\section{One-sided transitivity does not imply percolation}\label{sec:one-sided} Theorem~\ref{thm:classification} requires transitivity on both vertex parts. The corresponding statement for just one part is false, even for regular edge-transitive graphs. Coregliano suggested this one-sided analogue in \cite[Section~7]{CoreLeft}. We disprove it with the graph in Theorem~\ref{thm:one-sided}, without using the structural classification. \begin{definition}\label{def:left-percolation} A bigraph \emph{left-cut-percolates} if some word of oriented folds maps its whole left part to one vertex. Its \emph{minimum left image size} is the minimum of $|t(A)|$ over all such words $t$, including the empty word. Reversing the word identifies left cut-percolation with spreading one marked left vertex to the whole left part by successive inverse images \cite[Definition~2.4]{CoreLeft}. \end{definition} \paragraph{Idea.} Separate the binary coordinates into a protected part $D$ and an ordinary part $B$. There are four fold directions. Each fold reads one bit of $B$ and copies by a translation that may also change $D$. Products of the fold involutions generate every translation, so their group is transitive on points. But points with the same $B$ coordinates receive the same change in $D$. Their difference survives every copying operation. This leaves eight unavoidable point images. The main task is to prove that these are all the folds. \subsection{The binary-coset construction}\label{sec:binary-construction} All vector additions are over $\mathbb F_2=\{0,1\}$, with addition and multiplication modulo two. Let $D=\mathbb F_2^3$, choose its standard basis $a_1,a_2,a_3$, and put $a_4=a_1+a_2+a_3$. Their only nonempty linear dependence is $a_1+a_2+a_3+a_4=0$. Let $b_1,b_2,b_3,b_4$ be the standard basis of $B=\mathbb F_2^4$, and set $X=D\oplus B$. Write $x=(d,b)$ and $b=\sum_{i=1}^4 b_i(x)b_i$, where $b_i(x)\in\mathbb F_2$ is an ordinary coordinate. For each $i\in\{1,2,3,4\}$, put \[ U_i=\langle a_i,b_i\rangle. \] A coset $x+U_i$ has the four points $x,x+a_i,x+b_i,x+a_i+b_i$. Let $H_4$ be the incidence bigraph between the $128$ points of $X$ and all these cosets. Each direction has $128/4=32$ cosets, giving $128$ right vertices as well. Distinct directions give different point sets, so no repeated coset vertices occur. \begin{figure}[htbp] \centering \input{figures/one-sided-counterexample-drawing} \caption{Minimum left images for the binary-coset bigraph $H_4$. Only point images are shown; the coset vertices are omitted. Each row is a fiber $D\times\{b\}$, with its four ordinary bits $b$ on the left. The eight protected coordinates $d\in D$ are distinguished by color and by filled or hollow markers. Orienting translations by $b_1,b_2,b_3,b_4$ successively toward zero leaves $128,64,32,16,8$ point images. Any two points in one row receive the same translation under every oriented fold, so their difference in $D$ cannot disappear. The last eight images are unavoidable, even though the fold involutions generate a point-transitive group (Theorem~\ref{thm:binary-family}).} \label{fig:one-sided-counterexample} \end{figure} \begin{theorem}\label{thm:binary-family} The bigraph $H_4$ is connected, twin-free, $C_4$-free, four-regular, and edge-transitive, with $256$ vertices. Its fold involutions are exactly the eight translations by $b_i$ and $b_i+a_i$, $1\le i\le4$. Its fold group is the full translation group of $X$, transitive on the $128$ points and with four orbits on the coset part. Its minimum left fold image size is eight. \end{theorem} Connectedness follows since the $U_i$ span $X$. Distinct points have different incident cosets because $U_i\cap U_j=0$ for $i\ne j$, and distinct cosets have distinct neighborhoods; thus the graph is twin-free. Two cosets in distinct directions meet in at most one point, while cosets in one direction are disjoint; this also proves $C_4$-freeness. Each point is on one coset in each direction. Each coset has four points, proving four-regularity. Translations act on the graph. Every permutation of $a_1,\ldots,a_4$ extends to a linear automorphism of $D$, since the displayed dependence is unique; permuting the $b_i$ in the same way permutes the directions. These automorphisms act transitively on flags, hence on edges. \subsection{All automorphisms} We determine the part-preserving automorphisms before classifying folds. This prevents a choice of coordinates from hiding additional separating symmetries. Call two points \emph{collinear} if they belong to a common coset vertex. A \emph{parallel class} is the