\section{Some consequences of the classification}\label{app:consequences}
The classification turns questions about degrees and short cycles
into restrictions on finite Coxeter diagrams. We record a rank
bound, a finiteness criterion, the intrinsic fold group, and
three explicit lists. The positive
examples in the lists were already known to Conlon and Lee
\cite{CL}; the new deduction is that the lists are exhaustive
under their stated hypotheses. These statements concern
cut-percolating bigraphs, not all weakly norming bigraphs.

We use the Coxeter diagram and rank from
Section~\ref{sec:reflection-bigraphs}. The finite irreducible
diagrams have the following standard names \cite{Humphreys}.
The diagram $A_r$ is a path with all labels three; $B_r$ is
a path with one terminal label four; $D_r$ has three paths of
lengths $1,1,r-3$ meeting at one node, with all labels three.
The diagrams $E_6,E_7,E_8$ instead have path lengths $1,2,r-4$.
The diagram $F_4$ is a four-node path with labels $3,4,3$;
$H_3,H_4$ have a terminal label five and all other labels three.
Finally, $I_2(m)$ consists of two nodes joined with label $m$.
An isolated node is $A_1$.

The standard parabolic intersection identity is
$W_I\cap W_J=W_{I\cap J}$ \cite{Humphreys}. The subgroup index $[U:V]$ is the
number of cosets of $V$ in $U$. Our reflection presentations have
$S=I\cup J$, as required in Definition~\ref{def:reflection}.
Any entire diagram component in $I\cap J$
cancels from both coset spaces. We call a presentation with
these components removed \emph{effective}.

\subsection{Degree bounds and finiteness}\label{sec:degree-finiteness}
Degrees $(a,b)$ mean that every left vertex has degree $a$ and
every right vertex has degree $b$.

\Needspace{7\baselineskip}
\begin{lemma}\label{lem:parabolic-min-index}
If $(W,S)$ is irreducible of rank $r$ and $T\subsetneq S$, then
$[W:W_T]\ge r+1$.
\end{lemma}
\begin{proof}
In the real reflection representation, choose a nonzero point
$x$ in the closed fundamental chamber with stabilizer exactly
$W_T$: take a point on the walls indexed by $T$ and strictly
inside all other walls. Its orbit has $[W:W_T]$ points.
The linear span of the orbit is a nonzero invariant subspace,
so it is all of $\mathbb R^r$ by irreducibility. The sum of the
orbit points is invariant under $W$ and hence zero. Therefore
these spanning points are linearly dependent, and their number
is at least $r+1$.
\end{proof}

\begin{theorem}[Rank bound and degree finiteness]\label{thm:degree-finiteness}
Let $a,b\ge1$ be fixed integers.
\begin{enumerate}
\item Every connected cut-percolating bigraph of degrees $(a,b)$
has an effective reflection presentation of rank at most $a+b-2$.
\item There are finitely many such bigraphs up to isomorphism if
and only if at least one of $a,b$ is odd.
\end{enumerate}
\end{theorem}
\begin{proof}
Choose an effective presentation with $S=I\cup J$ and put
$K=I\cap J$. The degrees are
\[
 a=[W_I:W_K],\qquad b=[W_J:W_K].
\]
Call a component of the diagram induced on $I$ \emph{active}
if it contains a node outside $K$. If these active components
have ranks $r_1,\ldots,r_t$ and relative parabolic indices
$d_1,\ldots,d_t$, then
\[
 a=\prod_{i=1}^t d_i,\qquad
 \sum_{i=1}^t r_i\le\sum_{i=1}^t(d_i-1)\le a-1.
\]
The first inequality follows from Lemma~\ref{lem:parabolic-min-index}.
The second follows by repeatedly using
$(u-1)+(v-1)\le uv-1$ for $u,v\ge1$.
The analogous sum of active ranks for $J$ is at most $b-1$.
Every node of $I\setminus K$ or $J\setminus K$ belongs to an
active component. Every component of the diagram on $K$ must
touch one of these nodes, since otherwise it would be an entire
component of $S$ discarded from the effective presentation.
Thus the active $I$- and $J$-components together cover $S$,
giving $|S|\le a+b-2$.

