\section{From a microstalk to a rational trace}
\label{fl:sec:conservation}

We now have a chart $Y$ at the canonical point, with a fixed
section $x_*$ and a test function $f$ of weight two
(Section~\ref{fl:sec:graded-morse-chart}); we know that the
automorphic family is ULA over $S$ and, away from $x_*$, along $f$
(Corollary~\ref{fl:cor:phase-universal-local-acyclicity}); and we
know the vanishing cycles of $f$ at $x_*$ on the generic fibre
(Proposition~\ref{fl:lem:vacuum-microstalk-comparison}). This section
turns that information into the statement we need,
Proposition~\ref{fl:prop:composite-trace-nonzero}: in
characteristic $p$, the member of the family with coefficient
$P_{\rm reg}$ has nonzero Frobenius trace at some bundle defined
over $\mathbb F_q$. There are three steps.
\begin{enumerate}
\item \emph{Comparing the two ends of $S$}
(Theorem~\ref{fl:thm:scaling-character-conservation}). Vanishing
cycles along $f$ and specialization over $S$ need not commute. We
show that the discrepancy carries the trivial action of the scaling
group, so that it disappears on the part of the vanishing cycles on
which $-1\in\mathbb G_m$ acts by $-1$.
\item \emph{Frobenius}
(proof of Proposition~\ref{fl:prop:composite-trace-nonzero}). The
comparison of step (1) is an isomorphism of functors of the
coefficient $P$. By
Lemma~\ref{fl:lem:scalar-coefficient-covariance} the two Frobenius
structures then differ by one scalar, and the nonzero supertrace of
Proposition~\ref{fl:prop:puncture-exact-trace} survives.
\item \emph{From vanishing cycles to a stalk}
(Theorem~\ref{fl:thm:homogeneous-detector}). A nonzero trace on
vanishing cycles at $x_*$ is not yet a nonzero value of the trace
function. A trace formula for homogeneous functions shows that it
forces a nonzero stalk trace at some $\mathbb F_q$-rational point
of the chart.
\end{enumerate}
The nonzero number which is being followed sits on a different
object at each stage:
\begin{center}
\small
\begin{tabular}{@{}p{0.6\textwidth}p{0.32\textwidth}@{}}
\toprule
the trace of Frobenius is known to be nonzero on & by\\
\midrule
$\Gamma(\mathfrak F_\rho,P_{\rm reg})$, on the spectral side &
Proposition~\ref{fl:prop:puncture-exact-trace}\\
the vanishing cycles of $f$ at $x_*$ on the generic fibre, up to
one scalar &
Proposition~\ref{fl:lem:vacuum-microstalk-comparison} and step (2)\\
their minus part on the special fibre &
Theorem~\ref{fl:thm:scaling-character-conservation}\\
the stalk at a rational point of the chart &
Theorem~\ref{fl:thm:homogeneous-detector}\\
the stalk at a bundle defined over $\mathbb F_q$ &
Proposition~\ref{fl:prop:composite-trace-nonzero}\\
\bottomrule
\end{tabular}
\end{center}

\subsection{Scaling actions and the minus part}

Let $a:\mathbb G_m\times Y\to Y$ be an action. A complex
$\mathcal K$ on $Y$ is \emph{strongly equivariant} if it is
equipped with an isomorphism
$a^*\mathcal K\simeq\operatorname{pr}_Y^*\mathcal K$ that restricts
to the identity along the unit section and satisfies the cocycle
condition on $\mathbb G_m\times\mathbb G_m\times Y$. At a fixed
point of the action, the group $\mathbb G_m$ then acts on the
stalk of $\mathcal K$.

\begin{setup}\label{rw:setup:scaling}
Let $S$ be the spectrum of an absolutely integrally closed
valuation ring in which $2$ and $\ell$ are invertible. Let $Y/S$ be
separated and of finite presentation, with an action of
$\mathbb G_m$, a fixed section $x_*:S\to Y$, and a function
$f:Y\to\mathbb A^1_S$ of weight two that vanishes along $x_*$. Put
$Z=f^{-1}(0)$. Let $P$ be a strongly equivariant complex on $Y$.
\end{setup}

Since $f$ has weight two, the element $\epsilon=-1$ of
$\mathbb G_m$ preserves $f$, and it therefore acts on the vanishing
cycles of $f$ at the fixed point. For a geometric point $b$ of $S$
define
\[
 \mu_b(P)=(R\Phi_{f_b}P_b)_{x_{*,b}}[-1],
 \qquad \mu_b^-(P)=\frac{1-\epsilon}{2}\,\mu_b(P).
\]
The coefficients are rational, so the projector makes sense also
when $\ell=2$. The action of $\epsilon$ comes from the geometry of
$Y$. The monodromy of the vanishing cycles, which comes from the
inertia group at $0\in\mathbb A^1$, is a different action.

