\section{Spectral classification}\label{sec:structure}
We prove the difficult implication in Theorem~\ref{thm:classification}:
transitivity on the two vertex parts gives a reflection representation.
The proof has two stages. First, the lowest nonconstant Laplacian
eigenspace turns the fold group into a geometric reflection group.
Second, we recover the vertices that this representation identifies.
They form a smaller tensor factor to which we apply induction.

\subsection{Atomic folds on the Fiedler eigenspace}\label{sec:spectral-setup}
Let $\Gamma=\Fold(H)$. The \emph{combinatorial Laplacian} is the
symmetric matrix $\mathsf L=\mathsf D-\mathsf A$, where $\mathsf D$ is diagonal with
$\mathsf D_{vv}=\deg_H(v)$ and $\mathsf A_{uv}$ is one for an edge
and zero otherwise. Equivalently, for a real function $\phi$ on vertices,
\begin{equation}\label{eq:laplacian}
 (\mathsf L\phi)(v)=\sum_{w\in N_H(v)}(\phi(v)-\phi(w)).
\end{equation}
It commutes with every graph automorphism. An eigenvalue $\theta$
has eigenspace $\ker(\mathsf L-\theta I)$, the functions satisfying
$\mathsf L\phi=\theta\phi$; $\spec(\mathsf L)$ denotes the set of
these eigenvalues, and $I$ is the identity matrix.
Let $\lambda>0$ be the smallest positive eigenvalue and
$\mathcal E=\ker(\mathsf L-\lambda I)$ its \emph{Fiedler eigenspace}
\cite{Fiedler}.
Let $P$ be the orthogonal projection onto $\mathcal E$: it is the
identity on $\mathcal E$ and zero on its orthogonal complement.
Define the
\emph{spectral vertex map}
\begin{equation}\label{eq:spectral-map}
 p:V(H)\longrightarrow\mathcal E,\qquad p(v)=P\mathbf e_v.
\end{equation}
Equivalently, choose an orthonormal eigenbasis of $\mathcal E$
and use its values at $v$ as the coordinates of $p(v)$.
The permutation action of $\Gamma$ on $\RR^{V(H)}$ restricts to
an \emph{orthogonal representation} $\rho:\Gamma\to O(\mathcal E)$:
it assigns each group element a linear isometry, compatibly with
composition. Here $O(\mathcal E)$ is the group of linear isometries
of $\mathcal E$, and $p(gv)=\rho(g)p(v)$.

A function is \emph{$f$-odd} if $\phi(fv)=-\phi(v)$.
For $C\subseteq V(H)$, the \emph{Dirichlet Laplacian} on $C$ is
the principal submatrix $\mathsf L_{C,C}$, obtained by retaining
the rows and columns in $C$; it is the Laplacian quadratic form
restricted to functions zero outside $C$.
The \emph{Rayleigh quotient} of a nonzero function is
$\langle\phi,\mathsf L\phi\rangle/\langle\phi,\phi\rangle$.
Connectedness gives its variational characterization
\begin{equation}\label{eq:rayleigh}
 \lambda=\min_{\substack{\phi\ne0\\\sum_v\phi(v)=0}}
 \frac{\langle\phi,\mathsf L\phi\rangle}{\langle\phi,\phi\rangle}.
\end{equation}
The sum condition says that $\phi$ has \emph{mean zero}.

On one moved component of an atom, Perron--Frobenius gives a
positive lowest Dirichlet eigenvector unique up to scale.
It will supply the one-dimensional normal direction of a reflection.
Using the lowest nonconstant eigenvalue of the whole graph ensures
that no other odd direction can appear.

\begin{lemma}[An atomic fold acts as a reflection or the identity]\label{lem:atomic}
For each atomic fold involution $f$, either $\rho(f)=1$ or $\rho(f)$ is an orthogonal reflection. In the latter case, with reflecting hyperplane $M_f\subset\mathcal E$,
\[
 p^{-1}(M_f)=\Fix(f).
\]
Moreover no graph edge crosses $M_f$: its endpoints lie in a common closed halfspace.
\end{lemma}
\begin{proof}
Let $C,C'$ be the two nonfixed components, exchanged by $f$. An $f$-odd function on $V(H)$ is determined by its restriction $u$ to $C$: it is $-u\circ f$ on $C'$ and zero on $\Fix(f)$. Under this identification the Laplacian on odd functions is exactly the principal Dirichlet matrix $\mathsf L_{C,C}$. At a fixed vertex the contributions from $C$ and $C'$ cancel in pairs, so every Dirichlet eigenvector gives an actual odd Laplacian eigenvector on $H$.

