\section{Genus-two compact test classes}\label{sec:10-genus-two-detectors}

We construct cohomological test maps on compact moduli spaces that distinguish
the parallel and Picard Gysin images in the genus-five coefficient calculation
and determine the auxiliary kernel.

\begin{setup}\label{setup:g5-detector-row}
For a finite marking set $F$, choose distinct new flags $a,b,c,d$ and put
\[
I=F\sqcup\{c,d\},\qquad J=I\sqcup\{a,b\},\qquad
t=(c\,d),\qquad s=(a\,c)(b\,d).
\]
Let $\nu_{ab}:\Mbar_{3,J}\to\Mbar_{4,I}$ glue $a,b$, and set
\[
\mathscr D_F=
\{y\in H^{15}(\Mbar_{3,J}):(a\,b)^*y=(c\,d)^*y=y,\ s^*y=-y\},
\qquad
q_F=(1-s^*)\nu_{ab}^*:N_{5,F}\longrightarrow\mathscr D_F.
\]
Use the compact row and quotient $\mathscr Q_{5,F}$ of
Definition~\ref{def:compact-boundary-row}; write $\delta_F$ for its
map $\delta$, $\xi_F:\Mbar_{4,I}\to\Mbar_{5,F}$ for its map $\xi$
gluing $c,d$, and $\beta_{5,F}$ for the compact row map at $(5,F)$.
Order the double-loop determinant as
$\ell_{ab}\wedge\ell_{cd}$. For $\mu\vdash|F|$, put
$M_\mu(V)=\operatorname{Hom}_{\Sn_F}(V_\mu,V)$ and
\[
N_\mu=M_\mu(N_{5,F}),\qquad
\mathscr Q_\mu=M_\mu(\mathscr Q_{5,F}),\qquad
J_\mu=M_\mu(\xi_F^*)(M_\mu(\mathcal I_{5,F})).
\]
Write $\delta_\mu=M_\mu(\delta_F)$. The equality $\beta\alpha=0$
gives $J_\mu\subseteq\ker\delta_\mu$, and hence a map
\[
\overline\delta_\mu:N_\mu/J_\mu\longrightarrow\mathscr Q_\mu.
\]

Throughout, all graph classes and Gysin maps use the ordinary
quotient-stack normalization of Definition~\ref{def:stable-graphs};
the displayed sources include their Tate twists.
All sources and tests commute with relabelling of
$F$; hence every map below is $\Sn_F$-equivariant, and by
Set-up~\ref{setup:standing} every injectivity or kernel statement
passes to the exact functors $M_\mu$ and to invariants under finite
groups of labelled graph automorphisms, such as $\langle t\rangle$.
The group of order eight generated by the pair reversals $(a\,b)$,
$(c\,d)$ and by $s$ projects onto $\mathscr D_F$ by the average with
positive loop reversals and negative pair exchange; this projector
leaves each pairing with a class of $\mathscr D_F$ unchanged and
retains the order $\ell_{ab}\wedge\ell_{cd}$, so the tests below
introduce no additional normalization factor.
\end{setup}

\begin{lemma}\label{lem:g5-double-loop-coordinate}
For every finite $F$, projection to the double-loop coordinate induces
a map $\operatorname{pr}_{\mathrm{dl}}:\mathscr Q_{5,F}\to \mathscr D_F$ satisfying
\[
 \operatorname{pr}_{\mathrm{dl}}\delta_F=q_F.
\]
\end{lemma}
\begin{proof}
Both contractions defining the double-loop coordinate are the two
nonseparating contractions in the compact row. Hence the projection
annihilates the complete separating image in the quotient, and on the
remaining coordinate it is exactly $(1-s^*)\nu_{ab}^*$.
\end{proof}

\begin{definition}\label{def:g5-genus-two-moments}
On $\Mbar_{2,\{q\}}$ let $\gamma_{\mathrm{F}},\gamma_{\mathrm{E}},\gamma_{\mathrm{H},q},\gamma_{\mathrm{H},\varnothing},\gamma_{\mathrm{G}}$ be the five ordinary
pure two-edge classes of
Theorem~\ref{thm:cited-polito-projected-presentation}(2).

For a finite set $B$ disjoint from $q$, put
$Y_B=\Mbar_{2,B\sqcup\{q\}}$, let $\pi_B:Y_B\to \Mbar_{2,\{q\}}$ forget
$B$, and put $\vartheta_B=\prod_{i\in B}\psi_i$.
Let $\mathcal N_B\subseteq H^4(Y_B)$ be the sum of the following spaces:
the images $j_{T*}H^2(\operatorname{dom}j_T)$ for all nonempty
$T\subseteq B$, with
\[
j_T:\Mbar_{0,T\sqcup\{q,x\}}\times \Mbar_{2,(B\setminus T)\sqcup\{y\}}
\longrightarrow Y_B;
\]
the spans of pure two-edge trees with exactly two rational vertices;
the spans of trees with a rational $q$-center and two elliptic leaves;
and $\psi_qH^2(Y_B)$.
For $T\subseteq B$, let $\zeta_T$ be the ordinary parallel class
with rational vertex carrying $q\sqcup T$ and elliptic vertex
carrying $B\setminus T$.
\end{definition}

\begin{proposition}\label{prop:g5-compact-moments}
There exists a rational Hodge morphism
$\Lambda:H^4(\Mbar_{2,\{q\}})\to\mathbb Q(-2)$ with
\[
 (\Lambda(\gamma_{\mathrm{F}}),\Lambda(\gamma_{\mathrm{E}}),\Lambda(\gamma_{\mathrm{H},q}),\Lambda(\gamma_{\mathrm{H},\varnothing}),\Lambda(\gamma_{\mathrm{G}}))
 =(-12,1,0,12,0).
\]
For any such $\Lambda$ and every finite $B$, define
$\Lambda_B(u)=\Lambda(\pi_{B*}(u\vartheta_B))$.
Then
\[
 \Lambda_B(\mathcal N_B)=0,\qquad
 \Lambda_B(\zeta_T)=|T|!(|B\setminus T|+1)!
 \quad(T\subseteq B).
\]
These maps commute with relabelling $B$ and transport of $q$.
\end{proposition}
\begin{proof}
\emph{The one-marked functional.}
Theorem~\ref{thm:cited-polito-projected-presentation}(2) gives independence of
the five pure classes. The prescribed values therefore define a
functional on their Tate span. A rational polarization splits off this
sub-Hodge structure, extending the functional by zero on its orthogonal
complement. The genus-one identities of
Proposition~\ref{prop:genus-three-picard-input}, applied on the two
genus-two divisor normalizations, give
\[
\psi_q\delta_{\rm irr}=\gamma_{\rm F}/6+2\gamma_{\rm E}+\gamma_{\rm H,q},
\qquad
\psi_q\delta_1=\gamma_{\rm H,q}/12+2\gamma_{\rm G}.
\]
The first double-loop coefficient is $8/(2\cdot2\cdot12)=1/6$:
the two loop normalizations each contribute $1/2$, whereas the final
graph has automorphism order eight. The two rational-tail choices
give $2\gamma_{\rm E}$. In the second identity the two unmarked
elliptic leaves give $2\gamma_{\rm G}$. Thus these are identities
with the ordinary graph normalizations. Their images under $\Lambda$
vanish, as does the expression for $\psi_q^2$ displayed in
Theorem~\ref{thm:cited-polito-projected-presentation}(2).
Theorem~\ref{thm:cited-divisor-generation} and
Proposition~\ref{prop:genus-three-picard-input} show that
$\psi_q,\delta_{\rm irr},\delta_1$ generate the one-marked Picard
group. Consequently $\Lambda(\psi_q H^2(\Mbar_{2,\{q\}}))=0$.

\emph{Weighted forgetting.}
For a rational tail carrying $T\ne\varnothing$, the factor
$\vartheta_B$ contains $|T|$ cotangent classes on a rational factor of
dimension $|T|-1$, hence annihilates its full $H^2$ pushforward. The same
dimension argument annihilates every two-rational-vertex tree class. On a
$q$-centered tree, weighted forgetting leaves a multiple of
$\gamma_{\rm G}$, which is annihilated by $\Lambda$. The cotangent identity
under forgetting differs from $\pi_B^*\psi_q$ by the same $q$-containing
tails, so $\Lambda_B(\psi_qH^2)=0$.

The parallel quotient has the same simultaneous edge-exchange stabilizer
before and after forgetting, so no additional factor is introduced.
Finally, on a compact genus-$h$ factor with $n_0$ retained markings,
Proposition~\ref{prop:external-stable-forgetting} gives
\[
\pi_*\!\left(\prod_{r=1}^m\psi_{z_r}\right)
=(2h-2+n_0)_m,
\]
where $(x)_m=x(x+1)\cdots(x+m-1)$. Indeed, each section correction
has its forgotten cotangent on a rigid three-pointed rational tail,
where it vanishes; the remaining factors are removed successively by
the projection formula. Applying this to the two factors of $\zeta_T$ gives
$|T|!(|B\setminus T|+1)!$ times $\gamma_{\rm E}$, proving the claim.
\end{proof}

\begin{definition}\label{def:g5-parallel-source}
For $h\geq3$ and a finite set $L$, put
\[
 \mathcal R^{\mathrm{par}}_h(L)=\bigoplus_{S\subseteq L}
 H^{11}(\Mbar_{1,S\sqcup\{u,v\}})^{\langle(uv)\rangle}(-2).
\]
The partner factor, carrying the unit class, is $\Mbar_{h-2,(L\setminus S)\sqcup\{u',v'\}}$.
Let $p_h:\mathcal R^{\mathrm{par}}_h(L)\to H^{15}(\Mbar_{h,L})$ be twice the ordinary Gysin
map from the quotient by simultaneous exchange $(uv)(u'v')$.
For $h=3$ the two possible odd K\"unneth placements on the equal-genus
product are retained.
Put $\mathcal R^{\mathrm{par}}_F=\mathcal R^{\mathrm{par}}_4(I)^t$, and let $\mathcal R^{\mathrm{par}}_{0,F}$ and $\mathcal R^{\mathrm{par}}_{+,F}$ be its
allocation subspaces with $S\cap\{c,d\}$ empty and nonempty.
Let $\operatorname{Sep}_{h,L}$ be the complete ordinary separating
one-edge Gysin image in $H^{15}(\Mbar_{h,L})$, with full degree-thirteen
coefficients, and put $B_{\mathrm{sep},F}=\operatorname{Sep}_{4,I}^t$.