\subsection{Vanishing cycles at the two ends of the base}

The comparison rests on a result of Saito on nearby cycles over a
base of dimension two (Input~\ref{rw:in:saito}), which is stated
for noetherian schemes and torsion coefficients. Our base is a
valuation ring and our coefficients are rational, so two reductions
are needed. The first lemma provides the integral structure, and
the second carries out both reductions.

\begin{lemma}[A lattice for both acyclicity conditions]
\label{rw:lem:simultaneous-ula-lattice}
Let $R$ be an absolutely integrally closed valuation ring in which
$\ell$ is invertible, and $S=\operatorname{Spec}R$. Let $Y$ be
separated and of finite presentation over $S$, let $x_*:S\to Y$ be
a closed section, and let $f:Y\to\mathbb A^1_S$. Put $U=Y-x_*(S)$. Let $P$ be a bounded
constructible complex with rational coefficients that is ULA over
$S$ and such that $P|_U$ is ULA over $\mathbb A^1_S$. Then, for a
suitable finite extension $E/\mathbb Q_\ell$, there is a
constructible complex $P^+$ of $\mathcal O_E$-modules with
$P^+[1/\ell]\simeq P$ that has the same two properties.
\end{lemma}
\begin{proof}
\step{1}{a lattice on $U$} The ring $R$ is integrally closed, hence
so are $R[t]$ and its localizations. Thus $\mathbb A^1_S$ is normal
and irreducible, in particular geometrically unibranch. By
\cite[Proposition~6.11]{HS}, $P|_U$ admits a lattice
$\mathcal L_U$ that is ULA over $\mathbb A^1_S$. Since
$\mathbb A^1_S\to S$ is smooth, $\mathcal L_U$ is also ULA over
$S$ \cite[\S3, discussion following Proposition~3.3]{HS}.

\step{2}{extension over the generic fibre} The generic fibre
$Y_\eta$ is a noetherian scheme of finite type over a field. Write
$j_\eta:U_\eta\to Y_\eta$ and $i_\eta:x_{*,\eta}\to Y_\eta$ for the
inclusions. The complex $P_\eta$ is determined by
$j_\eta^*P_\eta$, by $i_\eta^*P_\eta$, and by the gluing morphism
\[
 i_{\eta,*}i_\eta^*P_\eta
       \longrightarrow j_{\eta,!}j_\eta^*P_\eta[1].
\]
On the open part take the lattice $j_\eta^*\mathcal L_U$ of step 1,
and on the closed part take any lattice of $i_\eta^*P_\eta$, a
bounded complex of finite free modules. By
\cite[Corollary~2.4]{HS}, morphisms between the rational complexes
are the morphisms between the lattices with $\ell$ inverted. So
after multiplying the lattice on the closed part by a large power
of a uniformizer, the gluing morphism is defined integrally. Its
fibre is a lattice $\mathcal L_\eta$ in $P_\eta$ whose restriction
to $U_\eta$ is $j_\eta^*\mathcal L_U$.

\step{3}{extension over $S$} By \cite[Theorem~4.1]{HS} with
integral coefficients, $\mathcal L_\eta$ has a unique extension
$P^+$ to $Y$ that is ULA over $S$. On $U$, both $P^+|_U$ and
$\mathcal L_U$ are ULA over $S$ with the same generic fibre, so the
full faithfulness in that theorem gives
$P^+|_U\simeq\mathcal L_U$. Hence $P^+|_U$ is ULA over
$\mathbb A^1_S$. By the same uniqueness, with rational
coefficients, $P^+[1/\ell]\simeq P$. The reductions of $P^+$ modulo
powers of the uniformizer of $\mathcal O_E$ retain both properties,
by the criterion for change of coefficients in the proof of
\cite[Proposition~3.8]{HS}.
\end{proof}

\begin{lemma}[Successive specializations]
\label{rw:lem:valuation-successive-cycles}
In the situation of Lemma~\ref{rw:lem:simultaneous-ula-lattice},
assume $f\circ x_*=0$ and put $Z=f^{-1}(0)$. Choose a
specialization $s\leftarrow\eta$ of geometric points of $S$, and
compatible geometric generic points of the strict localizations of
$\mathbb A^1$ at $0$ over $s$ and over $\eta$. Let
\[
 \mathcal D_P=\operatorname{Cone}\left(
 P_{x_{*,s}}\longrightarrow
 (R\Psi_{Z/S}(P|_{Z_\eta}))_{x_{*,s}}\right).
\]
Then there is a distinguished triangle
\begin{equation}\label{fl:eq:sign-comparison-triangle}
 \mathcal D_P[-1]\longrightarrow\mu_s(P)
       \longrightarrow\mu_\eta(P)\longrightarrow\mathcal D_P.
\end{equation}
It is natural in $P$: the three terms are exact functors on the
stable $\infty$-category of complexes satisfying the hypotheses,
and the maps are natural transformations between them.
\end{lemma}
\begin{proof}
\step{1}{torsion coefficients on a noetherian model} In the strict
localization of $\mathbb A^1_S$ at $(0,s)$ consider the four points
\[
\begin{tikzcd}[column sep=large]
 (0,s) & (0,\eta)\arrow[l]\\
 (\bar t,s)\arrow[u] & (\bar t,\eta)\arrow[l]\arrow[u],
\end{tikzcd}
\]
where $\bar t$ denotes the chosen geometric generic points of the
line. The locus where $P$ is not ULA over $\mathbb A^1_S$ is
contained in $x_*(S)$, which is finite over its image, the zero
section of $\mathbb A^1_S$. On a noetherian model and with torsion
coefficients, \cite[Proposition~1.8(2), (1.17)]{SaitoCC}
applied along the top and right edges of the square gives a
distinguished triangle
\[
 A_0\longrightarrow A_1\longrightarrow\mu_\eta(P)[1],
 \qquad
 A_0=(R\Psi_{Z/S}P|_{Z_\eta})_{x_{*,s}}.
\]
Along the left and bottom edges, the generic point of the line
does not meet $x_*$, and the same proposition identifies $A_1$ with
$(R\Psi_{f_s}P_s)_{x_{*,s}}$. Both $A_0$ and $A_1$ receive the
canonical map from $P_{x_{*,s}}$, compatibly, and the cones of
these maps are $\mathcal D_P$ and $\mu_s(P)[1]$. The octahedral
axiom for the composite
$P_{x_{*,s}}\to A_0\to A_1$, shifted by $-1$, gives
\eqref{fl:eq:sign-comparison-triangle}.

