\section{The genus-three rational Gram formula}\label{sec:06-genus-three-formula}

We compute the quotient of the finite-dimensional source $A_n$ that gives
$H^{15}(\Mbar_{3,n})$. Its intersection matrix is determined by
recursions for descendant integrals and integrals containing two cusp classes.
We first evaluate these entries, then identify the radical and compute
the quotient character.

\begin{definition}\label{def:genus-three-polynomial-pullback}
Write $\mathscr P_L=\mathbb Q[\kappa_1,(\psi_i)_{i\in L}]$.
For a partition $L=A\sqcup B$ and distinct new node flags $p,q$, let
\[
 \Delta_{A|B}:\mathscr P_L\longrightarrow
 \mathscr P_{A\cup\{p\}}\otimes\mathscr P_{B\cup\{q\}}
\]
send $\psi_i$ to the same variable on the factor containing $i$ and send
$\kappa_1$ to $\kappa_1\otimes1+1\otimes\kappa_1$. This is the
pullback rule of Theorem~\ref{thm:genus-three-excess-cited} for a
one-edge separating graph, and the same retained-flag rule and sum over
the inverse vertices there define the polynomial pullback for a general
graph contraction. A hook decoration is restricted by iterating $r_T$
of Definition~\ref{def:rational-tail-hook-map}
(Lemma~\ref{lem:cited-eleven-hook-pullback}).
\end{definition}

\begin{definition}\label{def:genus-three-descendants}
For $h\ge0$, $m\ge1$, and integer exponents $d_i$, define
$T_h(d_1,\ldots,d_m)=0$ unless $2h-2+m>0$, every $d_i\ge0$, and
$\sum_i d_i=3h-3+m$. Negative-genus and unstable expressions are zero.
Set
\[
 T_0(0,0,0)=1,\qquad T_1(1)=\frac1{24},\qquad (-1)!!=1.
\]
These base cases take precedence. For other dimension-compatible stable
inputs with a zero exponent, use symmetry and the string recursion
\begin{equation}\label{eq:genus-three-string}
 T_h(0,d_2,\ldots,d_m)
 =\sum_{j=2}^mT_h(d_j-1,(d_i)_{i\ne1,j}).
\end{equation}
If all exponents are positive, use
\begin{equation}\label{eq:genus-three-dvv}
\begin{split}
 &(2d_1+1)!!T_h(d_1,\ldots,d_m)\\
 &\quad=\sum_{j=2}^m
 \frac{(2d_1+2d_j-1)!!}{(2d_j-1)!!}
 T_h(d_1+d_j-1,(d_i)_{i\ne1,j})\\
 &\qquad+\frac12\sum_{\substack{a,b\ge0\\a+b=d_1-2}}
 (2a+1)!!(2b+1)!!\left(
 T_{h-1}(a,b,d_2,\ldots,d_m)\right.\\
 &\hspace{105pt}\left.+
 \sum_{\substack{h_1,h_2\ge0, h_1+h_2=h\\
 I\sqcup J=\{2,\ldots,m\}}}^{\mathrm{stable}}
 T_{h_1}(a,(d_i)_{i\in I})T_{h_2}(b,(d_j)_{j\in J})\right).
\end{split}
\end{equation}
The inner sum runs over ordered partitions and only stable factors.
Each nonzero descendant term on the right-hand side has smaller $2h-2+m$.

For $a\ge0$ and a stable marked pair $(h,m)$, define
\begin{equation}\label{eq:genus-three-kappa-descendants}
 W_h(a;d_1,\ldots,d_m)=
 \sum_{k=0}^a\frac{(-1)^{a-k}}{k!}
 \sum_{\substack{a_1+\cdots+a_k=a\\a_i\ge1}}
 \frac{a!}{a_1!\cdots a_k!}
 T_h(d_1,\ldots,d_m,a_1+1,\ldots,a_k+1).
\end{equation}
There is one empty composition when $a=k=0$ and none when $a>0,k=0$.
Extend $W_h$ linearly to $\mathscr P_{[m]}$ by evaluating the monomial
$\kappa_1^a\prod_i\psi_i^{d_i}$ with this expression.
\end{definition}

\begin{proposition}[Descendant integral identities
{\cite[Theorem 1.2, Proposition 3.1, Lemma 2.2]{liu2009}}]
\label{prop:genus-three-descendants-cited}
For every stable pair $(h,m)$ with $m\ge1$ and nonnegative integers
$a,d_1,\ldots,d_m$, the recursions
\eqref{eq:genus-three-string}--\eqref{eq:genus-three-kappa-descendants}
evaluate the ordinary stack integrals
\[
 T_h(d_1,\ldots,d_m)=\int_{\Mbar_{h,m}}\prod_i\psi_i^{d_i},\qquad
 W_h(a;d_1,\ldots,d_m)=
 \int_{\Mbar_{h,m}}\kappa_1^a\prod_i\psi_i^{d_i}.
\]
\end{proposition}