For $A\subseteq I$ with $|A|\ge10$, put
$H_A=H^{11}(\Mbar_{1,A\sqcup\{p\}})(-2)$.
Let
\[
\rho_A:H^{11}(\Mbar_{1,A\sqcup\{u,v\}})^{(uv)}(-2)\longrightarrow H_A
\]
be rational-tail restriction collapsing $u,v$ to $p$.
For $A\subseteq S$, let $f_{S,A}:H_S\to H_A$ be induced by the
tenth exterior power of the augmentation-space map identifying
every label of $S\setminus A$ with $p$.
For $A\cap\{c,d\}\ne\varnothing$, define
\[
 \iota_A:\Mbar_{1,A\sqcup\{p\}}\times \Mbar_{2,(J\setminus A)\sqcup\{q\}}
 \to \Mbar_{3,J},\qquad
 \mathcal D_A=(1\otimes\Lambda_{J\setminus A})
                 \operatorname{pr}_{11,4}\iota_A^*.
\]
\end{definition}

\begin{proposition}\label{prop:g5-parallel-left-inverse}
For every finite $F$, each $\rho_A$ is an isomorphism. The formulas
\[
 \mathcal L_A(y)=
 \frac{\rho_A^{-1}\left(\mathcal D_A(y)-
 \sum_{i\in I\setminus A}f_{A\sqcup\{i\},A}
                      \mathcal D_{A\sqcup\{i\}}(y)\right)}
 {2(|J\setminus A|+1)!}
\]
define a map
$\mathcal L:\mathscr D_F\to \mathcal R^{\mathrm{par}}_{+,F}$ satisfying
\[
 \mathcal Lq_Fp_4|_{\mathcal R^{\mathrm{par}}_{+,F}}=\mathrm{id},\qquad
 \mathcal Lq_F(B_{\mathrm{sep},F})=0.
\]
Moreover, the map
\[
 e_F:\mathcal R^{\mathrm{par}}_F\to \mathscr Q_{5,F}/\delta_F(B_{\mathrm{sep},F}),\qquad
 r\mapsto[\delta_Fp_4(r)]
\]
has kernel $\mathcal R^{\mathrm{par}}_{0,F}$, and $\mathcal L\operatorname{pr}_{\mathrm{dl}}$
induces a left inverse for its restriction to $\mathcal R^{\mathrm{par}}_{+,F}$.
\end{proposition}
\begin{proof}
\emph{The hook restriction.}
On augmentation spaces the map identifying $u,v$ with $p$ has kernel
$\mathbb Q(e_u-e_v)$ and maps its swap-invariant complement
isomorphically onto the augmentation space on $A\sqcup\{p\}$.
The tenth exterior power decomposes into its invariant part and the
part containing $e_u-e_v$, on which the swap acts negatively.
Proposition~\ref{prop:eleven-hook-realizations} and
Lemma~\ref{lem:cited-eleven-hook-pullback} identify the invariant
restriction with $\rho_A$, proving it is an isomorphism.

\emph{The complete separating image.}
Theorems~\ref{thm:cited-low-odd-vanishing} and
\ref{thm:cited-thirteen-cohomology}, together with
Proposition~\ref{prop:eleven-hook-realizations}, reduce the full
degree-thirteen coefficients of a genus-three separating source to
three types: an elliptic $H^{13}$ tail; an elliptic $H^{11}$ tail
with any Picard coefficient on its partner; and an elliptic
$H^{11}$ center with two unit elliptic leaves. Indeed, the only
odd K\"unneth degrees are thirteen and eleven, and an $H^{13}$
class on any higher-genus factor is generated by elliptic-tail
Gysin classes. An outer flag on the unit side gives a
Picard-decorated tail. An outer positive-genus side attached to
the odd factor gives a centered path; an outer rational side
there gives an elliptic $H^{13}$ tail.
This also retains rational boundaries inside every compact coefficient.

Compare the decorated tail set $S$ with the test set $A$.
If $S=A$, a Picard-decorated tail has $(11,4)$ component
$-z\otimes\psi_q\eta$ from the bridge normal line; an $H^{13}$
tail has elliptic degree thirteen and does not contribute. If $S\subsetneq A$, the
internal separating Gysin raises the elliptic degree to at least
thirteen. If $S$ and $A$ are disjoint, the odd class is on the
second test factor. If $A\subsetneq S$, the second test factor
has a $q$-containing rational tail with its full degree-two
coefficient. These are the nested or disjoint possibilities for
the two genus-one sides in a stable tree. For a centered path,
a cut isolating a unit leaf places the odd class on the other factor; a cut
inside the decorated center leaves a rational $q$-center with
the two unit elliptic leaves. Additional rational subdivisions
give two-rational-vertex trees. The same statements apply after
expanding an elliptic $H^{13}$ coefficient into its rational-tail
Gysin generators.
Thus every surviving degree-four coefficient belongs to
$\mathcal N_{J\setminus A}$. Proposition~\ref{prop:g5-compact-moments}
proves $\mathcal D_A(\operatorname{Sep}_{3,J})=0$.

\emph{The parallel pullback and its moments.}
On a labelled normal-crossings chart, a node normalized in the
compact pullback of $p_h$ is a selected parallel edge or an
internal node of a vertex factor. For a selected edge its
normal class is $-(\psi_x+\psi_y)$, and the other parallel edge
becomes separating; this branch lies in the complete separating
image with its full coefficient. An internal nonseparating node
of the decorated elliptic factor gives genus-zero odd cohomology
and vanishes by Theorem~\ref{thm:cited-genus-zero-cycles}. Such a
node in the unit factor gives $p_{h-1}$ with both new labels on
that factor. An internal separating node can be nonseparating in
the glued curve only when it separates the two attaching flags.
Cutting it turns the two parallel edges into a separating path.
These cases exhaust the compact fiber product; distinct smoothing
parameters give coefficient one and retain all deeper degenerations.
Consequently, modulo $\operatorname{Sep}_{3,J}$,
\[
q_Fp_4(r)=p_3\bigl((1-s^*)H_{a,b}r\bigr),
\]
where $H_{a,b}$ places $a,b$ on the unit factor.
Neither selected parallel edge is a bridge, so a genus-one tail
in the test pullback must split off inside one factor. A cut
in the unit factor places the odd class on the second test factor. A cut in the
decorated factor must leave both attaching flags on its rational
complement; it therefore contributes only for $A\subseteq S$.
Its hook restriction is $f_{S,A}\rho_S$ and its other factor
is $\zeta_{S\setminus A}$. This treats both odd placements when
the two original vertices have genus one. In the exchanged term
the decorated vertex contains neither $c$ nor $d$, so the
necessary containment fails for every old-present $A$.
Base change and Proposition~\ref{prop:g5-compact-moments} now give
\[
\mathcal D_A(q_Fp_4(r_S))=
2|S\setminus A|!(|J\setminus S|+1)!\,f_{S,A}\rho_S(r_S).
\]
The augmentation maps satisfy $f_{T,A}f_{S,T}=f_{S,A}$ for
$A\subseteq T\subseteq S$.
The $|S\setminus A|$ summands in the subtraction have factorial
$d(d-1)!$ when $d=|S\setminus A|$; all proper supersets cancel, leaving
$2(|J\setminus A|+1)!\rho_A(r_A)$. This proves the left-inverse formula.

The factor two here is the factor in $p_h$: the simultaneous
edge-exchange quotient is unchanged by the rational tail cut, and
the bridge defining the cut is uniquely determined by its labelled
side, so it contributes no
further multiplicity.

\emph{Descent and the old-absent kernel.}
A selected separating bridge remains separating after a
nonseparating normalization, so
$\nu_{ab}^*\operatorname{Sep}_{4,I}\subseteq\operatorname{Sep}_{3,J}$.
This and the preceding vanishing prove
$\mathcal Lq_F(B_{\rm sep,F})=0$.
For $r_0\in\mathcal R^{\rm par}_{0,F}$, regard its coefficients
as an element of $\mathcal R^{\rm par}_5(F)$, placing $c,d$ on the
unit factor after pullback. Let $\xi_{cd}$ glue these two flags.
The genus-five instance of the
compact pullback formula gives
$\xi_{cd}^*p_5(r_0)-p_4(r_0)\in B_{\rm sep,F}$.
In the compact restriction row, all other coordinates of
$\alpha(p_5(r_0))$ are separating. Therefore
$\beta\alpha=0$ from Proposition~\ref{prop:compact-restriction-row}
gives $\delta_F(\xi_{cd}^*p_5(r_0))=0$ in $\mathscr Q_{5,F}$,
and hence $e_F(r_0)=0$.
Conversely, if $e_F(r)=0$, Lemma~\ref{lem:g5-double-loop-coordinate}
and the left-inverse formula
force every old-present coordinate of $r$ to vanish, so $r\in
\mathcal R^{\rm par}_{0,F}$.
\end{proof}

\begin{theorem}[Hodge integrals
{\cite[Theorem 1 and Section 0.5]{faberpandharipande2003}}]
\label{thm:cited-lambda-g-integrals}
Let $g\geq1$, $n\geq1$, and $d_i\geq0$ satisfy
$\sum_i d_i=2g+n-3$. With $\lambda_g=c_g(\pi_*\omega_\pi)$, put
\[
b_g=\int_{\Mbar_{g,1}}\lambda_g\psi_1^{2g-2}.
\]
Then
\[
\int_{\Mbar_{g,n}}\lambda_g\prod_{i=1}^n\psi_i^{d_i}
=\binom{2g+n-3}{d_1,\ldots,d_n}b_g,
\qquad
1+\sum_{g\geq1}b_gt^{2g}=\frac{t/2}{\sin(t/2)}.
\]
In particular, $b_1=1/24$ and $b_2=7/5760$.
\end{theorem}

\begin{definition}\label{def:g5-auxiliary-picard-source}
Use the summands and maps of Definition~\ref{def:ordinary-summands}
at $(4,I)$. Define the following sub-sums, over every allocation of $I$:
$\mathcal D_F=\mathsf D_{4,I}\otimes\Pmot$, the elliptic $H^{13}$
tails $H^{13}(\Mbar_{1,A\cup\{p\}})(-1)$ with genus-three unit
partners; $\mathcal T_F=\mathsf T_{4,I}\otimes\Pmot$, the
centered-path source
\[
\bigoplus_{A\sqcup B\sqcup C=I}
H^{11}(\Mbar_{1,A\cup\{p_1,p_2\}})(-2)
\otimes H^0(\Mbar_{1,B\cup\{q_1\}})
\otimes H^0(\Mbar_{2,C\cup\{q_2\}})
\]
of paths with elliptic $H^{11}$ center and unit leaves of genera one
and two; $\mathcal E_F\subseteq\mathsf W_{4,I}\otimes\Pmot$, the
summands whose partner coefficient is a positive separating divisor
of the genus-three partner, which by composition of Gysin maps are
the endpoint paths of genera $(1,h,3-h)$, $h=1,2$, with $H^{11}$ at
the first endpoint; and
$\mathcal V_F\subseteq\mathsf W_{4,I}\otimes\Pmot$, the summands
whose partner coefficient lies in
$\langle\delta_{\mathrm{irr}},\psi_i:i\in I\setminus A\rangle$, the
elliptic $H^{11}$ tails whose genus-three partner is decorated by
$\delta_{\mathrm{irr}}$ or an external $\psi_i$; the attaching
cotangent is excluded from the last family. Let
$\mathscr U_F\subseteq N_{5,F}$ be the $t$-invariant part of
$\Omega_{4,I}(\mathcal D_F\oplus\mathcal T_F\oplus\mathcal E_F\oplus\mathcal V_F)$.
Put $\mathscr Q^{\mathrm{Pic}}_F=\mathscr Q_{5,F}/\delta_F(\mathscr U_F)$.