Since $H[C]$ is connected, the least eigenvalue $\mu$ of
$\mathsf L_{C,C}$ is simple, meaning its eigenspace is
one-dimensional, and has a strictly positive eigenvector.
Indeed, for large $c$ the matrix $cI-\mathsf L_{C,C}$ is nonnegative
and irreducible: its nonzero off-diagonal entries connect $C$.
Perron--Frobenius says that its largest eigenvalue is simple with
a positive eigenvector. Its odd extension has mean zero, so
\eqref{eq:rayleigh} gives $\mu\ge\lambda$.

If $f$ acts nontrivially on $\mathcal E$, its $-1$ eigenspace there is nonzero, and hence $\lambda$ occurs in the Dirichlet spectrum. Then $\lambda\ge\mu$, so $\lambda=\mu$ and this odd eigenspace is one-dimensional. Thus $\rho(f)$ is a reflection.

Choose its odd normal vector $\psi\in\mathcal E$, positive on $C$, negative on $C'$ and zero on $\Fix(f)$. Since $\langle p(v),\psi\rangle=\psi(v)$, the inverse image of its orthogonal hyperplane is exactly $\Fix(f)$. No edge joins $C$ to $C'$, proving the last assertion.
\end{proof}

\subsection{The reflection quotient}\label{sec:reflection-quotient}
The preceding lemma controls each atom separately. We next need
control of \emph{every} mirror of the resulting reflection group:
two image endpoints must share a chamber. For this purpose it
matters that the atomic folds are closed under conjugation.

\begin{definition}\label{def:spectral-quotient}
The \emph{spectral quotient} $R$ identifies vertices within each
specified part when their images under $p$ agree. Two quotient
vertices are adjacent if some original edge joins their classes;
repeated edges are identified.
The \emph{vertex fiber} over $x$ is its preimage in the appropriate
part of $H$. The \emph{edge fiber} over $xy\in E(R)$ is the set
of original edges with endpoints in those two vertex fibers.
\end{definition}
Write $T(W)$ for the set of all orthogonal reflections in a finite
real reflection group $W\le O(U)$. A subset $T_0$ is
\emph{conjugacy-invariant} if $wT_0w^{-1}=T_0$ for every $w\in W$.
We use the following standard fact to control all mirrors.

\begin{lemma}\label{lem:all-reflections}
If a conjugacy-invariant subset $T_0\subseteq T(W)$ generates $W$,
then $T_0=T(W)$.
\end{lemma}
\begin{proof}
The \emph{Coxeter diagram} of $(W,S)$ has vertices
$S$ and an edge labeled $m_{st}$ when the order of $st$ is
$m_{st}\ge3$. Reflection conjugacy classes correspond to the
components remaining after the even-labeled edges are removed.
The \emph{abelianization}, the quotient obtained by requiring all
group elements to commute, is a product of two-element groups
with one independent generator for each such component.
A reflection maps to the generator of its class, so a generating
set must meet every class. Conjugacy invariance then gives all
reflections. See the proofs of Lemmas 3.6 and 3.7 of
Brady--McCammond--M\"uhlherr--Neumann~\cite{BMMN}, pp.\ 7--8.
These facts apply to the essential space defined in
Section~\ref{sec:reflection-bigraphs}; a common fixed subspace
can be omitted.
\end{proof}

\begin{figure}[htbp]
\centering
\input{figures/reflection-chambers-drawing}
\caption{The spectral vertex map $p$ from \eqref{eq:spectral-map}
for $C_6$, up to scale. Its two vertex parts
form two three-point orbits of the type-$A_2$ reflection group.
The three fold mirrors pass through pairs of opposite vertices.
The marked edge has endpoints $x,y$ in the closed fundamental
chamber $\overline C$; every other edge lies in a translate of this
chamber. Here $s_1,s_2$ are its wall reflections, as in
Section~\ref{sec:reflection-bigraphs}. No edge crosses a mirror.}
\label{fig:reflection-chambers}
\end{figure}