\step{2}{descent to a noetherian model} Take the lattice $P^+$ of
Lemma~\ref{rw:lem:simultaneous-ula-lattice} and put
$P_m=P^+\otimes^L_{\mathcal O_E}\mathcal O_E/\varpi^m$. The schemes,
the section and $f$ are of finite presentation and descend to a
noetherian model; a finite affine cover with its presentations
gives a bound on the relative dimension that persists under every
further base change. For each fixed $m$, the complex $P_m$ descends
after refining the model, and so do the two ULA conditions, by
\cite[Proposition~3.7]{HS} applied to $Y/S$ and to
$U/\mathbb A^1_S$. This is done for one $m$ at a time, on a model
which depends on $m$.

\step{3}{return to $S$ and passage to the limit} Apply step 1 on
the model. By the compatibility of Saito's proposition with base
change, the resulting terms, pulled back to $R$, are the nearby and
vanishing cycles of $P_m$ over $R$. Let $d$ be a common bound for the relative dimensions of the
models of $Y\to\mathbb A^1_S$ and of $Z\to S$. By base change the
stalks of the nearby cycles in the triangle are the cohomology
complexes of Milnor fibres. If $P^+$ has Tor-amplitude in $[a,b]$,
these stalks have Tor-amplitude in $[a,b+d]$ and commute with
change of coefficients
\cite[Lemma~1.6(1) and Corollary~1.10(1)]{SaitoCC}. The cones and
shifts in the triangle change the bounds by integers which do not
depend on $m$. The same
formula identifies the reduction of each term for $m+1$ with the
term for $m$, compatibly with the maps, which are the canonical
adjunction, specialization and cone maps. Different models give
the same maps, by passing to a common refinement. We can therefore
take the derived limit over $m$ on the original base and invert
$\ell$. By the uniform bound the terms of the limit are the nearby
cycles of $P^+$ and, after inverting $\ell$, of $P$; compare
\cite[Corollary~1.10(1)]{SaitoCC}. This gives the
triangle for $P$.

\step{4}{naturality} The maps in
\eqref{fl:eq:sign-comparison-triangle} are composites of the
canonical transformations of the six operations, and these are
defined for rational coefficients without reference to a lattice.
The lattice was used only to prove that the cones are as stated.
The naturality in $P$ follows; in particular, no automorphism of
$P$ is required to preserve a lattice.
\end{proof}

\begin{theorem}[The minus part is the same at both ends]
\label{fl:thm:scaling-character-conservation}
In Setup~\ref{rw:setup:scaling}, suppose that $P$ is ULA over $S$
and that $P|_{Y-x_*(S)}$ is ULA over $\mathbb A^1_S$. Then for a
specialization $s\leftarrow\eta$ of geometric points of $S$, and
choices as in Lemma~\ref{rw:lem:valuation-successive-cycles}, the
map of \eqref{fl:eq:sign-comparison-triangle} restricts to an
isomorphism
\[
 \mu_s^-(P)\simeq\mu_\eta^-(P),
\]
natural in $P$ in the sense of that lemma.
\end{theorem}
\begin{proof}
The triangle \eqref{fl:eq:sign-comparison-triangle} is compatible
with the action of $\epsilon$ on its three terms, since $\epsilon$
is an automorphism of the data $(Y,f,x_*,P)$. It therefore suffices
to show that $\epsilon$ acts trivially on the cohomology of
$\mathcal D_P$: the image of the projector $(1-\epsilon)/2$ on
$\mathcal D_P$ is then a direct summand with zero cohomology, hence
zero, and applying the projector to the triangle gives the
isomorphism.