family of cosets of one $U_i$. \begin{lemma}\label{lem:parallel-recovery} The parallel classes are intrinsic. More precisely, form a graph on coset vertices by joining two disjoint cosets when at least four pairs of their points are collinear. Its connected components are exactly the four parallel classes. \end{lemma} \begin{proof} For two disjoint cosets of the same $U_i$, every allowed translation in some $U_j$, $j\ne i$, taking one coset to the other supplies four collinear point pairs. All such translations generate $X/U_i$, so these joins connect the whole parallel class. For different directions, translate one coset to $U_i$ and write the other as $x+U_j$. Disjointness means $x\notin U_i+U_j$. Each collinear pair corresponds uniquely to a vector in $(x+U_i+U_j)\cap U_k$ for $k\ne i,j$, because $U_i\cap U_j=0$. Any three of the $a_i$ are independent, so each of the two remaining $U_k$ has trivial intersection with $U_i+U_j$. Each contributes at most one pair. Thus different parallel classes have at most two collinear pairs and never reach the threshold four. \end{proof} \begin{lemma}\label{lem:binary-aut} Every part-preserving automorphism is generated by translations, simultaneous permutations of the $a_i,b_i$, and the independent shears \[ b_i\longmapsto b_i+a_i. \] In particular, the full part-preserving automorphism group has order $2^{11}4!=49152$. \end{lemma} \begin{proof} An automorphism permutes the parallel classes. Delete all coset vertices of direction $i$. The remaining graph has exactly two components, distinguished by the bit $b_i(x)$: the other $U_j$ span the codimension-one space $\ker b_i$, since $a_i=\sum_{j\ne i}a_j$. Thus an automorphism acts on the $b$ coordinates by a coordinate permutation and independent bit flips. Compose with the corresponding known automorphism so it preserves each direction and every $b$ fiber. Within a fiber, collinearity in direction $i$ is the matching $d\leftrightarrow d+a_i$. Preserving all these named matchings forces the action on that fiber to be a translation $d\mapsto d+\delta_b$, because the $a_i$ span $D$. Across two fibers differing only in bit $i$, the coset incidences force \[ \delta_{b+b_i}+\delta_b\in\langle a_i\rangle. \] For distinct $i,j$, a mixed difference of $\delta$ therefore lies both in $\langle a_i\rangle$ and $\langle a_j\rangle$, whose intersection is zero. Hence the difference in direction $i$ does not depend on the other bits. Consequently \[ \delta_b=\delta_0+\sum_i\varepsilon_i b_i(x)a_i, \qquad\varepsilon_i\in\mathbb F_2. \] This is a translation followed by the asserted shears. All these choices are distinct: the action on $b$ determines the signed coordinate permutation, and then the four differences and the value at $b=0$ determine the shear bits and $\delta_0$. Counting gives $2^7\cdot2^4\cdot4!$. \end{proof} \subsection{All fold involutions} Let $N$ be the point--coset incidence matrix. Its product $NN^{\mathsf T}$ is the convolution operator $f(x)\mapsto\sum_i\sum_{u\in U_i}f(x+u)$. The real characters of $X$ are the functions $x\mapsto(-1)^{\ell(x)}$ for linear functionals $\ell:X\to\mathbb F_2$. A character's eigenvalue is four times the number of $U_i$ on which $\ell$ vanishes. The constant eigenvalue is $16$. The largest nonconstant eigenvalue is $12$, with eigenspace spanned exactly by the four characters $(-1)^{b_i(x)}$. Indeed, a functional nonzero on $D$ is nonzero on at least two $a_i$, since their sum is zero. A functional zero on $D$ attains three vanishing subspaces exactly when it is one coordinate $b_i$. For the combinatorial Laplacian, the smallest positive eigenvalue is $\theta=4-2\sqrt3$, the smaller root of $(4-\theta)^2=12$. All other positive eigenvalue branches are larger. Thus the Fiedler vertex map on points, up to a common positive scale, is \[ (d,b)\longmapsto\big((-1)^{b_1},\ldots,(-1)^{b_4}\big). \] On a direction-$i$ coset its $i$-th coordinate is zero and all other coordinates are the corresponding signs, multiplied by the common positive factor $4/(4-\theta)$. The spectral quotient is the subdivision of the four-dimensional cube, obtained by inserting one new vertex in every cube edge: its point vertices are ordinary bit strings, and its coset vertices join the two strings differing in one coordinate. A point fiber is $D$, while a direction-$i$ coset fiber is $D/\langle a_i\rangle$. Thus the quotient remembers the ordinary bits but loses the protected coordinates. The automorphism action on this eigenspace is the group of signed coordinate permutations. Its reflections are coordinate sign