\section{Passing to isogenous groups while preserving the parameter}
\label{fl:sec:arithmetic-central}

The papers \cite{HuII} and \cite{HuIII} realize
parameters for simply connected groups. This section transfers the
result to the other groups with the same root system: to a quotient
$G=H_0/C$ of a semisimple group by a finite central subgroup
(Theorem~\ref{fl:thm:arithmetic-central-bridge}), and from the
derived group of a reductive group to the group itself
(Lemma~\ref{fl:lem:derived-extension}). As a consequence
Theorem~\ref{fl:thm:semisimple-elliptic} reduces to simply connected
simple groups (Corollary~\ref{fl:cor:simple-group-reduction}).

The difficulty is the following. Let
$\mathsf a:\widehat G\to\widehat{H_0}$ be the isogeny dual to
$H_0\to G$. A parameter $\sigma$ of $G$ determines the parameter
$\mathsf a\sigma$ of $H_0$, but several parameters of $G$ can have
the same image, and a cusp form on $H_0$ with parameter
$\mathsf a\sigma$ does not single one of them out. The proof
proceeds in three steps: embed $H_0$ in a group
$G^{\mathrm{env}}$ which maps onto $G$ with kernel a central
torus;
extend the cusp form to $G^{\mathrm{env}}$ and twist it by a
character so that its parameter is the prescribed one; and descend
to $G$. That the descended form has parameter $\sigma$, and not
only a parameter with the same image under $\mathsf a$, rests on
the surjectivity of a restriction map between rings of invariant
functions (Lemma~\ref{fl:lem:central-envelope-invariants}).

All groups in this section are split. For a parameter $\sigma$ we
write $\mathcal A_{\mathrm{cusp}}(G)_\sigma$ for the excursion
summand of $\sigma$ in the space of cusp forms on
$G(F)\backslash G(\mathbb A)$; functions on this set of a given
level are identified with functions on the points of the stack of
bundles with level structure as in \cite[Remark~8.21]{VLa18}. The
cusp forms of a given level which are invariant under a given
central lattice form a space of finite dimension
\cite[Theorems~0.1 and 12.3]{VLa18}. We use the notation
$T_G$, $Q_G$, $\alpha_G$, $\beta_G$ of
\eqref{fl:eq:two-central-maps}, and write $\pi_a:G\to Q_G$ for
the quotient map. For a homomorphism
$\psi:\Gamma_F\to Z(\widehat G)^\circ=\widehat Q_G$ which
has finite image or is of the form $c^{\deg_F}$, let
$\chi_\psi=\operatorname{CFT}_{Q_G}(\psi)$, normalized
as in Section~\ref{fl:sec:parameters-parameters}. As in
Proposition~\ref{fl:prop:compact-central-twists}, $T_\psi$
is the multiplication by this character:
\begin{equation}\label{fl:eq:central-twist-operator}
 (T_\psi h)(g)=\chi_\psi(\pi_a(g))h(g).
\end{equation}

\subsection{Functoriality and descent of all invariant functions}

\begin{lemma}[Pullback along a central map]
\label{fl:lem:arithmetic-central-pullback}
Let $v:G_1\to G_2$ be a homomorphism whose image is normal and
whose kernel is central, as a group scheme, and let
$\widehat v:\widehat G_2\to\widehat G_1$ be the dual
homomorphism. For compatible levels and central lattices,
\[
 v^*\mathcal A_{\mathrm{cusp}}(G_2)_\sigma
 \subseteq
 \mathcal A_{\mathrm{cusp}}(G_1)_{\widehat v\sigma}.
\]
The kernel of $v$ is not assumed to be smooth. The map $v^*$
commutes with right translations.
\end{lemma}

\begin{proof}
\step{1}{cuspidality} Let $P_1=P_{G_1}(\lambda)$ be the
parabolic subgroup of $G_1$ defined by a cocharacter $\lambda$,
and $P_2=P_{G_2}(v\lambda)$. Since the kernel of $v$ is central,
$v$ maps each root subgroup of $G_1$ isomorphically onto a root
subgroup of $G_2$. Since the image of $v$ is normal, the other
roots of $G_2$ belong to simple factors which commute with the
image, and they are orthogonal to $v\lambda$. So $v$ maps the
unipotent radical of $P_1$ isomorphically onto that of $P_2$,
and $P_2$ is proper if $P_1$ is. Consequently the constant term
of $v^*h$ along $P_1$ is the pullback of the constant term of
$h$ along $P_2$, and $v^*$ preserves cusp forms.