\Needspace{10\baselineskip}
\begin{theorem}[Spectral reflection quotient]\label{thm:quotient}
Suppose $\Gamma$ acts transitively on each of $A$ and $B$, and
let $R$ be the spectral quotient of Definition~\ref{def:spectral-quotient}.
Then:
\begin{enumerate}
\item $W=\rho(\Gamma)$ is a finite real reflection group, allowing the trivial group.
\item $R$ is a connected reflection bigraph for $W$. More precisely, after choosing a suitable fundamental chamber and corresponding simple system $S$, there are $I,J\subseteq S$ with $I\cup J=S$ such that
\[
 R\cong \mathcal R(W;I,J).
\]
\end{enumerate}
\end{theorem}
\begin{proof}
Atomic folds generate $\Gamma$, so Lemma~\ref{lem:atomic} shows that their nontrivial images generate $W$ by reflections. Their image set $T_0$ is invariant under conjugation by $W$, because atomic folds are invariant under conjugation by $\Gamma$ and $\rho$ is onto $W$. Lemma~\ref{lem:all-reflections} gives $T_0=T(W)$.
Consequently every reflecting hyperplane of $W$ is the image of an atomic fold as in Lemma~\ref{lem:atomic}, and no edge of $H$ crosses any of them.

Fix an edge $ab$ with $a\in A,b\in B$, and write $x=p(a),y=p(b)$. The points $x,y$ belong to a common closed fundamental chamber: for each reflection hyperplane their signed coordinates have nonnegative product. Any point $tx+(1-t)y$, $0<t<1$, has the same nonzero signs as either endpoint and lies on a hyperplane only when both endpoints do; every closed chamber containing that combination therefore contains both endpoints. The fixed ambient subspace, if present, can be factored off when making this choice.

Let $S$ be the simple system of this chamber, and let $I,J$ index its walls containing $x,y$, respectively. The standard stabilizer theorem for a finite reflection group (Humphreys~\cite{Humphreys}, Theorem 1.12(a)--(d), pp.\ 22--23) gives
\[
 \Stab_W(x)=W_I,\qquad \Stab_W(y)=W_J.
\]
Transitivity on each of $A,B$ implies that the two parts of $R$ are exactly
the two $W$-orbits of $x,y$, represented by $W/W_I$ and $W/W_J$.

Every image edge has its endpoints in some common closed chamber.
Move that chamber to the chosen one by an element of $W$. The resulting
endpoints must be $x,y$, since a $W$-orbit meets a closed fundamental
chamber in exactly one point. Thus the edges of $R$ are exactly the orbit
of $(x,y)$, namely the pairs $(wW_I,wW_J)$ for $w\in W$.
The quotient is connected because it is the image of the connected
graph $H$ under the quotient homomorphism. Thus $I\cup J=S$, and
this is the reflection bigraph of Definition~\ref{def:reflection};
original edge transitivity was not assumed. Keeping the two
parts distinguished handles the possibility that the same point occurs
in both orbit sets.
\end{proof}

An \emph{atomic lift} of a reflection in $W$ is an atomic fold
whose image under $\rho$ is that reflection. The proof above shows
that every reflection has an atomic lift. The following additional
fiber symmetry is useful, although the tensor splitting below does
not require it.

\begin{remark}[Symmetry within vertex fibers]\label{rem:kernel-fibers}
The \emph{representation kernel}
$K=\ker\rho=\{g\in\Gamma:\rho(g)=1\}$ consists of the group
elements acting trivially on $\mathcal E$. Under the hypotheses
of Theorem~\ref{thm:quotient}, it is transitive on every vertex fiber.
Indeed, take $v,v'$ in the same part with $p(v)=p(v')$ and choose
$g\in\Gamma$ with $gv=v'$. Then $\rho(g)$ belongs to the parabolic
subgroup $\Stab_W(p(v))$, generated by reflections fixing $p(v)$.
By Lemma~\ref{lem:atomic}, each atomic lift of such a reflection
fixes every vertex above $p(v)$ pointwise. A product $h$ of these
lifts therefore fixes $v$ and satisfies $\rho(h)=\rho(g)$.
Thus $gh^{-1}\in K$ sends $v$ to $v'$.
\end{remark}