The subscheme $Z=f^{-1}(0)$ is stable under $\mathbb G_m$, and
$P|_{Z_\eta}$ is strongly equivariant. Its extension
$Rj_{Z,*}(P|_{Z_\eta})$ is ULA over $S$ by \cite[Theorem~4.1]{HS},
and the restriction of this extension to $Z_s$ is
$R\Psi_{Z/S}(P|_{Z_\eta})$. The action and projection maps
$\mathbb G_m\times Z\to Z$ are smooth, so the pullbacks of the
extension along them are ULA over $S$; they are identified on the
generic fibre by the equivariant structure, and the full
faithfulness of \cite[Theorem~4.1]{HS} extends the identification,
with its unit and cocycle conditions, over $S$. Hence
$R\Psi_{Z/S}(P|_{Z_\eta})$ is strongly equivariant, compatibly with
the map from $P|_{Z_s}$, and $\mathbb G_m$ acts on $\mathcal D_P$.

For a strongly equivariant complex and a fixed point, the action
of $\mathbb G_m$ on each cohomology group of the stalk is locally
constant on $\mathbb G_m$ and is the identity at the unit. As
$\mathbb G_m$ is connected, every element acts trivially; in
particular $\epsilon$ does.
\end{proof}

When the data carry a Weil structure that commutes with the
scaling action, Frobenius commutes with $\epsilon$ and preserves
$\mu_s^-(P)$.

The comparison is useful only if the generic microstalk is not
killed by the projector. The next lemma arranges this.

\begin{lemma}[Changing the sign]\label{fl:lem:sign-square}
Suppose that the generic fibre has characteristic zero and that
$\epsilon$ acts on the functor $\mu_\eta$ by the scalar $+1$.
Replace
\[
 (Y,f,P)\quad\text{by}\quad
 (Y\times\mathbb A^1_y,\ f+y^2,\ \operatorname{pr}_Y^*P[1]),
\]
where $\mathbb G_m$ acts on $y$ with weight one. Then the two ULA
conditions of Theorem~\ref{fl:thm:scaling-character-conservation}
still hold, $\mu_\eta$ is tensored with a fixed graded line, and
$\epsilon$ acts on it by $-1$. If $f$ is smooth, so is $f+y^2$.
\end{lemma}
\begin{proof}
The pullback of $P$ is ULA over $S$ as a smooth pullback. For the
condition along the new function, cover the complement of
$(x_*,0)$ by two open sets. Where $y\ne0$, the map
$(w,y)\mapsto(w,f(w)+y^2)$ to $Y\times\mathbb A^1$ is \'etale, as
$2$ is invertible, and the pullback of $P$ to $Y\times\mathbb A^1$
is ULA over the second factor. Where
$w\ne x_*$, the pullback of $P$ is ULA over $\mathbb A^2$ with
respect to $(f,y)$ by hypothesis, and the map
$(t,y)\mapsto t+y^2$ is smooth.

On the generic fibre, the Thom--Sebastiani theorem
\cite[Theorem~4.5 and Remark~4.11(a)]{IllusieTS2016}
identifies the vanishing cycles of $f+y^2$, with the normalization
given by the shift, with the local convolution of those of $f$ and
those of $y^2$. The generic fibre lies over an algebraically closed
field of characteristic zero. There the monodromy is tame, and
local convolution is isomorphic to the tensor product, as a functor
of the two complexes, once a compatible system of trivializations
of the convolution kernels has been chosen
\cite[Corollaries~5.25 and~5.26, Remark~5.27]{IllusieTS2016}. We
choose one. Both functors are defined by kernels, and the
isomorphism is an isomorphism between the kernels
\cite[Proposition~5.18 and Corollary~5.25]{IllusieTS2016}; so it
applies to complexes as well as to sheaves. The resulting
isomorphism is natural in $P$, which is
what the next steps use; its compatibility with Frobenius is not
needed, because Frobenius is brought in afterwards
(Proposition~\ref{fl:prop:composite-trace-nonzero}, step~3).
The nearby fibre of $y^2$ consists of two points,
its reduced cohomology is a line, and $y\mapsto-y$ exchanges the
points and acts on the line by $-1$. This changes the sign of the
action of $\epsilon$. Finally $df$ is a component of $d(f+y^2)$, so
smoothness is preserved.
\end{proof}

\subsection{From vanishing cycles to a rational point}
\label{fl:sec:trace-detector}

We now work over $\mathbb F_q$, and write $\operatorname{Fr}_q$ for
the $q$-power map on points and $\operatorname{Frob}_q$ for
geometric Frobenius on cohomology. The theorem of this subsection
must produce a point over $\mathbb F_q$ itself; a point over an
extension would give an automorphic function over the wrong field.