\step{2}{excursion operators} For every function $f$ on
$\widehat G_1^I$ invariant under the diagonal translations,
\[
 S^{G_1}_{I,f,\boldsymbol\gamma}v^*
 =
 v^*S^{G_2}_{I,f\circ\widehat v^{\,I},
                 \boldsymbol\gamma}
\]
by \cite[Proposition~12.5]{VLa18}, where the corresponding
morphism of stacks of shtukas is shown to be compatible with
creation, with the action of the Galois groups and with
annihilation. By the comparison of roots in step~1, the
Schubert varieties which correspond have the same dimension, so
the Satake sheaves have the same normalization. For a central isogeny
$v$ and a coweight $\lambda$ of $G_1$, the map from the reduced
Schubert variety of $\lambda$ to that of $v\lambda$ is a finite
universal homeomorphism. To see this, work over an algebraically
closed extension of the residue field. The two positive loop
groups have the same root subgroups, and each is generated by its
root subgroups and by the integral points of the split torus. The
torus fixes the point of the coweight and normalizes the root
subgroups, so the map is surjective on the open orbits. It is
injective on geometric points: $v$ is finite, so a loop of $G_1$
whose image in $G$ is integral is integral. On the closures of the
orbits the map is proper with dense image, hence surjective; being
proper and injective on geometric points, it is finite and
radicial. The \'etale site is invariant under such a map, so the
intersection complexes correspond, with the same dimensions and
twists. For the simply connected cover this map is discussed in
\cite[Introduction, (1.1)]{HLRNormality}.

It follows that $v^*$ transforms a polynomial in the operators
$S^{G_2}_e-\nu_\sigma(e)$ into the corresponding polynomial in
the operators $S^{G_1}_{e'}-\nu_{\widehat v\sigma}(e')$, which
gives the inclusion. The compatibility with right translations
is \cite[Remark~12.4]{VLa18}.
\end{proof}

\begin{proposition}[Products]
\label{fl:prop:cuspidal-products}
If the parameters $\sigma_j$ have nonzero cuspidal excursion
summands on the groups $G_j$, the parameter $\prod_j\sigma_j$ has
a nonzero cuspidal excursion summand on $\prod_jG_j$.
\end{proposition}

\begin{proof}
A product of nonzero cusp forms on the factors is a nonzero cusp
form on the product. For a given set of legs, the stack of
shtukas of the product is the fibre product of the stacks of
shtukas of the factors over the space of the legs, and for an
exterior tensor product of representations the complex of
coefficients is the exterior product of those of the factors. By
the K\"unneth formula for direct images with compact supports,
the creation maps, the action of the Galois groups and the
annihilation maps are, on the tensor product of the parts of
degree zero, the tensor products of the corresponding maps for the
factors. Only these parts enter an excursion operator: creation
starts in degree zero on each factor, and the three operations
preserve the degree on each factor. A function on
$\prod_j\widehat G_j^I$ invariant under the diagonal translations
is a sum of products of such functions on the factors, so every
excursion operator of the product is a sum of tensor products of
excursion operators of the factors. On the tensor product of the
excursion summands of the $\sigma_j$ it is therefore the sum of
the scalar $\nu_{\prod\sigma_j}(e)$ and of a nilpotent operator.
No condition on the images of the $\sigma_j$ is needed.
\end{proof}

\begin{lemma}[Invariant functions restrict surjectively]
\label{fl:lem:central-envelope-invariants}
Let $H\hookrightarrow H^{\mathrm{env}}$ be a closed immersion of
reductive groups over $k$ such that
$H^{\mathrm{env}}=Z(H^{\mathrm{env}})^\circ H$.
For every $m\ge0$ the restriction map
\[
 k[(H^{\mathrm{env}})^m]^{H^{\mathrm{env}}}
       \longrightarrow k[H^m]^H
\]
between the rings of functions invariant under simultaneous
conjugation is surjective.
\end{lemma}

