\section{Marking supports in genus five}\label{sec:09-marking-supports}

We identify the open degree-fifteen quotient in genus five with the
direct sum of the images of its eleven-marking pullbacks. Injectivity of the sum of
these maps follows from rational-tail restriction and weighted forgetting. Compact boundary
restriction and the primitive marking quotient then reduce generation
to seven symplectic coefficient groups.

\begin{theorem}[Independence of marking supports]\label{thm:marking-support-independence}
Let $g\geq2$ and $d\geq2$, and suppose $\mathcal O_{g,d-1}=0$.
For every finite set $F$ with $|F|\geq d$, the sum of
forgetful pullbacks is injective:
\[
\bigoplus_{\substack{S\subseteq F\\|S|=d}}\mathcal O_{g,S}
\longrightarrow\mathcal O_{g,F},\qquad
(x_S)_S\longmapsto\sum_S\pi_S^*x_S.
\]
Here $\pi_S$ forgets the markings outside $S$.
\end{theorem}

\begin{proof}
For any marking set $A$, write
\[
r_A:H^{15}(\Mbar_{g,A})\longrightarrow\mathcal O_{g,A}.
\]
By Proposition~\ref{prop:primitive-degree-fifteen-quotient}, this map
is surjective, its kernel is $\mathcal I_{g,A}$, and
$W_{13}H^{13}(\Mopen_{g,A})=0$. Denote the sum in the statement by
$q_F$, and put $n=|F|$.

\emph{Rational-tail restriction.}
For distinct $i,j\in F$ and a new label $u$, put
$F'=(F\setminus\{i,j\})\sqcup\{u\}$. Let
\[
s_{ij}:\Mbar_{g,F'}\longrightarrow\Mbar_{g,F}
\]
attach at $u$ the three-pointed rational curve carrying $i,j$.
Its image is the divisor $D_{ij}$, with normalization
$\Mbar_{g,F'}\times\Mbar_{0,3}=\Mbar_{g,F'}$.
We claim that $r_{F'}s_{ij}^*$ annihilates $\mathcal I_{g,F}$.
It suffices to test the complete one-edge Gysin generators of that
space. Over $\Mopen_{g,F'}$, the image of $s_{ij}$ has just the node
defining $D_{ij}$, so a class supported on any other boundary divisor
pulls back to zero there. For a class $s_{ij*}\eta$ from $D_{ij}$ itself,
the normal class is $-\psi_u$ because $\psi$ vanishes on $\Mbar_{0,3}$, and hence
\[
\left.s_{ij}^*s_{ij*}\eta\right|_{\Mopen_{g,F'}}
=-\psi_u\smile\left.\eta\right|_{\Mopen_{g,F'}}=0.
\]
Indeed, $\eta\in H^{13}(\Mbar_{g,F'})$ restricts into the zero group
$W_{13}H^{13}(\Mopen_{g,F'})$. The labelled rational tail has no flag-exchange
factor. The claim gives a unique map
\[
c_{ij}:\mathcal O_{g,F}\longrightarrow\mathcal O_{g,F'},
\qquad c_{ij}r_F=r_{F'}s_{ij}^*.
\]

For a $d$-element support $S\subset F$, one computes
$c_{ij}\pi_S^*x_S$ by stabilizing the composite $\pi_Ss_{ij}$.
If $S$ avoids $i,j$, the result is $\pi_S^*x_S$ on $F'$, with
support avoiding $u$. If $S$ contains exactly one of them, say $v$,
the result is the pullback from
$(S\setminus\{v\})\sqcup\{u\}$ of $x_S$ relabelled by $v\mapsto u$.
If $S$ contains both, the composite factors through the collision map
with target $\mathcal O_{g,(S\setminus\{i,j\})\sqcup\{u\}}=0$.