\begin{definition}\label{def:genus-three-rational-integrals}
For $|L|\ge3$, define a linear functional $Z_L:\mathscr P_L\to\mathbb Q$,
zero outside degree $|L|-3$, with $Z_L(1)=1$ at $|L|=3$.
For larger $L$, select a factor $q$ of a monomial $qf'$.
If $q=\psi_z$, choose distinct $x,y\in L\setminus\{z\}$ and set
\[
 \mathcal A_q=\{A\subset L:z\in A, x,y\notin A,
                  |A|\ge2, |L\setminus A|\ge2\},\qquad c_q(A)=1.
\]
If $q=\kappa_1$, choose distinct $x,y\in L$ and set
\[
 \mathcal A_q=\{A\subset L:x,y\notin A,
                  |A|\ge2, |L\setminus A|\ge2\},\qquad c_q(A)=|A|-1.
\]
For each $A$ write
\(\Delta_{A|L\setminus A}f'=\sum_b f_{A,b}\otimes f_{B,b}\).
The recursion is
\[
 Z_L(qf')=\sum_{A\in\mathcal A_q}c_q(A)
 \sum_b Z_{A\cup\{p\}}(f_{A,b})
        Z_{(L\setminus A)\cup\{q'\}}(f_{B,b}).
\]
Here $p,q'$ are distinct new node flags used by $\Delta$.
Both new label sets have smaller cardinality than $L$.
\end{definition}

\begin{definition}\label{def:genus-three-cusp-recursion}
For $|I|\ge1$, define
$\mathcal C_I:K_I\times K_I\times\mathscr P_I\to\mathbb Q$,
bilinear in its first two arguments and linear in its last.
Set it to zero for $|I|<11$ and outside polynomial degree $|I|-11$.
For $|I|=11$, choose an ordering of $I$ and let $\epsilon_I:K_I\to\mathbb Q$
send its increasing eleven-symbol to one. For constant $f$ put
\[
 \mathcal C_I(k,l;f)=f\epsilon_I(k)\epsilon_I(l).
\]
For $|I|>11$, select a factor $q$ in a monomial $qf'$ and put
\[
\begin{array}{c|c|c}
q&\mathcal T_q&c_q(T)\\ \hline
\psi_i&\{T\subseteq I:i\in T, |T|\ge2\}&1\\
\kappa_1&\{T\subseteq I:|T|\ge2\}&|T|-1.
\end{array}
\]
For $A=I\setminus T$, expand
$\Delta_{A|T}f'=\sum_b f_{1,b}\otimes f_{0,b}$ with new flags $p,q'$.
Define
\begin{equation}\label{eq:genus-three-cusp-recursion}
 \mathcal C_I(k,l;qf')=
 \sum_{T\in\mathcal T_q}c_q(T)\sum_b
 \mathcal C_{A\cup\{p\}}(r_Tk,r_Tl;f_{1,b})
 Z_{T\cup\{q'\}}(f_{0,b}).
\end{equation}
The term $T=I$ is allowed: the elliptic factor has its single node
marking and is stable, and its hook is zero. Every elliptic term on the
right-hand side has $|A|+1<|I|$.
\end{definition}

\begin{proposition}[Vertex evaluation]\label{prop:genus-three-evaluation}
For every stable marked genus-zero space and every $f\in\mathscr P_L$,
\[
 Z_L(f)=\int_{\Mbar_{0,L}}f.
\]
For every $|I|\ge1$, $k,l\in K_I$, $u,v\in S$, and
$f\in\mathscr P_I$,
\[
 \int_{\Mbar_{1,I}}\omega_I(k,u)\omega_I(l,v)f
 =\mathcal C_I(k,l;f)J(u,v).
\]
The recursion values are independent of the selected factors, auxiliary
labels, and orderings. The functional $\mathcal C_I$ is symmetric in $k,l$
and equivariant under relabelling.
\end{proposition}

\begin{proof}
We identify the genus-zero recursion with integration, and then reduce
the paired-cusp integral to rational-tail divisors.

\paragraph{Genus-zero integration.}
In top degree and for $|L|>3$, a monomial has a positive-degree factor
$q$. Proposition~\ref{prop:genus-three-picard-input} gives
$q=\sum_{A\in\mathcal A_q}c_q(A)\delta_{0,A}$ with exactly the
coefficients in Definition~\ref{def:genus-three-rational-integrals}.
The auxiliary labels distinguish the complementary component, so each
divisor is counted once. Both components have labelled ordinary
markings, and the graph has automorphism group one. Its labelled
clutching pushforward of the unit is therefore $\delta_{0,A}$.
The remaining polynomial pulls back to
$\Delta_{A|L\setminus A}f'$ by
Theorem~\ref{thm:genus-three-excess-cited}. The projection formula gives
\[
 \int_{\Mbar_{0,L}}qf'
 =\sum_{A\in\mathcal A_q}c_q(A)\sum_b
 \left(\int_{\Mbar_{0,A\cup\{p\}}}f_{A,b}\right)
 \left(\int_{\Mbar_{0,(L\setminus A)\cup\{q'\}}}f_{B,b}\right).
\]
The two factors have fewer markings than $L$. Together with
$\int_{\Mbar_{0,3}}1=1$ and the degree convention, induction proves
the first assertion for every polynomial.