\begin{proof}
The restriction $k[(H^{\mathrm{env}})^m]\to k[H^m]$ is surjective,
and it commutes with conjugation by $H^{\mathrm{env}}$, which
normalizes $H$. A reductive group in characteristic zero is
linearly reductive, so the map remains surjective on invariants
of $H^{\mathrm{env}}$. The centre of $H^{\mathrm{env}}$ acts
trivially on $H^m$, so the invariants of $H^{\mathrm{env}}$ in
$k[H^m]$ are the invariants of $H$.
\end{proof}

\begin{lemma}[Descent along a quotient by a central torus]
\label{fl:lem:cuspidal-central-quotient}
Let $\pi_c:G^{\mathrm{env}}\to G$ be the quotient by a split
central torus $\mathsf T$, and
$\widehat{\pi_c}:\widehat G\to\widehat{G^{\mathrm{env}}}$ the
dual homomorphism. Every cusp form on $G^{\mathrm{env}}$ which is
invariant under $\mathsf T(\mathbb A)$ and under a central
lattice, and lies in the excursion summand of
$\widehat{\pi_c}\sigma$, is the pullback of a cusp form on $G$,
invariant under a central lattice, in the excursion summand
of $\sigma$. Joint
eigenfunctions correspond to joint eigenfunctions, the values of
Whittaker functionals are the same, and irreducible cuspidal
subrepresentations correspond to each other.
\end{lemma}

\begin{proof}
By Hilbert's Theorem~90 the map $\pi_c$ is surjective on points
with values in $F$ and in the fields $F_v$. At all but
finitely many places the exact sequence extends to the split models
over $\mathcal O_v$, and it is surjective on points with values in
$\mathcal O_v$ because $\operatorname{Pic}(\mathcal O_v)=0$. Hence
$G^{\mathrm{env}}(\mathbb A)\to G(\mathbb A)$ is surjective with
kernel $\mathsf T(\mathbb A)$. So $\pi_c^*$ is a bijection from
functions on $G(F)\backslash G(\mathbb A)$ to functions on
$G^{\mathrm{env}}(F)\backslash G^{\mathrm{env}}(\mathbb A)$
invariant under $\mathsf T(\mathbb A)$. The unipotent radicals of
corresponding parabolic subgroups are identified by $\pi_c$, so
constant terms and Whittaker coefficients correspond, and the
action of $G^{\mathrm{env}}(\mathbb A)$ on the invariant
functions is the action of $G(\mathbb A)$.

Next we choose compatible central lattices. Let $\pi_c^*h$ be invariant
under $\mathsf T(\mathbb A)$ and under a central lattice
$\Xi'$ of $G^{\mathrm{env}}$. The centre of $G$ is the quotient of
the centre of $G^{\mathrm{env}}$ by $\mathsf T$, so $\pi_c$ maps
$T_{G^{\mathrm{env}}}$ onto $T_G$ with kernel $\mathsf T$, and $h$
is invariant under the image of $\Xi'$. Choose elements of
$\Xi'$ whose degrees in $X_*(T_G)$ are a basis of a subgroup of
finite index. Their images generate a central lattice
$\Xi_G$ of $G$, under which $h$ is invariant; and these elements,
together with a central lattice of $\mathsf T$, generate a
central lattice of $G^{\mathrm{env}}$ which maps onto $\Xi_G$
and under which $\pi_c^*h$ is invariant.

It remains to compare excursion operators. By
Lemma~\ref{fl:lem:arithmetic-central-pullback}, applied with
these lattices, for a function
$f^{\mathrm{env}}$ on $(\widehat{G^{\mathrm{env}}})^I$,
\[
 S^{G^{\mathrm{env}}}_{I,f^{\mathrm{env}},\boldsymbol\gamma}\,
    \pi_c^*
 =\pi_c^*\,S^{G}_{I,f,\boldsymbol\gamma},
 \qquad f=f^{\mathrm{env}}\circ\widehat{\pi_c}^{\,I}.
\]
We claim that every function $f$ on $\widehat G^I$ invariant
under the diagonal translations is of this form. For
$I=\varnothing$ there are only constants. Otherwise choose
$i_0\in I$. The map $(g_i)\mapsto(g_ig_{i_0}^{-1})_{i\ne i_0}$
identifies the functions invariant under the diagonal
translations with the functions on $\widehat G^{I-\{i_0\}}$
invariant under simultaneous conjugation, and likewise for
$\widehat{G^{\mathrm{env}}}$. The homomorphism
$\widehat{\pi_c}$ is a closed immersion, and
\[
 \widehat{G^{\mathrm{env}}}
  =Z(\widehat{G^{\mathrm{env}}})^\circ\,
     \widehat{\pi_c}(\widehat G).
\]
So the claim follows from
Lemma~\ref{fl:lem:central-envelope-invariants}.