\subsection{Exact tensor splitting}\label{sec:tensor-splitting}
The quotient need not distinguish all vertices of $H$. Knowing the
quotient and the sizes of the fibers from
Definition~\ref{def:spectral-quotient} would not determine $H$:
we must also recover the incidence pattern inside each edge fiber.

We do this with one common folding word. It moves all image points
into a chosen chamber and identifies each original vertex fiber
bijectively with one of the two vertex fibers above a chosen edge,
called the \emph{base fibers}. The same word identifies
each whole edge fiber with the base edge fiber. This compatibility
is what gives an exact tensor product.

\begin{theorem}[The spectral quotient splits as a tensor factor]\label{thm:tensor}
Under the hypotheses of Theorem~\ref{thm:quotient}, there is a connected bigraph $F$ such that
\[
 H\cong R\otimes F,
\]
where $\otimes$ is the tensor product of Definition~\ref{def:tensor}.
The graph $F$ is the subgraph of $H$ induced by the two vertex fibers
above any quotient edge whose image points lie in a chosen closed
fundamental chamber. We call $F$ the \emph{residual factor}.
\end{theorem}
\begin{proof}
Retain the base edge and closed fundamental chamber $\overline C$ chosen in Theorem~\ref{thm:quotient}. Put $A_0=p^{-1}(x)\cap A$ and $B_0=p^{-1}(y)\cap B$ for its two vertex fibers. For a simple reflection $s$, let $T_s$ be the geometric fold that fixes the closed halfspace containing $\overline C$ and reflects the other halfspace.
There is a finite sequence $T_{s_m}\cdots T_{s_1}$ taking every point
of $p(V(H))$ into $\overline C$. Choose $c$ in the interior of the
essential chamber. If a current point lies outside $\overline C$,
choose a simple wall for which its coordinate is negative and fold
toward the chamber. The scalar product with $c$ strictly increases
for every moved point and stays unchanged for the others.
Each point remains in its finite $W$-orbit, so it can move only
finitely many times. Each step moves at least one current point;
the procedure therefore terminates with every point in
$\overline C$.

For each selected $s_i$ choose an atomic lift $f_i$ and orient its graph fold $t_i$ toward the positive halfspace. Lemma~\ref{lem:atomic} gives $p\circ t_i=T_{s_i}\circ p$. Define
\[
 q=t_m\cdots t_1:H\longrightarrow H.
\]
Each $t_i$ is a graph endomorphism, so $q$ is one as well.
Every $W$-orbit meets $\overline C$ in exactly one point (Humphreys~\cite{Humphreys}, Theorem 1.12). Consequently $q$ maps each left vertex fiber bijectively onto the base left fiber $A_0$ and each right vertex fiber bijectively onto the base right fiber $B_0$. The asserted bijectivity holds at each step: on a whole vertex fiber, $t_i$ is either the identity or the restriction of the automorphism $f_i$. For a fiber lying on the wall, $f_i$ fixes it pointwise by Lemma~\ref{lem:atomic}.

More strongly, on each whole edge fiber, $t_i$ is either the identity or the restriction of the same full automorphism $f_i$ at both endpoints. Indeed, the two image endpoints never have opposite nonzero signs with respect to the wall. If one lies on the wall, $f_i$ fixes its entire vertex fiber pointwise. Thus each $t_i$, and hence $q$, maps any complete edge fiber bijectively onto the corresponding image edge fiber. After all steps this image is the base edge fiber.

Let $F=H[A_0\cup B_0]$. The map $q$ fixes every vertex of $F$ because its image point already lies in $\overline C$, so $q$ is a \emph{retraction} onto $F$: a homomorphism to $F$
that is the identity on $F$. In particular $F$ is connected as the image of connected $H$.