\paragraph{Paired-cusp integration.}
Let $x\in H^{11}(\Mbar_{1,I})$. The labelled nonseparating
normalization is $\Mbar_{0,I\cup\{h,h'\}}$, whose degree-eleven
cohomology is zero by Proposition~\ref{prop:eleven-hook-realizations}.
The projection formula, with
$\delta_{\mathrm{irr}}=\tfrac12\xi_{\mathrm{irr}*}1$, gives
$\delta_{\mathrm{irr}}x=0$. The genus-one identities in
Proposition~\ref{prop:genus-three-picard-input} give
$qx=\sum_{T\in\mathcal T_q}c_q(T)\delta_{0,T}x$ with exactly the
coefficients in Definition~\ref{def:genus-three-cusp-recursion}.
The rational side distinguishes the two vertices, so these separating
graphs have automorphism group one, including $T=I$.

For $A=I\setminus T$ let
\[
 j_T:\Mbar_{1,A\cup\{p\}}\times\Mbar_{0,T\cup\{q'\}}
       \longrightarrow\Mbar_{1,I}
\]
be the labelled gluing map. Proposition~\ref{prop:eleven-hook-realizations}
identifies its pullback of $\omega_I(k,u)$ with
$\omega_{A\cup\{p\}}(r_Tk,u)\boxtimes1$; the same applies to the
second hook. Select a factor in $f=qf'$ and apply the divisor identity
to the first hook. Projection along each $j_T$, followed by the
polynomial pullback $\Delta_{A|T}$, expresses the integral as
\[
 \sum_{T\in\mathcal T_q}c_q(T)\sum_b
 \left(\int_{\Mbar_{1,A\cup\{p\}}}
 \omega_{A\cup\{p\}}(r_Tk,u)
 \omega_{A\cup\{p\}}(r_Tl,v)f_{1,b}\right)
 \left(\int_{\Mbar_{0,T\cup\{q'\}}}f_{0,b}\right).
\]
All polynomial factors have even cohomological degree, so this formula
preserves the order of the odd classes without a sign change.
The elliptic factor has fewer than $|I|$ markings. If $|I|<11$, its
hook is zero. At $|I|=11$, the ordered identification with
$\Mbar_{1,[11]}$ gives $f\epsilon_I(k)\epsilon_I(l)J(u,v)$.
Induction and the genus-zero formula give exactly
\eqref{eq:genus-three-cusp-recursion}. Degree mismatches vanish on
both sides.

These identifications with integration prove independence of the
recursive choices. Changing the eleven-flag ordering changes both
orientation signs and preserves their product. Relabelling commutes
with pullback, the divisor identities, and integration. Interchanging
the two degree-eleven classes changes the integral by $-1$, while
$J(v,u)=-J(u,v)$. Their rational coefficient is therefore symmetric
in $k,l$.
\end{proof}

\begin{definition}\label{def:genus-three-gram}
On a graph $\Gamma$, evaluate a tensor monomial containing two restricted
hook decorations as follows. Its value is zero if a hook restriction is
zero or the hooks occur on different vertices. If both occur at $v$, put
\[
 \operatorname{Ev}_\Gamma
 \left(k_v,l_v;\bigotimes_w f_w\right)=
 \mathcal C_{F(v)}(k_v,l_v;f_v)
 \prod_{w\ne v}W_{g(w)}(f_w).
\]
Extend linearly, retaining the first hook before the second. Here $W_h(f)$
means the linear polynomial evaluation in
Definition~\ref{def:genus-three-descendants}. In every nonzero term the
other vertices have genus at most two and at least one flag.

For the basis vector $a_r$, let $\Gamma_r$ and $\alpha_r$ be its graph and
formal hook decoration from Definition~\ref{def:genus-three-source}.
For $n\ge10$ define the rational matrix $C_n$ by
\begin{equation}\label{eq:genus-three-gram-entries}
\begin{split}
 (C_n)_{rs}={}&\frac1{|\Aut(\Gamma_r)||\Aut(\Gamma_s)|}
 \sum_{[\Gamma,f,g]\in\mathcal R(\Gamma_r,\Gamma_s)}
 \operatorname{Ev}_\Gamma\left(
 f^*\alpha_r\,g^*\alpha_s\right.\\
 &\left.\hspace{25pt}\cdot
 \prod_{\{h,h'\}\in H(f,g)}(-\psi_h-\psi_{h'})
 \left(\sum_{v\in V(\Gamma)}\kappa_{1,v}\right)^{n-9}
 \right).
\end{split}
\end{equation}
The evaluation acts on the polynomial expansion of the displayed formal
decorations. There is no further $|\Aut(\Gamma)|$ factor in this sum over
isomorphism classes of specified generic structures. Every refinement
here has at most four edges.
\end{definition}

\begin{theorem}[Ampleness of the logarithmic kappa class
{\cite[Theorem 2.2]{cornalba1993}}, {\cite[Theorem 1, specialized to $r=d=0$]{cornalba1995}}]
\label{thm:genus-three-ample-cited}
Let $(h,m)$ be stable, put $\lambda=c_1(\pi_*\omega_\pi)$, and let
$\delta$ be the total boundary divisor with ordinary stack normalization.
Then the logarithmic class $\kappa_1$ descends to an
ample rational divisor on the coarse space $\overline M_{h,m}$.
It is the first Chern class of the fractional line given by the Deligne
pairing of $\omega_\pi(\sum_iD_i)$ with itself and satisfies
\[
 \kappa_1=12\lambda-\delta+\sum_i\psi_i.
\]
\end{theorem}

