The geometry of Section~\ref{sec:p0-aes} is developed on the basic model, which
is complete and unpunctured: $\mathfrak E_0$ has no point at which the
arithmetic structure fails.  (The one-puncture model $\mathfrak E_1$ of
Example~\ref{ex:p0-e1-model} already leaves the regular category, and is the
case this chapter generalizes.)  This chapter develops the opposite situation through explicit punctured
models and the vocabulary proposed for them.  Its exact local constructions
are kept distinct from a register of six computed witnesses, each with its
own status.  The finite-type construction and reconstruction interface follow
in Section~\ref{sec:p0-compactification}.  The interpretive reading of those
witnesses --- the language of tearing --- is presented as a structural proposal,
and Subsection~\ref{subsec:p0-holed-nonclaims} lists what is not established.

\subsection{Punctured arithmetic expression spaces}
\label{subsec:p0-holed-definition}

\begin{definition}[Punctured arithmetic expression space]
\label{def:p0-punctured-aes}
Let $\overline M$ be an oriented smooth surface, let $S\subset\overline M$ be a
finite set of points, and put $M=\overline M\setminus S$.  Suppose given a
smooth Riemannian metric $g$ on $M$, a function $a:M\to\mathbb R$ that is smooth
on $M$, and constants $\mu\ne0$, $\lambda$, such that
\begin{enumerate}
  \item $(M,g,a;\mu,\lambda)$ is a regular AES in the sense of
  Definition~\ref{def:regular-aes}, and
  \item for every $p\in S$, no neighbourhood $U$ of $p$ in $\overline M$ admits
  extensions of $g|_{U\cap M}$ and $a|_{U\cap M}$ to smooth data
  $\widetilde g,\widetilde a$ on $U$ for which
  $(U,\widetilde g,\widetilde a;\mu,\lambda)$ is a regular AES.
\end{enumerate}
Then $(\overline M,S,g,a;\mu,\lambda)$ is called a \emph{punctured arithmetic
expression space}, and the points of $S$ are its \emph{punctures}.  When
$|S|=k$ we write $\mathfrak E_k$ for it, in the descriptive sense of
Convention~\ref{conv:p0-model-labels}.
\end{definition}

The ambient surface and the data are deliberately separated: the regular AES is
given only on the punctured surface $M$, so that condition (ii) is a genuine
requirement rather than a contradiction.  A puncture is not any point that has
been deleted, but a point at which the regular structure genuinely cannot be
continued.  This is the finite-point case of the singular arithmetic expression
space of Paper~I; the general singular theory, the classification of singular
germs, and everything concerning tubes, braids, and knot invariants belong to
Paper~I and Paper~III.  The notion is specialized here for one reason only: the
arithmetic motions of Subsection~\ref{subsec:p0-point-paths} are defined on the
punctured surface, and the question of what happens to them at a puncture is a
question about AES.

The first example is the one-puncture model of Example~\ref{ex:p0-e1-model}: the
unit disc with the complete Poincar\'e metric and the radial assignment
$a_{\mathbb D}=\frac{2\mu}{\lambda}\rho/(1-\rho^2)$, whose metric extends
smoothly across the centre while the assignment does not.  There $S=\{0\}$ and
the regular locus is $\mathbb D\setminus\{0\}$.  Two features of that example
are worth recording here, because they are typical of punctured models.  First,
the extended metric is complete on $\mathbb D$ while its restriction to
$\mathbb D\setminus\{0\}$ is \emph{not} complete: the puncture is at finite
distance from every interior point, so the incompleteness is caused by the
deletion and not by the geometry.  Second, the assignment has a finite limit at
the puncture (here $0$) but no $C^1$ extension, so the failure is a failure of
differentiability rather than of continuity.  Both statements are verified in
Paper~I together with the model.

\subsection{The local shape of a puncture}
\label{subsec:p0-holed-tangent-cone}

A puncture is not merely a missing point; at least in the example that has been
computed in detail, it carries a definite local shape.  The following statement
is exact.

\begin{proposition}[Tangent cone of the four limiting circles]
\label{prop:p0-puncture-tangent-cone}
Fix $\rho>0$ and consider the four circles through the origin
\[
  C_\pm:\ x^2+y^2\mp2\rho y=0,
  \qquad
  D_\pm:\ x^2+y^2\mp2\rho x=0 .
\]
Write $S_2=x^2+y^2$ and let
\[
  P=\bigl(S_2^2-4\rho^2y^2\bigr)\bigl(S_2^2-4\rho^2x^2\bigr)
\]
be the defining polynomial of their union.  Then
\[
  P=S_2^4-4\rho^2S_2^3+16\rho^4x^2y^2 ,
\]
so the lowest-degree homogeneous part of $P$ is $16\rho^4x^2y^2$.  Consequently
the tangent cone of the union at the origin is
\[
  (xy)^2=0 ,
\]
that is, the two coordinate axes, each with multiplicity two.
\end{proposition}

\begin{proof}
The four circles are $S_2=\pm2\rho y$ and $S_2=\pm2\rho x$, whose defining
polynomials are $S_2^2-4\rho^2y^2$ and $S_2^2-4\rho^2x^2$ respectively; the
product is $P$.  Expanding,
\[
  \bigl(S_2^2-4\rho^2y^2\bigr)\bigl(S_2^2-4\rho^2x^2\bigr)
  =S_2^4-4\rho^2S_2^2(x^2+y^2)+16\rho^4x^2y^2
  =S_2^4-4\rho^2S_2^3+16\rho^4x^2y^2 ,
\]
since $S_2^2(x^2+y^2)=S_2^3$.  The terms have degrees eight, six, and four, so
the lowest-degree part is $16\rho^4x^2y^2$, whose zero set is $xy=0$ counted
twice.
\end{proof}

Two remarks, both used below.  First, the two circles of the same family are
tangent at the origin, so the origin is a degenerate point of each family, while
circles of different families cross there with tangent lines $x=0$ and $y=0$;
the tangent cone is the union of those two lines.  Second, the tangent cone is
\emph{reducible with multiplicity}: it is the square of a product of two distinct
linear forms.  The computation is recorded in the exploration register of the
programme \cite{Yuan2026AEGHoleRegister}; the identity above is verified here
directly.