At bounded rank, the only unbounded parameter in the finite
Coxeter classification is the label $m$ in a dihedral component
$I_2(m)$. Consider its two nodes. If one lies in $I\setminus J$
and the other in $J\setminus I$, this component contributes a
factor two to each of $a,b$. In every other effective assignment
with $I\cup J$ containing both nodes, one parabolic contains
both nodes and the other contains zero or one. The nontrivial
local degree is then $2m$ or $m$, so $m\le\max\{a,b\}$.
If at least one degree is odd, the unbounded assignment is
impossible. There are then only finitely many effective diagrams
of bounded rank and bounded labels, and only finitely many choices
of $I,J$ in each. This proves finiteness.

If both degrees are even, start with $C_{2m}$ and replace every
left vertex by $b/2$ twins and every right vertex by $a/2$ twins.
The resulting bigraph has degrees $(a,b)$. It is a reflection
bigraph: the cycle uses $I_2(m)$, and uniform twin duplication
is obtained by taking its tensor product with the complete
bigraph $K_{b/2,a/2}$, itself a reflection bigraph for a product
of symmetric groups. Products of Coxeter systems realize these
tensor products. The number of vertices grows with $m$, giving
infinitely many pairwise nonisomorphic examples.
\end{proof}

\subsection{The intrinsic fold group}\label{sec:intrinsic-fold-group}
The reflection representation in Theorem~\ref{thm:classification}
need not use the entire fold group. The following refinement
identifies that group once twins are removed. Atomic folds
are the reflections themselves, rather than merely lifts of
reflections in a quotient.

\begin{theorem}\label{thm:intrinsic}
Let $H$ be a finite connected twin-free cut-percolating bigraph
with an edge. Then $\Fold(H)$ has a faithful finite real reflection
representation whose full reflection set is exactly the set of
atomic fold involutions. Moreover, there is a Coxeter system
$(\Fold(H),S)$ and $I,J\subseteq S$ realizing the given vertex
and edge action of $H$ as the reflection bigraph
$\mathcal R(\Fold(H);I,J)$.
\end{theorem}

\begin{lemma}\label{lem:product-connected}
The part-respecting tensor product of two connected bigraphs
with an edge is connected. Every fold of one factor lifts to a
fold of the tensor product by fixing the other coordinate.
If both factors are twin-free, their atomic folds lift to atomic
folds of the product.
\end{lemma}
\begin{proof}
For any two product vertices choose paths in the factors.
Their lengths have the same parity. Insert backtracks along an
edge into the shorter path until the lengths agree, and then
walk in both coordinates simultaneously. This proves connectivity.

For a fold $f$ of one factor, use the involution $(v,x)\mapsto
(v,fx)$. The two exchanged sides are the products with the
original sides; there is no edge between them. Hence each
component of the moved induced subgraph is sent to a different
component. If $f$ is atomic, each of its two moved components
contains an edge: an isolated moved vertex and its mate would
be twins, because all their neighbors are fixed. The connectivity
argument shows that their products with the other factor are
connected, so the lifted fold is atomic.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:intrinsic}]
Induct on the number of edges. For one edge the fold group and
its reflection representation are trivial. Otherwise
Theorem~\ref{thm:tensor} gives $H=R\otimes F$, where $R$ is the
Fiedler quotient and $F$ the residual factor. Both are connected,
both are twin-free (a twin pair in either factor would yield a
twin pair in $H$), and $F$ cut-percolates by Theorem~\ref{thm:inherit}.
The quotient has at least two edges, so $F$ is smaller.
Write $W$ for the image of $\Fold(H)$ on the Fiedler eigenspace.
It is a finite real reflection group. Its action on $R$ is
faithful because the spectral vertex images span that eigenspace.

Let $f$ be an atomic fold of $H$. If it is visible on the
Fiedler eigenspace, the proof of Theorem~\ref{thm:inherit} gives
\[
 f(v,x)=(\alpha v,\tau_vx),
\]
where $\alpha$ is a reflection in $W$ and every $\tau_v$ is a
twin permutation of $F$. Twin-freeness gives $\tau_v=1$.
Every reflection in $W$ has such an atomic lift, so these folds
generate $W\times1$ on the tensor coordinates.

If $f$ is invisible, it has the form $f(v,x)=(v,\tau_vx)$.
For every edge $vw$ of $R$, $(\tau_v,\tau_w)$ is an automorphism
of $F$. If $v,v'$ share the neighbor $w$, composing these
pairs yields an automorphism of $F$ fixing the opposite part
pointwise. Such an automorphism only permutes twins, so
$\tau_v=\tau_{v'}$. Connectedness propagates this equality:
all left $\tau_v$ agree, and all right $\tau_w$ agree. Thus
$f=1\times\beta$ for one automorphism $\beta$ of $F$.
Restricting $f$ to an edge fiber, as in Theorem~\ref{thm:inherit}, shows that $\beta$ is a genuine fold.
Moreover it is atomic; otherwise its distinct moved component
pairs would lift to more than one pair for $f$.