\emph{Weighted forgetting.}
Let $\pi_i:\Mbar_{g,F}\to\Mbar_{g,F\setminus\{i\}}$ forget $i$.
The degree-preserving map
\[
z\longmapsto\pi_{i*}(\psi_i\smile z)
\]
sends $\mathcal I_{g,F}$ into $\mathcal I_{g,F\setminus\{i\}}$.
To see this, let $\jmath$ be a normalized one-edge clutching map.
The projection formula gives
\[
\pi_{i*}(\psi_i\smile\jmath_*\eta)
=(\pi_i\jmath)_*(\jmath^*\psi_i\smile\eta).
\]
If the node survives forgetting $i$, this class remains supported on
the boundary. The only node that can disappear is the rational tail
with ordinary labels $i,k$ for some $k$. On its normalization,
$\jmath^*\psi_i=0$, since the cotangent line comes from
$\Mbar_{0,3}$. Thus there is a descended map
\[
T_i:\mathcal O_{g,F}\longrightarrow\mathcal O_{g,F\setminus\{i\}},
\qquad T_ir_F=r_{F\setminus\{i\}}\pi_{i*}(\psi_i\smile-).
\]
Proposition~\ref{prop:external-stable-forgetting} and the projection
formula imply
\begin{equation}\label{eq:marking-supports-diagonal-transfer}
T_i\pi_i^*=(2g+n-3)\operatorname{id}.
\end{equation}

\emph{The mixed transfer identity.}
Fix $k\neq i$, and write
\[
\begin{aligned}
\mathcal X&=\Mbar_{g,F},&
\mathcal Y_i&=\Mbar_{g,F\setminus\{i\}},\\
\mathcal Y_k&=\Mbar_{g,F\setminus\{k\}},&
\mathcal Z&=\Mbar_{g,F\setminus\{i,k\}}.
\end{aligned}
\]
Let $f_k:\mathcal Y_i\to\mathcal Z$ and
$f_i:\mathcal Y_k\to\mathcal Z$ forget the indicated markings.
On all rational cohomology, one has
\begin{equation}\label{eq:marking-supports-cycle-action}
\pi_{i*}\pi_k^*=f_k^*f_{i*}.
\end{equation}
To prove this identity, we compare the corresponding algebraic cycles. Both $f_i,f_k$ are
representable proper flat universal curves. Their fiber product
$\mathcal W=\mathcal Y_i\times_{\mathcal Z}\mathcal Y_k$ has the
expected dimension, and its fundamental cycle represents the
composition of the graph correspondences for $f_{i*}$ and $f_k^*$.
By \cite[Proposition 3.10, diagram (22)]{bae2022}, the induced map
$\tau=(\pi_i,\pi_k):\mathcal X\to\mathcal W$ is proper and
birational, and it satisfies $\tau_*[\mathcal X]=[\mathcal W]$. Flatness excludes components
of excess dimension, and the degree is one over the locus of smooth
curves with distinct markings. Pushing this equality to
$\mathcal Y_i\times\mathcal Y_k$ gives equality of the two
correspondence cycles, and hence of their cycle classes. Acting by
cup product and proper pushforward gives
\eqref{eq:marking-supports-cycle-action} in every degree.

Suppose now that $\mathcal O_{g,n-2}=0$. For
$x_k\in\mathcal O_{g,F\setminus\{k\}}$, choose a compact lift $z_k$.
The psi identity and \eqref{eq:marking-supports-cycle-action} give
\[
\pi_{i*}(\psi_i\smile\pi_k^*z_k)
=f_k^*f_{i*}(\psi_i^{\mathcal Y_k}\smile z_k)
 +\pi_{i*}([D_{ik}]\smile\pi_k^*z_k),
\]
where $\psi_i^{\mathcal Y_k}$ is the cotangent class on $\mathcal Y_k$.
The first term restricts through $\mathcal O_{g,n-2}$ and vanishes.
On $D_{ik}$, both forgetful maps contract the rational component and
are isomorphisms onto their targets, with $i,k$ interchanged.
The second term is therefore the relabelled class, with coefficient one. Writing $(i\,k)$ for
this relabelling, we obtain
\begin{equation}\label{eq:marking-supports-mixed-transfer}
T_i\pi_k^*x_k=(i\,k)x_k\qquad(k\neq i).
\end{equation}