For a quotient vertex $v$, let $V_v$ be its vertex fiber. The map
\[
 u\longmapsto (p(u),q(u))
\]
(with the part label included in the first coordinate) is a bijection from $V(H)$ onto the appropriate part-respecting Cartesian product of vertex sets, since $q|_{V_v}$ is a bijection onto $A_0$ or $B_0$. For every quotient edge $vw$, its restriction maps the entire edge fiber bijectively onto $E(F)$ using exactly those endpoint bijections. Hence adjacency in $H$ is equivalent to adjacency in both coordinates, proving $H\cong R\otimes F$.
\end{proof}

\begin{example}\label{ex:complete-bigraph}
The residual factor can be nontrivial even for a complete bigraph.
Let $H=K_{a,b}$ with $1\le a<b$. Its Fiedler eigenvalue is $a$,
and its Fiedler eigenspace consists of functions supported on the
$b$-vertex part with sum zero. All vertices of the $a$-vertex part
therefore have image zero, while the other images form a regular
simplex, with all pairwise distances equal. The spectral quotient is $K_{1,b}$ and the residual factor
is $K_{a,1}$, giving $K_{a,b}=K_{1,b}\otimes K_{a,1}$. Permutations
of twins in the $a$-vertex part are invisible on the first Fiedler
eigenspace and are recovered by induction on the residual factor.
\end{example}

\begin{figure}[htbp]
\centering
\input{figures/tensor-fibers-drawing}
\caption{A spectral quotient can lose a whole tensor factor.
For $K_{2,3}$, the Fiedler map identifies $a_0,a_1$ and keeps
the three right vertices distinct, giving $R=K_{1,3}$.
The colored edges above the selected quotient edge form the
entire residual graph $F=K_{2,1}$. Each other quotient edge has
the same edge fiber. The tensor splitting proves that these
copies have compatible endpoint identifications in general;
see Definition~\ref{def:spectral-quotient} and
Theorem~\ref{thm:tensor}.}
\label{fig:tensor-fibers}
\end{figure}

\begin{example}[Fibers need not be twin classes]\label{ex:non-twin-fibers}
The \emph{bipartite adjacency matrix} $M$ has rows indexed by the left
part and columns by the right part, with entry one exactly at an
edge. Its singular values are the square roots of the eigenvalues
of $M^{\mathsf T}M$. A matrix tensor product has entries equal to
the products of the entries at the corresponding coordinate pairs.
Take $H=C_{10}\otimes C_6$. The bipartite adjacency matrices of
$C_{10}$ and $C_6$ have largest singular value $2$ and
second singular values $2\cos(\pi/5)$ and $1$, respectively.
Since the bipartite adjacency matrix of $H$ is their tensor
product and $H$ is $4$-regular, meaning every vertex has degree four, its Fiedler
eigenvalue is
$4-4\cos(\pi/5)$. The corresponding functions depend only on
the $C_{10}$ coordinate. Thus its spectral quotient is $C_{10}$
and its residual factor is $C_6$. A vertex fiber contains three
vertices with different neighborhoods.
\end{example}

To continue by induction, we must show that $F$ still has
fold-group transitivity. Atoms \emph{invisible} on the Fiedler eigenspace, with $\rho(f)=1$,
restrict to folds inside its edge fibers. \emph{Visible} atoms, with $\rho(f)\ne1$, require a
different argument: their action on the $F$ coordinate consists
only of permutations of twins. The next lemma gives this argument.

