\subsection{A fixed section before a deep Taub visit}
\label{sec:deep-taub-crossings}

We locate a crossing of wall height $h$ before a deep Taub visit
by controlling a transverse radius backwards along the final part
of the approach. The estimate remains valid when the two weak walls
are equal or the shear vanishes.

Write $n=N_1$, $b=N_2$, $c=N_3$, and set
\begin{equation}
 \mathfrak r^2=s^2+(b-c)^2,\qquad
 \mathcal L=\tfrac12(\mathfrak r^2)'
       =ss'+(b-c)(b'-c').
 \label{eq:taub-transverse-radius}
\end{equation}
The zero set of $\mathfrak r^2$ is $\mathcal T_1$.
On the constraint, substitution of the equations gives the polynomial
identity
\begin{equation}
 \begin{split}
 (1-p)\mathcal L={}&-p(b-c)^2\mathfrak r^2\\
 &+n\bigl[(n-2(b+c))((1-p)\mathfrak r^2-(b-c)^2)
                       +(1-p)\sqrt3s(b-c)\bigr].
 \end{split}
 \label{eq:taub-radial-polynomial}
\end{equation}

\begin{lemma}[Backward control of the transverse radius]
\label{lem:taub-radius-backward}
On an orbit interval where $p\leq-1/2$, $n\geq0$ and
$n^2+b^2+c^2\leq2$, the function
$\mathfrak r^2e^{-8n}$ is nondecreasing. If its endpoints are $a\leq b_0$
and $n(a)\leq h\leq1/16$, then
\begin{equation}
 \mathfrak r^2(a)\leq2\mathfrak r^2(b_0).
 \label{eq:taub-backward-factor}
\end{equation}
\end{lemma}

\begin{proof}
Put $A_0=(1-p)\mathfrak r^2-(b-c)^2\geq0$.
The wall bound gives $b+c\leq2$, so $n-2(b+c)\geq-4$.
Also $\sqrt3s(b-c)\geq-\mathfrak r^2$, since the corresponding
quadratic form is nonnegative. Therefore the bracket in
\eqref{eq:taub-radial-polynomial} is at least
$-5(1-p)\mathfrak r^2$. Its first term is nonnegative, and division
by $1-p>0$ yields $\mathcal L\geq-5n\mathfrak r^2$.
The dominant wall rate satisfies
\[
 \lambda_1=-p^2-s^2+2p\leq-5/4.
\]
Consequently
\[
 (\mathfrak r^2e^{-8n})'
   =(2\mathcal L-8\lambda_1n\mathfrak r^2)e^{-8n}\geq0.
\]
Between the endpoints this gives
$\mathfrak r^2(a)\leq\mathfrak r^2(b_0)e^{8(n(a)-n(b_0))}$.
The exponential is at most $e^{1/2}\leq2$, proving the bound.
\end{proof}

\begin{proposition}[Crossing before a deep visit]
\label{prop:deep-taub-crossing}
Let $0<h\leq1/16$, $0<d<h/8$, and $0<\epsilon<d/4$.
Suppose an orbit segment $[a,b_0]$ in $\M$ satisfies
\[
 \mathcal P_{d/4}(\phi_tq)\quad(a\leq t\leq b_0),\qquad
 \mathfrak r^2(\phi_aq)>4\epsilon^2,\qquad
 \dist(\phi_{b_0}q,T_1)\leq\epsilon.
\]
There is $c_0\in(a,b_0)$ with
\begin{align}
 N_1(c_0)&=h,&p(c_0)&<-1/2,&|s(c_0)|&<d,\notag\\
 N_2(c_0),N_3(c_0)&<d,&
 \mathfrak r^2(c_0)&\leq2\mathfrak r^2(b_0),&
 N_1'(c_0)&\leq-5h/4.
 \label{eq:deep-taub-crossing}
\end{align}
\end{proposition}

\begin{proof}
The distance identity
\begin{equation*}
 \dist(q,T_1)^2=(1+p)^2+\mathfrak r^2+n^2+2bc
\end{equation*}
implies at $b_0$ that $p<-1/2$, $n<h$, and
$\mathfrak r^2\leq\epsilon^2$. Consider the last closed interval
$[c_0,b_0]$ on which $p\leq-1/2$ and $n\leq h$. More explicitly,
apply continuity to $f(t)=\max(p(t)+1/2,n(t)-h)$ and take the
left endpoint of its final nonpositive interval. Either $c_0=a$
or $f(c_0)=0$.

Equation~\eqref{eq:pair-product-state-bound} bounds the sum of wall
squares by two throughout the segment. Lemma~\ref{lem:taub-radius-backward}
therefore gives
\[
 \mathfrak r^2(c_0)\leq2\mathfrak r^2(b_0)\leq2\epsilon^2.
\]
The initial radius condition rules out $c_0=a$.
At $c_0$ the constraint reads
\[
 p^2+\mathfrak r^2+n^2-2n(b+c)=1.
\]
If $p=-1/2$, its left side is at most
$1/4+2\epsilon^2+h^2<1$, a contradiction. Thus $p<-1/2$ there,
and the boundary condition forces $n=h$.

The radius bound gives $|s|<d$ and $(b-c)^2\leq2\epsilon^2$.
The product bound gives $4bc\leq d^2/16$. Hence
\[
 (b+c)^2=(b-c)^2+4bc
           \leq2\epsilon^2+d^2/16<d^2.
\]
Both weak walls are less than $d$. Finally
$n'=\lambda_1h\leq-5h/4$, which also makes this a transverse
downward crossing of the fixed main wall.
\end{proof}

\begin{corollary}
\label{cor:late-fixed-axis-crossings}
Suppose $q\in\M$ has $\mathfrak r^2(q)>0$, satisfies
\eqref{eq:pair-product-decay}, and approaches $T_1$ arbitrarily
closely at arbitrarily late times. For every $0<h\leq1/16$,
$0<d<h/8$, $R\in\R$ and $\zeta>0$, there is a time
$c_0>\max(R,0)$ satisfying~\eqref{eq:deep-taub-crossing}, with
$\mathfrak r^2(c_0)<\zeta$.
\end{corollary}

\begin{proof}
Choose $a>\max(R,0)$ so late that $\mathcal P_{d/4}$ holds from $a$
onwards. The radius at $a$ is positive: its zero set is the fixed
point set of $\mathsf S$, and flow uniqueness makes that set invariant
in both time directions. Choose
\[
 0<\epsilon\leq\min\{d/8,\mathfrak r(a)/4,\sqrt\zeta/4\}.
\]
Take $b_0\geq a+1$ with $\dist(\phi_{b_0}q,T_1)<\epsilon$ and apply
Proposition~\ref{prop:deep-taub-crossing}. Its radius estimate is
at most $2\epsilon^2<\zeta$, and its crossing time is greater than
$a$.
\end{proof}