\begin{proposition}[The Gram identity and its radical]
\label{prop:genus-three-radical}
For every $n\ge10$, the matrix $C_n$ is symmetric and satisfies
\[
 \rho_n(\sigma)^tC_n\rho_n(\sigma)=C_n\quad(\sigma\in\Sn_n),
\]
and for all $a,b\in A_n$ and $u,v\in S$,
\[
 \int_{\Mbar_{3,n}}\Phi_n(a,u)\Phi_n(b,v)\kappa_1^{n-9}
 =C_n(a,b)J(u,v).
\]
As a map of Hodge structures,
\[
 \ker\Phi_n=(\ker C_n)\otimes\Pmot,\qquad
 H^{15}(\Mbar_{3,n})\cong(A_n/\ker C_n)\otimes\Pmot.
\]
\end{proposition}

\begin{proof}
We identify the matrix with the pulled-back intersection form, then
apply Hard Lefschetz to determine its radical.

\paragraph{The intersection entries.}
Lemma~\ref{lem:genus-three-formula-odd-excess} computes the product of
two boundary Gysin images in their displayed order. Under a contraction, the
inverse graph of a genus-one vertex either has a unique genus-one
vertex and rational trees, or consists of rational vertices with first
Betti number one. In the first case the hook pullback is obtained by
iterating Proposition~\ref{prop:eleven-hook-realizations}; in the
second it is zero, since each rational factor has only even cohomology
by Theorem~\ref{thm:cited-genus-zero-cycles}. Thus
Definition~\ref{def:genus-three-polynomial-pullback} gives the actual
coefficient pullbacks. The logarithmic convention and
Theorem~\ref{thm:genus-three-excess-cited} give
\[
 \xi_\Gamma^*\kappa_1^{n-9}
       =\left(\sum_{v\in V(\Gamma)}\kappa_{1,v}\right)^{n-9}.
\]
If the surviving hooks occur at different vertices, each such vertex
carries an odd-degree class multiplied by even-degree decorations.
Its integral is zero. If they occur at the same vertex,
Proposition~\ref{prop:genus-three-evaluation} evaluates that factor as
$\mathcal C_{F(v)}(k_v,l_v;f_v)J(u,v)$. Every other vertex has genus at
most two and at least one flag: total genus is three, and the refinement
is connected with an edge. Its polynomial integral is $W_{g(w)}(f_w)$
by Proposition~\ref{prop:genus-three-descendants-cited}. K\"unneth
integration gives precisely \eqref{eq:genus-three-gram-entries},
including its two input automorphism factors, and proves the asserted
Gram identity.

Every entry is a finite rational expression
(Definitions~\ref{def:genus-three-descendants}--\ref{def:genus-three-gram}).

The ambient intersection form is alternating because its arguments
have degree fifteen. Since $J$ is alternating and nondegenerate, the
Gram identity gives $C_n^t=C_n$. Relabelling preserves $\kappa_1$ and
integration and commutes with $\Phi_n$. Applying the identity after
relabelling both inputs gives
$\rho_n(\sigma)^tC_n\rho_n(\sigma)=C_n$.

\paragraph{Detection of the kernel.}
The coarse space of $\Mbar_{3,n}$ is a projective orbifold of complex
dimension $d=n+6$. Its rational cohomology agrees with that of the
stack: on a finite quotient chart the positive-degree rational
cohomology of each stabilizer vanishes, so the coarse-space map has
no higher rational direct images. By
Theorem~\ref{thm:genus-three-ample-cited}, a positive integral multiple
of this logarithmic $\kappa_1$ is ample and hence K\"ahler. For $n\ge10$
one has $15\le d$, and ordinary Hard Lefschetz for the projective
orbifold $\overline M_{3,n}$ with this K\"ahler class
\cite[Theorem 2.2(1)]{fernandez2006} gives
\[
 \kappa_1^{n-9}:H^{15}(\Mbar_{3,n};\mathbb Q)
        \xrightarrow{\sim}H^{2d-15}(\Mbar_{3,n};\mathbb Q).
\]
That theorem gives the isomorphism after extension to $\mathbb R$;
its map is rational, so it is already an isomorphism over $\mathbb Q$.
Poincar\'e duality makes $\int xy\kappa_1^{n-9}$ nondegenerate on
$H^{15}$.