Now let $\pi_c^*h$ lie in the excursion summand of
$\widehat{\pi_c}\sigma$. For an excursion datum $e=(I,f,
\boldsymbol\gamma)$ of $G$ choose $f^{\mathrm{env}}$ as in the
claim and let $e^{\mathrm{env}}$ be the corresponding datum.
Then $\nu_{\widehat{\pi_c}\sigma}(e^{\mathrm{env}})
=\nu_\sigma(e)$, and
\[
 \pi_c^*\prod_e\bigl(S^G_e-\nu_\sigma(e)\bigr)h
   =\prod_e\bigl(S^{G^{\mathrm{env}}}_{e^{\mathrm{env}}}
       -\nu_{\widehat{\pi_c}\sigma}(e^{\mathrm{env}})\bigr)
        \pi_c^*h
\]
for every finite product. Since $\pi_c^*$ is injective, the same
power of the ideal $\mathfrak m_\sigma$ annihilates $h$; for the
first power this is the statement about eigenfunctions.
\end{proof}

\subsection{The central character of a cuspidal summand}

\begin{proposition}[The central character of a cuspidal summand]
\label{fl:prop:connected-central-character}
Let $\Xi$ be a central lattice for $G$. On the cusp forms
invariant under $\Xi$ which lie in the excursion summand of
$\sigma$, the group $T_G(\mathbb A)$ acts by the character
\begin{equation}\label{fl:eq:extended-centre-label}
 \omega_\sigma=\operatorname{CFT}_{T_G}(\alpha_G\sigma).
\end{equation}
If there is a nonzero cusp form of this kind,
$\alpha_G\sigma$ has finite image.
\end{proposition}

\begin{proof}
Fix a level $N$ such that $\sigma$ is unramified outside $N$ and
$\omega_\sigma$ is trivial on $T_G(\mathbb A)\cap K_N$. On
functions of level $N$ invariant under $\Xi$ the group
$T_G(\mathbb A)$ acts through a finite quotient, and the action
commutes with the excursion operators. So the summand is the
direct sum of its eigenspaces for $T_G(\mathbb A)$, and it
suffices to show that a character $\omega$ which occurs is
$\omega_\sigma$.

Let $v\notin N$ and let $\lambda$ be a cocharacter of $T_G$. The
translation by $\lambda(\varpi_v)$ is the Hecke operator at $v$
of the character $\alpha_G^*\lambda$ of $\widehat G$
(Lemma~\ref{fl:lem:central-hecke-units}), hence an excursion
operator, and on the excursion summand of $\sigma$ its only
eigenvalue is $\omega_\sigma(\lambda(\varpi_v))$, by the
normalization of Section~\ref{fl:sec:parameters-parameters}. So
$\omega$ and $\omega_\sigma$ agree on these elements. Both
characters are trivial on $T_G(F)$ and on
$T_G(\mathbb A)\cap K_N$, and by the proof of
Lemma~\ref{fl:lem:central-hecke-units} the elements
$\lambda(\varpi_v)$ with $v\notin N$ generate $T_G(\mathbb A)$
modulo these two subgroups. Hence $\omega=\omega_\sigma$.