For $A\subseteq I$ meeting $\{c,d\}$, put $B=I\setminus A$, $k=|B|$,
and choose fresh attaching flags $p,q$. Define
\[
 E_A=\Mbar_{1,A\sqcup\{p\}},\qquad Y_A=\Mbar_{2,B\sqcup\{q,a,b\}},
 \qquad v_A:Y_A\to \Mbar_{3,B\sqcup\{q\}},
\]
\[
 i_A:E_A\times \Mbar_{3,B\sqcup\{q\}}\to \Mbar_{4,I},\qquad
 j_A:E_A\times Y_A\to \Mbar_{3,J}.
\]
The map $v_A$ glues $a,b$, and $i_A,j_A$ glue $p,q$.
On the genus-three partner put
\[
 Z_B=\langle\kappa_1,\psi_q,\delta_{0,T}:T\subseteq B,\ |T|\ge2\rangle.
\]
Define
\[
 \mathcal R^{\mathrm{Pic}}_F=
 \left(\bigoplus_{\substack{A\subseteq I,\ A\cap\{c,d\}\ne\varnothing\\
                             |I\setminus A|\le3}}
 H^{11}(E_A)(-1)\otimes Z_B\right)^t,
 \qquad
 \Phi^{\mathrm{Pic}}_F((z_A\otimes\eta_A)_A)
       =\sum_A i_{A*}(z_A\otimes\eta_A).
\]
For $\theta\in H^{2k+8}(Y_A)$ define
\[
 \mathcal T_A^\theta(y)=
 \left(1\otimes\left[u\mapsto\int_{Y_A}u\theta\right]\right)
       \operatorname{pr}_{11,4}j_A^*y.
\]
The original
divisor $\eta_A$ is on genus three; its tested pullback is $v_A^*\eta_A$.
\end{definition}

\begin{lemma}\label{lem:g5-basic-hodge-moments}
On every pointed genus-two space let $\lambda_i=c_i(\pi_*\omega_\pi)$.
The pullback of $\lambda_1\lambda_2$ to every irreducible or positive separating normalization
is zero, and it pulls back under forgetting markings. Put
\[
 I_0=\int_{\Mbar_{2,\{q\}}}\lambda_1\lambda_2\psi_q=\frac1{2880},\qquad
 \mathcal J_n(d_1,\ldots,d_n)=\int_{\Mbar_{2,[n]}}\lambda_1\lambda_2\prod_i\psi_i^{d_i}.
\]
With omitted exponents zero, the following identities hold in their
indicated ranges:
\[
 \begin{aligned}
 \mathcal J_n(n)&=I_0 &&(n\ge1),\\
 \mathcal J_n(n-1,1)&=(n+1)I_0 &&(n\ge2),\\
 \mathcal J_n(n-2,2)&=\left(\frac{n(n+1)}2-2\right)I_0 &&(n\ge2),\\
 \mathcal J_n(n-2,1,1)&=n(n+1)I_0 &&(n\ge3).
 \end{aligned}
\]
\end{lemma}
\begin{proof}
\emph{Hodge restrictions.}
Write $\mathbb E_h$ for the genus-$h$ Hodge bundle.
The residue sequence on a nonseparating normalization is
$0\to\mathbb E_1\to\nu^*\mathbb E_2\to\mathcal O\to0$; hence
$\nu^*\lambda_1=\lambda:=c_1(\mathbb E_1)$ and $\nu^*\lambda_2=0$. On a positive
separating normalization the Hodge bundle is the direct sum of the two
elliptic Hodge lines. Each line pulls back from $\Mbar_{1,1}$,
so its first Chern class has square zero. Thus
$(\lambda_1\lambda_2)|_{\rm irr}=(\lambda_1\lambda_2)|_{\rm sep}=0$.

\emph{The one-point integral.}
Writing $D_0$ for the rational-tail divisor sum and $D_1$ for the positive
separating divisor sum, Proposition~\ref{prop:genus-three-picard-input}
and Theorem~\ref{thm:genus-three-ample-cited} give
\[
5\kappa_1=5\sum_i\psi_i+\delta_{\rm irr}-5D_0+7D_1,\qquad
\kappa_1=12\lambda_1-\delta_{\rm irr}-D_0-D_1+\sum_i\psi_i.
\]
Subtracting these expressions for $5\kappa_1$ yields
$60\lambda_1=6\delta_{\rm irr}+12D_1$. Hence
$10\lambda_1=\delta_{\rm irr}+2D_1$ and
$\kappa_1=\sum_i\psi_i+2\lambda_1-D_0+D_1$.
Pairing the first identity with $\lambda_2\psi_q$ gives
\[
10I_0=2\left(\int_{\Mbar_{1,\{q,u\}}}\lambda\psi_q\right)
          \left(\int_{\Mbar_{1,\{v\}}}\lambda\right)
       =2(1/24)^2.
\]
The two genus-one integrals are
Theorem~\ref{thm:cited-lambda-g-integrals} with respectively
exponents $(1,0)$ and $(0)$. There is no factor $1/2$ because
$q$ distinguishes the two factors.
Thus $I_0=1/2880$.

\emph{Forgetting recursions.}
Forgetting preserves the Hodge bundle: contracting unstable rational
components does not change regular differentials. Under forgetting
one mark, Proposition~\ref{prop:external-stable-forgetting} gives
$\psi_i=f^*\psi_i+D_{it}$,
$D_{it}^2=-f^*\psi_iD_{it}$, and $\psi_tD_{it}=0$; also
$f_*\psi_t=2g-2+n$. Since $\lambda_1\lambda_2$ pulls back, these
are the string and dilaton recursions. Repeated string gives
$\mathcal J_n(n)=I_0$; one dilaton gives
$\mathcal J_n(n-1,1)=(n+1)I_0$; the base value at $n=2$ and the
string increment $n$ give
$\mathcal J_n(n-2,2)=(n(n+1)/2-2)I_0$; two dilaton steps give
$\mathcal J_n(n-2,1,1)=n(n+1)I_0$.
\end{proof}

\begin{definition}\label{def:g5-node-test}
For the sets in Definition~\ref{def:g5-auxiliary-picard-source}, with
no restriction on $k$, put
\[
 b_k=\frac{4(k+7)}{3(3k^2+15k+14)},\quad
 a_k=-k(k+4)b_k-2(k+2),\quad e_k=b_k+\frac23,
\]
\[
 \theta_{\psi,B}=\lambda_1\lambda_2\psi_q^k\left(
 a_k\psi_q+b_k\sum_{i\in B}\psi_i+\psi_a+\psi_b
 -\frac43D_{ab}+e_k\sum_{i\in B}D_{abi}\right).
\]
Here $D_T=\delta_{0,T}$ on $Y_A$; subscripts list sets of markings.
Put
\[
 \mathcal W_F=\left(\bigoplus_{A\cap\{c,d\}\ne\varnothing}
                   H^{11}(E_A)(-2)\right)^t,\qquad
 \Phi_{\psi,F}((z_A)_A)=\sum_Ai_{A*}(z_A\otimes\psi_q),
\]
and let $\mathcal T^\psi=(\mathcal T_A^{\theta_{\psi,B}})_A$.
\end{definition}

\begin{proposition}\label{prop:g5-node-left-inverse}
For every finite $F$,
\[
 \mathcal T^\psi q_F(\mathscr U_F)=0,\qquad
 \mathcal T^\psi q_F\Phi_{\psi,F}=-\frac1{1080}\mathrm{id}_{\mathcal W_F}.
\]
Consequently $-1080\mathcal T^\psi\operatorname{pr}_{\mathrm{dl}}$
induces a left inverse for
$\mathcal W_F\to \mathscr Q^{\mathrm{Pic}}_F$, $z\mapsto[\delta_F\Phi_{\psi,F}(z)]$.
\end{proposition}
\begin{proof}
\emph{The local moments.}
For rational numbers $\alpha_0,\beta_0,\chi_0,\epsilon_0$, put
\[
L_0=\alpha_0\psi_q+\beta_0\sum_{i\in B}\psi_i+\psi_a+\psi_b
       +\chi_0D_{ab}+\epsilon_0\sum_{i\in B}D_{abi},
\qquad
[D]_0=I_0^{-1}\int_{Y_A}\lambda_1\lambda_2\psi_q^{k+1}D L_0
\]
for a divisor $D$ on $Y_A$, and put
$\Xi=\psi_a+\psi_b-\sum_{\varnothing\ne S\subseteq B}(D_{aS}+D_{bS})$.
Lemma~\ref{lem:g5-basic-hodge-moments} gives
\[
\begin{aligned}
[\psi_q]_0={}&\alpha_0+k(k+4)\beta_0+2(k+4)+\chi_0,\\
[\psi_i]_0={}&(k+4)\alpha_0+
 \bigl((k-\tfrac12)(k+3)(k+4)-2\bigr)\beta_0\\
 &+2(k+3)(k+4)+(k+3)\chi_0+\epsilon_0,\\
[D_{ab}]_0={}&\alpha_0+k(k+3)\beta_0-(k+3)\chi_0+k\epsilon_0,\\
[D_{abi}]_0={}&\beta_0+2+\chi_0-\epsilon_0,\\
[\Xi]_0={}&8\bigl(\alpha_0+k(k+4)\beta_0+2(k+2)\bigr).
\end{aligned}
\]
Here $i\in B$ in the indexed rows. To see the boundary coefficients,
on $D_{ab}$ the two new cotangents vanish and its normal class is
$-\psi_v$ on the genus-two factor. Each $D_{abi}$ restricts there
to $D_{vi}$, contributing one. On $D_{abi}$ the rational factor is
$\Mbar_{0,4}$; its four contributions from $\psi_i,\psi_a+\psi_b,
D_{ab},D_{abi}$ are $\beta_0,2,\chi_0,-\epsilon_0$.
For the last row the contribution of $\psi_a+\psi_b$ is
\[
2(k+4)\alpha_0+2k(k+3)(k+4)\beta_0
 +3(k+3)(k+4)-4+2k\epsilon_0.
\]
The $2k$ tails with $|S|=1$ each contribute
$\alpha_0+(k+3)(k-1)\beta_0+k+3+\epsilon_0$.
The $k(k-1)$ tails with $|S|=2$ each contribute $2\beta_0+1$.
For larger $S$, the degree on the rational factor is below its dimension. Subtraction gives
the row for $\Xi$.

Specialize to $\alpha_0=a_k,\beta_0=b_k,\chi_0=-4/3,
\epsilon_0=e_k$, and denote the resulting bracket by $[D]_\psi$.
The $D_{abi}$ and $\Xi$ rows vanish by the definitions of $e_k,a_k$.
The $D_{ab}$ row reduces to $-4-3\chi_0=0$, and the $\psi_i$ row
reduces to
$-(3k^2+15k+14)b_k/2+2(k+7)/3=0$. The remaining row is
\[
[\psi_q]_\psi=4-\frac43=\frac83.
\]
This also covers $k=0$, when the indexed rows are absent.

\emph{The endpoint boundary.}
Theorem~\ref{thm:cited-nonseparating-divisor-pullbacks} gives
\[
v_A^*\delta_{\rm irr}
=\delta_{\rm irr}^{Y_A}-(\psi_a+\psi_b)
 +\sum_{\text{separating types splitting }a\text{ from }b}\delta_{\rm type}.
\]
The irreducible and positive separating terms are annihilated by
$\lambda_1\lambda_2$. A rational tail containing $q$ and exactly
one new flag has dimension at most $k$, so $\psi_q^{k+1}$ restricts to zero there.
The remaining terms are precisely $-\Xi$, and hence
$[v_A^*\delta_{\rm irr}]_\psi=0$.
For $C\subseteq B$, the rational factor of $D_{abC}$ has
dimension $|C|$. Only $|C|=0,1$ can contribute with the single divisor factor in
$L_\psi=L_0|_{(a_k,b_k,-4/3,e_k)}$, and those are the two zero
boundary rows above. The rational factor of $D_{qabC}$ has
dimension $|C|+1$. If $C\ne B$, $\psi_q^{k+1}$ restricts to zero there; if
$C=B$, every term of $L_\psi$ restricts to that rational factor,
and the total degree is $k+2>k+1$. Thus both families vanish.