Conversely, every fold of $F$ lifts by
Lemma~\ref{lem:product-connected}. Therefore
\[
 \Fold(H)=W\times\Fold(F).
\]
The two factors commute and intersect trivially. The induction
hypothesis supplies a faithful reflection representation of
$\Fold(F)$ with atomic folds exactly its reflections. Take its
direct sum with the representation of $W$. A reflection in this
direct sum acts as a reflection in exactly one factor and as the
identity in the other. The visible lifts above and the atomic
lifts from $F$ show that these are precisely the atomic folds
of $H$.

Finally, $R$ is a reflection bigraph for $W$ with its given
permutation action, and the induction hypothesis says the same
for $F$ and $\Fold(F)$. The product of the two Coxeter systems
and the union of the respective parabolics give exactly the
action on $R\otimes F=H$.
\end{proof}

\paragraph{Adding twins.}
Use the uniform twin blowup of Definition~\ref{def:twins}. The symmetric group $\mathfrak S_a$ consists of all
permutations of $a$ elements. A \emph{semidirect product}
$N\rtimes Q$ below means that $Q$ permutes the independent
symmetric-group factors of $N$ by its action on twin classes.

\begin{lemma}\label{lem:twin-fold}
An atomic fold in a connected bigraph either transposes two
twins and fixes all other vertices, or descends to an atomic
fold of its twin quotient, fixing every vertex in each fixed
quotient class pointwise.
\end{lemma}
\begin{proof}
Let $f$ be atomic, and suppose a twin class $T$ is preserved
by $f$ but not fixed pointwise. Choose $u\in T$ with $fu\ne u$.
If a neighbor $w$ of $u$ were moved, twinhood gives the two
edges $uw$ and $(fu)w$ inside the moved induced subgraph.
They connect $u$ to $fu$, forcing that component to be
preserved by $f$, contrary to the fold condition. Hence all
neighbors of $T$ are fixed. The moved vertices of $T$ are
isolated components. Atomicity now forces $f$ to transpose
only $u,fu$, which are twins.

Otherwise every quotient-fixed class is fixed pointwise, and
every moved class is exchanged with a different twin class.
The moved subgraph is a blowup of the moved subgraph in the
quotient. Components containing an edge lift to connected
components, since all adjacencies between classes are complete.
An isolated moved quotient vertex would have an isolated
exchanged mate with the same fixed neighborhood, contradicting
twin-freeness of the quotient. Thus all moved components
contain an edge, and the two components exchanged by $f$
project to exactly two quotient components exchanged by its
induced involution. This is an atomic quotient fold.
\end{proof}

\begin{corollary}[Uniform twins]\label{cor:all-twins}
Let $G=(A_0,B_0;E_0)$ be a finite connected twin-free bigraph
with an edge. For every $a,b\ge1$, with the natural permutation
action,
\[
 \Fold(G(a,b))\cong
 \left(\mathfrak S_a^{A_0}\times\mathfrak S_b^{B_0}\right)
 \rtimes\Fold(G).
\]
Every finite connected cut-percolating bigraph $H$ with an edge
has the form $G(a,b)$ for a twin-free cut-percolating quotient
$G$. In this case, modulo independent permutations inside twin
classes, the full fold group is the finite Coxeter group
in Theorem~\ref{thm:intrinsic}.
\end{corollary}
\begin{proof}
By Lemma~\ref{lem:twin-fold}, every atomic fold of $G(a,b)$
induces a quotient fold or the identity. All transpositions
inside a twin class are atomic folds, so the full product of
symmetric groups lies in $\Fold(G(a,b))$. Conversely, choose
clone indices consistently and lift every fold of $G$ by
retaining its clone index. A quotient fold's fixed classes
are fixed pointwise, and its moved components lift to
exchanged components. Thus these lifts belong to the fold
group and give a group section. The quotient-map kernel is
precisely the independent twin permutations, proving the formula.

