\subsection{An \(S_3\)-parameter and the determinant \(-64\)}
\label{fl:sec:low-rank-example}

In the following example the parameter, the group $Z_\rho$ and the
supertrace of Proposition~\ref{fl:prop:puncture-exact-trace} are
computed by hand. The rest of the construction is not explicit: the
automorphic function is obtained from these data by the results of
Sections~\ref{fl:sec:geometrization}--\ref{fl:sec:exceptional-assembly}.

Take \(F=\mathbb F_5(t)\), \(G=\mathrm{SL}_2\), and
\(\widehat G=\mathrm{PGL}_2(k)\), with \(\ell\ne5\).
The splitting field of \(z^3-z-t\) has geometric and arithmetic
Galois group \(S_3\). To see this directly, the map
\(t=x^3-x\) of projective lines has separable degree three, so
the cubic is geometrically irreducible. Its discriminant
\(4-27t^2=4-2t^2\) has two simple zeros and is not a square.
The geometric Galois group is therefore \(S_3\). The arithmetic
group lies between it and \(S_3\), so it is \(S_3\) as well, and
the splitting field has constant field \(\mathbb F_5\).

Use the projective images of
\[
 r=\begin{pmatrix}0&-1\\1&-1\end{pmatrix},
 \qquad s=\begin{pmatrix}0&1\\1&0\end{pmatrix}.
\]
They satisfy \(r^3=s^2=1\) and \(srs=r^{-1}\), and give the
parameter
\(\sigma:\Gamma_F\twoheadrightarrow S_3\hookrightarrow
\mathrm{PGL}_2(k)\).
Its ramification has order two at \(t^2=2\) and order three at
infinity, hence is tame at five. Its centralizer is
\begin{equation}\label{fl:eq:rank-one-centralizer}
 Z_\sigma=\{1,c\},\qquad
 c=\left[\begin{pmatrix}1&-2\\2&-1\end{pmatrix}\right].
\end{equation}
Indeed, \(c^2=[-3I]=1\); it commutes with \(r\) and commutes
projectively with \(s\). Conversely, diagonalizing \(r\) gives
its torus centralizer, on which \(s\) acts by inversion. The
fixed subgroup is \(\mu_2\). Thus \(\sigma\) is elliptic,
although its centralizer is not trivial.

The point \(t=0\) has identity Frobenius, since the cubic splits
there as \(z(z-1)(z+1)\). In Section~\ref{rw:sec:example-curve} the
divisor $D_{\rm aux}$ is taken to be one rational point of the
auxiliary component above \(t=0\). So $d_{\rm aux}=1$, and the
Frobenius element at the puncture is $g_D=1$. It is also shown
there that the geometric monodromy group of the local system $\rho$
on the smooth curve is \(S_3\); hence
\(Z_\rho=\Pi_\rho=\mu_2\), by \eqref{fl:eq:rank-one-centralizer}.
For any representation \(V\) of this group, the mutually inverse
maps
\[
 V\longrightarrow (k[\mu_2]\otimes V)^{\mu_2},
 \quad v\longmapsto e_1\otimes v+e_c\otimes cv,
 \qquad
 e_1\otimes v_1+e_c\otimes v_c\longmapsto v_1
\]
commute with Frobenius; this is step~1 of the proof of
Proposition~\ref{fl:prop:puncture-exact-trace}. Therefore the
cohomology of \(\Gamma(\mathfrak F_\rho,P_{\mathrm{reg}})\) is the
exterior algebra $\Lambda$ of that proof, with its grading. Here
$\mathsf B_\rho$ is the conormal space \(\mathfrak{pgl}_2^*\) at
the one puncture, which has dimension three. The arithmetic
operator of step~3 of that proof, which is defined by transport,
acts by \(5\,\id\) on $\mathsf B_\rho$, in degree \(-1\), because
\(q=5\), $d_{\rm aux}=1$ and \(g_D=1\). Its supertrace is
\begin{equation}\label{fl:eq:rank-one-nonzero-trace}
 1-3\cdot5+3\cdot25-125=(1-5)^3=-64.
\end{equation}
The geometric operator is its inverse, with supertrace
\((1-5^{-1})^3=64/125\). Both numbers are nonzero in the
coefficient field.

In the later sections this number is transported to the automorphic
side. By Proposition~\ref{fl:lem:vacuum-microstalk-comparison} and
Lemma~\ref{fl:lem:scalar-coefficient-covariance}, the supertrace of
geometric Frobenius on the vanishing cycles studied in
Section~\ref{fl:sec:conservation} is \(64c/125\), where the scalar
\(c\in k^\times\) accounts for the comparison of the two Frobenius
structures and for the graded line in that lemma. The scalar is not
determined, and it need not be: the supertrace is not zero.
Proposition~\ref{fl:prop:composite-trace-nonzero} then gives a
bundle on the special fibre of $C_0$, defined over $\mathbb F_5$,
at which the Frobenius supertrace of the sheaf is not zero.

There is also an elementary check on the \emph{parameter}.
At \(t=0,1,2\) the cubic has respectively three, one and no
roots in \(\mathbb F_5\). Its Frobenius classes have orders
one, two and three. The adjoint traces are respectively
\(3,-1,0\). These are predicted Hecke eigenvalues of any
realization of \(\sigma\), not computed values of
\(\varphi_\sigma\).

For this parameter automorphy can also be proved directly. The
matrices $r$ and $s$ define a linear representation of $S_3$ of
dimension two, which is irreducible: $r$ has two distinct
eigenspaces, and $s$ exchanges them. Its determinant has finite
order, so by the theorem of L.~Lafforgue it is the parameter of a
cuspidal representation of $\mathrm{GL}_2$. Take a cusp form in
the corresponding summand and translate it on the right so that
its value at the identity is not zero. Its restriction to
$\mathrm{SL}_2(\mathbb A_F)$ is a nonzero cusp form, and it lies
in the summand of $\sigma$ by
\cite[Proposition~12.5 and Remark~12.13]{VLa18}. This summand, at
a finite level, contains a joint eigenvector of the excursion
operators.