In particular $\omega_\sigma$ factors through a finite group, and
by class field theory $\alpha_G\sigma$ has finite image.
\end{proof}

\subsection{Extension from the derived group}

Let $G$ be split connected reductive. Recall from
\eqref{fl:eq:two-central-maps} that
$\ker\beta_G=\widehat Q_G=Z(\widehat G)^\circ$ and that the
restriction of $\alpha_G$ to $\widehat Q_G$ is an isogeny.

\begin{lemma}[Extension by zero from the derived group]
\label{fl:lem:cuspidal-zero-extension}
Let $\varphi_0$ be a nonzero cusp form on $G_{\mathrm{der}}$.
There are a central lattice $\Xi$ of $G$ and a cusp form on
$G$, invariant under $\Xi$, whose restriction to
$G_{\mathrm{der}}(\mathbb A)$ is $\varphi_0$.
\end{lemma}

\begin{proof}
\step{1}{extension by zero} Let $K_{\mathrm{der}}$ be an open
compact subgroup under which $\varphi_0$ is invariant. Since
$G_{\mathrm{der}}$ is semisimple, the support of $\varphi_0$ is
compact \cite[Chapter~I, Proposition~5.3]{Morris1982}. The group
$Q_G(F)$ is discrete in $Q_G(\mathbb A)$, so there is an open
compact subgroup $K_G\subset G(\mathbb A)$ with
\[
 \pi_a(K_G)\cap Q_G(F)=\{1\},\qquad
 K_G\cap G_{\mathrm{der}}(\mathbb A)\subset K_{\mathrm{der}}.
\]
We claim that the map
\[
 G_{\mathrm{der}}(F)\backslash G_{\mathrm{der}}(\mathbb A)
     /(K_G\cap G_{\mathrm{der}}(\mathbb A))
 \longrightarrow G(F)\backslash G(\mathbb A)/K_G
\]
is injective. Indeed, let $h_2=\gamma h_1\kappa$ with
$h_1,h_2\in G_{\mathrm{der}}(\mathbb A)$, $\gamma\in G(F)$ and
$\kappa\in K_G$. Applying $\pi_a$ we get
$\pi_a(\gamma)\pi_a(\kappa)=1$, hence
$\pi_a(\gamma)=\pi_a(\kappa)=1$ by the choice of $K_G$, and
$\gamma\in G_{\mathrm{der}}(F)$,
$\kappa\in G_{\mathrm{der}}(\mathbb A)$. Let $\varphi^c$ be the
function on $G(F)\backslash G(\mathbb A)/K_G$ which is
$\varphi_0$ on the image of this map and zero elsewhere. It has
compact support, contained in the image of
$\mathcal S=G(F)G_{\mathrm{der}}(\mathbb A)K_G$.

\step{2}{cuspidality} The unipotent radical $U$ of a parabolic
subgroup of $G$ is contained in $G_{\mathrm{der}}$, and
$\mathcal S$ is stable under left multiplication by
$G_{\mathrm{der}}(\mathbb A)$, which is normal. So the constant
term of $\varphi^c$ vanishes at points outside $\mathcal S$. At a
point $\gamma h\kappa$ of $\mathcal S$ we have
\[
 \int_{U(F)\backslash U(\mathbb A)}
     \varphi^c(u\gamma h\kappa)\,du
 =\int_{U'(F)\backslash U'(\mathbb A)}\varphi_0(u'h)\,du',
 \qquad U'=\gamma^{-1}U\gamma ,
\]
and $U'$ is the unipotent radical of a proper parabolic subgroup
of $G_{\mathrm{der}}$. The integral is zero because $\varphi_0$
is cuspidal.

\step{3}{the lattice} Since $r_G:T_G\to Q_G$ is an isogeny, there
is a central lattice $\Xi$ of $G$ on which the degree of the
image in $Q_G$ is injective. Put
$\varphi(g)=\sum_{\xi\in\Xi}\varphi^c(g\xi)$. The image in $Q_G$
of a point of $\mathcal S$ has degree zero, so at most one term
of the sum is not zero, and for
$g\in G_{\mathrm{der}}(\mathbb A)$ it is the term $\xi=1$. Thus
$\varphi$ is a cusp form invariant under $\Xi$ which restricts
to $\varphi_0$.
\end{proof}

\begin{lemma}[From the derived group to the group]
\label{fl:lem:derived-extension}
Let $\sigma:\Gamma_F\to\widehat G(k)$ be continuous,
finitely ramified and semisimple, and suppose that
$\beta_G\sigma$ has a nonzero cuspidal excursion summand on
$G_{\mathrm{der}}$.
\begin{enumerate}
\item There is a nonzero cuspidal joint eigenfunction of the
excursion operators in $\mathcal A^{\mathrm{cont}}(G)$ with
parameter $\sigma$.
\item If $\alpha_G\sigma$ has finite image, there is a full
central lattice $\Xi$ such that the excursion summand of $\sigma$
in the cusp forms invariant under $\Xi$ is not zero.
\end{enumerate}
\end{lemma}