flips and signed transpositions of two coordinates. \begin{lemma}\label{lem:binary-atoms} The atomic folds are exactly translation by $b_i$ or $b_i+a_i$. \end{lemma} \begin{proof} First rule out invisible folds, whose Fiedler action is the identity. By Lemma~\ref{lem:binary-aut}, such an involution fixes $b$ and acts by $d\mapsto d+\delta_b$. Fix a moved point $(d,b)$, so $\delta_b\ne0$. A direction-$j$ coset through it is fixed exactly when $\delta_b\in\langle a_j\rangle$. If $\delta_b$ is none of the $a_j$, every direction is moved, and their $a$-vectors span $D$. If $\delta_b=a_i$, every direction except $i$ is moved, and $a_i=\sum_{j\ne i}a_j$. In either case, add a suitable sequence of $a$-vectors from moved directions to connect $(d,b)$ to $(d+\delta_b,b)$. All intermediate points keep $b$ and remain moved; the cosets used along the path are moved as well. Thus a moved component contains a point and its image and is preserved by the involution, contrary to the fold condition. An atom visible on the Fiedler space must be one of its reflections. A signed transposition of coordinates $i,j$ fixes all spectral points with the corresponding equal or opposite bits. The atomic Fiedler lemma would require it to fix every point of each such $D$ fiber. But its action on that fiber has linear part exchanging $a_i,a_j$, which is nontrivial, followed by a translation. It cannot fix the whole fiber. This rules out these reflections. For a coordinate flip $i$, the mirror contains exactly the direction-$i$ coset vertices. The lemma requires all these cosets fixed individually. In the normal form of Lemma~\ref{lem:binary-aut}, this forces $\varepsilon_j=0$ for $j\ne i$ and $\delta_0\in\langle a_i\rangle$. The involution condition also forces $\varepsilon_i=0$. The only possibilities are translation by $b_i$ and by $b_i+a_i$. Both are atoms. Their fixed vertices are exactly the direction-$i$ cosets, and deleting these leaves the two connected components distinguished by $b_i(x)$, which the translation exchanges. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:binary-family}] Every atom moves the entire point part. A fold cannot decompose into two or more atoms, whose moved supports must be disjoint. Thus every genuine fold is one of the listed atoms. These translations generate every $a_i,b_i$ and hence all of $X$. Translations are transitive on points and on each direction class of cosets, and never change direction. Each oriented fold reads only bit $b_i(x)$ and adds one of $b_i,b_i+a_i$ when copying that side. Points differing by a vector in $D$ have identical bits and receive identical additions. Their difference is therefore preserved by every folding word. The eight points in any one $D$ fiber can never merge. Conversely, for each $i$ orient translation by $b_i$ toward $b_i(x)=0$. These four folds send all points to $D\oplus0$, attaining exactly eight images. \end{proof} \paragraph{A stable two-color witness.} Choose a nonzero linear functional $\ell:D\to\mathbb F_2$ and color every edge at point $(d,b)$ by $\ell(d)$. This is a nonconstant two-color star coloring: every edge at a point has its color. It is fold-stable in the sense of Definition~\ref{def:fold-stable}, even though the fold group is point-transitive. For an oriented atom in direction $i$, the new point color is $\ell(d+g(b_i)a_i)$, where $g$ is either zero or the indicator of the copied bit. The map $(d,b)\mapsto(d+g(b_i)a_i,b)$ is an automorphism from Lemma~\ref{lem:binary-aut}; it identifies the new colored bigraph with the original one. All folds of $H_4$ are atoms, so this covers every oriented fold. The witness records the protected coordinate rather than a point orbit of the fold group. \paragraph{Analytic scope.} Unlike a counterexample lacking edge transitivity, this graph passes Sidorenko's basic necessary symmetry condition. Nevertheless $H_4$ is not weakly norming. Appendix~\ref{app:local-obstruction} proves this by a local distance obstruction and the deleted-root profile inequality of Lemma~\ref{lem:profile-psd}. \begin{proof}[Proof of Theorem~\ref{thm:one-sided}] Take $H_4$. Theorem~\ref{thm:binary-family} gives the combinatorial assertions, and Corollary~\ref{cor:binary-not-norming} gives the analytic one. \end{proof} \paragraph{Relation to a known graph.} The graph $H_4$ agrees with the $256$-vertex graph on line $687$ of the semisymmetric tetravalent catalogue \cite{GraphSym,PotocnikWilson}. Our statement concerns its complete fold classification and the failure of the one-sided criterion. The proof uses no catalogue computation.