\subsection{An explicit punctured model with four zero branches}
\label{subsec:p0-four-circle-model}

The review of this paper observed, correctly, that the tangent cone above was a
statement about a curve configuration with no arithmetic expression space
attached to it.  The gap is now closed: the configuration is exactly the zero
locus of an explicit punctured AES.  We need one general fact first.

\begin{lemma}[The gradient never vanishes in a regular AES]
\label{lem:p0-gradient-nonvanishing}
In a regular AES $\mathfrak E=(M,g,a;\mu,\lambda)$ one has $\nabla a\ne0$
everywhere on $M$; indeed $|\nabla a|_g^2=\mu^2+\lambda^2a^2\ge\mu^2>0$.  Hence,
when $M$ is boundaryless, \emph{every} level set of $a$ is a smooth embedded
curve.  The regular-zero theorem of Paper~I is the case of level zero.
\end{lemma}

\begin{proof}
Both sides of the eikonal identity are non-negative and the right-hand side is at
least $\mu^2>0$, so $\nabla a$ does not vanish.  The regular-value theorem then
applies at every level, not only at $0$.
\end{proof}

The next theorem gives a conditional answer to the canonicality question.
Both smooth extension of the assignment and extension of the conformal
background are required.  The latter hypothesis repairs the statement printed
in the September 22 version: constructing a new AES metric does not by itself
extend the prescribed metric.

\begin{theorem}[Critical punctures with an extending conformal background]
\label{thm:p0-punctures-are-critical}
Let $(\overline M,S,g,a;\mu,\lambda)$ be a punctured AES in the sense of
Definition~\ref{def:p0-punctured-aes}, and suppose that $a$ extends to a smooth
function $\widetilde a$ on $\overline M$.  Suppose in addition that, near each
$p\in S$, the given metric $g$ is conformal on $U\setminus S$ to a smooth
Riemannian metric $g_0$ on a neighbourhood $U$ of $p$, with $g_0$ non-degenerate
at $p$.  Then
\begin{equation}\label{eq:p0-punctures-are-critical}
  S=\bigl\{p\in\overline M:\ d\widetilde a(p)=0\bigr\} .
\end{equation}
Within this conformal-background class, the puncture set is determined by the
assignment and its cardinality is the number of critical points of
$\widetilde a$ in $\overline M$.
\end{theorem}

\begin{proof}
For the inclusion $S\subseteq\{d\widetilde a=0\}$, let $p\in S$ and suppose
$d\widetilde a(p)\ne0$.  Use the extending background metric $g_0$ supplied by the hypothesis on a
neighbourhood $U$ of $p$.  Writing $g=f g_0$ on $U\setminus S$, the eikonal
identity forces
\[
  f=\frac{|d\widetilde a|_{g_0}^2}{\mu^2+\lambda^2\widetilde a^2}.
\]
Thus the extension constructed below agrees with the \emph{given} metric.
After shrinking $U$ we have $|d\widetilde a|^2_{g_0}>0$ on $U$, because
$|d\widetilde a|^2_{g_0}$ is continuous and positive at $p$, and also
$\mu^2+\lambda^2\widetilde a^2>0$ on $U$, because $\mu\ne0$.  Define
\[
  \widetilde g=\frac{|d\widetilde a|_{g_0}^2}{\mu^2+\lambda^2\widetilde a^2}\,g_0
  \qquad\text{on }U .
\]
Then $\widetilde g$ is a smooth Riemannian metric on $U$, and
\[
  |d\widetilde a|^2_{\widetilde g}
  =\frac{\mu^2+\lambda^2\widetilde a^2}{|d\widetilde a|_{g_0}^2}
   \,|d\widetilde a|_{g_0}^2
  =\mu^2+\lambda^2\widetilde a^2 ,
\]
so $(U,\widetilde g,\widetilde a;\mu,\lambda)$ is a regular AES extending
$(g,a)$ across $p$.  This contradicts condition (ii) of
Definition~\ref{def:p0-punctured-aes}.  Hence $d\widetilde a(p)=0$.

