geom}}$
has finite index in $\Gamma_F^{\mathrm{geom}}$. Let
$H'_{\mathrm{ar}}$ and $H'_{\mathrm{geom}}$ be the monodromy groups
of the restriction of $\sigma$ to $\Gamma_{F'}$.

The closure of a subgroup of finite index has the same identity
component as the closure of the group: finitely many translates of
the smaller closure cover the larger one. Together with
Lemma~\ref{fl:lem:arithmetic-monodromy-quotient} this gives
\[
 (H'_{\mathrm{ar}})^\circ=H_{\mathrm{ar}}^\circ
   =H_{\mathrm{geom}}^\circ=(H'_{\mathrm{geom}})^\circ,
\]
so that $H'_{\mathrm{ar}}/H'_{\mathrm{geom}}$ is finite.

Now let $\chi$ be an algebraic character of $H'_{\mathrm{ar}}$.
The character $\chi\circ\sigma$ of $\Gamma_{F'}$ is continuous and
finitely ramified, so by Lemma~\ref{fl:lem:torus-geometric-image},
applied to $F'$, it has finite image on
$\Gamma_{F'}^{\mathrm{geom}}$. By density $\chi(H'_{\mathrm{geom}})$
is finite, and since $H'_{\mathrm{geom}}$ has finite index in
$H'_{\mathrm{ar}}$, so is $\chi(H'_{\mathrm{ar}})$. The
representation $r$ of the reductive group $H'_{\mathrm{ar}}$ is
semisimple, its irreducible constituents are those of
$r\sigma|_{\Gamma_{F'}}$, and the determinant of each is a
character $\chi$ as above.
\end{proof}

\begin{remark}\label{rw:rem:centre-after-extension}
The proof shows that $H_{\mathrm{ar}}^\circ$ is semisimple. Indeed,
take $F'$ so large that $H'_{\mathrm{ar}}$ is connected, that is,
equal to $H_{\mathrm{ar}}^\circ$. Every character of this
connected group has finite image, hence is trivial, and a
connected reductive group without nontrivial characters is
semisimple. Consequently the centre of $H'_{\mathrm{ar}}$ is finite
for every finite extension $F'$: it centralizes the semisimple
group $H_{\mathrm{ar}}^\circ$, whose centralizer in
$H_{\mathrm{ar}}$ is finite. For a general algebraic group with
finite centre this can fail after passing to a subgroup of finite
index: the group generated by a torus and the inversion has finite
centre, and its identity component is the torus. Such a group is
excluded here by class field theory, through
Lemma~\ref{fl:lem:torus-geometric-image}.
\end{remark}

\subsection{Purity}

\begin{definition}[Strong purity]
\label{fl:def:strong-purity}
A parameter with values in a semisimple group is \emph{strongly
pure of weight zero} if, for every finite extension $F'$ of $F$ and
every algebraic representation of the group, the eigenvalues of the
geometric Frobenius elements of $\Gamma_{F'}$ at the unramified
places are algebraic numbers all of whose complex absolute values
are one.
\end{definition}

\begin{theorem}[Purity from ellipticity]
\label{fl:thm:arithmetic-strong-purity}
Every continuous, finitely ramified, semisimple, elliptic parameter
with values in a semisimple group over $k$ is strongly pure of
weight zero.
\end{theorem}

\begin{proof}
Let $F'/F$ be a finite extension and $r$ an algebraic
representation. By
Lemma~\ref{fl:lem:arithmetic-finite-coefficients} the
representation $r\sigma|_{\Gamma_{F'}}$ and its irreducible
constituents are defined over a finite extension of
$\mathbb Q_\ell$. By
Proposition~\ref{fl:prop:arithmetic-finite-determinants} each
constituent has a determinant of finite order. So
Input~\ref{rw:in:llafforgue}(2)
\cite[Theorem~VII.6(i)--(ii), p.~198]{LLa02} applies to each
constituent, on the open subset of the curve where it is
unramified: its eigenvalues of Frobenius are algebraic numbers of
absolute value one. The eigenvalues of $r\sigma|_{\Gamma_{F'}}$
are those of its constituents.
\end{proof}

\begin{remark}
Ellipticity cannot be omitted. The parameter
$\gamma\mapsto\operatorname{diag}(q^{\deg_F(\gamma)},q^{-\deg_F(\gamma)})$
with values in $\mathrm{SL}_2(k)$ is well defined and continuous,
because $q$ is an $\ell$-adic unit; it is unramified and
semisimple, and its eigenvalues of Frobenius do not have absolute
value one. Its centralizer contains the diagonal torus.
\end{remark}

\subsection{Separating the central character}
\label{fl:sec:common-central-normalization}

For a reductive group, the image of the parameter in the dual of
the connected central torus need not be pure. Its restriction to
$\Gamma_F^{\mathrm{geom}}$ has finite image, by
Lemma~\ref{fl:lem:torus-geometric-image}. We remove a character of
the Galois group of the field of constants, after which the whole
image in the dual of the central torus is finite.