\emph{All compact allocations.}
The selected ordinary edges are bridges and remain separating under
degeneration. Thus $\nu_{ab}$ cuts an internal node of one source
factor with transverse coefficient one. Cutting the decorated
elliptic factor gives zero odd cohomology on genus zero. In a common
refinement with $j_A$, an original edge internal to $E_A$ raises its
odd degree from at least eleven to at least thirteen; an odd factor
outside $E_A$ cannot supply its degree eleven. Consequently a surviving
decorated allocation $A'$ contains $A$. An exchanged branch has
both $c,d$ outside its decorated factor, so none survives.
For an endpoint source with $S=A'\setminus A\ne\varnothing$, the
intervening rational factor has markings $S,q$ and its original
attaching flag, and dimension $|S|-1\le k-1$. The restriction of
$\psi_q^k$ to this factor is zero. For $A'=A$, stability excludes an intervening unmarked
rational tree with two attachments, and the bridges coincide.
An $H^{13}$ endpoint then has elliptic degree thirteen and does not contribute. An
$H^{11}$ endpoint contributes $-z_A\otimes\psi_q v_A^*\eta_A$.
The external cotangents and irreducible divisor vanish by the
local rows. The two cuts of an endpoint path give a positive divisor
and either $D_{abC}$ or $D_{qabC}$, whose pairings have already been shown to vanish.

A centered path has an intervening rational factor with markings
$S,q$ and two leaf attachments, of dimension $|S|$. Only $S=B$
can survive $\psi_q^k$. Then the original decorated factor carries
all of $I$. Cutting the genus-two leaf leaves two elliptic leaves,
on which $\lambda_1\lambda_2$ vanishes. Cutting the genus-one leaf
leaves a rigid rational $ab$ leaf and a genus-two leaf. The $q$ and
$B$ cotangents lie on the dimension-$k$ center, the $a,b$ cotangents
are zero, $D_{ab}$ is minus the attaching cotangent on the center,
and $D_{abi}$ restricts to a boundary divisor on the center.
Therefore every term of $L_\psi$ lies on the center, and
$\psi_q^kL_\psi$ has degree $k+1$ there. Thus the remaining
compact centered term vanishes, including $k=0$.

\emph{The node scalar and descent.}
Only the matching tail decorated by the node cotangent class contributes, with value
$-z_A\int_{Y_A}\theta_{\psi,B}\psi_q^2
=-(8/3)I_0z_A=-z_A/1080$.
Nonempty decorated marking sets distinguish the ordinary graph vertices;
the new pair distinguishes a lowered leaf. Their transverse
coefficients are one, and by Set-up~\ref{setup:g5-detector-row} the
order-eight projector onto $\mathscr D_F$ introduces no additional
normalization factor. Lemma~\ref{lem:g5-double-loop-coordinate} gives
the claimed left inverse on the full quotient.
\end{proof}

\begin{definition}\label{def:g5-picard-tests}
Assume $k=|B|\le3$. Let $\mathcal C_B$ have coordinates $x_T$ for
two-element subsets of $B$ and $y_U$ for three-element subsets.
For a vector of these coordinates define
\[
 X=\sum_Tx_T,\quad Y=\sum_Uy_U,\quad
 X_i=\sum_{T\ni i}x_T,\quad Y_i=\sum_{U\ni i}y_U,
\]
\[
 d_H=3k^2+15k+14,\quad m_H=(k^2+7k+14)/2,\quad
 S_H=\frac{6(Y-(k+2)X)}{d_H},
\]
\[
 r_i=\frac{-X-(k+3)X_i+Y_i-(k+4)S_H}{m_H},\qquad
 a_0=-(k+4)S_H-X,
\]
\[
 L_H=a_0\psi_q+\sum_{i\in B}r_i(\psi_i+D_{abi})
           +\sum_Tx_TD_T+\sum_Uy_UD_U,\qquad
 \theta_{H,B}(x,y)=\lambda_1\lambda_2\psi_q^kL_H.
\]
Define a second test by the rational coefficients
\[
 \begin{gathered}
 d_\lambda=21k^2+105k+50,\quad b_\lambda=-28(k+3)/d_\lambda,\quad
 c_\lambda=-19/7-(5k/14)b_\lambda,\quad e_\lambda=b_\lambda+2+c_\lambda,\\
 f_\lambda=56/25,\quad h_\lambda=96/25-(k/2)b_\lambda,\quad
 a_\lambda=-16/5-k(k+4)b_\lambda-2(k+4)-c_\lambda,
 \end{gathered}
\]
\[
 \begin{aligned}
 L_\lambda={}&a_\lambda\psi_q+b_\lambda\sum_{i\in B}\psi_i
                  +\psi_a+\psi_b+c_\lambda D_{ab}
                  +e_\lambda\sum_{i\in B}D_{abi}\\
 &+f_\lambda E_\varnothing+(b_\lambda+f_\lambda)\sum_{i\in B}E_i
                  +h_\lambda(E_a+E_b),\qquad
 \theta_{\lambda,B}=\lambda_2\psi_q^{k+1}L_\lambda.
 \end{aligned}
\]
Here $E_S$ is the genus-one/genus-one separating divisor with one
side carrying $S$; complementary descriptions designate the same
divisor and are counted once. Put
$\eta_B^0=\kappa_1+\sum_{T\subseteq B,\ |T|\ge2}\delta_{0,T}\in Z_B$.
Let $\mathcal K_F\subseteq\mathcal R_F^{\mathrm{Pic}}$ have
each divisor coefficient in $\langle\psi_q,\eta_B^0\rangle$,
and let $\mathcal R_{F,\psi}^{\mathrm{Pic}}$ be the subspace with divisor coefficient in the span of $\psi_q$.
\end{definition}

\begin{proposition}\label{prop:g5-picard-injection}
For every finite $F$, the families
$\mathcal T^H=(\mathcal T_A^{\theta_{H,B}(x,y)})_{A,(x,y)}$
and $\mathcal T^\lambda=(\mathcal T_A^{\theta_{\lambda,B}})_A$
annihilate $q_F(\mathscr U_F)$ and induce maps on $\mathscr Q^{\mathrm{Pic}}_F$.
On the source $\mathcal R_F^{\mathrm{Pic}}$,
\[
 \ker(\mathcal T^Hq_F\Phi_F^{\mathrm{Pic}})=\mathcal K_F.
\]
The family $\mathcal T^\lambda q_F\Phi_F^{\mathrm{Pic}}$ is diagonal
in $A$, is zero on $\mathcal R_{F,\psi}^{\mathrm{Pic}}$, and has
value $z_A/75$ on a coefficient $z_A\otimes\eta_B^0$.
In particular the map
\[
 \mathcal R_F^{\mathrm{Pic}}\longrightarrow \mathscr Q^{\mathrm{Pic}}_F,\qquad
 r\longmapsto[\delta_F\Phi_F^{\mathrm{Pic}}(r)]
\]
is injective.
\end{proposition}
\begin{proof}
\emph{The first family annihilates the auxiliary classes.}
For a divisor $D$ on $Y_A$, put
\[
[D]_H=I_0^{-1}\int_{Y_A}\theta_{H,B}(x,y)\psi_qD.
\]
Summing the defining numerators of $r_i$, using
$\sum_iX_i=2X$, $\sum_iY_i=3Y$, and
$m_H+k(k+4)=d_H/2$, gives $\sum_i r_i=S_H$.
Boundary restriction and Lemma~\ref{lem:g5-basic-hodge-moments} now give
\[
\begin{aligned}
[\psi_q]_H=[D_{ab}]_H&=a_0+(k+4)S_H+X=0,\\
[D_{abi}]_H&=r_i-r_i=0,\\
[\psi_i]_H&=-(k+4)S_H-X-(k+3)X_i+Y_i-m_Hr_i=0.
\end{aligned}
\]
For the last row, a pair divisor not containing $i$ contributes
$(k+3)x_T$, a pair containing $i$ contributes zero, and a triple
containing $i$ contributes $y_U$; the other triple terms vanish by
dimension. The cotangent terms contribute
\[
(k+4)a_0+(k+3)(k+4)S_H
 -\left(\frac{(k+3)(k+4)}2+2\right)r_i,
\]
and the $D_{abi}$ correction adds $r_i$. These are all the terms
in the displayed row.

Use the full irreducible pullback $v_A^*\delta_{\rm irr}$ of
Theorem~\ref{thm:cited-nonseparating-divisor-pullbacks}, as in the
proof of Proposition~\ref{prop:g5-node-left-inverse}. Its irreducible and
positive terms are annihilated by $\lambda_1\lambda_2$, and its
$q$-containing rational tails by $\psi_q^{k+1}$. For the remaining
class $\Xi$ defined there, the cotangent and $D_{abi}$ terms contribute
$8(a_0+(k+4)S_H)$. Each pair correction contributes $8x_T$:
the contribution $2(k+3)x_T$ of $\psi_a+\psi_b$ loses
$2(k-2)x_T$ from disjoint singleton tails and $2x_T$ from
the two tails with subset $T$. Triple corrections contribute zero.
Thus $[\Xi]_H=8(a_0+(k+4)S_H+X)=0$.
For $D_{abC}$ only $|C|\le1$ can contribute; these are the two
zero boundary rows. For $D_{qabC}$ the cotangent power at $q$ vanishes for every
proper $C\subset B$. If $C=B$, all terms of $L_H$, including its
pair and triple divisors, lie on the rational factor of dimension
$k+1$, where $\psi_q^{k+1}L_H$ vanishes.

The common-refinement argument of
Proposition~\ref{prop:g5-node-left-inverse} therefore applies to all
endpoint families. For centered paths the only additional case has
all original markings on the decorated center. The cut giving two
elliptic leaves is annihilated by $\lambda_1\lambda_2$. For the cut giving
a rigid rational $ab$ leaf, each added $D_T,D_U$ has rational side
contained in $B$, so it restricts to the rational center or to zero.
It cannot include the positive-genus leaf. Thus all of $L_H$ lies
on the dimension-$k$ center and $\psi_q^kL_H=0$ there.
It follows that $\mathcal T^Hq_F$ annihilates every summand of
$\mathscr U_F$, and that its Picard evaluation is diagonal in $A$,
with matching value $-I_0z_A[v_A^*\eta_A]_H$.

\emph{The first kernel.}
For a pair $T=\{i,j\}\subseteq B$, restriction to its rigid rational
tail gives
\[
[D_T]_H=a_0+(k+3)(S_H-r_i-r_j)-(k+3)x_T
 +\sum_{\substack{|T'|=2\\T'\cap T=\varnothing}}x_{T'}
 +\sum_{\substack{|U|=3\\U\supset T}}y_U.
\]
The normal class gives $-(k+3)x_T$, disjoint pairs give two rigid
tails, and containing triples give a rigid tail on the contracted
core. All crossing cases are incompatible. On a triple $U$ the
rational factor is $\Mbar_{0,4}$, giving
\[
[D_U]_H=\sum_{i\in U}r_i+\sum_{\substack{T\subset U\\|T|=2}}x_T-y_U.
\]
The terms are its cotangents, boundary points, and normal class.
There are no tail columns for $k=0,1$. For $k=2$, writing $x$
for the only pair parameter gives $S_H=-3x/7$, $r_i=-3x/14$,
$a_0=11x/7$, and $[D_B]_H=-24x/7$. For $k=3$, put
$y=y_B$. Substitution yields
\[
[D_T]_H=\frac{308y+94X-2064x_T}{473},\qquad
[D_B]_H=\frac{28X-40y}{43}.
\]
On the two-dimensional space $X=y=0$ the pair part is
$-48/11$ times the identity. On equal pair parameters
$x_{12}=x_{13}=x_{23}=x$ and triple parameter $y$, the matrix is
\[
\frac1{43}\begin{pmatrix}-162&28\\84&-40\end{pmatrix},
\qquad \det=\frac{96}{43}.
\]
This proves invertibility of the full labelled tail matrix.