For the reverse inclusion, let $p\in\overline M$ with $d\widetilde a(p)=0$ and
suppose $p\notin S$.  Then $p\in M$ and the eikonal identity at $p$ reads
\[
  0=|da(p)|_g^2=\mu^2+\lambda^2a(p)^2\ge\mu^2>0 ,
\]
a contradiction.  Hence $p\in S$.
\end{proof}

\begin{remark}[The hypothesis is active]
\label{rem:p0-puncture-hypothesis}
The theorem requires the assignment to extend smoothly.  The isolated-zero disc
model imported from Paper~I (Example~\ref{ex:p0-e1-model}) does not satisfy the
hypothesis: there the metric extends smoothly across the centre while the
assignment is only continuous, and not $C^1$, so $d\widetilde a$ does not exist
at the puncture and \eqref{eq:p0-punctures-are-critical} has no meaning.  The two
verified punctured models are therefore of different kinds: in the disc model the
\emph{assignment} fails at the puncture, while in the model below the assignment is
smooth everywhere and the \emph{metric} is what cannot be non-degenerately extended.
The critical-set conclusion applies to the second kind only with the theorem's
extending-conformal-background hypothesis.  Without it even a smoothly
extending assignment need not determine the punctures.  For example, on a
punctured disc, $a=x$, $\mu=1$, $\lambda=0$, and
\[
  g=dx^2+(x^2+y^2)^{-1}dy^2
\]
satisfy $|da|_g^2=1$, while $g$ cannot extend at the origin and $da$ never
vanishes.  In every smoothly extending case only
$\operatorname{Crit}(\widetilde a)\subseteq S$ holds without further hypotheses.
\end{remark}

\begin{theorem}[A punctured AES whose zero locus is the four-circle configuration]
\label{thm:p0-four-circle-model}
Fix $\rho>0$ and $0<r<\rho$, let
\[
  \overline M=D_r=\{(x,y):x^2+y^2<r^2\},
  \qquad
  S=\{0\},
  \qquad
  M=\overline M\setminus S ,
\]
let $P$ be the union polynomial of
Proposition~\ref{prop:p0-puncture-tangent-cone}, put $a=P|_M$, fix
$\mu\ne0$ and $\lambda$, and define the conformal metric
\begin{equation}\label{eq:p0-four-circle-metric}
  g=\frac{|\nabla a|^2}{\mu^2+\lambda^2a^2}\,
    \bigl(dx^2+dy^2\bigr)
  \qquad\text{on }M .
\end{equation}
Then:
\begin{enumerate}
  \item $g$ is a smooth Riemannian metric on $M$, and $(M,g,a;\mu,\lambda)$ is a
  regular AES;
  \item the zero locus of $a$ in $\overline M$ is exactly the union of the four
  circles, so the configuration is the zero set of the assignment and not an
  auxiliary curve family;
  \item the origin is locally essential: no neighbourhood of $0$ admits a smooth
  non-degenerate extension of $(g,a)$ to a regular AES;
  \item consequently $(\overline M,S,g,a;\mu,\lambda)$ is a punctured arithmetic
  expression space in the sense of Definition~\ref{def:p0-punctured-aes}, whose
  zero locus has four branches meeting at the puncture;
  \item the metric is incomplete on $M$, and the puncture lies at finite distance.
\end{enumerate}
\end{theorem}

\begin{proof}
Throughout, write $S_2=x^2+y^2$ and
\[
  f(t)=S_2^2\bigl(S_2-3\rho^2\bigr)+4\rho^4t^2 ,
\]
so that, differentiating $P=S_2^4-4\rho^2S_2^3+16\rho^4x^2y^2$,
\begin{equation}\label{eq:p0-union-polynomial-gradient}
  \partial_xP=8x\,f(y),
  \qquad
  \partial_yP=8y\,f(x).
\end{equation}

\emph{(1)} The denominator $\mu^2+\lambda^2a^2$ is bounded below by $\mu^2>0$ on
$M$ and $|\nabla a|^2$ is smooth, so $g$ is smooth and positive definite.  Its
eikonal identity is immediate:
\[
  |\nabla a|_g^2
  =\frac{\mu^2+\lambda^2a^2}{|\nabla a|^2}\,|\nabla a|^2
  =\mu^2+\lambda^2a^2 .
\]
We must also check that $\nabla a$ does not vanish on $M$, so that the conformal
factor is strictly positive there.  Suppose $\nabla P=0$ at a point of
$\overline M$.  If $x=0$ then $S_2=y^2>0$ at a punctured point and
$f(0)=y^4(y^2-3\rho^2)$, which is nonzero because $y^2=S_2<r^2<\rho^2<3\rho^2$;
then $\partial_yP=8y\,f(0)\ne0$, a contradiction.  The case $y=0$ is symmetric.
If $x\ne0$ and $y\ne0$, then \eqref{eq:p0-union-polynomial-gradient} forces
$f(x)=f(y)=0$.  Here $S_2<3\rho^2$, so $f$ is strictly increasing in $t^2$ for
fixed $S_2$; hence $f(x)=f(y)$ gives $x^2=y^2$, so $S_2=2x^2$ and
\[
  0=f(x)=4x^2\Bigl[x^2\bigl(2x^2-3\rho^2\bigr)+\rho^4\Bigr],
\]
that is $2u^2-3u+1=0$ for $u=x^2/\rho^2$, whose roots are $u=1$ and
$u=\tfrac12$.  These give $S_2=2\rho^2$ and $S_2=\rho^2$, both at least $r^2$,
so no such point lies in $\overline M$.  Therefore $\nabla P\ne0$ on $M$.

\emph{(2)} $P$ factors as $\bigl(S_2^2-4\rho^2y^2\bigr)\bigl(S_2^2-4\rho^2x^2\bigr)$,
and each quadratic factor splits into two circles through the origin:
\[
  S_2=2\rho y \iff x^2+(y-\rho)^2=\rho^2,
  \qquad
  S_2=-2\rho y \iff x^2+(y+\rho)^2=\rho^2 ,
\]
and similarly with $x$ and $y$ exchanged.  Inside $\overline M$ the only common
point of any two of these circles is the origin: two circles of the same family
are tangent there, and two of different families meet again only at radius
$\sqrt2\,\rho>r$, as one checks by solving $S_2=2\rho y$, $S_2=-2\rho x$ or
$S_2=2\rho y$, $S_2=2\rho x$.  Hence the zero locus in $\overline M$ consists of
four arcs, two tangent to the $x$-axis and two to the $y$-axis, meeting only at
the origin --- which is exactly the tangent cone computed in
Proposition~\ref{prop:p0-puncture-tangent-cone}.