The map $\Phi_n$ is surjective by
Proposition~\ref{prop:genus-three-generation}. A vector in the radical
of its pulled-back form has image pairing to zero with the whole
target and therefore has zero image. Conversely, each kernel vector
lies in that radical. The pulled-back form is $C_n\otimes J$, whose
radical is $(\ker C_n)\otimes S$ because $J$ is nondegenerate.
Restoring the common Tate factor gives the stated kernel and quotient
as Hodge structures.
\end{proof}

\begin{definition}\label{def:genus-three-character}
Put $\En_{3,n}=A_n/\ker C_n$ for $n\ge10$, and put $\En_{3,n}=0$
for $0\le n\le9$.
For $n\ge10$ with $N_n>0$, put
\[
 Q_n=\sum_{\sigma\in\Sn_n}\rho_n(\sigma)^t\rho_n(\sigma),\qquad
 M_n=Q_n^{-1}C_n^tQ_n^{-1}C_n,
\]
and define, for $\sigma\in\Sn_n$,
\begin{equation}\label{eq:genus-three-character}
 e_{3,n}(\sigma)=\operatorname{tr}\rho_n(\sigma)
 -[z^0]\frac{z\operatorname{tr}\left(
 \rho_n(\sigma)\operatorname{adj}(zI_{N_n}+M_n)\right)}
 {\det(zI_{N_n}+M_n)}.
\end{equation}
The coefficient is the Laurent constant at $z=0$, and
$\operatorname{adj}$ is the classical adjugate. Set $e_{3,n}=0$ when
$N_n=0$ or $0\le n\le9$. For a partition $\mu$ let $m_j(\mu)$ count its
parts equal to $j$, put $z_\mu=\prod_jj^{m_j(\mu)}m_j(\mu)!$ and
$p_\mu=\prod_i p_{\mu_i}$, and let $\chi_\lambda(\mu)$ be the ordinary
character of $V_\lambda$ on that cycle type. Define
\[
 F_3=\sum_{n\ge10}\sum_{\mu\vdash n}
 \frac{e_{3,n}(\mu)}{z_\mu}p_\mu,
 \qquad
 m_{3,n}^{\lambda}=\sum_{\mu\vdash n}
 \frac{\chi_\lambda(\mu)e_{3,n}(\mu)}{z_\mu}.
\]
\end{definition}

\begin{lemma}[Restriction of elliptic degree-eleven classes]
\label{lem:elliptic-eleven-open-restriction}
For every nonempty finite marking set $L$, restriction induces an
$\Sn_L$-equivariant isomorphism
\[
H^{11}(\Mbar_{1,L};\mathbb Q)
\xrightarrow{\ \sim\ }W_{11}H^{11}(\Mopen_{1,L};\mathbb Q).
\]
\end{lemma}

\begin{proof}
We apply the ordinary weight sequence to the full normalized boundary,
keeping every quotient-stack component. On an \'etale normal-crossing
chart of $\Mbar_{1,L}$, its incoming degree-eleven term is the direct
sum over all stable one-edge graphs $\Gamma$ of type $(1,L)$:
\[
 \bigoplus_{[\Gamma]}
 \left(H^9(X_\Gamma;\mathbb Q)\otimes\Lmot\right)^{\Aut_L(\Gamma)}
 \longrightarrow H^{11}(\Mbar_{1,L};\mathbb Q)
 \longrightarrow W_{11}H^{11}(\Mopen_{1,L};\mathbb Q)
 \longrightarrow0.
\]

For a nonseparating graph, the normalization is a genus-zero moduli
stack, so its degree-nine cohomology vanishes by
Theorem~\ref{thm:cited-genus-zero-cycles}. For a separating graph, the
vertex genera are one and zero. K\"unneth decomposes $H^9(X_\Gamma)$
into tensor products whose degrees add to nine. At least one factor
has odd degree at most nine, and that factor vanishes by
Theorem~\ref{thm:cited-low-odd-vanishing}; the genus-zero factor also
has no odd cohomology by the same genus-zero theorem. Thus every
$H^9(X_\Gamma)$ is zero, before taking automorphism invariants.

The normalized boundary and its branch-order local system descend from
the charts by finite quotient groups. Taking invariants is exact over
$\mathbb Q$, and the quotient-stack Gysin maps commute with this
descent. Consequently the displayed source is zero on the stack as
well. The weight sequence therefore identifies restriction with an
isomorphism onto $W_{11}H^{11}(\Mopen_{1,L})$. All constructions commute
with relabelling of $L$, which proves $\Sn_L$-equivariance.
\end{proof}

\begin{proposition}[The genus-three threshold]
\label{prop:genus-three-threshold}
For $0\le n\le8$, nonseparating pullback is injective from
$H^{15}(\Mbar_{3,n})$ into $H^{15}(\Mbar_{2,n+2})$, and the latter
group is zero. Also $H^{15}(\Mbar_{3,9})=0$.
For $n\ge10$, write $\operatorname{Std}_{[n]}=
\ker(\mathbb Q^{[n]}\xrightarrow{\sum}\mathbb Q)$.
Clutching two unmarked elliptic tails to an elliptic component carrying
all ordinary labels, with the tail-exchange quotient, induces an injection
\[
 \left(\bigwedge^{10}\operatorname{Std}_{[n]}
 \oplus\bigwedge^9\operatorname{Std}_{[n]}\right)\otimes\Pmot
 \lhook\joinrel\longrightarrow H^{15}(\Mbar_{3,n}).
\]
\end{proposition}

\begin{proof}
We prove vanishing below arity nine by duality, prove that every boundary
Gysin image vanishes at arity nine, and detect the equal-tail image in
every larger arity.