The proof uses hyperbolic restriction, and at one point exchanges
it with the cohomology of an inertia group. We isolate that
exchange. It is also the source of the factor $1-q$ in the trace
computation.

\begin{lemma}[Inertia and hyperbolic restriction]
\label{rw:lem:inertia-hyperbolic-exchange}
Let $I_{\rm in}$ be the inertia group of a trait of residue characteristic
$p\ne\ell$, and let $\mathcal A$ be a bounded constructible complex
with a continuous action of $I_{\rm in}$ on a scheme of finite type over
$\overline{\mathbb F}_q$. Suppose the scheme carries an action of
$\mathbb G_m$ commuting with that of $I_{\rm in}$ on $\mathcal A$, let $x$
be a rational fixed point, and let
$\operatorname{Hyp}\mathcal A$ be the cohomology with compact
supports of the attractor of $x$, the set of points $y$ with
$\lim_{t\to0}ty=x$, with coefficients in $\mathcal A$. Then
\[
 \operatorname{Hyp}R\Gamma(I_{\rm in},\mathcal A)
       \simeq R\Gamma(I_{\rm in},\operatorname{Hyp}\mathcal A),
\]
compatibly with geometric Frobenius and with any further symmetry
of finite order. If $I_{\rm in}$ acts trivially on the cohomology of
$\operatorname{Hyp}\mathcal A$, then
\begin{equation}\label{rw:eq:inertia-trace-factor}
 \operatorname{Str}\bigl(\operatorname{Frob}_q;
       R\Gamma(I_{\rm in},\operatorname{Hyp}\mathcal A)\bigr)
  =(1-q)\operatorname{Str}\bigl(\operatorname{Frob}_q;
       \operatorname{Hyp}\mathcal A\bigr).
\end{equation}
Both statements hold for the eigenspaces of such a symmetry.
\end{lemma}
\begin{proof}
\step{1}{the exchange} Work first with torsion coefficients. The
action of wild inertia factors, on a finite stratification, through
a finite $p$-group. Since $p$ is invertible in the coefficients,
its invariants are the image of the averaging idempotent, and there
is no higher cohomology. The part of tame inertia of order prime to
$\ell$ is treated in the same way. These idempotents commute with
pullback and with pushforward with compact supports. The remaining
quotient of $I_{\rm in}$ is $\mathbb Z_\ell(1)$. For a topological generator
$\tau$, its continuous cohomology is computed by the complex
\[
            [\,\mathcal B\xrightarrow{\tau-1}\mathcal B\,]
\]
in degrees $0$ and $1$, where $\mathcal B$ is the invariant part
just described. Hyperbolic restriction is a pullback followed by a
pushforward with compact supports, so it commutes with this cone.
This gives the comparison for torsion coefficients. For integral
and rational coefficients, use the compatible system of torsion
complexes: the two geometric operations have finite cohomological
dimension and the complex above has length one, so the bounds are
uniform and the derived limits give the comparison. A different
choice of $\tau$ changes the complex but not the functor it
computes, so the isomorphism is compatible with the action of
Frobenius on $I_{\rm in}$ by conjugation and with commuting symmetries.

\step{2}{the trace} There is a spectral sequence, equivariant for
Frobenius,
\[
 H^a\bigl(I_{\rm in},H^b(\operatorname{Hyp}\mathcal A)\bigr)
       \Longrightarrow H^{a+b}\bigl(R\Gamma(I_{\rm in},
                                  \operatorname{Hyp}\mathcal A)\bigr).
\]
If $I_{\rm in}$ acts trivially on $H^b(\operatorname{Hyp}\mathcal A)$, its
only nonzero columns are $H^b(\operatorname{Hyp}\mathcal A)$ for
$a=0$ and $H^b(\operatorname{Hyp}\mathcal A)(-1)$ for $a=1$.
Geometric Frobenius acts on $k(-1)$ by $q$, and the alternating
trace of the abutment equals that of the $E_2$ page, which is
$(1-q)$ times the alternating trace on
$\operatorname{Hyp}\mathcal A$. Passing to an eigenspace of a
commuting symmetry is another idempotent, to which the same
argument applies.
\end{proof}

\begin{theorem}[A nonzero microstalk trace gives a rational point]
\label{fl:thm:homogeneous-detector}
Let $p\ne2$. Let $Y/\mathbb F_q$ be smooth and affine with an
action of $\mathbb G_m$, let $f:Y\to\mathbb A^1$ be smooth of
weight two, and let $x_*\in Y(\mathbb F_q)$ be a fixed point. Let
$P$ be a bounded constructible Weil complex on $Y$, strongly
equivariant compatibly with its Weil structure, such that the
vanishing cycles of $f$ on $f^{-1}(0)$ are supported at $x_*$. If
\[
 \operatorname{Str}(\operatorname{Frob}_q;\mu^-(P))\ne0,
\]
then there is $x\in Y(\mathbb F_q)$ at which the stalk of $P$ has
nonzero Frobenius supertrace.
\end{theorem}

The action of Frobenius on vanishing cycles depends on a lift of
Frobenius to the Galois group of the generic point of the
henselization of $\mathbb A^1$ at the origin, and two lifts differ
by an element of the inertia group. In the theorem, and wherever
$\operatorname{Frob}_q$ acts on $\mu^-$, the lift is taken in the
subgroup of index two attached to the covering $t=u^2$. Step~3 of
the proof shows that the inertia group of that covering acts
trivially on the cohomology of $\mu^-(P)$; so the supertrace does
not depend on the lift within this subgroup.