\emph{(3)} Let $U$ be a neighbourhood of the origin and let $\widetilde g$,
$\widetilde a$ be smooth extensions to $U$ of $g|_{U\cap M}$ and $a|_{U\cap M}$
for which $(U,\widetilde g,\widetilde a;\mu,\lambda)$ is a regular AES.  Since
$\widetilde a$ is continuous on $U$ and agrees with $a$ off the origin, while $a$
extends continuously to the origin with value $P(0)=0$, we have
$\widetilde a(0)=0$, and $\nabla\widetilde a(0)=\nabla a(0)=0$ because $a$ is
smooth on $\overline M$.  Evaluating the eikonal identity at the origin by
continuity gives
\[
  0=\widetilde g^{ij}(0)\,\partial_i a(0)\,\partial_j a(0)
   =\mu^2+\lambda^2a(0)^2=\mu^2\ne0 ,
\]
a contradiction.  So no such extension exists, which is condition (ii) of
Definition~\ref{def:p0-punctured-aes}.  (Equivalently: by
Lemma~\ref{lem:p0-gradient-nonvanishing} a regular AES has no critical point of
its assignment, whereas the origin is a critical point of $P$.)

\emph{(4)} Conditions (i) and (ii) of Definition~\ref{def:p0-punctured-aes} are
(1) and (3), so the tuple is a punctured AES.  By (2) its zero locus is the
four-circle configuration, and by the local structure computed there the four
arcs meet only at the puncture.

\emph{(5)} On the $x$-axis and away from the origin, $\partial_yP=0$ and
$f(0)=x^4(x^2-3\rho^2)$, so $|\nabla P|=8|x|\,|f(0)|=8|x|^5|3\rho^2-x^2|$, which
is $24\rho^2|x|^5+O(|x|^7)$ as $x\to0$.  Hence along the axis
\[
  \sqrt{g_{xx}}\;dx
  =\frac{|\nabla P|}{\sqrt{\mu^2+\lambda^2P^2}}\,dx
  =\frac{24\rho^2}{|\mu|}\,|x|^5\,dx\,(1+o(1)),
\]
whose integral from $0$ to $\varepsilon$ is finite; so the puncture lies at
finite distance and the metric on $M$ is incomplete.  The distance is not zero:
on the annulus $\varepsilon/2\le S_2^{1/2}\le r$ the conformal factor is bounded
below by a positive constant, so every path from a point of radius $\varepsilon$
to the puncture has length bounded below.
\end{proof}

The theorem answers the question the review posed, and it is worth noting what it
does \emph{not} do.  Its puncture set is not stipulated: by
Theorem~\ref{thm:p0-punctures-are-critical}, applied to the extending Euclidean
background of this construction, the set $S$ must equal the critical set, and the proof above computes that critical set on
$\overline M=D_r$ to be $\{0\}$ alone, so the model's puncture is forced by $a$.
What the theorem does not do is attach the
configuration to the arithmetic motions rather than to the assignment, or treat
more than one puncture.  Finally it makes the interface with
Paper~I visible rather than implicit: the zero locus of a regular AES is a smooth
curve (Lemma~\ref{lem:p0-gradient-nonvanishing}), the complete-splitting theorem
of Paper~I forbids branching on a connected complete boundaryless regular space,
and here branching occurs exactly at the puncture --- with the metric incomplete,
which is one of the ways the theorem's hypotheses can fail.

\begin{corollary}[Puncture counts in the conformal-background class]
\label{cor:p0-puncture-count}
Under all the hypotheses of Theorem~\ref{thm:p0-punctures-are-critical},
$|S|=|\operatorname{Crit}(\widetilde a)|$.  Conversely, if a smooth function
$\widetilde a$ on an oriented smooth surface $\overline M$ has a finite critical
set $S$, then for every smooth background metric $g_0$ the formula
\[
 g=\frac{|d\widetilde a|_{g_0}^2}{\mu^2+\lambda^2\widetilde a^2}\,g_0
 \quad\text{on }\overline M\setminus S
\]
constructs a punctured AES with precisely that declared puncture set.
\end{corollary}

\begin{proof}
The equality follows from the theorem with its conformal hypothesis.  For the
converse, the factor is positive off $S$ and the eikonal identity follows by
inverting the factor.  At each point of $S$ a non-degenerate regular extension
would give $0=\mu^2+\lambda^2\widetilde a^2$, which is impossible.  This proves
essentiality for the constructed metric, not a classification of arbitrary
prescribed metrics.
\end{proof}

\begin{remark}[A construction rather than a classification]
\label{rem:p0-multi-puncture}
The corollary supplies models from assignments with finitely many critical
points.  It does not make an arbitrary AES metric extendible and does not
identify all punctures with critical points.  Section~\ref{sec:p0-compactification}
constructs compactified carriers with every positive finite puncture count;
the point at infinity in that family is essential because the assignment
diverges, so it is not an additional critical point of a smooth extension.
\end{remark}