For cut-percolating $H$, part transitivity makes the twin
classes on each specified part have a common size, say $a,b$.
Thus $H=G(a,b)$. The formula implies that $\Fold(G)$ is
transitive on each quotient part; the classification gives
cut-percolation of $G$. Apply Theorem~\ref{thm:intrinsic}.
\end{proof}

\begin{corollary}[Atomic generators bounded by the degrees]\label{cor:degree-generators}
Let $H$ be a finite connected cut-percolating bigraph of degrees
$d_A,d_B$. Its full fold group is generated by at most
$d_A+d_B-2$ atomic fold involutions. Consequently a positive
recognition certificate needs at most
$\min\{|V(H)|-2,d_A+d_B-2\}$ atoms.
\end{corollary}
\begin{proof}
Write $H=G(a,b)$ as in Corollary~\ref{cor:all-twins}, and let
$G$ have degrees $r,s$. By Theorem~\ref{thm:intrinsic}, the
intrinsic Coxeter system of $\Fold(G)$ realizes $G$.
It is effective: any diagram component in both vertex
stabilizers would act trivially, contradicting faithfulness;
and connectedness requires the union of the two parabolics
to contain the simple system. The rank argument in the proof of
Theorem~\ref{thm:degree-finiteness} gives at most $r+s-2$ simple
generators, all atoms.
Their clone-index-preserving lifts are atomic by the proof of Lemma~\ref{lem:twin-fold}, since a moved quotient component
contains an edge. Add $a-1$ transpositions connecting the
clones in one left twin class and $b-1$ such transpositions
in one right twin class. The quotient group is transitive
on each part. Conjugating these transpositions by its lifts
gives the corresponding generators in every twin class.
Thus they generate the full semidirect product, using at most
\[
 r+s+a+b-4\le br+as-2=d_A+d_B-2
\]
atoms. The inequality is equivalent to
$(b-1)(r-1)+(a-1)(s-1)\ge0$. The $|V(H)|-2$ certificate
bound follows by adding atoms that merge two current vertex
orbits until the two part orbits remain.
\end{proof}

\paragraph{Why the twin-free hypothesis is needed.}
For $H=C_6(3,1)$, the group above is
$\mathfrak S_3^3\rtimes\mathfrak S_3$, where the last factor
permutes the three left twin classes. Its order is $1296$.
Its subgroup $C_3^3\rtimes C_3$ of order $81$ is nonabelian;
here $C_3$ is a cyclic group of order three. It is a Sylow
three-subgroup, meaning its order is the largest power of
three dividing the full group order. In contrast, every
finite Coxeter group of order $1296$ has abelian Sylow
three-subgroups. Indeed, the only irreducible finite types
whose orders can divide $1296=2^4 3^4$ are $A_1,A_2,A_3,B_2,B_3$
and dihedral types; all have cyclic Sylow three-subgroups.
Their direct products have abelian Sylow three-subgroups.
Thus $\Fold(H)$ itself need not be a Coxeter group before
twins are removed, even for this twelve-vertex example.


\subsection{Odd prime degree}\label{sec:prime-degree}
The following small part of the finite Coxeter classification is
enough for a complete list at every odd prime degree.

\Needspace{9\baselineskip}
\begin{lemma}\label{lem:prime-index}
Let $(W,S)$ be a finite irreducible Coxeter system and let
$T\subsetneq S$ have odd prime index $p$. Then one
of the following holds:
\begin{enumerate}
\item $W$ has type $A_{p-1}$ and $T$ omits an endpoint of its
path diagram;
\item $W$ has type $I_2(p)$ and $T$ contains one of its two nodes.
\end{enumerate}
The alternatives coincide for $p=3$.
\end{lemma}
\begin{proof}
The parabolic must be maximal among standard parabolics. Otherwise
an intermediate standard parabolic factors its index into two
integers greater than one. It therefore omits a single node.
In type $A_{n-1}$, the resulting indices are $\binom nk$ for
$1\le k<n$. For $2\le k\le n-2$, this number exceeds $n$.
If it were prime $p$, the identity
$k\binom nk=n\binom{n-1}{k-1}$ would force $p$ to divide the
positive integer $\binom{n-1}{k-1}=(k/n)p<p$, a contradiction.
Only endpoint omissions remain, of index $n=p$.
The indices in $B_r$ are $2^k\binom rk$ for $1\le k\le r$.
In $D_r$ they are $2^k\binom rk$ for $1\le k\le r-2$ and
$2^{r-1}$ at the two fork endpoints. All are even.
For the exceptional types, the distinct indices are:
\begin{center}\small
\begin{tabular}{ll}
\toprule
Type&Maximal standard parabolic indices\\\midrule
$E_6$&$27,72,216,720$\\
$E_7$&$56,126,576,756,2016,4032,10080$\\
$E_8$&$240,2160,6720,17280,60480,69120,241920,483840$\\
$F_4$&$24,96$\\
$H_3$&$12,20,30$\\
$H_4$&$120,600,720,1200$\\\bottomrule
\end{tabular}
\end{center}
None is prime. These integers follow directly by dividing the
standard finite Coxeter group orders by the orders after one
node is deleted \cite{Humphreys}. Finally, a maximal standard
parabolic in $I_2(m)$ has index $m$, giving $m=p$.
\end{proof}