\begin{proof}
The function is obtained in three moves: extend a cusp form from
$G_{\mathrm{der}}$ to $G$ by zero, choose in the extension a
component with a parameter $\sigma_{\mathrm{in}}$ which has the
right image in $\widehat{G_{\mathrm{der}}}$, and twist by the
character of $G/G_{\mathrm{der}}$ which corresponds to the
quotient of $\sigma$ by $\sigma_{\mathrm{in}}$. The point which
makes this work for the parameter itself, and not only for its
conjugacy class at each place, is in step~1: the two parameters
are made equal after projection, as homomorphisms, so that their
quotient is a homomorphism to the centre.

\step{1}{a parameter with the right image in the derived group}
Let $\varphi_0\ne0$ be in the excursion summand of
$\beta_G\sigma$, and let $\varphi$ be an extension as in
Lemma~\ref{fl:lem:cuspidal-zero-extension}. The cusp forms on $G$
of the level of $\varphi$ invariant under $\Xi$ are the direct
sum of their excursion summands, and each summand belongs to a
parameter; write $\varphi=\sum_\tau\varphi_\tau$ accordingly. By
Lemma~\ref{fl:lem:arithmetic-central-pullback}, applied to the
inclusion of $G_{\mathrm{der}}$, the restriction of
$\varphi_\tau$ to $G_{\mathrm{der}}$ lies in the excursion
summand of $\beta_G\tau$. The excursion summands on
$G_{\mathrm{der}}$ of parameters which are not conjugate are
independent, so $\varphi_0$ is the sum of the restrictions of
the $\varphi_\tau$ with $\beta_G\tau$ conjugate to
$\beta_G\sigma$. Choose such a $\tau$ with nonzero restriction
and call it $\sigma_{\mathrm{in}}$. After conjugating
$\sigma_{\mathrm{in}}$ by an element of $\widehat G$ which lifts
the conjugating element, we have
$\beta_G\sigma_{\mathrm{in}}=\beta_G\sigma$.

\step{2}{the difference of the two parameters} Put
\[
 z(\gamma)=\sigma(\gamma)\sigma_{\mathrm{in}}(\gamma)^{-1}
       \in\ker\beta_G=\widehat Q_G .
\]
Since $z$ has values in the centre, it is a homomorphism: in
$\sigma(\gamma\delta)\sigma_{\mathrm{in}}(\gamma\delta)^{-1}$
write $\sigma(\delta)=z(\delta)\sigma_{\mathrm{in}}(\delta)$ and
move $z(\delta)$ to the left. It is continuous and finitely
ramified. By Lemma~\ref{fl:lem:finite-coefficients}, applied to
the torus $\widehat Q_G$, we can write $z=z_0\psi$ with
$z_0$ of finite image and $\psi=c^{\deg_F}$.

\step{3}{twisting} By \eqref{fl:eq:transported-full-action} the
operator $T_{z_0}$ maps the excursion summand of
$\sigma_{\mathrm{in}}$ into that of $\sigma_{\mathrm{in}}z_0$,
after $\Xi$ has been replaced by its subgroup of finite index
$\Xi\cap\ker\chi_{z_0}$, which is again a central lattice.
(This can also be seen from
Lemma~\ref{fl:lem:arithmetic-central-pullback} and
Proposition~\ref{fl:prop:cuspidal-products}: $T_{z_0}h$ is the
pullback of the function $h\otimes\chi_{z_0}$ on $G\times Q_G$
by $g\mapsto(g,\pi_a(g))$, and the dual of this homomorphism is
the multiplication
$\widehat G\times\widehat Q_G\to\widehat G$.) In the same way
$T_\psi$ maps the excursion summand of
$\sigma_{\mathrm{in}}z_0$ to a space of functions in
$\mathcal A^{\mathrm{cont}}(G)$ of finite dimension on which the
operators $S_e^{\mathrm{fun}}-\nu_\sigma(e)$ are nilpotent, since
$\sigma_{\mathrm{in}}z_0\psi=\sigma$. Both twists preserve
cusp forms, because the characters are trivial on unipotent
radicals. The space obtained is not zero, so it contains a joint
eigenvector. This proves (1).