\emph{Independence with $d+1$ markings.}
The case $n=d$ is the identity map. For $n=d+1$, index a putative
relation by the omitted label:
\[
\sum_{i\in F}\pi_i^*x_i=0,
\qquad x_i\in\mathcal O_{g,F\setminus\{i\}}.
\]
Applying $c_{ik}$ and identifying its target with
$\mathcal O_{g,F\setminus\{i\}}$ gives
\[
x_i+(i\,k)x_k=0.
\]
All other terms vanish by the collision calculation, since their
supports contain both $i,k$ and $\mathcal O_{g,d-1}=0$.
Applying $T_i$ to the relation and using
\eqref{eq:marking-supports-diagonal-transfer} and
\eqref{eq:marking-supports-mixed-transfer}, we find
\[
0=(2g+n-3)x_i+\sum_{k\neq i}(i\,k)x_k
  =(2g+n-3-(n-1))x_i=(2g-2)x_i.
\]
Since $g\geq2$, every $x_i$ is zero.

\emph{Independence in larger arities.}
Suppose $n\geq d+2$ and independence is known on $n-1$ markings.
Let $q_F((x_S)_S)=0$, and fix one $d$-element subset $S\subset F$.
Choose distinct $i,j\in F\setminus S$ and apply $c_{ij}$.
The collision calculation expresses the result as a relation among
$d$-element supports in $F'$. In that relation the coefficient at
$S$, which avoids $u$, is exactly $x_S$: a support meeting $i,j$
either vanishes or becomes a support containing $u$. The induction
hypothesis makes every coefficient zero, in particular $x_S$.
This holds for each $S$, and proves the theorem.
\end{proof}

\begin{lemma}\label{lem:marking-supports-row-bounds}
For $0\leq n\leq12$, put
\[
r_n=\begin{cases}
9,&0\leq n\leq10,\\
8,&n=11,\\
7,&n=12.
\end{cases}
\]
Then the following hold.
\begin{enumerate}[label=(\roman*)]
\item For every stable $(g,n)$ with $g\geq1$ and every partition
$\mu\vdash n$ with $\ell(\mu)\leq8$, one has
\[
\operatorname{Hom}_{\Sn_n}(V_\mu,B^{(1)}_{g,n})=0.
\]
\item For every $0\leq n\leq12$ and every partition $\mu\vdash n$
with $\ell(\mu)<r_n$, one has
\[
\operatorname{Hom}_{\Sn_n}
\bigl(V_\mu,H^{15}(\Mbar_{5,n})\bigr)=0.
\]
\item For $h=2,3,4$, every $m$ with $(h,m)$ stable, and every partition
$\mu\vdash m$ with $\ell(\mu)\leq8$, one has
\[
\operatorname{Hom}_{\Sn_m}
\bigl(V_\mu,H^{15}(\Mbar_{h,m})\bigr)=0.
\]
\item For every stable $(g,F)$, the summand of $C^1_{\mathrm{sep}}$ in
Definition~\ref{def:compact-boundary-row} formed by the K\"unneth
sectors with degree pairs $(13,2)$, $(2,13)$, $(11,4)$, $(4,11)$ on the
separating normalizations has row bound nine (Lemma~\ref{lem:row-bound-calculus}).
\end{enumerate}
\end{lemma}

\begin{proof}
\emph{The source and the lower-genus targets.}
Assertion (i) is Lemma~\ref{lem:row-bound-calculus}(3), and every
ordinary one-edge image in degree fifteen inherits the bound. Consequently
Propositions~\ref{prop:genus-two-gysin-surjectivity}
and~\ref{prop:genus-three-generation}, together with
Theorem~\ref{thm:genus-four-base}, give at least nine rows for
$H^{15}(\Mbar_{h,m})$ for $h=2,3,4$ and every $m$, which is
assertion (iii).
Proposition~\ref{prop:genus-five-initial} gives the same bound for
$h=5$ and $m\leq10$. Theorem~\ref{thm:genus-one} and the primitive
term of Theorem~\ref{thm:motivic-types} give
\[
H^{15}(\Mbar_{1,m})=0\quad(1\leq m\leq12),
\qquad
H^{15}(\Mbar_{1,13})=\mathcal I_{1,[13]}.
\]
In particular the latter group also has at least nine rows.
The genus-zero degree-fifteen group is zero by
Theorem~\ref{thm:motivic-types}.