\subsection{Tearing across a puncture}
\label{subsec:p0-tearing-at-puncture}

The tearing of Definition~\ref{def:p0-tearing} is a defect between two
add--scale histories.  On a punctured space a further question arises: a
comparison of two motions may have to be performed around a puncture rather than
through it, and one may ask whether the tearing is then measured by an ACS area,
by a holonomy of a loop around the puncture, or by neither.  The answer proved
below is that it is measured by the ACS area in every case, that no canonical
holonomy exists in the AES data, and that what the puncture changes is the
admissibility of a \emph{comparison on the surface} rather than the value of the
tearing.

\begin{proposition}[The tearing does not see the puncture]
\label{prop:p0-tearing-puncture-independent}
Let $\gamma,\delta$ be scale-compatible add--scale histories over a field $K$,
with translation coordinates $\xi_\gamma,\xi_\delta$ as in
Section~\ref{sec:p0-affine-cocycles}.  Then
\[
  \tau(\gamma,\delta)=\xi_\gamma-\xi_\delta
\]
depends only on the two words.  Consequently $\tau(\gamma,\delta)$ is unchanged
by puncturing the space in which the histories are realized, by changing that
space, or by choosing a different realization of the same words; and the area
formula of Theorem~\ref{thm:torsion-stokes} applies verbatim on a punctured
model, with the filling taken in the charge plane rather than on the surface.
\end{proposition}

\begin{proof}
By Proposition~\ref{prop:affine-cocycle-formulas} the target-frame translation
$\xi$ is a function of the chronological word alone, and
Definition~\ref{def:relative-torsion} defines $\tau$ as the difference of the two
$\xi$'s.  Nothing in either statement mentions a surface, a metric, or a puncture.
The area formula is Theorem~\ref{thm:torsion-stokes}, whose data are the two
charge chains.
\end{proof}

The next statement explains why the charge plane is the canonical place for the
comparison, and it has nothing to do with punctures: on a surface the two orders
of a rectangle never close up.

\begin{proposition}[The surface never closes the comparison]
\label{prop:p0-surface-does-not-close}
Let $\mathfrak E$ be a regular AES with canonical frame $(X_u,X_v)$, let
$p\in M$, and let $h,k>0$ be such that the flows $\Phi_u^h$ and $\Phi_v^k$ of
$X_u$ and $X_v$ are defined at $p$ and along both compositions.  Then the two
chronological compositions end at points whose assignments differ by
\begin{equation}\label{eq:p0-surface-comparison-defect}
  a\bigl(\Phi_v^k\Phi_u^h(p)\bigr)-a\bigl(\Phi_u^h\Phi_v^k(p)\bigr)
  =\mu h\bigl(e^{\lambda k}-1\bigr)\ne0 ,
\end{equation}
whenever $\mu h\ne0$ and $\lambda k\ne0$.  Hence the two flow paths alone bound
no region of $M$: closing them requires a further path along which the
assignment changes by that amount.

Consequently the ACS comparison of Section~\ref{sec:p0-acs-tearing} is not a
device for avoiding punctures.  It is the canonical comparison for every regular
AES, punctured or not, and the puncture changes nothing about it.
\end{proposition}

\begin{proof}
By Theorem~\ref{thm:continuous-affine-flow}, the flows satisfy
$\frac{d}{ds}(a\circ\Phi_u^s)=\mu$ and
$\frac{d}{ds}(a\circ\Phi_v^s)=\lambda\,(a\circ\Phi_v^s)$; hence
$a\circ\Phi_u^h=a+\mu h$ and $a\circ\Phi_v^k=e^{\lambda k}a$.  Composing in the
two orders,
\[
  a\circ\Phi_v^k\circ\Phi_u^h=e^{\lambda k}(a+\mu h),
  \qquad
  a\circ\Phi_u^h\circ\Phi_v^k=e^{\lambda k}a+\mu h ,
\]
whose difference is $\mu h(e^{\lambda k}-1)$.  This is nonzero under the stated
hypotheses, so the two endpoints are distinct; two paths that do not share both
endpoints bound no region.
\end{proof}

\begin{proposition}[No canonical holonomy around a puncture]
\label{prop:p0-no-puncture-holonomy}
Let $\mathfrak E$ be a regular AES.
\begin{enumerate}
  \item The horizontal structure determined by the assignment is the foliation
  by the level sets of $a$, whose tangent distribution $\ker da$ is integrable;
  it is therefore flat, and it has no holonomy.
  \item The AES data determine no further connection on $M$.  A connection with
  curvature requires a choice of splitting beyond $(M,g,a;\mu,\lambda)$; the
  contact model of Section~\ref{sec:p0-contact} is one such choice, and relative
  to its added charge projection the curvature form $\omega$ of
  Proposition~\ref{prop:p0-curvature-form} is canonical.
  \item Since the assignment is a globally single-valued function on $M$, no loop
  around a puncture can change the value of $a$: there is no monodromy of the
  assignment.
\end{enumerate}
Hence no tearing is measured by a holonomy of a loop around a puncture.  The
natural holonomy in this theory lives in the contact model, over a loop in the
charge plane, where it is the closed horizontal drift
$T_{\mathrm{closed}}=e^{-\lambda k}T_{\mathrm{open}}$ of
Section~\ref{sec:p0-contact} --- a holonomy of the added structure, not of the
puncture.
\end{proposition}

\begin{proof}
(1) The level sets of a smooth function foliate the open set where its
differential does not vanish, and by
Lemma~\ref{lem:p0-gradient-nonvanishing} that is all of $M$; the tangent
distribution of a foliation is integrable by definition.  (2) The canonical
frame supplies two global directions, but a splitting of the tangent bundle is
not part of the data; §9 fixes one by declaring the coordinates $(u,v,a)$ and the
charge projection.  (3) A single-valued function has no monodromy along any loop.
\end{proof}

