\section{Proof of the main theorem}\label{sec:16-main-proof}

\begin{theorem}[Uniform high-genus character]\label{thm:uniform-high-genus}
Use the series of Definition~\ref{def:common-series}.
For $g\geq4$, let $t_g=\lfloor(g-2)/2\rfloor$ and let $\epsilon_g$ be $1$
for odd $g$ and $0$ for even $g$. Then
\[
B_g=E\frac{\partial U}{\partial p_1}+H(3E+(g-1)E^2)
 +t_g(H+G)E^2+\epsilon_g(Hh_2[E]+Ge_2[E]).
\]
The degree-$n$ part is $\ch(\En_{g,n})$,
which is nonzero exactly for $n\geq9$, with the stated realizations in cohomology.
The full ordinary Gysin map is surjective and
$W_{15}H^{15}(\Mopen_{g,n})=0$ for all $g\geq4$ and $n\geq0$.
\end{theorem}

\begin{proof}
Theorem~\ref{thm:genus-four-base} and
Corollary~\ref{thm:genus-five-base} give surjectivity of the complete ordinary
one-edge Gysin map and vanishing of the open weight piece in those two
genera.

\emph{Induction in the genus.}
Assume $g\geq6$ and that the complete ordinary map is surjective in genus
$g-1$ for every finite marking set.  Let $F$ be finite, choose new flags
$c,d$, and let
\[
\xi:\Mbar_{g-1,F\cup\{c,d\}}\longrightarrow\Mbar_{g,F}
\]
be the nonseparating clutching map.  For $z\in H^{15}(\Mbar_{g,F})$ put
$x=\xi^*z$.  Then $x$ is fixed by $t=(c\,d)$ and, by the induction
hypothesis, belongs to
$\mathcal I_{g-1,F\cup\{c,d\}}^{\langle t\rangle}$.

Since $\beta\alpha(z)=0$ (Proposition~\ref{prop:compact-restriction-row})
and the remaining coordinates of $\alpha(z)$ lie in $C^1_{\rm sep}$,
$\delta(x)=[\beta(0,x)]=0$ in $\mathscr Q_{g,F}$.
Theorem~\ref{thm:ordinary-lifting} therefore gives $y\in\mathcal I_{g,F}$
with $\xi^*y=x$.  Theorem~\ref{thm:compact-injectivity} implies $y=z$.
Thus $\mathcal I_{g,F}=H^{15}(\Mbar_{g,F})$, completing the induction.

Proposition~\ref{prop:primitive-degree-fifteen-quotient}
identifies the cokernel of the complete ordinary map with
$W_{15}H^{15}(\Mopen_{g,F})$.  Hence the established surjectivity gives
the asserted vanishing for every $g\geq4$.

\emph{Character and realizations.}
Adding the four characteristic series displayed in
Theorem~\ref{thm:ordinary-image} gives the displayed formula for $B_g$.
Since the ordinary
image is the full compact cohomology group, Proposition~\ref{prop:h13-input} gives
\[
H^{15}(\Mbar_{g,n})\cong\En_{g,n}\otimes\Pmot
\]
as pure Hodge structures and, for every prime $\ell$,
\[
H^{15}_{\mathrm{et}}(\Mbar_{g,n,\overline{\mathbb Q}};\mathbb Q_\ell)
\cong(\En_{g,n}\otimes_{\mathbb Q}\mathbb Q_\ell)\otimes\Pmot_\ell,
\]
as a $\Sn_n\times\Gal(\overline{\mathbb Q}/\mathbb Q)$-representation before semisimplification.
Here $\Pmot_\ell$ denotes the $\ell$-adic realization of
$\Pmot$ supplied by that proposition.

It remains to check the first nonzero arity.  The series $H$ starts in
degree ten, $G$ in degree nine, and $U$ in degree twelve, so
$E\partial U$ starts in degree eleven.  Thus $B_g$ has no term of degree
less than nine.  In degree nine, $G$ contributes $e_9$, while the constant
term of $h_2[E]$ is one and that of $e_2[E]$ is zero; consequently the
degree-nine part is $t_g e_9\neq0$.  For $n\geq9$, the product
$t_g e_9h_{n-9}$ is a nonzero Schur-positive summand.  Every other term is
Schur-positive (the derivative and products encode restriction and
induction, and $h_2[E],e_2[E]$ encode symmetric and exterior squares), so
there can be no cancellation.
\end{proof}

\begin{theorem}[Degree-fifteen determination]\label{thm:main}
This is Theorem~\ref{thm:introduction-degree-fifteen}.
For every stable $(g,n)$, there are $\Sn_n$-equivariant isomorphisms
of pure Hodge structures and compatible $\ell$-adic realizations before semisimplification
\[
H^{15}(\Mbar_{g,n})\cong
\bigoplus_{\lambda\vdash n}V_\lambda\otimes\Pmot^{\oplus [s_\lambda]F_g}
\;\oplus\;\delta_{g,1}\Khook_n\otimes\Sonesix,
\]
where $F_0=0$ and $F_1,F_2,F_3,F_g$ for $g\geq4$ are specified in
Theorems~\ref{thm:genus-one}, \ref{thm:genus-two},
\ref{thm:genus-three}, and~\ref{thm:uniform-high-genus}, respectively.
The cohomology group is nonzero precisely in the following cases:
\[
g=0:\ \varnothing;\quad g=1:\ n\geq13;\quad g=2:\ n\geq11;\quad
g=3:\ n\geq10;\quad g\geq4:\ n\geq9.
\]
\end{theorem}

\begin{proof}
Theorem~\ref{thm:motivic-types} reduces every stable arity to the
$\Pmot$-multiplicity module $\En_{g,n}$ and, only in genus one, the
primitive summand $\Khook_n\otimes\Sonesix$; in genus zero it gives
$F_0=0$.  The four theorems named in the statement give
$\ch_n(\En_{g,n})=[F_g]_n$, so $[s_\lambda]F_g$ is the $V_\lambda$
multiplicity of $\Pmot$, and their nonzero thresholds $n\geq13$, $11$,
$10$, $9$ make the five displayed cases exhaustive.  The same four
theorems give the realizations before semisimplification; for
$g\geq4$ this is the $\ell$-adic splitting displayed in the proof of
Theorem~\ref{thm:uniform-high-genus}.
\end{proof}

\begin{corollary}[Hodge numbers]\label{cor:expert-questions}
Let $(g,n)$ be stable and let $e_{g,n}=\dim\En_{g,n}$. The only possibly nonzero
Hodge numbers in degree fifteen are
\[
h^{13,2}=h^{2,13}=e_{g,n},\qquad
h^{15,0}=h^{0,15}=\delta_{g,1}\binom{n-1}{14},
\]
where the binomial coefficient is zero for $n<15$. In particular,
$\dim H^{15}(\Mbar_{g,n})=2e_{g,n}+2\delta_{g,1}\binom{n-1}{14}$.
\end{corollary}

\begin{proof}
The motivic pieces have Hodge types $(13,2)$ and $(2,13)$ for $\Pmot$,
and $(15,0)$ and $(0,15)$ for $\Sonesix$, by
Notation~\ref{def:motivic-pieces}.  The $\Pmot$-multiplicity space has
dimension $e_{g,n}$, while the only possible $\Sonesix$ summand is
$\Khook_n=V_{(n-14,1^{14})}$.  The hook-length formula gives
\[
\dim V_{(n-14,1^{14})}=\binom{n-1}{14}
\]
for $n\geq15$, and it is zero otherwise.  The Hodge numbers follow.
\end{proof}