\paragraph{The four families.}
The \emph{hypercube} $Q_p$ has vertices $\{0,1\}^p$, with an
edge when two strings differ in one coordinate; its parts are
the strings of even and odd coordinate sum. The \emph{crown}
on $2q$ vertices is $K_{q,q}$ with one perfect matching removed.
The \emph{middle-layer inclusion bigraph} $M_p$ has parts
$\binom{[2p-1]}{p-1}$ and $\binom{[2p-1]}p$, with adjacency
by containment. Here $[n]=\{1,\ldots,n\}$ and $\binom Xk$
denotes the family of all $k$-element subsets of $X$.

\begin{theorem}[Odd prime degree]\label{thm:prime-degree}
For every odd prime $p\ge5$, the connected $p$-regular
cut-percolating bigraphs are exactly
\[
 K_{p,p},\qquad Q_p,\qquad
 K_{p+1,p+1}\setminus\text{a perfect matching},\qquad M_p.
\]
For $p=3$, the complete list is $K_{3,3}$, $Q_3$, $M_3$
(also called the Desargues graph), and the $192$-vertex edge--triangle
incidence bigraph of the regular $24$-cell, described in
Section~\ref{sec:large-girth}. The crown with
four vertices in each part coincides with $Q_3$.
\end{theorem}
\begin{proof}
Take an effective connected presentation and put $K=I\cap J$.
Since $[W_I:W_K]=[W_J:W_K]=p$, each of $I,J$ has exactly one
active component. Lemma~\ref{lem:prime-index} shows that each
is either $A_{p-1}$ or $I_2(p)$ and that
\[
 I\setminus J=\{s\},\qquad J\setminus I=\{t\}.
\]
The active component on $I$ consists of $s$ and exactly one
component $L$ of $K$: either $L=A_{p-2}$ with $s$ joined to an
endpoint by an edge labeled three, or $L=A_1$ with its edge
to $s$ labeled $p$. Similarly $t$ attaches to one component
$L'$ of $K$. No other component of $K$ remains in an effective
presentation.

For $p\ge5$, suppose that an active component is $I_2(p)$.
If $L=L'$, the other active component would also have to be
$I_2(p)$, because its part in $K$ must be the same one-node
component. Two edges labeled $p\ge5$ meeting at that node are
incompatible with positive definiteness of the Coxeter Gram
matrix, the matrix of inner products of the unit normals to
its simple walls; see Lemma~\ref{lem:large-labels}.
If $L\ne L'$ and $s,t$ were joined, the whole diagram
would be irreducible. For $p\ge7$, no finite irreducible diagram
of rank at least three contains an edge labeled $p$. For $p=5$,
the only possibilities are $H_3,H_4$; a second $I_2(5)$ gives
two edges labeled five, while an $A_4$ component gives total
rank six. Neither is possible. Hence $s,t$ commute, the two
active components are separate, and the bigraph is $K_{p,p}$.

It remains to consider two active components of type $A_{p-1}$.
If $L=L'$, then $s,t$ attach to the same or opposite endpoints
of one $A_{p-2}$ path. They cannot be joined directly, since
finite Coxeter diagrams have no cycle. Attachment to the same
endpoint gives $D_p$, and attachment to opposite endpoints
gives $A_p$. These presentations yield respectively $Q_p$
and the crown with $p+1$ vertices in each part.
For the first identification, $D_p$ acts by coordinate
permutations and changes of an even number of coordinate signs.
Its two vertex orbits are the sign vectors in $\{-1,1\}^p$
with an even or odd number of negative coordinates, and
adjacency changes one coordinate.
For the second, $A_p$ is the permutation group on $p+1$ points:
the two parts are singletons and $p$-subsets, with containment
as adjacency. Identifying a $p$-subset with its missing point
gives the crown.