The role of the scaling action can be seen on points. In the
argument below there appear points $x$ of $Y$ that are fixed not by
$\operatorname{Fr}_q$ but by $\epsilon\operatorname{Fr}_q$. Choose
$a\in\overline{\mathbb F}_q^\times$ with $a^{q-1}=-1$. Then
\[
 \operatorname{Fr}_q(ax)=a^q\operatorname{Fr}_q(x)
 =(-a)(\epsilon x)=ax,
\]
so $ax$ is a rational point, and the equivariant structure of $P$,
being compatible with the Weil structure, identifies the trace of
$\epsilon\operatorname{Frob}_q$ on the stalk at $x$ with the trace
of $\operatorname{Frob}_q$ on the stalk at $ax$. Twisted fixed
points thus contribute values of the trace function at rational
points, and no extension of $\mathbb F_q$ is needed.

\begin{proof}
Steps~1 to~4 compute one hyperbolic restriction in two ways. One
gives the vanishing cycles at $x_*$, the other the cohomology with
compact supports of the part of a fibre of $f$ which the action
contracts to $x_*$, and the outcome is the equality
of supertraces \eqref{fl:eq:detector-finite-trace}. This is the
content of the proof. Step~5 is then the trace formula for the
fibre.

\step{1}{a function of weight one} Make the base change $t=u^2$:
put $Y'=Y\times_{\mathbb A^1_t}\mathbb A^1_u$ and
$P'=\operatorname{pr}_Y^*P$. For the action $c(x,u)=(cx,cu)$, the
function $u$ on $Y'$ has weight one. The involution
$\epsilon'(x,u)=(\epsilon x,u)$ preserves $u$. Since $f$ is smooth,
$Y'$ is smooth. The nearby cycles of $u$ for $P'$, as a complex
with Weil structure, are those of $f$ for $P$; only the inertia
group is replaced by its subgroup of index two
\cite[\S1.1(b)]{Illusie}.

\step{2}{hyperbolic restriction} Let $Z=(u=0)$, with inclusion
$i:Z\to Y'$ and complement $j:Y'-Z\to Y'$, and let $\mathsf F$ be
the fixed locus of $\mathbb G_m$ in $Y'$. It lies in $Z$, because
$u$ has weight one. Hyperbolic restriction takes values in complexes on $\mathsf F$;
the point $x_*$ need not be isolated in $\mathsf F$, and stalks
at $x_*$ are taken at the end. For $X=Y'$ or $X=Z$, let $X^+$ and $X^-$ be the
attractor and the repeller, with their inclusions
$p^\pm:X^\pm\to X$ and the limit maps $q^\pm:X^\pm\to\mathsf F$,
and put
\[
 \operatorname{Hyp}^+_X=q^+_!(p^+)^*,\qquad
 \operatorname{Hyp}^-_X=q^-_*(p^-)^! .
\]
Braden's theorem gives an isomorphism
$\operatorname{Hyp}^+_XK\simeq\operatorname{Hyp}^-_XK$ of complexes
on $\mathsf F$ for every equivariant complex $K$ on $X$
\cite[Theorem~1]{Braden}\cite[Theorem~2.6]{Richarz}. Because $u$
has positive weight, the repeller of $Y'$ lies in $Z$, so that
$Z^-=(Y')^-$ and $p^-_{Y'}=i\circ p^-_Z$. For an equivariant
complex $Q$ on $Y'-Z$ we claim
\begin{equation}\label{fl:eq:braden-recollement}
 \operatorname{Hyp}^+_Z i^*Rj_*Q
 \simeq\operatorname{Hyp}^+_{Y'}j_!Q[1]
\end{equation}
as complexes on $\mathsf F$. Indeed
\[
 \operatorname{Hyp}^+_Zi^*Rj_*Q
 \simeq\operatorname{Hyp}^-_Zi^!j_!Q[1]
 =\operatorname{Hyp}^-_{Y'}j_!Q[1]
 \simeq\operatorname{Hyp}^+_{Y'}j_!Q[1] .
\]
The first isomorphism is Braden's theorem on $Z$ for the
equivariant complex $i^*Rj_*Q\simeq i^!j_!Q[1]$, the equality holds
because $(p^-_{Y'})^!=(p^-_Z)^!i^!$, and the last isomorphism is
Braden's theorem on $Y'$ for $j_!Q$. Both complexes are
equivariant.

We now take the stalk at $x_*$ of both sides of
\eqref{fl:eq:braden-recollement}. By base change for the direct
image with compact supports, the stalk
at $x_*$ of $\operatorname{Hyp}^+_XK$ is the cohomology with
compact supports, with coefficients in $K$, of the fibre of $q^+$
over $x_*$, which is the attractor of $x_*$ in $X$. We write
$\operatorname{Hyp}_X$ for this functor; it is the one of
Lemma~\ref{rw:lem:inertia-hyperbolic-exchange}. All maps commute with Frobenius
and with $\epsilon'$. We apply the resulting isomorphism
$\operatorname{Hyp}_Zi^*Rj_*Q\simeq\operatorname{Hyp}_{Y'}j_!Q[1]$ to
$Q=P'|_{Y'-Z}$, take the part on which $\epsilon'$ acts by $-1$,
and compute the Frobenius supertrace of both sides.