For a split connected reductive group $G$ put
$T_G=(Z_G^\circ)_{\mathrm{red}}$ and $Q_G=G/G_{\mathrm{der}}$. The
restriction $r_G:T_G\to Q_G$ of the quotient map is an isogeny. On
the dual side there are an exact sequence and a homomorphism
\begin{equation}\label{fl:eq:two-central-maps}
\begin{gathered}
 1\longrightarrow\widehat Q_G\xrightarrow{\ j_G\ }\widehat G
   \xrightarrow{\beta_G}\widehat{G_{\mathrm{der}}}\longrightarrow1,
 \qquad \widehat Q_G=Z(\widehat G)^\circ,\\
 \alpha_G:\widehat G\longrightarrow\widehat T_G,
 \qquad \alpha_Gj_G=\widehat r_G ,
\end{gathered}
\end{equation}
so the restriction of $\alpha_G$ to the connected centre of
$\widehat G$ is an isogeny. Recall from
Section~\ref{fl:sec:parameters-parameters} the central character
$\omega_\sigma$ of a parameter, and that with the normalization
fixed there the character attached to $c^{\deg_F}$ is
$t\mapsto c^{-\deg(t)}$.

\begin{definition}[Twists by the field of constants]
\label{fl:def:constant-field-central-homomorphism}
A point $c$ of a torus over $k$ is a \emph{unit point} if the
values of all characters at $c$ are $\ell$-adic units. For a unit
point $c$ of $Z(\widehat G)^\circ$ we denote by $c^{\deg_F}$ the
continuous homomorphism $\Gamma_F\to Z(\widehat G)^\circ(k)$
obtained by extending $n\mapsto c^n$ from $\mathbb Z$ to
$\widehat{\mathbb Z}$ and composing with $\deg_F$.
\end{definition}

The extension exists and is unique: in coordinates, $c$ is a tuple
of units of a finite extension $E$ of $\mathbb Q_\ell$, and the
group $\mathcal O_E^\times$ is profinite, so $n\mapsto c^n$ extends
uniquely to a continuous homomorphism on $\widehat{\mathbb Z}$.
The homomorphism $c^{\deg_F}$ is trivial on
$\Gamma_F^{\mathrm{geom}}$, in particular on every inertia group.

\begin{lemma}[Central normalization]
\label{fl:lem:finite-coefficients}
Every continuous, finitely ramified homomorphism
$\sigma:\Gamma_F\to\widehat G(k)$ can be written as
\[
 \sigma=\psi\,\sigma^{\mathrm{fin}},\qquad
 \psi=c^{\deg_F},
\]
with $c$ a unit point of $Z(\widehat G)^\circ$ and
$\alpha_G\sigma^{\mathrm{fin}}$ of finite image. The parameters
$\sigma$ and $\sigma^{\mathrm{fin}}$ have the same image under
$\beta_G$ and the same restriction to every inertia group, and one
is semisimple, or elliptic, if and only if the other is. If
$\alpha_G\sigma$ has finite image one can take $\psi=1$.
\end{lemma}

\begin{proof}
Choose $\gamma_0\in\Gamma_F$ of degree one. By
Lemma~\ref{fl:lem:torus-geometric-image} the homomorphism
$\alpha_G\sigma$ is finite on $\Gamma_F^{\mathrm{geom}}$. Lift the
point $\alpha_G\sigma(\gamma_0)$ of $\widehat T_G$ to a point $c$
of $Z(\widehat G)^\circ$ through the isogeny $\alpha_G$. The point
$\alpha_G\sigma(\gamma_0)$ is a unit point, because the image of
$\alpha_G\sigma$ is compact, and then so is $c$: the valuations of
the character values of a point and of its image under an isogeny
determine each other. Put $\psi=c^{\deg_F}$ and
$\sigma^{\mathrm{fin}}=\psi^{-1}\sigma$. The homomorphism
$\alpha_G\sigma^{\mathrm{fin}}$ is trivial at $\gamma_0$ and finite
on $\Gamma_F^{\mathrm{geom}}$. Since the powers of $\gamma_0$ are
dense in $\Gamma_F/\Gamma_F^{\mathrm{geom}}$, its image is finite.

The twist $\psi$ has values in the kernel of $\beta_G$ and is
trivial on inertia, which gives the second assertion. Multiplying
by a central element does not change centralizers, so ellipticity
is preserved. For semisimplicity, let $M$ be the closure of the
image of $(\sigma,\psi)$ in
$\widehat G\times Z(\widehat G)^\circ$. The unipotent radical of
$M^\circ$ has trivial image in both factors, because the closures
of the images of $\sigma$ and of $\psi$ have reductive
identity components; so it is trivial. The closure of the image
of $\sigma^{\mathrm{fin}}$ is the image of $M$ under
$(g,z)\mapsto z^{-1}g$, and its identity component is reductive.
The same argument with $\psi^{-1}$ gives the converse.
\end{proof}