If $L\ne L'$, then $s,t$ are endpoints of two disjoint
$A_{p-1}$ paths. If they commute, the bigraph is again
$K_{p,p}$. If they are joined by a label three, the diagram is
$A_{2p-2}$ and the two coset spaces are the $(p-1)$- and
$p$-subsets of a $(2p-1)$-point set, yielding $M_p$.
A larger joining label would be internal in a path of rank
$2p-2\ge8$, impossible in the finite Coxeter classification.
This exhausts $p\ge5$.

When $p=3$, the two alternatives in
Lemma~\ref{lem:prime-index} coincide as $A_2=I_2(3)$.
The common component $L=L'$ consists of one node, giving $A_3$
and $Q_3$, also the four-by-four crown. For $L\ne L'$, the
joining labels two, three, and four give respectively
$A_2\times A_2$, $A_4$, and $F_4$. Labels at least five do
not give a finite diagram of the form $3$--$m$--$3$.
The three resulting graphs are $K_{3,3}$, $M_3$, and the
$24$-cell edge--triangle incidence graph.
All listed presentations are reflection bigraphs and hence
cut-percolating by Conlon--Lee \cite{CL}.
\end{proof}

\subsection{Graphs without short cycles}\label{sec:large-girth}
The \emph{girth} of a graph is the length of a shortest cycle, with
girth infinity if it has no cycle. The \emph{subdivision} $S(G)$
replaces each edge $uv$ of $G$ by a path $u,x_{uv},v$ of two edges.
Its specified parts are the original vertices and the new edge
vertices. The $d$-dimensional \emph{hypercube} $Q_d$ has vertex set
$\{0,1\}^d$ and joins two binary strings differing in one coordinate.
The dodecahedral graph uses the twenty vertices and thirty edges
of the regular dodecahedron. The $120$-cell graph uses the
vertices and edges of the regular four-dimensional polytope
with $120$ dodecahedral three-dimensional faces. The $24$-cell is the convex hull
of the twenty-four coordinate permutations of $(\pm1,\pm1,0,0)$.
The \emph{edge--triangle
incidence bigraph of the $24$-cell} has one vertex for each edge
and one for each triangular face, with adjacency meaning containment.

\begin{theorem}\label{thm:large-girth}
Up to exchanging the specified parts, the connected cut-percolating
bigraphs of minimum degree at least two and girth at least eight
are exactly the following:
\begin{center}
\begin{tabular}{lrrl}
\toprule
Bigraph & Vertices & Degrees in the two parts & Girth\\
\midrule
$C_{2m}$, $m\ge4$ & $2m$ & $(2,2)$ & $2m$\\
$S(Q_d)$, $d\ge3$ & $(d+2)2^{d-1}$ & $(d,2)$ & $8$\\
$S(\text{dodecahedral graph})$ & $50$ & $(3,2)$ & $10$\\
$S(\text{$120$-cell graph})$ & $1800$ & $(4,2)$ & $10$\\
Edge--triangle incidence of the $24$-cell & $192$ & $(3,3)$ & $8$\\
\bottomrule
\end{tabular}
\end{center}
\end{theorem}

\Needspace{5\baselineskip}
\begin{corollary}\label{cor:large-girth-min-degree}
The edge--triangle incidence bigraph of the $24$-cell is the only
connected cut-percolating bigraph of minimum degree at least three
and girth at least eight. Every other connected cut-percolating
bigraph of minimum degree at least three contains a four-cycle or
a six-cycle.
\end{corollary}

\begin{figure}[htbp]
\centering
\input{figures/coxeter-large-girth-drawing}
\caption{The effective Coxeter diagrams for girth at least eight.
Blue marks $I\setminus J=\{s\}$, orange marks
$J\setminus I=\{t\}$, and gray nodes belong to $I\cap J$.
The two distinguished nodes must be joined by a label at least
four. The finite Coxeter classification leaves just these five
possibilities; see Theorem~\ref{thm:large-girth}.
Each marked endpoint is omitted from the opposite parabolic.
Unlabeled edges have label three.}
\label{fig:coxeter-large-girth}
\end{figure}

Barnhill's angle formula for standard parabolic subgroups
\cite[Proposition~6.8]{Barnhill} gives the exact girth
$2\min\{m_{st}:s\in I\setminus J,\ t\in J\setminus I\}$
of a reflection bigraph when neither parabolic contains the other.
For the exhaustion argument, we only need the following elementary
upper bound, which we prove directly.