\step{3}{the left side} We have
$i^*Rj_*Q=R\Gamma(I_{\rm in},R\Psi_uP')$, where $I_{\rm in}$ is the inertia group of
$\mathbb A^1_u$ at $0$. By
Lemma~\ref{rw:lem:inertia-hyperbolic-exchange} the left side of
\eqref{fl:eq:braden-recollement} is
$R\Gamma(I_{\rm in},\operatorname{Hyp}_ZR\Psi_uP')$.

We first identify the minus part of
$\operatorname{Hyp}_ZR\Psi_uP'$. On $Z$ the involution $\epsilon'$
is the action of $-1\in\mathbb G_m$; hence it is trivial on the
cohomology of $\operatorname{Hyp}_Zi^*P'$. Indeed, the action by
scaling is a family of automorphisms of the attractor,
parametrized by $\mathbb G_m$, together with an isomorphism of the
complex of coefficients. The direct image with compact supports
along the projection to $\mathbb G_m$ is a constant complex with
an endomorphism, which is the identity at the unit element, and
therefore on the cohomology at every point of the connected group
$\mathbb G_m$. The triangle relating
$i^*P'$, nearby cycles and vanishing cycles therefore gives
\[
 (\operatorname{Hyp}_ZR\Psi_uP')^-
 \simeq(\operatorname{Hyp}_ZR\Phi_uP')^-=\mu^-(P)[1];
\]
the last equality holds because the vanishing cycles are supported
at $x_*$, and the attractor of $x_*$ meets this support in the
point $x_*$ alone; so here hyperbolic restriction is the stalk.

Next, $I_{\rm in}$ acts trivially on the cohomology of
$(R\Psi_uP')_{x_*}$. To see this, consider on
$Y'\times\mathbb G_m$ the function $\lambda u$ and the pullback
$P'^\lambda$ of $P'$ from $Y'$. Verdier
\cite[\S7, Proposition~7.1(a)]{Verdier1983} proves that the nearby
cycles $R\Psi_{\lambda u}(P'^\lambda)$ form a monodromic complex in
the direction of $\mathbb G_m$, that wild inertia acts trivially on
them, and that tame inertia acts by the inverse of the canonical
monodromy of this complex. On the other hand the automorphism
$\varphi(y,\lambda)=(\lambda y,\lambda)$ of
$Y'\times\mathbb G_m$ carries the function $u$, pulled back from
$Y'$, to $\lambda u$, and by strong equivariance it carries the
pullback of $P'$ to $P'^\lambda$. Hence
\[
 R\Psi_{\lambda u}(P'^\lambda)\simeq a^*R\Psi_u(P'),
\]
compatibly with the action of $I_{\rm in}$, where $a$ is the action map. In
particular wild inertia acts trivially on $R\Psi_u(P')$. Along
$\{x_*\}\times\mathbb G_m$ the right side is constant, because
$x_*$ is a fixed point, so its canonical monodromy is trivial there,
and tame inertia acts trivially on the cohomology of the stalk at
$x_*$. The isomorphisms of the
preceding paragraph are compatible with $I_{\rm in}$, and on the minus part
the stalk of the vanishing cycles at $x_*$ is identified with that
of the nearby cycles. So $I_{\rm in}$ acts trivially on the cohomology of
$(\operatorname{Hyp}_ZR\Psi_uP')^-$.

Formula~\eqref{rw:eq:inertia-trace-factor}, with the shift by one,
now shows that the supertrace of Frobenius on the minus part of the
left side of \eqref{fl:eq:braden-recollement} is
\[
 -(1-q)\operatorname{Str}(\operatorname{Frob}_q;\mu^-(P)).
\]

\step{4}{the right side} The attractor of $x_*$ in $Y'-Z$ is, by
homogeneity, the product of $\mathbb G_m$ and its fibre over
$u=1$. Let $C_1$ be the cohomology with compact supports of that
fibre with coefficients in $P'$. The involution $\epsilon'$ acts
trivially on the factor $\mathbb G_m$, whose cohomology with
compact supports has Frobenius supertrace $q-1$. With the shift,
the supertrace on the minus part of the right side is
$-(q-1)\operatorname{Str}(\operatorname{Frob}_q;C_1^-)$. Comparing
with step 3 and dividing by $1-q$,
\begin{equation}\label{fl:eq:detector-finite-trace}
 \operatorname{Str}(\operatorname{Frob}_q;\mu^-(P))
 =-\operatorname{Str}(\operatorname{Frob}_q;C_1^-).
\end{equation}