\paragraph{Arities at most eight.}
The dimension of $\Mbar_{3,n}$ is $n+6$, so the Poincar\'e-dual degree
to fifteen is $2n-3$. For $n=0,1$ this degree is negative. For
$2\le n\le6$ it is one of $1,3,5,7,9$, and
Theorem~\ref{thm:cited-low-odd-vanishing} applies. At $n=7$ it is
eleven, which vanishes in genus three by
Proposition~\ref{prop:eleven-hook-realizations}. At $n=8$ it is
thirteen, which vanishes by Theorem~\ref{thm:cited-thirteen-cohomology}:
the defining sum for $U_{3,[8]}$ is empty. Thus the genus-three source
of the asserted pullback is zero. Its genus-two target is zero by
Theorem~\ref{thm:genus-two}, since $n+2\le10$.

\paragraph{The separating source at arity nine.}
Proposition~\ref{prop:genus-three-generation} gives its complete
ordinary one-edge boundary source. By
Theorem~\ref{thm:cited-low-odd-vanishing}, a degree-thirteen class on
a separating normalization can only have degrees
$(13,0),(11,2),(2,11),(0,13)$. A degree-eleven elliptic factor would
need ten ordinary labels in addition to the attaching flag, so both
mixed sectors are zero. A rational component has no odd cohomology;
if it carries a stable rational tail, the complementary genus-three
factor has at most eight markings and has zero $H^{13}$. On a
genus-one factor there are at most ten markings, so $U_{1,[m]}=0$.
Theorem~\ref{thm:cited-thirteen-cohomology} therefore leaves only
\[
 H^{13}(\Mbar_{2,[9]\cup\{p\}})
       \boxtimes H^0(\Mbar_{1,\{p'\}}).
\]
The same theorem describes the first factor by its unique allocation
of ten labels to an elliptic degree-eleven component. Attaching the
outer unmarked elliptic tail produces the two-edge star with an
elliptic center carrying all nine ordinary labels and two unmarked
elliptic tails. Its center has eleven flags, so its hook is the sign
representation. The graph automorphism group has order two, generated
by exchanging the tails and their central flags. This exchange acts
by $-1$ on the hook and by $+1$ on the unit decorations. If
$\gamma$ is such a decoration and $\tau$ is the exchange, then
\[
 \xi_{\Gamma*}\gamma
   =\xi_{\Gamma*}\tau^*\gamma=-\xi_{\Gamma*}\gamma=0.
\]
Each of the two separating one-edge maps used in this factorization
has graph automorphism group one; their composite is the labelled
$\xi_{\Gamma*}$, twice the normalized star class. Thus its vanishing
also proves that the full separating source maps to zero.

\paragraph{The nonseparating source at arity nine.}
Write $I=[9]\cup\{h,h'\}$. The remaining source is
$H^{13}(\Mbar_{2,I})^{\langle(h\,h')\rangle}$, and its outer Gysin
map is $\tfrac12\xi_{\mathrm{irr}*}$. By
Theorem~\ref{thm:cited-thirteen-cohomology}, the inner elliptic
allocations have size ten or eleven. We may check their labelled
generators before averaging under $(h\,h')$.

If a size-ten allocation omits an ordinary label, gluing $h,h'$ makes
a loop on the decorated elliptic vertex. Reversal of the loop
transposes two of its eleven hook flags. If it omits $h$ or $h'$,
the resulting graph has two parallel edges between its elliptic
vertices; their exchange transposes two of the eleven flags on the
decorated vertex. In each case the graph automorphism group has
order two, the other factor carries the unit, and the displayed
anti-invariance calculation shows that the labelled pushforward vanishes.
The outer factor $1/2$ makes it exactly the normalized two-edge class.

For the size-eleven allocation the elliptic center carries all of
$I$ and an attaching flag $p$, and the unmarked elliptic tail has
flag $p'$. Let
\[
 \eta:\Mbar_{1,I\cup\{p\}}\longrightarrow
          \Mbar_{2,[9]\cup\{p\}},\qquad
 \iota:\Mbar_{2,[9]\cup\{p\}}\times\Mbar_{1,\{p'\}}
          \longrightarrow\Mbar_{3,9}
\]
glue $h,h'$ and attach the tail, respectively. For a degree-eleven
class $s$ at the center, transitivity of labelled Gysin maps writes
the two-edge contribution as
\[
 \frac12\xi_{\Gamma*}(s\boxtimes1)
       =\iota_*\left(\left(\frac12\eta_*s\right)\boxtimes1\right).
\]
The factor $1/2$ on both sides is the loop-reversal normalization;
the outer separating graph has automorphism group one. The class
$\tfrac12\eta_*s$ belongs to
$H^{13}(\Mbar_{2,[9]\cup\{p\}})$, which is the image of the unique
ten-label elliptic allocation already treated. Applying $\iota_*$
therefore gives the equal-tail star class, which vanishes. All one-edge sectors
have now been exhausted, and surjectivity of the sum of their Gysin maps proves
$H^{15}(\Mbar_{3,9})=0$.

\paragraph{The equal-tail source.}
Let $n\ge10$, and choose four distinct new flags $p,q,p',q'$.
Put $L=[n]\cup\{p,q\}$ and let $\tau$ exchange $p,q$ and the two
tail factors in
\[
 \mathcal D_n=
 \left[\Mbar_{1,L}\times\Mbar_{1,\{p'\}}
       \times\Mbar_{1,\{q'\}}\big/\langle\tau\rangle\right].
\]
Let $\zeta_n:\mathcal D_n\to\Mbar_{3,n}$ glue $p$ to $p'$ and
$q$ to $q'$. Thus its degree-eleven source is
$K_L^{\langle(p\,q)\rangle}\otimes S$, with Tate factor $\Lmot^2$.

