\section{Homology, no shear, and the common Lagrangian}
\label{sec:homology}

The geometric maps now have a common boundary surface on which their action
can be read.  This section records the embedded meridian and dual-longitude
transports, computes the blocks of the homology action, and extracts the Lagrangian data
needed for the Walker lift.

The homological calculation uses the global maps \(\Fglob e\) from
\cref{prop:global-extension}.  A cut map or quotient map has neither the
domain nor the exterior data required in this section.

For each edge \(e:v\to w\), define
\begin{equation}
 h_e=p_w(\Fglob e|_{\partial_+W})p_v^{-1}:Y\longrightarrow Y.
 \label{eq:outgoing-map-section12}
\end{equation}
Let
\[
 U=\Span_{\Q}\{m_i,m_j,m_k\},
 \qquad
 U^\vee=\Span_{\Q}\{\lambda_i,\lambda_j,\lambda_k\}.
\]
The vertex parameterizations send the embedded meridian and
product-framed dual-longitude rows to the corresponding fixed homology
labels on \(Y\).

\subsection{Embedded transports}

\begin{proposition}[Geometric row transport]
\label{prop:geometric-row-transport}
The same global maps \(\Fglob e\) used in \eqref{eq:outgoing-map-section12}
induce the following transports of embedded product-framed rows:
\begin{equation}
\begin{array}{c|c|c}
e&\text{meridians}&\text{dual longitudes}\\ \hline
0B&(i_0,j_0,k_0)\mapsto(i_0,J,k_0)
 &(\ell_i,\ell_j,\ell_k)\mapsto(\ell_i,\ell_j,K_{kj})\\
B\star&(i_0,J,k_0)\mapsto(I_\star,J,k_0)
 &(\ell_i,\ell_j,K_{kj})\mapsto(\ell_i,L_{ji},K_{kj})\\
0A&(i_0,j_0,k_0)\mapsto(I_A,j_0,k_0)
 &(\ell_i,\ell_j,\ell_k)\mapsto(\ell_i,L_{ji},\ell_k)\\
AAC&(I_A,j_0,k_0)\mapsto(I_\star,j_0,k_0)
 &(\ell_i,L_{ji},\ell_k)\mapsto(\ell_i,L_{ji},K_{ki})\\
AC\star&(I_\star,j_0,k_0)\mapsto(I_\star,J,k_0)
 &(\ell_i,L_{ji},K_{ki})\mapsto(\ell_i,L_{ji},K_{kj}).
\end{array}
\label{eq:embedded-row-transports}
\end{equation}
The literal spectator pairs, in the same edge order, are
\begin{equation}
 (i_0,\ell_i),\quad(k_0,K_{kj}),\quad(k_0,\ell_k),\quad
 (j_0,L_{ji}),\quad(I_\star,\ell_i).
 \label{eq:spectator-row-pairs}
\end{equation}
Every meridian image has zero dual-longitude component, and every
dual-longitude image has zero meridian component.
\end{proposition}

\begin{proof}
On each active local branch, the positive two-site theorem
\cref{thm:positive-two-site-input} and the selected compatible
representative transport the two embedded mover/carrier rows.
On the spectator branch, the coordinate identity transports the pair in
\eqref{eq:spectator-row-pairs} with its complete product framing.
\Cref{prop:pair15} identifies the two signed representatives with the same
normal coordinate, so these cut-level transports descend to embedded
curves in the quotient.  The boundary identity
\eqref{eq:global-boundary-conjugacy} and the parameterizations \(p_v,p_w\)
then give the displayed global transports.  In the terminal \(AC\star\)
row, the exceptional-ribbon isotopy identifies the raw product-framed
\(\widetilde K\) ribbon of \eqref{eq:raw-sharp-longitude} with
\(K_{kj}\); its product normal and zero turn are part of that
construction.