\step{4}{the case of a finite central character} By
Proposition~\ref{fl:prop:connected-central-character},
$\alpha_G\sigma_{\mathrm{in}}$ has finite image. If
$\alpha_G\sigma$ has finite image, so has $\alpha_Gz$. The
restriction of $\alpha_G$ to $\widehat Q_G$ has finite kernel, so
$z$ has finite image. Then we may take $\psi=1$, and the
twist by $z_0=z$ gives cusp forms invariant under a central
lattice. This proves (2).
\end{proof}

\begin{corollary}[Genericity is preserved]
\label{fl:cor:generic-derived-extension}
In Lemma~\ref{fl:lem:derived-extension}(2), suppose that the
Whittaker functional does not vanish on the excursion summand of
$\beta_G\sigma$ on $G_{\mathrm{der}}$. Then the excursion summand
of $\sigma$ on $G$ contains a globally generic irreducible
cuspidal representation.
\end{corollary}

\begin{proof}
Choose $\varphi_0$ with nonzero Whittaker coefficient. The
maximal unipotent subgroups of $G$ and of $G_{\mathrm{der}}$ are
the same, so the extension $\varphi$ has the same Whittaker
coefficient at the identity. In step~1 of the preceding proof the
sum of the $\varphi_\tau$ with $\beta_G\tau$ conjugate to
$\beta_G\sigma$ restricts to $\varphi_0$, so one of them has a
nonzero Whittaker coefficient, and we take it for
$\sigma_{\mathrm{in}}$. The twist by $z$ multiplies the Whittaker
coefficient at the identity by a nonzero scalar. The resulting
cusp form lies in the excursion summand of $\sigma$ and is
invariant under a central lattice. The representation which
it generates under right translation consists of cusp forms
invariant under this lattice, in the same summand. By
Section~\ref{fl:sec:unitary-extraction} and
Lemma~\ref{fl:lem:unitary-admissible-semisimplicity} it contains
an irreducible subrepresentation on which the Whittaker
functional is not zero.
\end{proof}

\subsection{Finite central quotients and the prescribed lift}

\begin{theorem}[Transfer to a central quotient]
\label{fl:thm:arithmetic-central-bridge}
Let $H_0$ be split semisimple, let $C\subset Z_{H_0}$ be a
finite diagonalizable group scheme, and put $G=H_0/C$. Let
$\mathsf a:\widehat G\to\widehat{H_0}$ be the dual isogeny. If
$\mathsf a\sigma$ has a nonzero cuspidal excursion summand on
$H_0$, then $\sigma$ has a nonzero cuspidal excursion summand on
$G$. If the first summand contains a globally generic irreducible
cuspidal representation, so does the second.
\end{theorem}

\begin{proof}
Embed $C$ in a split torus $\mathsf T$ and put
\begin{equation}\label{fl:eq:central-envelope}
 G^{\mathrm{env}}
   =(H_0\times\mathsf T)/\{(c,c^{-1}):c\in C\},
 \qquad \mathsf Q=\mathsf T/C.
\end{equation}
The derived group of $G^{\mathrm{env}}$ is $H_0$, and $\mathsf T$
is its maximal central torus. The centre of
$G^{\mathrm{env}}$ is $(Z_{H_0}\times\mathsf T)/C$, and the quotient
of this centre by $\mathsf T$ is $Z_{H_0}/C$. So the centre is not
connected in general, and in characteristic $p$ it can be
nonreduced; we use only the torus $\mathsf T$. There
are exact sequences
\[
 1\to\mathsf T\to G^{\mathrm{env}}\xrightarrow{\pi_c}G\to1,
 \qquad
 1\to H_0\to G^{\mathrm{env}}\to\mathsf Q\to1.
\]
The dual sequences are
\begin{equation}\label{fl:eq:dual-central-envelope}
\begin{split}
 1&\to\widehat G\xrightarrow{\widehat{\pi_c}}
          \widehat{G^{\mathrm{env}}}
          \xrightarrow{\alpha}\widehat{\mathsf T}\to1,\\
 1&\to\widehat{\mathsf Q}\to\widehat{G^{\mathrm{env}}}
          \xrightarrow{\beta}\widehat{H_0}\to1,
 \qquad \beta\,\widehat{\pi_c}=\mathsf a ,
\end{split}
\end{equation}
where $\alpha$ and $\beta$ are the maps
$\alpha_{G^{\mathrm{env}}}$ and $\beta_{G^{\mathrm{env}}}$ of
\eqref{fl:eq:two-central-maps}.