\begin{lemma}[Fiber automorphisms with one fixed fiber]\label{lem:twins}
Let $R,F$ be connected bigraphs. Suppose an automorphism of $R\otimes F$ induces an automorphism $\alpha$ on the $R$ coordinate, so that it has the form
\[
 (v,x)\longmapsto (\alpha v,\tau_v x),
\]
where $\tau_v$ is a permutation of the appropriate part of $F$. If $\tau_{v_0}$ is the identity for at least one quotient vertex $v_0$, then every $\tau_v$ permutes vertices only within their twin classes in $F$.
\end{lemma}
\begin{proof}
For each edge $vw\in E(R)$, the pair $(\tau_v,\tau_w)$ is a part-preserving automorphism of $F$, because it bijects the full edge fiber over $vw$ onto the full edge fiber over $\alpha(vw)$. A permutation $\tau$ of one part of $F$ extends to an automorphism fixing the other part pointwise if and only if it preserves every twin class.
Call a permutation preserving each twin class a \emph{twin permutation}.
Thus if $\tau_v$ is a twin permutation, compose $(\tau_v,\tau_w)$ with $(\tau_v^{-1},1)$ to deduce that $\tau_w$ is a twin permutation too. Starting from $v_0$ and following paths in connected $R$ proves the assertion.
\end{proof}

\begin{theorem}[Fold transitivity passes to the residual tensor factor]\label{thm:inherit}
If $\Fold(H)$ is transitive on each part, then $\Fold(F)$ is transitive
on each part for the factor $F$ in Theorem~\ref{thm:tensor}.
\end{theorem}
\begin{proof}
Work in the tensor coordinates supplied by Theorem~\ref{thm:tensor}.
We show that every atomic fold of $H$ preserves the $\Fold(F)$-orbit of
the $F$-coordinate of each vertex and each edge of $H$.
Let $f$ be such an atomic fold.
If $\rho(f)=1$, then $f$ preserves every vertex fiber and every edge
fiber individually. On the subgraph formed by any edge fiber it is
either the identity or a fold involution: intersect its fixed set
and its two exchanged open sides with that edge fiber.
The intersections still have no edge between the open sides and
are exchanged by the restricted involution. Transporting this
restriction by the fiber isomorphism to $F$ gives the identity
or a fold involution of $F$. Thus it preserves the claimed vertex
and edge orbits. Every vertex lies in an incident edge fiber because
$H$ has no isolated vertices.

If $\rho(f)\ne1$, its image is a reflection $\alpha$ on $R$.
It has the form $f(v,x)=(\alpha v,\tau_vx)$.
Since $H$ is connected and $f$ is a nonidentity fold,
$\Fix(f)$ is nonempty: otherwise the sole component of
$H-\Fix(f)$ would be preserved setwise.
Choose a vertex of $H$ fixed by $f$, and let $v_0$ be its quotient
vertex. Its spectral image lies on the reflecting hyperplane.
Lemma~\ref{lem:atomic} says that $f$ fixes \emph{the entire fiber}
over $v_0$ pointwise, so $\tau_{v_0}=1$.

Lemma~\ref{lem:twins} now implies that every $\tau_v$ is a twin permutation of $F$. A transposition of twins is a fold involution: after its fixed vertices are removed, the two moved vertices are separate components. Hence all twin permutations belong to $\Fold(F)$. For any quotient edge $vw$, the pair $(\tau_v,\tau_w)$ is therefore an element of $\Fold(F)$, proving the edge-orbit assertion also in the visible case.

Atomic folds generate $\Fold(H)$. If $\Fold(F)$ had more than one vertex
orbit in a part, the corresponding $F$-coordinate classes of vertices of
$H$ would be invariant under $\Fold(H)$, contradicting transitivity in
that part. The same argument also shows that edge transitivity descends
whenever it is assumed.
\end{proof}

\subsection{Completion of the classification}
The quotient and the residual factor now have the same symmetry
property with which we started, and the quotient is reflection.
We finish by checking that the residual factor is strictly smaller.
We also recall the geometric proof of the known reverse implication,
so that all four equivalences can be followed directly.

\begin{lemma}[Reflection bigraphs cut-percolate]\label{lem:reflection-percolates}
Every reflection bigraph is cut-percolating.
\end{lemma}
\begin{proof}
We recall a geometric proof of the result of Conlon--Lee~\cite{CL};
an algebraic proof on Coxeter coset spaces is given in
\cite[Section 3.1]{BCL}. Choose a fundamental closed chamber $\overline C$
for $W$, and points $x,y$ in its faces whose stabilizers are exactly
$W_I,W_J$. Embed the two parts as the distinguished orbits $Wx,Wy$.
Every edge has endpoints in some $w\overline C$, so no edge has endpoints
on opposite strict sides of a reflection hyperplane. A reflection fixes
exactly the vertices on its hyperplane; its positive and negative vertex
sets are exchanged unions of components. If the reflection acts
nontrivially on the graph, it therefore gives a genuine fold.
Repeatedly fold across simple walls toward $\overline C$. The finite-orbit
argument in Theorem~\ref{thm:tensor} gives a common finite word
taking every vertex into $\overline C$. The unique orbit representatives
there are $x$ and $y$, so the word sends every edge to the base edge $xy$.
Reverse the word and take inverse images to obtain cut-percolation.
\end{proof}