\begin{lemma}[A cycle from a cross pair]\label{lem:dihedral-cycle}
For $s\in I\setminus J$ and $t\in J\setminus I$,
$\mathcal R(W;I,J)$ contains a cycle of length $2m_{st}$.
\end{lemma}
\begin{proof}
The rank-two parabolic group $D=W_{\{s,t\}}$ is dihedral of order
$2m_{st}$. By the parabolic intersection property,
\[
 D\cap W_I=\langle s\rangle,\qquad
 D\cap W_J=\langle t\rangle.
\]
Consequently, the left and right cosets of these two order-two
subgroups embed in the two vertex parts of $\mathcal R(W;I,J)$.
The edges indexed by $D$ are distinct: their common stabilizer
is $D\cap W_{I\cap J}=\{1\}$. They form the rank-two reflection
bigraph, which is the cycle $C_{2m_{st}}$. Extra edges in the
ambient graph do not remove this cycle.
\end{proof}

\begin{lemma}[No two large labels meet]\label{lem:large-labels}
In a finite Coxeter diagram, no vertex is incident with two
edges whose labels are at least four.
\end{lemma}
\begin{proof}
Choose unit inward normals to the simple mirrors. Their Gram
matrix is positive definite and has entries
$-\cos(\pi/m_{st})$ off the diagonal. Suppose a vertex has two
neighbors with labels at least four. Put
$a,b\ge1/\sqrt2$ for the corresponding cosines and
$c\ge0$ for the cosine between those neighbors. The relevant
three-by-three principal determinant is
\[
 1-a^2-b^2-c^2-2abc\le0,
\]
contrary to positive definiteness.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:large-girth}]
Choose a connected reflection presentation with $S=I\cup J$,
and put $K=I\cap J$. Neither $I\setminus J$ nor
$J\setminus I$ is empty: otherwise one part would have degree
one. Lemma~\ref{lem:dihedral-cycle} and girth at least eight
give $m_{st}\ge4$ for every pair
$s\in I\setminus J$, $t\in J\setminus I$.
Lemma~\ref{lem:large-labels} therefore forces
\[
 I\setminus J=\{s\},\qquad J\setminus I=\{t\}.
\]
These two nodes are joined by a label at least four. All other
nodes belong to $K$. Every irreducible component avoiding $s,t$
lies in both parabolics and cancels from both coset spaces, so
it can be discarded.

The finite irreducible Coxeter diagrams containing an edge labeled
at least four are precisely $I_2(m)$ for $m\ge4$, $B_d$ for
$d\ge3$, $F_4$, $H_3$, and $H_4$. In every case this edge is
unique. Our parabolics omit its opposite endpoints:
\[
 I=S\setminus\{t\},\qquad J=S\setminus\{s\}.
\]
For $I_2(m)$ this gives $C_{2m}$. For $B_d$, marking the
terminal edge labeled four gives the vertex--edge incidence of
the $d$-cube, namely $S(Q_d)$. The diagrams $H_3,H_4$ have a
terminal edge labeled five; they give respectively vertex--edge
incidence of the dodecahedron and of the $120$-cell. The middle
edge labeled four in $F_4$ gives edge--triangle incidence of the
$24$-cell. The rank-two case $B_2$ is already $I_2(4)$.

All these incidence graphs are reflection bigraphs and hence
cut-percolating by Conlon--Lee. Their girths are as listed.
For subdivisions, girth doubles: the hypercube has girth four,
and the dodecahedral and $120$-cell graphs have girth five.
The edge--triangle incidence of the $24$-cell has girth eight.
The geometric verification of these girths is given below.
\end{proof}

\paragraph{Girths of the positive examples.}
The dodecahedron and the $120$-cell have pentagonal two-faces.
The directions of the incident edges at a vertex form a simplex,
called its \emph{vertex figure}. Thus any two incident edges
lie in a pentagonal face and meet at angle $108^\circ$.
In a cycle of length three or four, let $u_1,\ldots,u_k$ be its
successive unit displacement vectors. Adjacent vectors have scalar
product $c=\cos72^\circ>0$, and their sum is zero. For $k=3$ this
would give $0=3+6c$. For $k=4$ it would give
\[
 0=4+8c+2(\langle u_1,u_3\rangle+\langle u_2,u_4\rangle)
 \ge8c>0.
\]
Thus neither skeleton has a triangle or a square; pentagonal faces
give girth five. Their subdivisions have girth ten. A hypercube
has no triangle and has square faces, giving girth four and eight
before and after subdivision.