The tuple presentation identifies $K_L$ with
$\bigwedge^{10}\operatorname{Std}_L$: the symbol
$[b_1,\ldots,b_{11}]$ maps to
\[
 (\mathbf e_{b_1}-\mathbf e_{b_{11}})\wedge\cdots\wedge
 (\mathbf e_{b_{10}}-\mathbf e_{b_{11}}),
\]
where $\mathbf e_b$ is the permutation-basis vector. Exterior
multilinearity gives the alternating tuple relations, and the anchor
symbols map to an exterior basis. Under $\Sn_n\times\langle(p\,q)\rangle$
there is a decomposition
\[
 \operatorname{Std}_L=\operatorname{Std}_{[n]}
 \oplus\mathbb Q\left(\mathbf e_p+\mathbf e_q
                 -\frac2n\sum_{i=1}^n\mathbf e_i\right)
 \oplus\mathbb Q(\mathbf e_p-\mathbf e_q).
\]
The last line is anti-invariant and the other two summands are fixed
by $(p\,q)$. Taking the tenth exterior power and its invariants gives
exactly $\bigwedge^{10}\operatorname{Std}_{[n]}
\oplus\bigwedge^9\operatorname{Std}_{[n]}$.

\paragraph{Detection by self-intersection.}
Restrict the central factor to $\Mopen_{1,L}$, and write
\[
 \widetilde{\mathcal U}_n=\Mopen_{1,L}\times\Mbar_{1,\{p'\}}
                   \times\Mbar_{1,\{q'\}},\qquad
 \mathcal U_n=[\widetilde{\mathcal U}_n/\langle\tau\rangle].
\]
Remove from $\Mbar_{3,n}$ the image under $\zeta_n$ of the locus
where the central factor is singular, obtaining an open substack
$\mathcal V_n$. This removed image is closed because the full
clutching map is proper. It is disjoint from the image of
$\mathcal U_n$: a curve in that image has a unique smooth elliptic
component carrying every ordinary label, and the two separating
nodes cutting off unmarked genus-one subcurves are determined by
the curve. This remains true when either one-pointed tail is nodal.
Consequently $\zeta_n$ restricts to a closed regular embedding
$i:\mathcal U_n\hookrightarrow\mathcal V_n$ of codimension two.
In local deformation coordinates its equations are the smoothing
parameters of these two nodes. The quotient accounts for their
exchange, so there are no additional branches of this embedding.

For $s\in H^{11}(\Mbar_{1,L})^{\langle(p\,q)\rangle}$, let $r(s)$
be its restriction to $\Mopen_{1,L}$. By
Lemma~\ref{lem:elliptic-eleven-open-restriction}, this map is injective
on the weight-eleven source, so $s\ne0$ implies $r(s)\ne0$.
Pulling the self-intersection identity for $i$ back to the ordered
cover gives exactly
\[
 (r(s)\boxtimes1\boxtimes1)
             (\psi_p+\psi_{p'})(\psi_q+\psi_{q'}).
\]
The two normal lines have first Chern classes
$-\psi_p-\psi_{p'}$ and $-\psi_q-\psi_{q'}$; their product has the
displayed positive sign. Its K\"unneth component of degrees $(11,2,2)$ is
\[
 r(s)\boxtimes\psi_{p'}\boxtimes\psi_{q'}.
\]
For $s\ne0$ this is nonzero: $r(s)\ne0$, and each tail cotangent
class has integral $1/24$ by
Proposition~\ref{prop:genus-three-descendants-cited}. The other three
terms have different K\"unneth degrees. Pullback to the ordered cover
is injective, since its trace composed with pullback is multiplication
by two. For the quotient-normalized Gysin map the displayed
self-intersection has coefficient one: the class descends with
pullback $s\boxtimes1\boxtimes1$, while its global pushforward is
$\tfrac12$ of the ordered pushforward. Thus every nonzero invariant
$s$ has nonzero global image. Restoring the codimension-two twist
proves the asserted equivariant injection. Finally,
$\dim\bigwedge^9\operatorname{Std}_{[n]}=\binom{n-1}{9}>0$ and
$S\ne0$, so the injected module is nonzero for every $n\ge10$.
\end{proof}