\paragraph{What the puncture does change.}  It changes which comparisons can be
performed \emph{on the surface}.  In the model of
Theorem~\ref{thm:p0-four-circle-model} the local picture is computed as follows.

\begin{proposition}[The zero-set germ and the metric at the puncture]
\label{prop:p0-puncture-local-picture}
In the model of Theorem~\ref{thm:p0-four-circle-model}, with $0<\varepsilon<\rho$:
\begin{enumerate}
  \item the circle $S_2=\varepsilon^2$ meets the zero locus in exactly eight
  points, four on the pair of circles tangent to the $x$-axis and four on the
  pair tangent to the $y$-axis;
  \item the assignment changes sign at each of them, and the punctured disc is
  divided into eight sectors, four with $a>0$ and four with $a<0$; the negative
  sectors are the cusps $|y|<x^2/(2\rho)$ and $|x|<y^2/(2\rho)$ along the
  coordinate axes;
  \item $g$ is conformal to the Euclidean metric with conformal factor vanishing
  at the puncture, so the \emph{conformal class} of $g$ extends across the puncture
  although the metric does not; angles at the puncture are therefore well
  defined and equal those of the tangent cone of
  Proposition~\ref{prop:p0-puncture-tangent-cone};
  \item the puncture lies at finite $g$-distance from every interior point, so a
  path that approaches it does so with finite $g$-length, and $M$ is incomplete.
\end{enumerate}
\end{proposition}

\begin{proof}
(1) On the circle $S_2=\varepsilon^2$ the union polynomial restricts to
\[
  P=\varepsilon^4\bigl(\varepsilon^2-4\rho^2y^2\bigr)
              \bigl(\varepsilon^2-4\rho^2x^2\bigr),
\]
whose zeros are $y=\pm\varepsilon^2/(2\rho)$ and $x=\pm\varepsilon^2/(2\rho)$,
each giving two points of the circle since $\varepsilon<\rho$ implies
$\varepsilon^2/(2\rho)<\varepsilon$.  (2) The sign of $P$ on a small punctured
neighbourhood is the sign of the product of the two factors
$S_2^2-4\rho^2y^2$ and $S_2^2-4\rho^2x^2$, and near the origin
$P=S_2^3(S_2-4\rho^2)+16\rho^4x^2y^2$ is dominated by $-4\rho^2S_2^3$ exactly
where $|y|<x^2/(2\rho)$ and by $-4\rho^2S_2^3$ with the roles of $x$ and $y$
exchanged exactly where $|x|<y^2/(2\rho)$.  (3) and (4) are parts (1) and (5) of
Theorem~\ref{thm:p0-four-circle-model}, the conformal factor being
$|dP|^2/(\mu^2+\lambda^2P^2)$.
\end{proof}

So the answer to the question of this subsection is the following.  A tearing is
always the ACS area of Theorem~\ref{thm:torsion-stokes}, and its value is
independent of the model and of its punctures; the admissible fillings are the
oriented two-chains of the charge plane with the given boundary, and by that
theorem the value does not depend on which filling is chosen.  A comparison that
would have to pass through a puncture is simply not available on the surface, and
in the computed model the puncture is exactly where the surface stops being
metric (though not conformal) and where the zero-set germ has eight
sign-changing branches.  These scalar statements do not determine whether an assignment-preserving
word moves the full surface state; Section~\ref{sec:p0-compactification}
retains that distinction explicitly.

\subsection{Six computed witnesses}
\label{subsec:p0-holed-witnesses}

The exploratory work recorded in \cite{Yuan2026AEGHoleRegister} produced six
separately computed statements about punctured configurations.  They are
reproduced here in the register's own order, each with its exact content and its
status.  None of them is promoted by appearing in this paper: the register
carries no authoritative status, and the items below are classified according to
the labels of this repository.

\begin{enumerate}
  \item \emph{A topological witness.}  Cut a disc carrying four right-angle
  sectors along the two diagonal seams and the four coordinate rays; eight
  triangular pieces result, each with a left radial edge, an outer arc, and a
  right radial edge.  Regluing is a permutation $\sigma$ of the eight left edges
  onto the eight right edges.  If $C$ is the number of cycles of $\sigma$, then
  the centre vertex has $C$ orbits, the outer vertices form eight orbits, and
  $V=C+8$, $E=16$, $F=8$, whence
  \[
    \chi=V-E+F=C=b ,
  \]
  the number of boundary components.  Consequently every gluing of this family
  produces either one disc or a disjoint union of $C$ discs, and the compact
  three-holed sphere ($\chi=-1$ with three boundary components) is unreachable
  in this family.  A surface of Euler characteristic $-1$ is obtained instead by
  removing two points from the disc: it has one boundary component, two
  punctures, and hence three ends, so its Euler characteristic agrees with that
  of the compact pair of pants while its boundary structure does not.  An
  exhaustive check of all $40320$ permutations confirms the
  distribution $5040,13068,13132,6769,1960,322,28,1$ for $C=1,\ldots,8$.
  \emph{Status:} the identity $\chi=C$ is elementary and provable as above; the
  enumeration is a computationally verified example.  The topological
  conclusion holds for the stated gluing rule, not for arbitrary gluings of the
  eight pieces.