For the $24$-cell, use the $24$ coordinate permutations of
$(\pm1,\pm1,0,0)$, joining pairs at squared distance two. Its
triangular faces are precisely its three-vertex cliques, and the
graph induced on the neighbors of any vertex is a cube $Q_3$.
Two different edges belong together to at most one triangular
face, ruling out a four-cycle in edge--triangle incidence. A
six-cycle would use three distinct edges that pairwise lie in
three distinct triangles. These three edges are either the sides
of a single triangle, which would repeat that face, or have a
common endpoint. In the latter case their other endpoints form
a triangle in its cube neighborhood, also impossible. A square
in that cube neighborhood gives four incident edges and four
triangular faces, hence an eight-cycle in the incidence bigraph.
This proves its girth is eight.

\subsection{Odd degree without four-cycles}
The same diagram argument gives a sharper conclusion at odd
regular degree, with no primality assumption. Recall that $M_d$
is the inclusion bigraph between the $(d-1)$- and $d$-subsets
of $[2d-1]$, defined in Section~\ref{sec:prime-degree}.

\begin{theorem}\label{thm:odd-no-four-cycles}
For every odd integer $d\ge3$, the connected $d$-regular
cut-percolating bigraphs without a four-cycle are exactly $M_d$,
and, when $d=3$, the edge--triangle incidence bigraph of the
$24$-cell.
\end{theorem}
\begin{proof}
Choose an effective connected reflection presentation, with
$S=I\cup J$ and $K=I\cap J$.
By Lemma~\ref{lem:dihedral-cycle}, every node of $I\setminus J$
is adjacent in the diagram to every node of $J\setminus I$:
a nonadjacent pair would have label two and give a four-cycle.
Finite irreducible Coxeter diagrams are trees. The complete
bipartite subgraph between these two nonempty exclusive sets
therefore has a singleton side. Suppose $I\setminus J=\{s\}$.

If $J\setminus I$ has at least two nodes, they are neighbors
of $s$. If $s$ has no other neighbor, its active component in
$I$ is $A_1$, giving degree two, contrary to oddness.
Thus $s$ has degree three in the diagram, with two neighbors
in $J\setminus I$ and one in $K$. It is the unique branch
node of a diagram of type $D$ or $E$.
Let $\ell_1,\ell_2$ be the lengths of the two arms starting
in $J\setminus I$, and let $\ell_3$ be the length of the
arm entirely in $K$. The local degrees are
\[
 [W_I:W_K]=\ell_3+2,
 \qquad [W_J:W_K]=(\ell_1+1)(\ell_2+1).
\]
For their common value to be odd, both $\ell_1,\ell_2$
must be even. A type-$D$ diagram has arms $1,1,r-3$,
so this is impossible. The type-$E$ arms are $1,2,L$,
where $L\in\{2,3,4\}$. The only possibility would use
the two even arms, leaving $\ell_3=1$; but then the two
degrees are $3$ and $3(L+1)$, unequal. We conclude that
$J\setminus I$ is also a singleton, say $\{t\}$.

The adjacent nodes $s,t$ are omitted from opposite maximal
parabolics. Regularity says these parabolics have equal
orders. In type $A_r$, adjacent omissions have indices
$\binom{r+1}i,\binom{r+1}{i+1}$; equality forces
$r=2i$, and the resulting graph is $M_{i+1}$.
In type $B_r$, number the nodes $0,\ldots,r-1$ with the
edge labeled four between $0$ and $1$. Omitting adjacent
nodes $k,k+1$ gives local degrees $2(k+1)$ and $r-k$;
the first is always even. In type $D_r$, number the long
arm edges starting at the branch node by $k=0,1,\ldots$.
Their local degrees are $2(k+3)$ and $r-k-2$, while a
fork edge has one local degree two. Thus neither type
contributes an odd regular example.
The maximal parabolic indices in $E_7,E_8,H_3,H_4$ are
all distinct. In $E_6$, the equal-index nodes are in opposite
arms and are not adjacent. The only adjacent equal-index
pair in $F_4$ is its middle edge, giving degree three and
the $24$-cell graph. Dihedral diagrams give degree two.
This exhausts the finite list.

Conversely, $M_d$ is a reflection bigraph with the two
adjacent central nodes of $A_{2d-2}$ distinguished. Its
girth is six by Barnhill's formula, or directly by subset
containment: two distinct $(d-1)$-subsets have at most
one common $d$-superset, excluding four-cycles, and the
three $(d-1)$-subsets extending a fixed $(d-2)$-subset by
three different points give a six-cycle. The $24$-cell
example was verified above. Both are cut-percolating.
\end{proof}