Logarithmic dualizing differentials identify $v_A^*\kappa_1$
with the logarithmic $\kappa_1$ on $Y_A$; also
$v_A^*\psi_q=\psi_q$ and $v_A^*\delta_{0,T}=D_T$ for $T\subseteq B$.
The genus-two Picard relation from
Lemma~\ref{lem:g5-basic-hodge-moments} gives
\[
[v_A^*\kappa_1]_H=-\sum_{\substack{T\subseteq B\\|T|\ge2}}[D_T]_H.
\]
Indeed, $\lambda_1^2\lambda_2$ is pulled back from a
codimension-four class on the three-dimensional unmarked genus-two
space, and vanishes. Positive boundary terms vanish as above.
The external and node cotangents vanish, the $q$-tails and $abC$
tails have already been shown to have zero pairing, and the full irreducible-pullback
equation cancels the new cotangents with the one-new-flag tails.
Only the negative sum of the base rational tails remains.
Consequently, for vector coefficients
$z_\kappa\kappa_1+z_\psi\psi_q+\sum_Tz_T\delta_{0,T}$, all first
tests vanish exactly when $z_T=z_\kappa$ for every $T$.
Together with the zero node column this is precisely $\mathcal K_F$.

\emph{The second family and its positive boundary terms.}
Put
\[
m_\lambda=k+4,\qquad
X_\lambda=a_\lambda+k(k+4)b_\lambda+2(k+4)+c_\lambda,\qquad
[D]_\lambda=5760\int_{Y_A}\lambda_2\psi_q^{k+2}L_\lambda D.
\]
Theorem~\ref{thm:cited-lambda-g-integrals} gives coefficient $7$ for
a genus-two factor and coefficient $5760(1/24)^2=10$ for two
genus-one factors. In particular $\lambda_2$ restricts to
$\lambda_1'\lambda_1''$ on a positive normalization; those
contributions must be included. Restriction gives the following table:
\[
\begin{aligned}
[\psi_q]_\lambda={}&7X_\lambda+10f_\lambda,\\
[\psi_i]_\lambda={}&7\bigl[m_\lambda a_\lambda+
 (k-\tfrac12)m_\lambda(m_\lambda-1)b_\lambda
 +2m_\lambda(m_\lambda-1)\\
 &\hspace{25mm}+(m_\lambda-1)c_\lambda+e_\lambda\bigr]
 +10\bigl[(m_\lambda-1)f_\lambda+(b_\lambda+f_\lambda)\bigr],\\
[D_{ab}]_\lambda={}&7\bigl[a_\lambda+k(m_\lambda-1)b_\lambda
 -(m_\lambda-1)c_\lambda+ke_\lambda\bigr]+10f_\lambda,\\
[D_{abi}]_\lambda={}&7(b_\lambda+2+c_\lambda-e_\lambda),\\
[E_\varnothing]_\lambda={}&10\bigl[a_\lambda+k(m_\lambda-1)b_\lambda
 +2(m_\lambda-1)+c_\lambda\\
 &\hspace{25mm}-(m_\lambda-2)f_\lambda
 +k(b_\lambda+f_\lambda)+2h_\lambda\bigr],\\
[E_i]_\lambda={}&10\bigl(b_\lambda-(b_\lambda+f_\lambda)+f_\lambda\bigr),\\
[E_a]_\lambda+[E_b]_\lambda={}&20(1-h_\lambda+f_\lambda).
\end{aligned}
\]
The cotangent entries are the multinomial integrals on the surviving
genus-two factor. The $D_{ab}$ and $D_{abi}$ entries use the rigid
tail and $\Mbar_{0,4}$ normalizations as in the node calculation,
now with the genus-two coefficient $7$. Their positive intersections
and those of the cotangents give the terms with coefficient $10$.
In the $E_\varnothing$ row its normal class contributes
$-10(m_\lambda-1)f_\lambda$. There is also a branch with all
markings on a rational component, contributing $+10f_\lambda$;
together these give $-10(m_\lambda-2)f_\lambda$.
The intersections with $E_i,E_a,E_b$ contribute
$10k(b_\lambda+f_\lambda)$ and $20h_\lambda$.
Each is a product of two marked genus-one Hodge integrals
(counting the self-incidence of the unmarked elliptic side once).

For the full irreducible pullback of
Theorem~\ref{thm:cited-nonseparating-divisor-pullbacks}, retain the
persistent normal, the one-new-flag rational tails, and the positive
splitting branches. To make their signs explicit, the class $\Xi$ from
the node proof has bracket
\[
[\,\Xi\,]_\lambda
=7\bigl(8a_\lambda+8k(k+4)b_\lambda+16k+36\bigr)
 +60f_\lambda+20h_\lambda.
\]
Indeed, the two new cotangents contribute
\[
\begin{aligned}
[\psi_a+\psi_b]_\lambda
={}&7\bigl(2m_\lambda a_\lambda
 +2k m_\lambda(m_\lambda-1)b_\lambda
 +3m_\lambda(m_\lambda-1)+2ke_\lambda\bigr)\\
 &+20(m_\lambda-1)f_\lambda+20h_\lambda.
\end{aligned}
\]
Each of the $2k$ singleton rational tails contributes
$7(a_\lambda+(k-1)(k+3)b_\lambda+k+3+e_\lambda)+10f_\lambda$;
each of the $k(k-1)$ pair tails contributes $7(1+2b_\lambda)$;
larger tails vanish by dimension. Their subtraction gives the
displayed $\Xi$ bracket. The only positive splitting sides
that can receive the remaining divisor degree are $\{a\}$ and
$\{b\}$. Consequently
\[
\begin{aligned}
[v_A^*\delta_{\rm irr}]_\lambda
 &=-[\Xi]_\lambda+[E_a]_\lambda+[E_b]_\lambda\\
 &=-7\bigl(8a_\lambda+8k(k+4)b_\lambda+16k+36\bigr)
   -40f_\lambda-40h_\lambda+20 .
\end{aligned}
\]
Substituting the coefficients gives
\[
\begin{gathered}
X_\lambda=-16/5,\qquad
[\psi_q]_\lambda=[\psi_i]_\lambda=[D_{abi}]_\lambda
 =[E_i]_\lambda=0,\\
[D_{ab}]_\lambda+[E_\varnothing]_\lambda=0,\qquad
[v_A^*\delta_{\rm irr}]_\lambda=0.
\end{gathered}
\]

The complete pullback of an endpoint divisor with elliptic
side marked by $C\subseteq B\sqcup\{q\}$ is $D_{abC}+E_C$.
If $q\in C$, its rational part has dimension at most $k+1$
and its elliptic part has dimension at most $k+2$ with one
Hodge class; both terms are annihilated by $\psi_q^{k+2}$.
If $q\notin C$ and $|C|\ge2$, the side away from $q$ needs
at least two further divisor degrees, while $L_\lambda$ supplies
one. The cases $C=\varnothing,\{i\}$ are the cancellations
in the table. This proves all endpoint vanishings.

\emph{The kappa scalar.}
Restriction to the remaining base rational tails gives
\[
[D_T]_\lambda=
\begin{cases}
7\bigl(-3(k+2)b_\lambda-2\bigr),&|T|=2,\\
21b_\lambda,&|T|=3,\\
0,&|T|\ge4 .
\end{cases}
\]
For a pair, its rigid tail removes its two cotangents.
The remaining genus-two terms and its intersection with
$E_\varnothing$ give
\[
7\bigl(a_\lambda+(k-2)(k+3)b_\lambda+2(k+3)+c_\lambda\bigr)
 +10f_\lambda
=7X_\lambda+10f_\lambda+7\bigl(-3(k+2)b_\lambda-2\bigr).
\]
The node row is zero, giving the first value. For a triple,
only its three rational cotangents have top degree on $\Mbar_{0,4}$,
giving $21b_\lambda$; larger tails vanish by dimension. Apply the full genus-two relation
$\kappa_1=\sum\psi_i+2\lambda_1-D_0+D_1$.
The zero rows just proved remove the old cotangents, node
cotangent and endpoint pairs; the full irreducible equation
removes the one-new-flag terms. Since
$[\lambda_1]_\lambda=2X_\lambda$ by
Lemma~\ref{lem:g5-basic-hodge-moments}, the remaining terms are
\[
[v_A^*\kappa_1]_\lambda
=2([E_\varnothing]_\lambda+[E_a]_\lambda+[E_b]_\lambda)
 +4X_\lambda-\sum_{\substack{T\subseteq B\\|T|\ge2}}[D_T]_\lambda .
\]
The positive rows simplify to
$[E_\varnothing]_\lambda=10(X_\lambda-2-2f_\lambda+2h_\lambda)$
and $[E_a]_\lambda+[E_b]_\lambda=20(1-h_\lambda+f_\lambda)$.
Their sum is $10X_\lambda=-32$.
It follows that
\[
[v_A^*\eta_B^0]_\lambda
=2(-32)+4(-16/5)=-384/5.
\]
Thus its local integral is $-1/75$, and the bridge minus sign
gives $+1/75$ on the coefficient.

The common-refinement argument for the first family applies verbatim,
$\psi_q^{k+1}$ restricting to zero on every intervening rational factor
including the centered one of dimension $k$; so $\mathcal T^\lambda q_F$
annihilates $\mathscr U_F$ and the node column and is diagonal in $A$
with matching value $-z_A\psi_qv_A^*\eta_A$.

\emph{Injectivity of the combined test maps.}
For a class in the common kernel, the first family forces its divisor coefficient to lie in
$\langle\psi_q,\eta_B^0\rangle$. The second, with scalar $1/75$,
forces its $\eta_B^0$ coefficient to vanish. The node test, with
scalar $-1/1080$ by Proposition~\ref{prop:g5-node-left-inverse},
then forces its node coefficient to vanish.
Lemma~\ref{lem:g5-double-loop-coordinate} gives descent through
the complete separating denominator, and the auxiliary
vanishings give descent to $\mathscr Q_F^{\rm Pic}$.
\end{proof}

\begin{definition}\label{def:g5-auxiliary-source}
Use the sets $F,I,J$ of Set-up~\ref{setup:g5-detector-row} and the
four families of Definition~\ref{def:g5-auxiliary-picard-source}.
Let $\mathcal D_F^+$, $\mathcal E_F^+$, $\mathcal V_F^+$ be the
sub-sums of $\mathcal D_F$, $\mathcal E_F$, $\mathcal V_F$ over the
allocations whose decorated vertex's marking set $A$ meets
$\{c,d\}$, and keep the full centered-path source $\mathcal T_F$. Let
\[
\mathcal A_F^{\rm aux}=
(\mathcal D_F^+\oplus\mathcal T_F\oplus
\mathcal E_F^+\oplus\mathcal V_F^+)^{\langle t\rangle},
\qquad
\Phi_F^{\rm aux}:\mathcal A_F^{\rm aux}\longrightarrow N_{5,F}
\]
be the sum of their Gysin maps, the restriction of $\Omega_{4,I}$.