Consider the parameter $\widehat{\pi_c}\sigma$ of
$G^{\mathrm{env}}$. Its image under $\beta$ is $\mathsf a\sigma$,
which has a nonzero cuspidal excursion summand on $H_0$, and its
image under $\alpha$ is trivial. By
Lemma~\ref{fl:lem:derived-extension}(2) there are nonzero cusp
forms on $G^{\mathrm{env}}$, invariant under a central
lattice, in the excursion summand of $\widehat{\pi_c}\sigma$. By
Proposition~\ref{fl:prop:connected-central-character} the group
$\mathsf T(\mathbb A)$ acts on them by the character
$\omega_{\widehat{\pi_c}\sigma}$, which is trivial. By
Lemma~\ref{fl:lem:cuspidal-central-quotient} they are pullbacks
of cusp forms on $G$ in the excursion summand of $\sigma$. The
statement on generic representations follows from
Corollary~\ref{fl:cor:generic-derived-extension} and the last
assertion of Lemma~\ref{fl:lem:cuspidal-central-quotient}.
\end{proof}

\begin{remark}
For $\mathrm{Sp}_6\to\mathrm{PGSp}_6$ one can take
$G^{\mathrm{env}}=\mathrm{GSp}_6$, and the dual groups are
$\mathrm{Spin}_7\subset\mathrm{GSpin}_7\to\mathrm{SO}_7$. Any
two lifts of the same $\mathrm{SO}_7$-valued homomorphism to
$\mathrm{GSpin}_7$ differ by a central character.
The proof does not stop there: it conjugates the parameter found
on $\mathrm{GSp}_6$ so that the two images in $\mathrm{SO}_7$ are
equal as homomorphisms, and then twists by the quotient $z$ of
the two parameters. So it produces a cusp form with the
prescribed parameter $\widehat{\pi_c}\sigma$, and not only one
with the same representation of dimension seven. In the descent
to $\mathrm{PGSp}_6$ it is the surjectivity of
Lemma~\ref{fl:lem:central-envelope-invariants} which shows that
no excursion operator is lost.
\end{remark}

\begin{corollary}[Reduction to simply connected simple groups]
\label{fl:cor:simple-group-reduction}
Let $G$ be split semisimple, with simply connected cover
$\prod_aG_a$, the $G_a$ simple, and let $\sigma$ be an elliptic
parameter of $G$. The images $\sigma_a$ of $\sigma$ in the groups
$\widehat G_a$ are elliptic, and if every $\sigma_a$ has a nonzero
cuspidal excursion summand on $G_a$, then $\sigma$ has one on
$G$. Thus Theorem~\ref{fl:thm:semisimple-elliptic} follows from
the same statement for simply connected simple groups, with the
bounds on the characteristic attached to their types. The central
subgroup by which the cover is divided need not be a product of
subgroups of the factors.
\end{corollary}

\begin{proof}
The dual of the covering map is a central isogeny
$\widehat G\to\prod_a\widehat G_a$. It induces an isomorphism of
Lie algebras compatible with the adjoint actions, so the Lie
algebra of the centralizer of $\sigma$ is the direct sum of the
Lie algebras of the centralizers of the $\sigma_a$. Hence each
$\sigma_a$ is elliptic. If each $\sigma_a$ has a nonzero cuspidal
excursion summand, so has the product parameter on $\prod_aG_a$
by Proposition~\ref{fl:prop:cuspidal-products}, and
Theorem~\ref{fl:thm:arithmetic-central-bridge}, applied once to
the product and the central subgroup, gives the summand of
$\sigma$ on $G$.
\end{proof}