\begin{theorem}[Genus-three decomposition]
\label{thm:genus-three}
For every integer $n\ge0$ and partition $\lambda\vdash n$, the numbers
$m_{3,n}^{\lambda}$ are
nonnegative integers and there are $\Sn_n$-equivariant isomorphisms
\[
 H^{15}(\Mbar_{3,n};\mathbb Q)
 \cong\bigoplus_{\lambda\vdash n}
 V_\lambda\otimes\Pmot^{\oplus m_{3,n}^{\lambda}},
 \qquad
 \operatorname{ch}_n(\En_{3,n})=[F_3]_n.
\]
The isomorphism is one of pure Hodge structures. For every prime $\ell$
the same multiplicities give an actual
$\Sn_n\times\Gal(\overline{\mathbb Q}/\mathbb Q)$-equivariant isomorphism
\[
 H^{15}_{\mathrm{et}}(\Mbar_{3,n,\overline{\mathbb Q}};\mathbb Q_\ell)
 \cong\bigoplus_{\lambda\vdash n}
 (V_\lambda\otimes\mathbb Q_\ell)\otimes
 \left(\mathbb Q_\ell(-2)\otimes S_\ell\right)^{\oplus m_{3,n}^{\lambda}},
\]
Moreover, $H^{15}(\Mbar_{3,n})\ne0$ if and only if $n\ge10$.
Every displayed multiplicity is determined by the finite graph sum and
terminating recursions in Definitions~\ref{def:genus-three-descendants}--
\ref{def:genus-three-character}.
\end{theorem}

\begin{proof}
We compute the character of the Gram quotient, identify its Hodge
realization, and then apply comparison to the sum of all ordinary
boundary Gysin maps.

\paragraph{The quotient character.}
Fix $n\ge10$ with $N_n>0$. For every nonzero real column $x$,
\[
 x^tQ_nx=\sum_{\sigma\in\Sn_n}
                   \|\rho_n(\sigma)x\|^2>0.
\]
Reindexing the sum gives
$\rho_n(\sigma)^tQ_n\rho_n(\sigma)=Q_n$. Together with the
corresponding identity for $C_n$ from
Proposition~\ref{prop:genus-three-radical}, this implies that
$Q_n^{-1}C_n$ commutes with every $\rho_n(\sigma)$. Its adjoint for
the $Q_n$-inner product is $Q_n^{-1}C_n^t$, so $M_n$ also commutes
with the action and is self-adjoint. Moreover,
\[
 x^tQ_nM_nx=(C_nx)^tQ_n^{-1}(C_nx)\ge0,
 \qquad \ker M_n=\ker C_n.
\]
The real spectral theorem diagonalizes $M_n$ with nonnegative
eigenvalues. Therefore the rational endomorphism
\[
 \Pi_n=[z^0]\,z(zI_{N_n}+M_n)^{-1}
\]
is the projection onto $\ker C_n$. On the zero eigenspace the
displayed function is one, while on an eigenspace with eigenvalue
$\nu>0$ the constant coefficient of $z/(z+\nu)$ is zero. The
projection is rational because the inverse matrix has rational
function entries, and it commutes with the group action.

The Laurent coefficient in \eqref{eq:genus-three-character} is
consequently $\tr(\rho_n(\sigma)\Pi_n)=\tr(\sigma\mid\ker C_n)$.
Subtracting it from the source character proves
that $e_{3,n}$ is the character of $A_n/\ker C_n$. When $N_n=0$
the quotient and the specified character are both zero.

Character orthogonality gives
$\operatorname{mult}(V_\lambda,A_n/\ker C_n)=m^\lambda_{3,n}$
(Definition~\ref{def:genus-three-character}).
These are nonnegative integers, and the Frobenius characteristic is
$\sum_{\mu\vdash n}e_{3,n}(\mu)p_\mu/z_\mu=[F_3]_n$.
Proposition~\ref{prop:genus-three-radical} identifies the corresponding
Specht decomposition tensored with $\Pmot$ with the full Hodge
structure $H^{15}(\Mbar_{3,n})$. For $n\le9$, the definitions give
$e_{3,n}=0$ and $m_{3,n}^{\lambda}=0$, and
Proposition~\ref{prop:genus-three-threshold} gives the same vanishing
for the cohomology. This proves the Hodge assertions in every arity.

\paragraph{The actual Galois realization.}
Use the complete map
$\Phi^{15}_{3,n,\mathrm B}:\mathcal D^{13}_{3,n,\mathrm B}
\to H^{15}(\Mbar_{3,n})$ of
Definition~\ref{def:degree-thirteen-boundary-source}.
It is surjective in every arity by
Proposition~\ref{prop:genus-three-generation}, and its Betti image has
just been identified with $\En_{3,n}\otimes\Pmot$.
The source identifications use the actual forgetful maps realizing
the hooks in Proposition~\ref{prop:eleven-hook-realizations}, the
elliptic separating maps in
Theorem~\ref{thm:cited-thirteen-cohomology}, and the algebraic divisor
classes in the mixed degree-eleven and degree-two sectors. Their
composites with the outer quotient-normalized clutching maps are
defined over $\mathbb Q$. In particular these are the geometric maps
of Proposition~\ref{prop:h13-input}, which commute with comparison
and every Specht projection. Its surjective-image assertion applies
and gives the actual equivariant isomorphism
\[
 H^{15}_{\mathrm{et}}(\Mbar_{3,n,\overline{\mathbb Q}};\mathbb Q_\ell)
 \cong(\En_{3,n}\otimes\mathbb Q_\ell)
                \otimes\mathbb Q_\ell(-2)\otimes S_\ell.
\]
The rational Specht decomposition established above gives exactly
the displayed multiplicities in this isomorphism.

The nonzero injected module of
Proposition~\ref{prop:genus-three-threshold} gives $H^{15}\ne0$ for
$n\ge10$; every multiplicity is determined by the finite formulas above.
\end{proof}