  \item \emph{A combinatorial witness.}  If three components interact pairwise,
  the interaction sites are the three unordered pairs, and each site carries two
  directions of exchange, giving six directed transitions in total.
  \emph{Status:} counting, given the definition of a hole as an interaction
  site.  The definition is a choice, not a theorem.

  \item \emph{A frame witness.}  Let the state of a three-component frame be
  $(c,\sigma,\varphi)$ with component label $c\in\mathbb Z_3$, direction
  $\sigma\in\{\pm1\}$, and phase $\varphi\in\mathbb Z_4$, and let one step act by
  \[
    (c,\sigma,\varphi)\longmapsto(c+\sigma,\sigma,\varphi+\sigma).
  \]
  Since $\gcd(3,4)=1$, the Chinese remainder theorem identifies
  $(c,\varphi)$ with a single counter $k$ modulo $12$, and the orbit of each
  direction has length $12$.  Within one such orbit the component label
  advances in the direction $\sigma$, so the transitions realized are the three
  pairs of that direction, each occurring four times --- once per phase class.
  The six directed transitions of the combinatorial witness therefore appear
  four times each only over the two orbits together, that is, over $24$ steps;
  a single $12$-step orbit cannot contain $24$ transitions.  Since one full
  phase period is four steps and $4\equiv1\pmod3$, one period of the phase
  returns the phase to its initial value while advancing the component label by
  \[
    c\longmapsto c+\sigma \pmod 3 ;
  \]
  for the fixed direction $\sigma=+1$ this is the advance by one used in the
  register, and for $\sigma=-1$ it is the advance by two.  In either case the
  phase loop is not closed in the total state: its monodromy is the generator
  $-\sigma$ of $\mathbb Z_3$.  \emph{Status:} exact arithmetic (Chinese
  remainder theorem), verified exhaustively on the two $12$-state orbits and by
  replay.

  \item \emph{An observer witness.}  Let a frame be described by an angle
  $\theta$ and three joint offsets $\delta_1,\delta_2,\delta_3$, and let the
  residuals available to an observer be
  \[
    r_i=\bigl|2\sin\delta_i\bigr| \quad (i=1,2,3),
    \qquad
    r_4=\bigl|2\sin\bigl(2\theta+\tfrac12(\delta_1+\delta_2+\delta_3)\bigr)\bigr| .
  \]
  All four residuals vanish exactly at the ideal frame
  $(\theta,\delta)=(\pi/2,0,0,0)$.  Suppose an observer sees only residuals of
  magnitude at least $\varepsilon$, and that an iterative procedure stops as
  soon as every visible residual is below that threshold.  At the halt one then
  has the following bound.  From $r_i<\varepsilon$ we get
  $|\delta_i|<\arcsin(\varepsilon/2)=\varepsilon/2+O(\varepsilon^3)$, and from
  $r_4<\varepsilon$ together with
  $r_4=\bigl|4(\theta-\pi/2)+(\delta_1+\delta_2+\delta_3)\bigr|+O(\lVert(\theta-\pi/2,\delta)\rVert^2)$
  we get
  \[
    \bigl|\theta-\pi/2\bigr|
    <\frac58\,\varepsilon+O(\varepsilon^2).
  \]
  The halting offset is therefore of order $\varepsilon$ and decreases linearly
  in $\varepsilon$; it is not zero, and it is the best that a procedure obeying
  this stopping rule can report.  \emph{Status:} the residual functions, their
  exact common zero at the ideal frame, and the bound are proved above.  The
  particular iteration counts and offsets recorded in the register depend on the
  implementation of the update step, which the register does not fix, and they
  are accordingly \emph{not} reproduced here.  The witness is \emph{not} an
  obstruction: nothing in the computation shows that no procedure can do better,
  and the exact zero of the residuals at the ideal frame shows the opposite.

  \item \emph{A semantic witness.}  Model the three interaction sites by
  vocabularies built from four terms common to all three components, four
  further control terms distributed over the pairs as two, one, and one, and
  three terms private to the individual components (which therefore enter no
  site vocabulary).  The three site vocabularies then have sizes
  \[
    6=4+2,\qquad 5=4+1,\qquad 5=4+1 .
  \]
  Filling the three sites independently --- one word chosen from
  each site's own vocabulary --- gives $6\cdot5\cdot5=150$ possibilities;
  filling them so that the same core word is used at all three sites gives
  $4^3=64$.  The difference $86$ is carried by the adapters at the sites.  On
  the control layer the pairwise intersections are non-empty while the common
  intersection is empty; on the full vocabularies the common intersection is the
  four-term core, so the emptiness statement is layer-dependent.
  \emph{Status:} the arithmetic is verified given the model; the model is a
  modelling choice, and the "Borromean" description is an analogy with no
  link-theoretic content.

  \item \emph{A phase-geometry witness.}  With components at phases
  $\{0,+1,-1\}$ modulo $n$, a forward motion of $k$ steps and a reverse motion of
  $m$ steps meet exactly when $k+m\equiv0\pmod n$.  A meeting phase different
  from the three component phases exists precisely for $n\ge4$; at $n=4$ it is
  the first case in which the meeting is simultaneously balanced ($k=m$) and
  antipodal ($\varphi=n/2$), namely $k=m=2$, $\varphi=2$.  \emph{Status:}
  computationally verified example over the stated finite model, exhaustive for
  $n=3,\ldots,8$.  The value $n=4$ is minimal, not unique: $n=6$ and $n=8$ also
  admit balanced antipodal meetings.
\end{enumerate}

\begin{remark}[Register bookkeeping]
\label{rem:p0-holed-register-bookkeeping}
The exploration register introduces these items as five classes while tabulating
six rows; the sixth row is the phase-geometry witness listed last above.  This
paper follows the table and counts six, recording the discrepancy rather than
resolving it silently \cite{Yuan2026AEGHoleRegister}.
\end{remark}