\emph{The mixed separating sectors.}
Let $(g,F)$ be stable. By
Theorem~\ref{thm:cited-low-odd-vanishing}, the only possible degree
pairs on a separating normalization are
\[
(15,0),\ (0,15),\ (13,2),\ (2,13),\ (11,4),\ (4,11).
\]
In the $(13,2)$ and $(2,13)$ terms,
Theorem~\ref{thm:cited-thirteen-cohomology} and the bound on $U$
give at least nine rows after fixing the node. In the $(11,4)$
and $(4,11)$ terms, Proposition~\ref{prop:eleven-hook-realizations}
places the odd class on an elliptic vertex; fixing its node leaves
at least ten rows. Induction and graph invariants preserve these
bounds regardless of the representation on the even factor. This
proves assertion (iv).

\emph{The complete compact boundary.}
For $n=11,12$, Proposition~\ref{prop:compact-restriction-row} gives
an injection
\[
H^{15}(\Mbar_{5,n})\lhook\joinrel\longrightarrow C^1_{5,[n]},
\]
since $15<2\cdot5-2+n$. The mixed separating terms have at least nine
rows by (iv).

The nonseparating term is
\[
H^{15}(\Mbar_{4,[n]\sqcup\{c,d\}})^{\langle(c\,d)\rangle}.
\]
It has at least eight rows by the nine-row genus-four bound (iii)
and Lemma~\ref{lem:row-bound-calculus}(1). The exchange fixes the single
edge, so its action on the edge determinant in $C^1$ is trivial.

Consider a separating $(15,0)$ term; the $(0,15)$ term is identical
with the vertices interchanged. If the degree-fifteen vertex has
genus $h<5$, it has at most $n+1\leq13$ flags. The lower-genus
results just proved give at least nine rows before fixing its
node, and hence at least eight afterward. This also covers $h=1$,
whose factor vanishes unless it has thirteen flags; the rational
case is zero.

If that vertex has genus five, its rational partner carries $b\geq2$
ordinary markings. The genus-five factor then has $m=n-b+1$ flags.
At $n=11$ one has $m\leq10$, so its nine-row bound leaves eight
rows after fixing the node. Every summand of $C^1_{5,[11]}$ has
therefore at least eight rows, which proves the bound for $n=11$.
At $n=12$ and $b\geq3$, the same argument applies. The remaining
case is $b=2$, $m=11$: the bound just obtained has eight rows before
fixing the node and seven afterward. Hence every summand of
$C^1_{5,[12]}$ has at least seven rows. The boundary injections
prove the remaining cases $n=11,12$ of assertion (ii).
\end{proof}

\begin{proposition}[The seven coefficient reduction]\label{prop:genus-five-support}
For every $0\leq r\leq14$ and every partition $\lambda\vdash r$
of length at most five, one has
\[
\begin{aligned}
W_{15}H^{15-r}(\Mopen_5,\mathbb V_\lambda)&=0
\quad\text{unless }r=11\text{ and }\lambda\in\mathscr R_{11},\\
\mathscr R_{11}&=\{(11),(10,1),(9,2),(9,1,1),(8,3),(8,2,1),(8,1,1,1)\}.
\end{aligned}
\]
Put $\mathcal C_\lambda=W_{15}H^4(\Mopen_5,\mathbb V_\lambda)$.
There is a canonical equivariant isomorphism
\[
\mathcal O_{5,11}\cong
\bigoplus_{\lambda\in\mathscr R_{11}}
\mathcal C_\lambda\otimes V_{\lambda^T}.
\]
For every $n\geq11$, the sum of eleven-marking forgetful
pullbacks gives an isomorphism
\[
\operatorname{Ind}^{\Sn_n}_{\Sn_{11}\times\Sn_{n-11}}[\mathcal O_{5,11}\boxtimes\mathbf1]\xrightarrow{\sim}\mathcal O_{5,n}.
\]
Equivalently, restriction to the smooth locus gives an exact sequence
\[
0\to\mathcal I_{5,[n]}\to H^{15}(\Mbar_{5,n})\to\mathcal O_{5,n}\to0
\]
with $\mathcal O_{5,n}$ the induced module displayed above.
\end{proposition}

\begin{proof}
Throughout we use the primitive quotient of
Proposition~\ref{prop:primitive-degree-fifteen-quotient} with $g=5$:
$\mathcal O_{5,n}/\mathcal F^{15}_{5,n}$ is the direct sum over
$\lambda\vdash n$ with $\ell(\lambda)\leq5$ of
$W_{15}H^{15-n}(\Mopen_5,\mathbb V_\lambda)\otimes V_{\lambda^T}$.