Let \(P_U\) and \(P_{U^\vee}\) denote the projections associated with
the active splitting \(U\oplus U^\vee\).  The meridian half of
\eqref{eq:embedded-row-transports} consists of boundaries of the
transported embedded meridian disks.  The transporting annuli join such
disk boundaries without entering a longitude track, and therefore
\[
 P_{U^\vee}h_{e*}|_U=0.
\]
Independently, every dual-longitude row is carried by its displayed
product-framed annulus.  These annuli have zero turn in the fixed product
normal and introduce no winding about a meridian core; the terminal
exceptional-ribbon reserve is compared to \(K_{kj}\) by a product-normal isotopy
with the same zero-turn property.  Consequently their classes are
linear combinations only of the displayed dual-longitude rows, so
\[
 P_Uh_{e*}|_{U^\vee}=0.
\]
The two zero mixed components are thus conclusions about two distinct
families of embedded annuli, proved before any matrix is introduced.
\end{proof}

\subsection{The block action}

The exterior identity in \eqref{eq:old-unused-identity} gives the splitting
\begin{equation}
 H_1(Y;\Q)=H_{\mathrm{old}}\oplus
 \bigoplus_{t\notin\{i,j,k\}}
   (\Q m_t\oplus\Q\ell_t)\oplus(U\oplus U^\vee).
 \label{eq:global-homology-splitting}
\end{equation}

\begin{corollary}[No-shear block form]
\label{cor:no-shear-block-form}
\leavevmode\par\nobreak
In the ordered active basis
\((m_i,m_j,m_k,\lambda_i,\lambda_j,\lambda_k)\),
\begin{equation}
 h_{e*}=I_{\mathrm{ext}}\oplus
 \begin{pmatrix}M_e&0_3\\0_3&N_e\end{pmatrix},
 \qquad N_e=M_e^{-T}.
 \label{eq:block-action-section12}
\end{equation}
The five keyed matrices and their route products are retained as repository
reproducibility data; the public argument uses only the block identities
displayed here.
\end{corollary}

\begin{proof}
The two geometric zero statements in
\cref{prop:geometric-row-transport} give the off-diagonal zero blocks.
Reading the meridian coefficients in
\eqref{eq:embedded-row-transports} gives \(M_e\), and reading the
independently transported dual-longitude coefficients gives \(N_e\).
Direct calculation gives \(N_e=M_e^{-T}\).  The old and unused summands are fixed by the actual
global map, giving \(I_{\mathrm{ext}}\).  No symplecticity argument is
used to create either zero block.
\end{proof}

The matrix arithmetic also gives
\begin{equation}
 M_{B\star}M_{0B}
 =M_{AC\star}M_{AAC}M_{0A}
 =\begin{pmatrix}1&0&0\\1&1&0\\1&1&1\end{pmatrix}.
 \label{eq:route-matrix-equality}
\end{equation}
This is a consistency check and a homological consequence of the embedded
rows.  It will not be used to produce the relative isotopy in
\cref{sec:rigidity-pentagon}.

\subsection{The common Lagrangian}

Set
\begin{equation}
 \Lambda=\lambda_g^Y\oplus
 \bigoplus_{t\notin\{i,j,k\}}\Q[m_t]\oplus U.
 \label{eq:common-lagrangian-section12}
\end{equation}
The old summand is Lagrangian by hypothesis, the unused meridians are the
disk-kernel summands of the unused handles, and \(U\) is the active
meridian summand.  Hence \(\Lambda\) is a rational Lagrangian in
\(H_1(Y;\Q)\).

\begin{proposition}[Two-sided preservation]
\label{prop:two-sided-lagrangian}
For every edge \(e\), the actual global map and its actual inverse satisfy
\begin{equation}
 h_{e*}\Lambda=\Lambda=h_{e*}^{-1}\Lambda.
 \label{eq:two-sided-lagrangian-section12}
\end{equation}
\end{proposition}

\begin{proof}
The global map fixes \(\lambda_g^Y\) and every unused meridian.  By
\cref{cor:no-shear-block-form}, it sends \(U\) into \(U\) by the
unimodular matrix \(M_e\); hence it sends \(U\) onto \(U\).  Therefore
\(h_{e*}\Lambda=\Lambda\).  The inverse of \(\Fglob e\) constructed in
\cref{prop:global-extension} induces the actual inverse \(h_e^{-1}\).
Applying \(h_{e*}^{-1}\) to the established equality gives
\(h_{e*}^{-1}\Lambda=\Lambda\).  This proves both directions and does not
infer invariance merely from dimensions.
\end{proof}

If \(N=3\), the unused direct sums in
\eqref{eq:global-homology-splitting} and
\eqref{eq:common-lagrangian-section12} are empty.  The proof then reduces
verbatim to the old and active summands; no additional case is needed.