Write $\mathcal T_F=\mathcal T_{0,F}\oplus\mathcal T_{1,F}
\oplus\mathcal T_{2,F}$, where $\mathcal T_{i,F}$ for $i=1,2$
has both $c,d$ on the genus-$i$ unit leaf, and $\mathcal T_{0,F}$
has all other allocations. On $\mathcal T_{1,F}$ define
\[
\sigma_T(A,B\cup\{c,d\},C;z)
=(A,C\cup\{c,d\},B;(p_1\,p_2)^*z),
\]
and let $\mathcal T_{1,F}^\pm$ be its eigenspaces.
For $(A,B,C\cup\{c,d\};w)\in\mathcal T_{2,F}$, glue the
decorated center along $p_1$ to
$\Mbar_{0,B\cup\{c,d,q_1\}}$. This gives a degree-thirteen
class on $\Mbar_{1,A\cup B\cup\{c,d,p_2\}}$; placing it in
the elliptic-tail source with genus-three unit partner marked by $C$
defines $L:\mathcal T_{2,F}\to(\mathcal D_F^+)^{\langle t\rangle}$.
\end{definition}

\begin{proposition}[The auxiliary kernel]\label{prop:g5-auxiliary-kernel}
For every finite $F$, in the coordinate order
$(\mathcal D_F^+,\mathcal T_{0,F},\mathcal T_{1,F},
\mathcal T_{2,F},\mathcal E_F^+,\mathcal V_F^+)$, one has
\[
\ker(q_F\Phi_F^{\rm aux})=
\{(Lw,0,u,w,0,0):u\in\mathcal T_{1,F}^+,\ w\in\mathcal T_{2,F}\}.
\]
The map $q_F\Phi_F^{\rm aux}$ is injective on the direct complement
\[
(\mathcal D_F^+)^{\langle t\rangle}\oplus
\mathcal T_{0,F}^{\langle t\rangle}\oplus\mathcal T_{1,F}^-
\oplus(\mathcal E_F^+)^{\langle t\rangle}
\oplus(\mathcal V_F^+)^{\langle t\rangle}.
\]
Moreover, for a source vector $v$ with $\mathcal V_F^+$ coefficient
$e_A\otimes\delta_{\rm irr}+\sum_{i\in B}d_{A,i}\otimes\psi_i$ at an
old-present allocation $A$, $B=I\setminus A$, the following hold.
\begin{enumerate}
\item[(a)] (External Picard tests.) For each $C\subseteq B$, restrict
$q_F\Phi_F^{\rm aux}(v)$ to the open of the positive path with odd
endpoint carrying $A$, middle vertex carrying
$(B\setminus C)\sqcup\{a\}$, and other endpoint carrying
$C\sqcup\{b\}$, and take the $(11,2,2)$ component of the
self-intersection. The result is
$-(10e_A+\sum_{i\in C}d_{A,i})|_{\Mopen_{1,A\sqcup\{p\}}}\otimes x\otimes y$,
with $x,y$ the nonzero cotangents on the two partial elliptic
factors; the other families of $\mathcal A_F^{\rm aux}$ and the
exchanged terms contribute nothing to these tests, which therefore
recover the $\mathcal V_F^+$ coefficient.
\item[(b)] (Positive support tests.) If the $\mathcal V_F^+$
coefficient of $v$ vanishes, the same tests on the positive paths
obtained by lowering the genus-two unit vertex of an endpoint path
recover each $\mathcal E_F^+$ coefficient with scalar $1$; if
moreover the $\mathcal E_F^+$ coefficient vanishes, lowering the
genus-two leaf of a centered path recovers each $\mathcal T_{0,F}$
coefficient with scalar $1$ and each $\mathcal T_{1,F}^-$ coefficient
with scalar $2$. The elliptic $H^{13}$ tails and all classes
supported on rational strata contribute nothing to these tests.
\end{enumerate}
\end{proposition}
\begin{proof}
Coefficients are recovered in the order
$\mathcal V_F^+,\mathcal E_F^+,\mathcal T_{0,F},\mathcal T_{1,F}^-,\mathcal D_F^+$.

\emph{Restrictions on the selected strata.}
A collection of stable graphs closed under contraction defines an
open substack: its complement is the finite union of the proper
clutching images of the excluded graphs. On charts with labelled nodes,
each selected bridge has its own smoothing parameter and normal
class $-\psi_h-\psi_{h'}$. A uniquely determined set of selected
bridges therefore defines a regular embedding, with the curve
stabilizers retained.

Let $\Gamma$ be a three-elliptic-vertex path in $\Mbar_{3,J}$ with
a specified odd vertex, replace its two unit vertices by rational
vertices with one loop each, and let $V_\Gamma$ be the open union
of the strata obtained by contracting edges of the resulting graph.
The normalization over this
open is
\[
\Mopen_{1,L_0}\times\mathscr U_{L_1}\times\mathscr U_{L_2},
\]
where $L_0,L_1,L_2$ are the respective marking and bridge-flag
sets of the odd vertex and of the two unit vertices, and
$\mathscr U_L$ is the open of
Lemma~\ref{lem:ordinary-image-cotangent-restrictions}, the smooth
or irreducible one-nodal genus-one curves with distinct smooth
markings.

If the odd vertex is the center, this is the situation of
Lemma~\ref{lem:positive-path-detection} with leaf genera $(1,1)$;
in every application below the two unit leaves carry different
external label sets. That lemma makes the normalization a closed
regular embedding of codimension two, gives the self-intersection
with $(11,2,2)$ component
$z|_{\Mopen_{1,L_0}}\otimes\psi_{q_1}\otimes\psi_{q_2}$, injective
in the odd coefficient $z$, and shows that $V_\Gamma$ meets no
other two-edge tree stratum and no two-edge tree stratum with a
rational vertex.

If the odd vertex is an endpoint, the same argument applies with
the roles of the vertices changed. The open meets only the
specified two-edge tree: a tree contraction must contract both
loops and retain both bridges. The bridges are determined by their
labelled sides. Two nested
bridges of the same type would enclose an unmarked rational tree
with two exterior attachments. On such a tree the sum of valences
minus two is zero, whereas stability makes every summand positive.
Thus there is no duplicate bridge.
Lemma~\ref{lem:elliptic-eleven-open-restriction} makes restriction
of the odd coefficient to $\Mopen_{1,L_0}$ injective. The
cotangents on each $\mathscr U_L$ are equal, since
Proposition~\ref{prop:genus-three-picard-input} identifies them
modulo rational-tail divisors with the elliptic Hodge class
$\lambda$, and nonzero by
Lemma~\ref{lem:ordinary-image-cotangent-restrictions}.
It follows that self-intersection on this path is injective on
the odd coefficient in the K\"unneth component of degree eleven
there and degree two on each unit vertex. The coefficient is
$+1$: from each bridge normal factor choose a cotangent on a
different unit vertex. If the odd vertex is an endpoint, choose
the middle cotangent at the first bridge and the other endpoint
cotangent at the second. All other choices have different
multidegrees. Odd decorations on a different vertex are excluded
by parity.

In both cases, by Theorem~\ref{thm:cited-thirteen-cohomology},
elliptic $H^{13}$ is generated by rational-tail Gysin classes, so
its tail sources are supported on trees with a rational vertex and
do not meet these positive-path opens.

\emph{The two kernel families.}
Every selected source edge is a bridge, which remains separating
through compact degenerations. A node normalized by $\nu_{ab}$
must therefore lie inside a source factor. Each such branch lowers
that factor's genus and puts $a,b$ on it, with transverse
coefficient one. Lowering an odd elliptic factor gives zero by
Theorem~\ref{thm:cited-genus-zero-cycles}. The compact factors
in these pullbacks already contain all deeper degenerations.

For $\mathcal T_{1,F}$ let $R_1$ denote the positive branch
obtained by lowering the genus-two leaf. Its leaves carry the
old pair and the new pair respectively. Pair exchange interchanges
these leaves and the two center flags, so on this source
\[
q_F\Phi_F^{\rm aux}=R_1(1-\sigma_T).
\]
The other branch lowers the genus-one leaf to a rational leaf
carrying all four flags, and cancels under $1-s^*$.
Exchanging ordinary bridge Thom classes has positive sign;
no determinant is imposed on these bridges. Thus
$\mathcal T_{1,F}^+$ is contained in the kernel.

For $w\in\mathcal T_{2,F}$ the positive branch has all four flags
on the lowered leaf and cancels. The remaining rational path has
$B\sqcup\{a,b\}$ on its rational leaf and $C\sqcup\{c,d\}$ on
its genus-two leaf, minus its pair exchange. The composite
Gysin defining $Lw$ gives the exchanged path. Composition
preserves complex orientations and the two Tate twists, so
\[
q_F\Phi_F^{\rm aux}(w)=-q_F\Phi_F^{\rm aux}(Lw).
\]
The rational factor in $L$ is stable also for $B=\varnothing$,
since it carries $c,d$ and its node. This is an equality of
compact classes with all $H^{13}$ relations imposed.
Subtracting $(Lw,0,u_+,w,0,0)$ from any source vector therefore
leaves its image unchanged and places it in the displayed complement.

\emph{Recovery of partner divisors.}
Use the path of (a).
Write $p,q$ for the first bridge flags and $u,v$ for the second,
and $x,y$ for the common nonzero cotangents on the middle and
unit endpoint partial elliptic factors.
After gluing $a,b$, the partner divisor is restricted along the
parallel gluing of those two elliptic factors. By
Theorem~\ref{thm:cited-nonseparating-divisor-pullbacks} at the two
persistent nonseparating nodes, its full
irreducible-boundary restriction is
\[
\delta_{\rm irr}^{(1)}+\delta_{\rm irr}^{(2)}
 -\psi_a-\psi_b-\psi_u-\psi_v
=12x+12y-2x-2y=10x+10y.
\]
Both persistent nonseparating nodes contribute their normal
lines, and an additional node on either factor contributes the
corresponding transverse Cartier branch. These are all branches
on the chosen partial product.

Write its coefficient as in (a).
The first bridge normal multiplies this by $-\psi_p-\psi_q$.
Its $(11,2,2)$ component is the class displayed in (a).
The $\psi_p$ term raises the odd degree, and the $x^2$ term
has multidegree $(11,4,0)$.
Only this unexchanged partner-divisor allocation contributes.
An elliptic tail can isolate either endpoint of the path, and
the other endpoint places the odd class in a different K\"unneth summand. An exchanged
odd endpoint has a new flag and no old flag, so cannot carry
$A$. Endpoint-path branches have $a,b$ together on the lowered
unit vertex, while this test splits them. Centered paths have
their odd degree at the middle vertex; rational trees miss the
open. Hence $C=\varnothing$ recovers $10e_A$, and $C=\{i\}$
then recovers $d_{A,i}$. Injectivity of smooth restriction and
the nonzero factors $x,y$ force all coefficients in the $\mathcal V_F^+$ summand to vanish.

\emph{Recovery of the positive paths.}
Lower the unique genus-two unit vertex of an endpoint path.
Its old-present endpoint and the vertex carrying $a,b$ recover
the original path and its marking allocation uniquely.
An exchanged endpoint is old-absent, centered paths place the
odd class on the middle vertex, and classes supported on rational strata restrict to zero on the test open.
The self-intersection just proved recovers every
$\mathcal E_F^+$ coefficient with scalar $1$.

After these coefficients have been shown to vanish, lower the genus-two leaf of a
$\mathcal T_{0,F}$ source. Raising the vertex carrying $a,b$
recovers the source uniquely. If an old flag lies on the
decorated center, no exchanged branch has that center. Otherwise
the old pair is split between the leaves, whereas an exchanged
branch has the old pair together. The same test has scalar $1$
and forces the coefficients in $\mathcal T_{0,F}$ to vanish.
The $\mathcal T_{1,F}^{-}$ targets have the old pair on one
leaf and the new pair on the other, with none on the center.
These allocations exclude the preceding targets and their
exchanges. The formula $R_1(1-\sigma_T)$ gives scalar $2$ and
recovers the entire negative eigenspace.

