\subsection{Strict expansion of the weighted volume}
\label{sec:strict-volume-expansion}

Expanding section measures provide a useful route to
almost-everywhere statements in homogeneous cosmology~\cite{Brehm}.
For the present argument the expansion must remain strict even
where the curvature polynomial $Q$ is negative. The factor
$1+4H$ in~\eqref{eq:section-volume-weight} achieves this
through a sharp algebraic lower bound for $Q$.

\begin{lemma}
\label{lem:curvature-product-lower-bound}
For $a,b,c>0$,
\begin{equation}
 Q(a,b,c)\geq-3(abc)^{2/3}.
 \label{eq:curvature-product-lower-bound}
\end{equation}
Consequently, on $\M$ the function
\begin{equation}
 \mathcal D(x)=\frac{2[Q(x)+4H(x)]}{1+4H(x)}
       \geq\frac{2H(x)}{1+4H(x)}>0
 \label{eq:positive-corrected-divergence}
\end{equation}
is continuous and strictly positive.
\end{lemma}

\begin{proof}
By symmetry assume $a\geq b\geq c>0$, and put
$r=(abc)^{1/3}$. The identity
\[
 (a+b+c)^3-27abc
  =9c\big[(a-b)^2+(a-b)(b-c)+(b-c)^2\big]
       +(a+b-2c)^3
\]
gives $a+b+c\geq3r$. A second identity is
\begin{equation*}
 \begin{split}
 (a+b+c)(Q+3r^2)
    &=(a-b)^2(a+b-c)+c(a-c)(b-c)\\
    &\quad+3r^2(a+b+c-3r).
 \end{split}
\end{equation*}
Every term on the right is nonnegative. Dividing by
$a+b+c>0$ proves~\eqref{eq:curvature-product-lower-bound}.
Apply this with $a,b,c=N_1,N_2,N_3$ and $r^2=H$.
The denominator in~\eqref{eq:positive-corrected-divergence}
is positive, and $Q+4H\geq H>0$.
\end{proof}

Let $F$ denote the polynomial vector field in
\eqref{eq:normalized-flow}. Its divergence in the five
original coordinates is
\begin{equation}
 \operatorname{div}F=2Q-3S^2,
 \qquad S^2=p^2+s^2.
 \label{eq:ambient-flow-divergence}
\end{equation}
Indeed, the first two diagonal derivatives are $Q,Q$,
and the last three are the three wall rates, whose sum
is $-3S^2$. Along a positive-wall solution,
\begin{equation}
 (N_1N_2N_3)'=-3S^2N_1N_2N_3,\qquad H'=-2S^2H.
 \label{eq:volume-weight-evolution}
\end{equation}
Hence the weight $\mathcal W=N_1N_2N_3(1+4H)$ satisfies
\[
 \frac{\mathcal W'}{\mathcal W}
        =-3S^2-\frac{8S^2H}{1+4H}.
\]
Using $S^2+Q=1$ yields the exact cancellation
\begin{equation}
 \operatorname{div}F-\frac{\mathcal W'}{\mathcal W}
       =2Q+\frac{8S^2H}{1+4H}
       =\mathcal D.
 \label{eq:weighted-divergence-identity}
\end{equation}

For $x\in\M$ and $t\in I(x)$ write
$P_t=D\phi_t(x)$, with derivative taken in the five
original initial coordinates. Set
\begin{equation}
 J(x,t)=\frac{\mathcal W(x)}{\mathcal W(\phi_t x)}
                                      \det P_t.
 \label{eq:weighted-flow-jacobian}
\end{equation}

\begin{proposition}
\label{prop:weighted-flow-expansion}
The multiplier in~\eqref{eq:weighted-flow-jacobian} is
\begin{equation}
 J(x,t)=\exp\left(\int_0^t\mathcal D(\phi_u x)\dd u\right).
 \label{eq:weighted-flow-exponential}
\end{equation}
For every positive time $t$, $J(x,t)>1$. On tangent
vectors to the constraint, the four-form $\Omega$ and
the flux form $\omega$ of Section~\ref{sec:small-section-flux}
obey
\begin{align}
 \Omega_{\phi_t x}(P_tv_1,\ldots,P_tv_4)
       &=J(x,t)\Omega_x(v_1,\ldots,v_4),
 \label{eq:constraint-volume-transport}\\
 \omega_{\phi_t x}(P_tv_1,P_tv_2,P_tv_3)
       &=J(x,t)\omega_x(v_1,v_2,v_3).
 \label{eq:section-flux-transport}
\end{align}
\end{proposition}

\begin{proof}
The ordinary flow differential satisfies
$P_t'=DF(\phi_t x)P_t$ and $P_0=\mathrm{Id}$.
Differentiating the determinant by row multilinearity
gives
\[
 (\det P_t)'=\operatorname{tr}(DF(\phi_t x))\det P_t.
\]
In this calculation, replacing a row by any other row
makes its determinant zero; only the diagonal entries of
$DF$ contribute. Thus the identity also holds without
assuming the matrix is invertible. Differentiating
\eqref{eq:weighted-flow-jacobian} and using
\eqref{eq:weighted-divergence-identity} gives
$J'=\mathcal D(\phi_t x)J$, $J(x,0)=1$.
The scalar integrating factor proves
\eqref{eq:weighted-flow-exponential}. On every compact
positive-time interval the integrand is continuous and
positive, so its integral is strictly positive.

To restrict this multiplier to the constraint, differentiate
the ordinary flow identity $\mathcal C\circ\phi_t=\mathcal C$.
It gives
\[
 d\mathcal C_{\phi_t x}P_t=d\mathcal C_x.
\]
In particular $P_t(x/2)$ is a normalized transverse vector
at $\phi_t x$, and $P_tv_i$ are tangent there. Use this
transverse vector in~\eqref{eq:constraint-volume-form}.
Multilinearity of the ambient determinant gives
\[
 \begin{split}
 \Omega_{\phi_t x}(P_tv_1,\ldots,P_tv_4)
  &=\frac{\det P_t}{\mathcal W(\phi_t x)}
                       \mathcal V(x/2,v_1,\ldots,v_4)\\
  &=J(x,t)\Omega_x(v_1,\ldots,v_4).
 \end{split}
\]
Finally, differentiating the flow identity in its time
shift gives $P_tF(x)=F(\phi_t x)$.
Insert $F(x)$ as the first tangent vector to obtain
\eqref{eq:section-flux-transport}.
\end{proof}

The strict inequality is pointwise in the initial state
and travel time. It will suffice for capture, even when
the multipliers approach one along a sequence of crossings.