\step{5}{conclusion} Suppose that the stalk of $P$ has zero
supertrace at every point of $Y(\mathbb F_q)$. By the trace formula
for cohomology with compact supports,
$\operatorname{Str}(\operatorname{Frob}_q;C_1)$ is a sum of such
stalk traces and vanishes. The same formula applied to
$\epsilon'\operatorname{Frob}_q$, that is, to the quadratic twist
of the fibre over $u=1$, expresses
$\operatorname{Str}(\epsilon'\operatorname{Frob}_q;C_1)$ as a sum
over points fixed by $\epsilon\operatorname{Fr}_q$, and by the
remark preceding the proof each term is a stalk trace at a rational
point. So this vanishes too. Half the difference of the two is the
supertrace on $C_1^-$, which contradicts
\eqref{fl:eq:detector-finite-trace}.
\end{proof}

\begin{remark}[The two factors]
The factors $1-q$ in step 3 and $q-1$ in step 4 have different
origins: the first comes from the cohomology of inertia, the second
from the cohomology with compact supports of a scaling orbit. Their
quotient is the sign in \eqref{fl:eq:detector-finite-trace}. The
argument needs the part on which $\epsilon$ acts by $-1$: on the
invariant part the contribution of $i^*P'$ in step 3 does not drop
out.
\end{remark}

\subsection{Nonvanishing for the automorphic family}
\label{fl:sec:morse-geometry}

\begin{proposition}[A rational bundle with nonzero trace]
\label{fl:prop:composite-trace-nonzero}
Let $P\mapsto\mathcal F_P$ be the family over $S$ of
Theorem~\ref{fl:thm:compact-automorphic-family}. The complex
$\mathcal F_{P_{\rm reg},s}$ on the special fibre has nonzero
Frobenius supertrace at some bundle on the curve
$C_0=X\cup_DC_2$ defined over $\mathbb F_q$.
\end{proposition}
\begin{proof}
If $G$ is trivial, the stack of bundles of each fibre is a point,
and the complex attached to $P_{\rm reg}$ is a graded line, on
which Frobenius is invertible.
Otherwise $p\nmid|W_G|$ implies $p\ne2$.

\step{1}{the hypotheses of the comparison} Pull the family back to
the chart $Y$ of Lemma~\ref{fl:lem:graded-morse-chart}, with the
shift by the relative dimension. The pullbacks are strongly
equivariant for the scaling action, which is induced by a change of
trivialization and does not change the underlying bundle. They are
ULA over $S$, and by
Corollary~\ref{fl:cor:phase-universal-local-acyclicity} they are
ULA over $\mathbb A^1_S$ away from $x_*$. The chart and the test
function depend only on the nilpotent cone, so this holds for every
perfect coefficient $P$ on the same chart.

\step{2}{the sign} By
Proposition~\ref{fl:lem:vacuum-microstalk-comparison}, the functor
$\mu_\eta$ on $\operatorname{Perf}(\mathfrak F_\rho)$ is
$\Gamma(\mathfrak F_\rho,-)\otimes L_{\rm shift}$. The involution
$\epsilon$ is a natural automorphism of this functor, hence a
scalar by Lemma~\ref{fl:lem:scalar-coefficient-covariance}, and the
scalar is $1$ or $-1$. If it is $1$, apply
Lemma~\ref{fl:lem:sign-square} to the whole family. In either case
$\mu_\eta=\mu_\eta^-$, and
Theorem~\ref{fl:thm:scaling-character-conservation}, applied along
the flag of $S$, gives an isomorphism of functors
\begin{equation}\label{fl:eq:puncture-special-microfunctor}
 \mu_s^-\simeq\mu_\eta
 \simeq\Gamma(\mathfrak F_\rho,-)\otimes L_{\rm shift}.
\end{equation}

\step{3}{Frobenius} Both sides of
\eqref{fl:eq:puncture-special-microfunctor} carry a Frobenius
covariance with respect to the same autoequivalence of
$\operatorname{Perf}(\mathfrak F_\rho)$: the left side from the
Weil structure of the family on the special fibre, with Frobenius
acting on vanishing cycles through a lift chosen as explained after
Theorem~\ref{fl:thm:homogeneous-detector}, the right side
from the action of Frobenius on $\mathfrak F_\rho$
(Lemma~\ref{fl:lem:coarse-covariance}). By
Lemma~\ref{fl:lem:scalar-coefficient-covariance} they differ by one
nonzero scalar. Evaluated at $P_{\rm reg}$, this identifies
geometric Frobenius on $\mu_s^-(\mathcal F_{P_{\rm reg}})$, up to
that scalar and the action on the line $L_{\rm shift}$, with the
inverse of arithmetic Frobenius on
$\Gamma(\mathfrak F_\rho,P_{\rm reg})$, including the inverse of
the permutation of the basis of $k[\Pi_\rho]$. Its supertrace is
not zero, by Proposition~\ref{fl:prop:puncture-exact-trace}, which
covers the inverse of Frobenius.

\step{4}{a rational point} The finitely many objects and maps
involving $P_{\rm reg}$ are defined over a finite extension of
$\mathbb Q_\ell$. On the special fibre of the chart, which is
defined over $\mathbb F_q$, the vanishing cycles of $f$ are
supported on $x_*$ by step 1. So
Theorem~\ref{fl:thm:homogeneous-detector} applies and gives an
$\mathbb F_q$-rational point of the chart at which the stalk of the
pullback has nonzero supertrace. The pullback from the stack of
bundles to the chart is smooth, and it changes stalk traces by a
sign and a power of $q$. The image of the point is the required
bundle.
\end{proof}