\emph{The first possible support.}
Proposition~\ref{prop:genus-five-initial} gives
$\mathcal O_{5,n}=0$ for $0\leq n\leq10$. For $r\leq10$ the primitive
quotient is therefore zero, and since its Specht factors are all
nonzero, $W_{15}H^{15-r}(\Mopen_5,\mathbb V_\lambda)=0$ for every
$\lambda\vdash r$ of length at most five. Thus
$\mathcal F^{15}_{5,11}=0$, and the primitive quotient identifies
$\mathcal O_{5,11}$ with the full sum of size-eleven coefficients.
It is a quotient of $H^{15}(\Mbar_{5,11})$, so
Lemma~\ref{lem:marking-supports-row-bounds} excludes every
$V_{\lambda^T}$ with $\ell(\lambda^T)=\lambda_1\leq7$, hence
\[
W_{15}H^4(\Mopen_5,\mathbb V_\lambda)=0
\quad\bigl(|\lambda|=11,\ \ell(\lambda)\leq5,\ \lambda_1\leq7\bigr).
\]
For first parts $11,10,9,8$, the remaining parts partition
$0,1,2,3$, respectively. Their $1,1,2,3$ possibilities are
exactly $\mathscr R_{11}$. This proves the first isomorphism.

\emph{No new support at twelve markings.}
The primitive quotient at $n=12$ is a subquotient of
$H^{15}(\Mbar_{5,12})$, so the same lemma gives
\[
W_{15}H^3(\Mopen_5,\mathbb V_\lambda)=0
\quad\bigl(|\lambda|=12,\ \ell(\lambda)\leq5,\ \lambda_1\leq6\bigr).
\]
For the complementary range $\lambda_1\geq7$,
Lemma~\ref{lem:low-degree-symplectic-vanishing}, applied with
$d=3$, $g=5=d+2$, and $\lambda_1>3$, gives the stronger
vanishing of the full group $H^3(\Mopen_5,\mathbb V_\lambda)$.
These two ranges cover every admissible partition of twelve.
Therefore $\mathcal O_{5,12}=\mathcal F^{15}_{5,12}$.

\emph{No new support in higher arities.}
If $\lambda\vdash n$ with $n=13,14$ and $\ell(\lambda)\leq5$, then
$\lambda_1\geq\lceil n/5\rceil=3>15-n$, so the same lemma with
$d=15-n$, for which $5\geq d+2$, gives
$H^{15-n}(\Mopen_5,\mathbb V_\lambda)=0$ for every such $\lambda$.
The primitive quotient therefore vanishes, and
$\mathcal O_{5,n}=\mathcal F^{15}_{5,n}$ at $n=13,14$.
For every $n\geq15$ this equality is the last assertion of
Proposition~\ref{prop:primitive-degree-fifteen-quotient}.
This proves the first assertion.

Iterating these equalities and composing the forgetful maps
gives
\[
\mathcal O_{5,n}
=\sum_{\substack{S\subseteq[n]\\|S|=11}}
\pi_S^*\mathcal O_{5,S}\qquad(n\geq11).
\]
Indeed, at each larger arity every eleven-element subset misses a
label, and forgetting that label followed by forgetting to $S$ is
precisely $\pi_S$.

\emph{The pullback isomorphism and the restriction kernel.}
Theorem~\ref{thm:marking-support-independence}, with $g=5$, $d=11$,
applies because $\mathcal O_{5,10}=0$. It makes the displayed sum
injective, while the preceding argument makes it surjective.
The stabilizer of a fixed eleven-element subset is
$\Sn_{11}\times\Sn_{n-11}$, and its second factor acts trivially
on the corresponding summand. Thus the labelled direct sum is the
induced representation in the statement.

The exact sequence follows, since the kernel of compact restriction
is $\mathcal I_{5,[n]}$ by
Proposition~\ref{prop:primitive-degree-fifteen-quotient}.
\end{proof}