The following elementary implications will be useful
\cite[Section 1]{Coregliano}.
Every oriented fold maps an edge either to itself or to its image under
the underlying fold involution. It therefore preserves each
$\Fold(H)$-edge orbit. If its compositions send all edges to one edge,
there can be only one such orbit. Thus cut-percolation implies
fold-group edge transitivity. Edge transitivity in turn implies
transitivity on each vertex part, since every vertex is incident
with an edge.

\begin{lemma}[Part transitivity forces a reflection representation]
\label{lem:transitive-reflection}
Let $H$ be a finite connected simple bigraph with at least one edge.
If $\Fold(H)$ is transitive on each vertex part, then $H$ is a
reflection bigraph. Consequently $H$ cut-percolates and $\Fold(H)$
is edge-transitive.
\end{lemma}
\begin{proof}
Induct on $|E(H)|$ to obtain the reflection representation.
The single-edge graph is reflection, using the trivial Coxeter group.
For $|E(H)|>1$, Theorems~\ref{thm:quotient}--\ref{thm:inherit} give $H\cong R\otimes F$, with $R$ reflection and $F$ connected and fold-group transitive on each part. We verify that $|E(R)|\ge2$, making $|E(F)|<|E(H)|$.

Transitivity on each part makes $H$ biregular, of degrees $r,s$ on the two parts. If $R$ had only one edge, every function in the Fiedler eigenspace would be constant on each part. The only Laplacian eigenvalues on that two-dimensional space are $0$ and $r+s$, so $\lambda=r+s$. But at least one part has two vertices $u,v$. They have the same degree $d\in\{r,s\}$ and are nonadjacent. The mean-zero test function $\mathbf e_u-\mathbf e_v$ has Rayleigh quotient $d$, whence $\lambda\le d<r+s$, a contradiction.

Thus the induction hypothesis makes $F$ reflection. The part-respecting tensor product of reflection bigraphs is reflection, realized by the direct product of their finite Coxeter systems. Therefore $H$ is reflection.
Lemma~\ref{lem:reflection-percolates} gives cut-percolation, and the
elementary orbit argument above gives fold-group edge transitivity.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:classification}]
The known implications form the chain
\textup{(iii)}$\Rightarrow$\textup{(iv)}$\Rightarrow$\textup{(ii)}
$\Rightarrow$\textup{(i)}: the first is
Lemma~\ref{lem:reflection-percolates}, and the other two are the
elementary implications just recalled.
Lemma~\ref{lem:transitive-reflection} proves
\textup{(i)}$\Rightarrow$\textup{(iii)}, closing the cycle.
\end{proof}

\begin{corollary}[A binary star obstruction]\label{cor:star-obstruction}
If a finite connected simple bigraph $H$ with an edge is not
cut-percolating, it has a nonconstant two-color edge coloring $c$ such that
$c\circ t=c$ for every oriented fold $t$ of $H$.
Moreover, all edges at each vertex of one specified part have the same
color: the coloring can be chosen by coloring the stars centered in
that part.
\end{corollary}
\begin{proof}
Theorem~\ref{thm:classification} says that $\Fold(H)$ has at least two
vertex orbits in one of the specified parts, say $A$. Choose a nonempty
proper union $U$ of these orbits, and color an edge red exactly when its
$A$-endpoint lies in $U$. There are edges of both colors since $H$ has no
isolated vertices. An oriented fold sends a vertex either to itself or
to its image under an element of $\Fold(H)$, so it preserves membership
in $U$ and therefore preserves the coloring pointwise.
\end{proof}