\subsection{What the witnesses do not establish}
\label{subsec:p0-holed-nonclaims}

The register's own summary sentence is that a hole "is not a defect to be
removed but the position that carries exchange".  That sentence is the
motivation for this chapter, and it is not proved by any of the six witnesses.
The following list of non-claims is part of the content of this section.

\begin{itemize}
  \item No witness shows that punctures are unavoidable in AES.  The witnesses
  are about chosen finite models: a chosen gluing family, a chosen frame, a
  chosen vocabulary, a chosen stopping rule.  Changing the model changes the
  count, and the register says so for its own semantic witness.
  \item No witness establishes an obstruction of AEG.  The monodromy of the
  frame witness is a monodromy of a stipulated finite flow; the calibration
  witness is a property of a stipulated halting rule; the gluing witness is a
  statement about one gluing family.
  \item The $\mathbb Z_3$ monodromy of the frame witness is not identified with
  any $\mathbb Z_4$ twist of sheet labels appearing elsewhere in the register.
  Those are different groups, and no computation in the register relates them;
  the identification asserted there is recorded as unproved.
  \item The word "hole" in this chapter means a puncture in the sense of
  Definition~\ref{def:p0-punctured-aes}, or an interaction site in the sense of
  the combinatorial witness.  It is not the "one-hole context" of
  Section~\ref{sec:p0-expansion}, which is a placeholder in an expression, and
  the two must not be conflated.
  \item The engineering study that accompanies the register is explicitly
  recorded by its author as an analogy and not as a mathematical assertion; one
  of its three language prototypes was never compiled.  It contributes
  motivation only.
  \item Nothing here changes the status of Paper~I's regular-zero theorems or of
  Paper~III's singular theory.  A punctured AES is a specialization of the
  singular notion, and its classification is not attempted.
\end{itemize}

\subsection{Open problems}
\label{subsec:p0-holed-open}

The register closes with the author's own verdict that this route is not yet
free, for three reviewable reasons: the family of cuts depends on a chosen
permutation, the covering sheets are defined by the cut, and the graded
description depends on an externally supplied frame.  It registers three
directions that have not been attempted, and they are the natural open problems
of this chapter.

\begin{enumerate}
  \item Replace the tangent cone of Proposition~\ref{prop:p0-puncture-tangent-cone}
  by the formal neighbourhood of the puncture: compute the successive jets at
  $p$ and test whether the description ceases to depend on the choice of tangent
  directions.
  \item Blow up the puncture and rewrite the cut-and-glue construction as a
  canonical construction on the exceptional divisor, so that the seam is
  replaced by a construction the data itself determines.
  \item Replace the cut-and-glue calculus by an operadic composition, in which
  the gluing of punctured surfaces is intrinsic and contains no chosen seam.
\end{enumerate}

Each of these is a programme, not a result.  Four further problems arise
directly from the discussion above; the first of them is the natural
generalization of the construction now proved, and the second is the comparison
question raised in Subsection~\ref{subsec:p0-tearing-at-puncture}.

\begin{enumerate}
  \item \label{op:p0-four-circle}\emph{Generalize the model of
  Theorem~\ref{thm:p0-four-circle-model}.}  The construction there produces one
  puncture and a zero locus with four branches from a prescribed polynomial
  assignment and the metric that the eikonal identity forces.  Two questions
  follow.  Which curve configurations arise as zero loci of punctured AES in this
  way?  The existence of some compactifiable model for each $k\ge1$ is settled
  in Theorem~\ref{thm:p0-arbitrary-k-punctures}; prescribed zero configurations,
  puncture locations, and conformal moduli remain separate questions.  And does the construction survive if the
  assignment is required to come from the arithmetic motions rather than being
  prescribed as a polynomial?  A negative answer to the last question would mean
  that the configuration is realizable as a zero locus but not as an expression.
  \item \label{op:p0-canonical-punctures}\emph{Classify full-state residue across
  punctured models.}  The scalar affine defect is word-dependent and hence
  puncture-independent, while Proposition~\ref{prop:p0-k4-motion-residue}
  exhibits a level-set displacement invisible to that scalar observation.
  Determine which residual data and admissible motion domains suffice for
  continuation across recorded collars, especially when the assignment does
  not extend to a puncture.  The scalar statement alone does not answer this
  full-state question.
  \item \emph{Canonicalize the punctures when the assignment does not extend.}
  For smoothly extending assignments with an extending conformal background,
  Theorem~\ref{thm:p0-punctures-are-critical} gives the critical set.  Outside
  that class, metric singularities may occur where $da\ne0$.  When the
  assignment is not differentiable at the puncture, as in the disc model of
  Paper~I, the critical set is not even defined there.  Determine useful
  intrinsic end data in both cases.  This is the first
  unfinished item of the exploration register's own programme.
  \item \emph{Carry ripple geometry across a puncture.}  Decide whether a
  punctured model carries a pullback structure, in the sense of
  Section~\ref{sec:p0-ripples}, that survives at the puncture; in the model of
  Theorem~\ref{thm:p0-four-circle-model} the pole would have to be placed at the
  critical point, where the conformal factor of the metric vanishes.
\end{enumerate}

A solution of any of them would turn the language of this chapter from a
description into a theory.