\emph{Recovery of elliptic degree-thirteen tails.}
Only $\mathcal D_F^+$ remains in a relation. Choose a nonzero
coefficient $z_A$ with $|A|$ minimal, and let
$W_A\subseteq\Mbar_{3,J}$ be the open of
Lemma~\ref{lem:minimal-elliptic-tail-detection} for the partner
genus $h=2$ and the decomposition $J=A\sqcup(J\setminus A)$, with
tail normalization
$\mathcal M^{\rm ct}_{1,A\sqcup\{p\}}\times
\Mopen_{2,(J\setminus A)\sqcup\{q\}}$.
By that lemma, any genus-one tail meeting $W_A$ has its external
labels contained in $A$. Strict subsets have zero coefficients by
minimality; an equal-sized set must be $A$. Exchanged terms contain
a new flag outside $A$. The only remaining $(13,2)$ component is
therefore
\[
-z_A|_{\mathcal M^{\rm ct}_{1,A\sqcup\{p\}}}
 \otimes\psi_q|_{\Mopen_{2,(J\setminus A)\sqcup\{q\}}},
\]
which depends injectively on $z_A$ by the same lemma, a
contradiction.

Nonempty decorated marking sets distinguish odd
endpoints, and the old and new pairs distinguish the relevant
unit leaves. Thus the recovery scalars are exactly $1,1,2$;
node-flip stabilizers remain inside the partial elliptic stacks.
\end{proof}

\begin{lemma}[Multiplicities of the genus-four sources]
\label{lem:genus-four-source-multiplicities}
Let $(F,\mu)$ be one of
$([11],(1^{11}))$, $([11],(2,1^9))$, $([12],(2,2,2,1^6))$,
$([11],(3,1^8))$. Use the sources of
Definitions~\ref{def:g5-auxiliary-source},
\ref{def:g5-auxiliary-picard-source}, and~\ref{def:g5-parallel-source};
the $\kappa_1$, $\psi_q$, and rational-tail columns give the three
summands of the Picard source, and the parallel column is
$M_\mu(\mathcal R^{\mathrm{par}}_{+,F})$. The multiplicities of
$\Pmot$ in these sources, each computed after taking invariants
under exchange of the old flags and applying the functor $M_\mu$,
are as follows; entries marked $-$ are not used.
\[
\begin{array}{c|rrrrrrr|rrr|r}
\mu&\mathcal D_F^+&\mathcal T_{0,F}&\mathcal T_{1,F}^+
 &\mathcal T_{1,F}^-&\mathcal T_{2,F}&\mathcal E_F^+&\mathcal V_F^+
 &\kappa_1&\psi_q&\text{tails}&\mathrm{par}\\\hline
(1^{11})&25&36&-&2&-&36&15&6&6&3&6\\
(2,1^9)&45&102&-&6&-&72&28&12&12&8&12\\
(2,2,2,1^6)&5&16&1&0&1&6&2&0&0&1&0\\
(3,1^8)&25&96&2&4&6&36&13&6&6&5&6
\end{array}
\]
For $\mu=(2,2,2,1^6)$ the rational-tail summand is supported precisely
on $|A\cap F|=10$, $|A\cap\{c,d\}|=1$. For such an allocation,
let $e$ be the unique old flag outside $A$ and write
$\{x,y\}=F\setminus A$. Then its partner coefficient is the line
$\mathbb Q\bigl(\delta_{0,\{e,x\}}-\delta_{0,\{e,y\}}\bigr)$.
For $\mu=(1^{11}),(2,1^9)$ the sources $S_{0,\mu},S_{e,\mu},R_{0,\mu}$
of Definition~\ref{def:g5-selected-separating-sources} have
multiplicities $114,9,6$ and $253,9,6$.

The multiplicity of $\Pmot$ in $N_\mu=M_\mu(N_{5,F})$ is
$179$, $349$, $32$, $205$, respectively. For $\rho=(2,2,2,1^6)$ and
$(3,1^8)$, the cohomological maps of the sources induce an isomorphism
\[
 M_\rho(\mathcal A_F^{\rm aux})\oplus
 M_\rho(\mathcal R_F^{\mathrm{Pic}})\oplus
 M_\rho(\mathcal R^{\mathrm{par}}_{+,F})
 \xrightarrow{\ \sim\ }N_\rho,
\]
whose first summand maps onto $M_\rho(\mathscr U_F)$.
\end{lemma}

\begin{proof}
\emph{The parent tables.}
By Theorems~\ref{thm:genus-four-base} and~\ref{thm:ordinary-image},
the genus-four compact multiplicity series is
\[
 E\partial U+3HE+4HE^2+GE^2.
\]
The multiplicity in $N_\mu$ is the coefficient of $s_\mu h_2$ in
this series: the two loop flags carry ordinary invariants.
The horizontal-two-strip parents of $(1^{11})$ are
$(3,1^{10})$ and $(2,1^{11})$; those of $(2,1^9)$ are
$(4,1^9)$, $(3,2,1^8)$, $(3,1^{10})$, $(2,2,1^9)$.
For $\rho=(2,2,2,1^6)$ the parents are
$(4,2,2,1^6)$, $(3,2,2,2,1^5)$, $(3,2,2,1^7)$, $(2,2,2,2,1^6)$, and for
$\rho=(3,1^8)$ they are $(5,1^8)$, $(4,2,1^7)$, $(4,1^9)$, $(3,3,1^7)$, $(3,2,1^8)$.
The required coefficients are
\[
\begin{array}{c|rrrrr}
 & (4,1^9)&(3,2,1^8)&(3,1^{10})&(2,2,1^9)&(2,1^{11})\\\hline
 E\partial U&12&8&17&9&9\\
 3HE&12&6&18&6&6\\
 4HE^2&40&32&72&36&36\\
 GE^2&30&24&18&9&3\\\hline
 \text{sum}&94&70&125&60&54
\end{array}
\]
The same series is $E\partial U+H(2E+3E^2)+(H+G)E^2+HE$, the four
summands $\mathsf D,\mathsf W,\mathsf T,\mathsf R$ of
Theorem~\ref{thm:ordinary-image} at genus four; grouped in this way,
the coefficients at the parents of the two nine-row partitions are
\[
\begin{array}{c|rrrr}
\theta&\mathsf D&\mathsf W&\mathsf T&\mathsf R\\\hline
(4,2,2,1^6)&1&0&6&0\\
(3,2,2,2,1^5)&0&0&0&0\\
(3,2,2,1^7)&4&9&12&0\\
(2,2,2,2,1^6)&0&0&0&0\\\hline
\text{sum}&5&9&18&0
\end{array}
\qquad
\begin{array}{c|rrrr}
\theta&\mathsf D&\mathsf W&\mathsf T&\mathsf R\\\hline
(5,1^8)&3&0&15&0\\
(4,2,1^7)&2&0&15&0\\
(4,1^9)&12&38&40&4\\
(3,3,1^7)&0&0&6&0\\
(3,2,1^8)&8&28&32&2\\\hline
\text{sum}&25&66&108&6.
\end{array}
\]
The table can be evaluated by the following finite rule. Removing corners gives
\[
 E\partial U=E^2\bigl(s_{(2,1^{10})}+e_{11}+s_{(2,1^9)}\bigr)
       +EE_{\ge2}H+EE_{\ge3}(H+G).
\]
For a contained core $\rho\subseteq\theta$, the coefficient from
$E^2s_\rho$ is $s_{\theta/\rho}(1,1)$, and that from
$EE_{\ge j}s_\rho$ is
\[
 \sum_{v=0}^{|\theta|-|\rho|-j}
 \langle s_{\theta/\rho},h_vh_{|\theta|-|\rho|-v}\rangle.
\]
An empty sum is zero; a single $E$ contributes precisely when the skew diagram is
horizontal. These rules give all the displayed entries by successive
horizontal strips. The lists of partitions are
exhaustive by $\theta_i\ge\rho_i\ge\theta_{i+1}$ and
$|\theta|-|\rho|=2$. In the first table, the second and fourth parents
have four boxes in column two, while every possible core with at most
ten rows acquires at most two additional boxes in that column from its
two complete-function factors. In the second table, the parent
$(3,3,1^7)$ contributes $1+2+3=6$ from the three nine-row cores in
$GE^2$; it is included in the count.
Thus the multiplicities in $N_\mu$ are
$125+54=179$, $94+70+125+60=349$, $5+9+18+0=32$, and
$25+66+108+6=205$.

\emph{The restriction rule.}
For a set with $u$ external labels, $r$ old flags and $v$ node flags,
let $\operatorname{Std}_u$ be the augmentation kernel of the
permutation module on the external labels. The elliptic hook
restricts as
\[
 \bigwedge^{10}\bigl(\operatorname{Std}_u\oplus\mathbb Q^{r+v}\bigr).
\]
This is the exterior-power form of the tuple presentation in
Definition~\ref{def:low-genus-eleven-hook}. When $r=2$, the old
transposition has one negative line among the auxiliary directions,
and invariants omit that line. When $r=1$, its two allocations form
one orbit and the invariant coefficient is one representative.
For induction over the remaining labels, the multiplicity is the
pairing of the corresponding skew Specht character with the unit
or divisor character on those labels.

\emph{The sign and ten-row partitions.}
For the old-present centered hook the auxiliary positive space has
dimension three in either old-pair case. After distributing all
remaining external labels over the two unit leaves, its counts
for a fixed old-pair pattern are
\[
\begin{array}{c|rrrr}
 u&8&9&10&11\\\hline
 (1^{11})&0&3&6&1\\
 (2,1^9)&2&13&12&3.
\end{array}
\]
There are two choices for the other old flag when just one is on
the center, and one choice when both are there. With neither old
flag on the center and with the pair split between the leaves,
the counts for $u=9,10,11$ are $(1,4,1)$ and $(4,6,2)$.
Hence the $\mathcal T_{0,F}$ counts are
$3(3+6+1)+(1+4+1)=36$ and
$3(2+13+12+3)+(4+6+2)=102$.

For clarity, the entry $13$ comes from three sign-nine endpoint
coefficients, each with leaf counts $1+1+2$, and one further
vertical-two-box skew character contributing on the split leaves.
The entry $2$ at $u=8$ comes from the two leaf sizes $(1,2)$ and
$(2,1)$. On $\mathcal T_{1,F}$, the leaf exchange also exchanges
the node flags. Its negative-eigenspace multiplicities at $u=9,10,11$ are
$(0,2,0)$ and $(2,3,1)$, respectively. At $u=9$ for the sign
partition the sign of exchanging the two singleton leaf labels
cancels the sign of exchanging the two exterior node directions;
at $u=10$ the two leaf locations pair off. For the ten-row
partition the split two-label skew character has both signs.
This gives $2$ and $6$ for $\mathcal T_{1,F}^-$.

For an old-present endpoint the auxiliary positive space has
dimension two. For each genus order and each fixed old-pair
pattern the $u=9,10,11$ counts are $(1,4,1)$ and $(4,6,2)$.
There are three such patterns and two genus orders, giving
$36$ and $72$ for $\mathcal E_F^+$.
For the external Picard source, its partner character is
$\mathbf1\oplus\mathbb Q[B]$. The contributions with one and two
old flags on the endpoint are
\[
\begin{array}{c|rrr|rrr}
 &\multicolumn{3}{c|}{\text{one old flag}}&
   \multicolumn{3}{c}{\text{two old flags}}\\
 u&9&10&11&9&10&11\\\hline
 (1^{11})&1&6&2&1&4&1\\
 (2,1^9)&4&9&4&3&6&2.
\end{array}
\]
For example, at $u=9$ with one old flag the partner consists of
one fixed old flag and two external labels. Its divisor character
is three trivial lines and one sign line for their transposition.
The sign partition uses the latter; the ten-row skew character
uses both. The totals are $15$ and $28$.

The full elliptic $H^{13}$ tail source has character
$E\partial U$. Its two-old-flag invariant coefficients were
computed in the parent tables above:
they are $17+9=26$ and $12+8+17+9=46$.
If neither old flag is on the elliptic factor, the threshold in
Definition~\ref{def:low-genus-thirteen-modules} forces all eleven
external labels onto it. The restriction of
$V_{(2,1^{10})}$ after fixing the node contains each of our two
partitions once. Removing this one old-absent copy leaves $25$
and $45$. We have therefore obtained
\[
\begin{array}{c|rrrrr|r}
 &\mathcal D_F^+&\mathcal T_{0,F}&\mathcal T_{1,F}^-&
   \mathcal E_F^+&\mathcal V_F^+&S_{0,\mu}\\\hline
 (1^{11})&25&36&2&36&15&114\\
 (2,1^9)&45&102&6&72&28&253.
\end{array}
\]

With no old flag on an endpoint, the hook is
$\bigwedge^{10}\operatorname{Std}_u\oplus
\bigwedge^9\operatorname{Std}_u$. For either partition, only
$u=11$ and $u=10$ contribute: once at $u=11$, and once for each
of the two unit placements of the remaining label at $u=10$.
Each of the three selected patterns in $\mathcal S_e$ therefore
contributes three, giving $S_{e,\mu}=\Pmot^9$.
The same restrictions give one copy for each of the two $m=0$
columns and each of the four $m=1$ columns, hence
$R_{0,\mu}=\Pmot^6$.

It remains to count $R_{p,\mu}$. The rooted divisor generators are
independent by Corollary~\ref{cor:ordinary-image-rooted-picard-basis}.
For either one or two old flags on the elliptic endpoint, its
invariant hook is
\[
 \bigwedge^{10}\operatorname{Std}_u
 \oplus2\bigwedge^9\operatorname{Std}_u
 \oplus\bigwedge^8\operatorname{Std}_u.
\]
For a constant partner coefficient the counts at $u=9,10,11$
are $(0,2,1)$ for the sign partition and $(1,3,2)$ for the
ten-row partition. Each of $\kappa_1$ and $\psi_q$ therefore
contributes $6$ and $12$, respectively, after the two old-pair
cases.
For rational-tail coefficients with one old flag on the endpoint,
$u=10$ gives the single divisor on the partner $\{d,i\}$, with
multiplicities $2$ and $3$. At $u=9$ the partner is
$\{d,i,j\}$ and its four divisors are
\[
 D_{di},\ D_{dj},\ D_{ij},\ D_{dij}.
\]
Under $(i\,j)$ they give three trivial lines and one sign line.
The sign partition contributes one and the ten-row partition four.
With both old flags on the endpoint, the partner $\{i,j\}$ has
one invariant tail divisor, contributing zero and one.
Thus the rational-tail counts are $3$ and $8$, and
\[
 R_{p,(1^{11})}=\Pmot^{6+6+3},\qquad
 R_{p,(2,1^9)}=\Pmot^{12+12+8}.
\]
All contributing endpoints have $u\ge9$, so their partners have
at most three flags. The stated restriction on the Picard source
loses no term.
Finally put
$R_{\mathrm{par},\mu}=M_\mu(\mathcal R^{\mathrm{par}}_{+,F})$.
By Proposition~\ref{prop:g5-parallel-left-inverse}, the maps $\rho_A$
identify its coefficients with the old-present single-node elliptic
hook, with a unit partner. The constant-partner count just given
therefore gives
\[
 R_{\mathrm{par},(1^{11})}=\Pmot^6,\qquad
 R_{\mathrm{par},(2,1^9)}=\Pmot^{12}.
\]

\emph{The nine-row partitions.}
Let now $\rho=(2,2,2,1^6)$ or $(3,1^8)$. For the character counts put
\[
 \mathcal H_j=\sum_{r\ge1}s_{(r,1^{j-1})}\quad(8\le j\le11).
\]
Thus $H=\mathcal H_{11}+\mathcal H_{10}$ and
$G=\mathcal H_{10}+\mathcal H_9$.

\emph{Allocations without an old flag.}
An elliptic endpoint with no old flag has only its attaching flag fixed.
On its ordinary labels its one-node hook has at least ten rows.
Every Littlewood--Richardson constituent contains that hook, so neither
of the nine-row partitions under consideration occurs, whatever the
partner coefficient. This excludes all such endpoint-path and Picard
allocations.
The parallel hook has the same property, since $\rho_A$ is an
isomorphism (Proposition~\ref{prop:g5-parallel-left-inverse}).

For the degree-thirteen endpoint, the only possible full flag counts
without an old flag are twelve and thirteen. By
Definition~\ref{def:low-genus-thirteen-modules} and
Theorem~\ref{thm:cited-thirteen-cohomology}, the corresponding characters
are
\[
 \ch(U_{1,[12]})=s_{(2,1^{10})},\qquad
 \ch(U_{1,[13]})=s_{(2,1^{10})}h_1+e_{10}h_3.
\]
Fixing the attaching flag applies $\partial$. All resulting terms
have at least ten rows except $e_9h_3$ in the second expression.
The latter cannot contain $(2,2,2,1^6)$, since the skew diagram above
$(1^9)$ is a vertical three-strip. At $|F|=11$ only the twelve-flag
case is possible. Thus these degree-thirteen allocations also vanish.
Consequently the full auxiliary multiplicity is accounted for by
$\mathcal A_F^{\rm aux}$, and the full parallel multiplicity comes from
allocations with at least one old flag on the decorated vertex.

\emph{Centered paths.}
Two ordered unit allocations are counted by $E^2$. If a hook core is
$\eta\subseteq\rho$, its contribution is $s_{\rho/\eta}(1,1)$.
The complete core calculations are
\[
\begin{array}{c|cc}
\rho &[s_\rho]\mathcal H_9E^2 &[s_\rho]\mathcal H_8E^2\\\hline
(2,2,2,1^6)&0+1=1&0+2=2\\
(3,1^8)&3+2+1=6&6+4+2=12.
\end{array}
\]
For the first row the skew Schur functions are $e_3,e_2$ over the nine-row cores,
and $e_3h_1,e_2h_1$ over the eight-row cores. For the second row
they are $h_2,h_1,1$ and $h_2h_1,h_1^2,h_1$.

If both old flags lie on the center, its augmentation representation
on the ordinary center labels is the standard representation plus
three trivial lines and one $t$-sign line. Invariants exclude the sign
line. The resulting hook character is
$\mathcal H_{11}+3\mathcal H_{10}+3\mathcal H_9+\mathcal H_8$.
The same character occurs when exactly one old flag is on the center
and the other is on a specified leaf. There are two such choices of
leaf; each $t$-orbit is counted once. With the old flags on different
leaves the character is
$\mathcal H_{11}+2\mathcal H_{10}+\mathcal H_9$.
Thus the $\mathcal T_{0,F}$ entries are
$3(3\cdot1+2)+1=16$ and $3(3\cdot6+12)+6=96$.
Both $\mathcal T_{1,F}$ and $\mathcal T_{2,F}$ have multiplicities
one and six, respectively in the two cases.

The node exchange acts by $-1$ on the nine-row center hook.
For $(2,2,2,1^6)$ the sole $\mathcal T_{1,F}$ contribution has ten
ordinary center labels; its complementary pair has the sign character
and its split-leaf allocation changes sign under leaf exchange.
Their product gives $\sigma_T=+1$.
For $(3,1^8)$, center sizes nine, ten, and eleven give respectively
$(1,2)$, $(1,1)$, and $(0,1)$ copies in the positive and negative
eigenspaces of $\sigma_T$. For size nine, the allocations with both
remaining labels on a single leaf form an exchanged pair, whereas the
trivial constituent of the split allocation has leaf-exchange sign
$+1$. For size ten the two possible locations of the remaining label
are exchanged. For size eleven only the center sign remains.
This proves the asserted $\mathcal T_{1,F}^\pm$ counts.

\emph{Endpoint and Picard generators.}
Write $a_0=|A\cap F|$. If exactly one old flag is on the endpoint,
its augmentation representation is
$\operatorname{Std}_{A\cap F}\oplus\mathbb Q^2$.
If both old flags are there, it is
$\operatorname{Std}_{A\cap F}\oplus\mathbb Q^2\oplus\mathbb Q_{\rm sign}$;
taking $t$-invariants again excludes the last line. In either case the
only nine-row term is $s_{(a_0-8,1^8)}$, once.

For $(2,2,2,1^6)$ only $a_0=9,10$ can contribute. At size nine the three
remaining ordinary labels require their sign representation. Neither
allocations to two unit vertices nor the permutation representations
on markings or subsets of three markings contain that sign. At size
ten the complementary pair requires sign. For endpoint paths it must
split between the unit vertices. There are four choices when one old
flag is on the endpoint, and two when both are there, giving six.
For the external cotangents the sign line is $\psi_x-\psi_y$, once
for each of these two old-flag patterns, giving two. The coefficients
$\kappa_1$ and $\psi_q$ are trivial and contribute zero. For rational
tails, $B=\{e,x,y\}$ has precisely the sign line
$\delta_{0,\{e,x\}}-\delta_{0,\{e,y\}}$.
When $B=\{x,y\}$ its only stable subset is fixed and contributes zero.
This proves the one-copy tail assertion; every contributing complement
has size three and hence occurs in $\mathcal R_F^{\mathrm{Pic}}$.
The invariant parallel source has a trivial partner coefficient, so
its multiplicity is zero.

For $(3,1^8)$ the sizes are $a_0=9,10,11$. The complementary skew
is horizontal and requires the trivial representation on the remaining
ordinary labels. Endpoint paths give $4\cdot6+2\cdot6=36$ copies,
using the centered-path core count above. In the non-node Picard
family, one old flag on the endpoint gives $3,3,2$ copies at these
sizes: $\delta_{\mathrm{irr}}$, the cotangent at the other old flag,
and, when present, the sum of the ordinary cotangents. Both old flags
on the endpoint give $2,2,1$ copies. The total is thirteen.
Each of $\kappa_1$ and $\psi_q$ gives one copy for every size and
old-flag pattern, hence six copies. For rational tails the numbers of
orbits of stable subsets of $B$ are
\[
\begin{array}{c|rrr}
\text{old flags on the endpoint}&a_0=9&a_0=10&a_0=11\\\hline
\text{one}&3&1&0\\
\text{both}&1&0&0.
\end{array}
\]
For example, when $B=\{e,x,y\}$ the three orbits are
$\{x,y\}$, the pair $\{e,x\},\{e,y\}$, and $B$.
This gives five tail copies. All these complements have size at most
three. The parallel collapse gives the same count as a single trivial
Picard coefficient, namely six.

\emph{Independence for the nine-row partitions.}
The endpoint, non-node Picard, and remaining Picard generating spaces
have total multiplicities $6+2+1=9$ and $36+13+17=66$.
They surject onto the corresponding $\mathsf W$ multiplicities, which
have exactly these dimensions by the tables. The generating maps are
therefore isomorphisms on these multiplicity spaces. This proves the required
Picard splitting without imposing any additional divisor relation.
The degree-thirteen entries are five and twenty-five, and the other
two ordinary summands have already been counted.
Together they give $N_\rho\cong\Pmot^{32}$ and $\Pmot^{205}$ and
the claimed source decomposition.
\end{proof}