\section{Cut-percolating incidence bigraphs of simple two-designs} \label{app:designs} \begin{definition}\label{def:design} A \emph{simple $2$-$(v,k,\lambda)$ design} $\mathcal D=(X,\mathcal B)$ has $v$ points and a family $\mathcal B$ of distinct $k$-element subsets, called \emph{blocks}, with each pair of distinct points in exactly $\lambda$ blocks. Its \emph{incidence bigraph} has parts $X,\mathcal B$ and joins $x$ to $V$ when $x\in V$. The design is \emph{complete} if $\mathcal B=\binom{X}{k}$, \emph{symmetric} if $|\mathcal B|=v$, and \emph{nontrivial} here if $2\le k0$ is a prescribed inner \emph{unary multiplier}, a factor depending only on the left state. Let $d$ be the degree of a right vertex in $G$, and choose \[ 00\}$, and let $e$ be the number of marked left clones beyond the ones forced by these $j$-clones. Up to a common conditioning factor, its power of $t$ is \begin{equation}\label{eq:marked-exponent} ad+a|N_G(T)\setminus N_G(v)|+e-c\sum_z k_z. \end{equation} If $T\ne\varnothing$, twin-freeness and equal right degrees imply $|N_G(T)\setminus N_G(v)|\ge1$. Relative to the configuration $k_v=b$, $T=\varnothing$, $e=0$, the exponent is larger by at least $a-cb(|Y|-1)>0$. If $T$ is empty, either missing a $j$-clone at $v$ or adding an unnecessary marked clone also strictly increases the exponent. Therefore the unique leading family has all $b$ root clones at $j$, no other right clone at $j$, and exactly the inner left clones marked. As $t\downarrow0$, the conditional model converges to the model with the whole root class deleted. Each inner clone acquires the factor $\ell_x$, since its $b$ incident root entries contribute $(\ell_x^{1/b})^b=\ell_x$. Outer activities remain $\mu_x$. Apply \eqref{eq:split-state} to this new right state, with $C_*,D_*$ the clone unions above $C,D$, and let the observable on marked left states be an arbitrary function $f$ of their old labels. Right transitivity of $G$ and clone symmetry make $\Pr(y_z=j)$ equal for every right source clone before taking the limit. Dividing out this positive common probability gives, in the limit, \begin{equation}\label{eq:deleted-raw} \sum_{v\in Y}\sum_{\substack{u\in N(v)\cap C\\w\in N(v)\cap D}} \EE_v\left[ \left(\sum_{i=1}^a f(x_{u,i})\right) \left(\sum_{i=1}^a f(x_{w,i})\right)\right]\ge0. \end{equation} The sum over the $b$ clones of each root contributes a common factor $b$, which has also been divided out. All $Z_v$ are equal, including when inner and outer activities differ. To see this, an automorphism of $G$ taking $v$ to $v'$ takes $N(v)$ to $N(v')$ and its complement to the corresponding complement. Lift it to the clones, retaining each clone index. It gives a weight-preserving bijection of assignments between the two deleted-root backgrounds: the kernel and right activities are common, and the left activity vectors are common within the inner and outer sets separately. The two vectors need not agree with each other. The source coloring occurs only in the observable sums, so this automorphism need not preserve $C,D$. Multiplying \eqref{eq:deleted-raw} by the common positive partition function gives its unnormalized version. \emph{Interpolating an arbitrary profile function.} For a positive deleted-root background, the conditional partition factor of one inner left clone with profile $h$ is \[ Z(h)=\sum_x\mu_x\ell_x\prod_{j\in\mathcal J}K(x,j)^{h_j}>0. \] Recall that a kernel row records its values at all right target states. Add finitely many \emph{rare} left rows, whose activities will tend to zero, with strictly positive row vectors $w_i=(w_{i,j})_{j\in\mathcal J}$, masses $\rho c_i$ with $c_i>0$, and inner multipliers one. In this proof $O(\rho^d)$ means a quantity bounded by a constant times $\rho^d$ as $\rho\downarrow0$, with the finite background and the chosen rare rows fixed. Let $f$ be zero on old rows and equal to $f_i\in\RR$ on rare row $i$. Write $w_i^h=\prod_j w_{i,j}^{h_j}$, $B(h)=\sum_i c_i f_iw_i^h$, and $S(h)=\sum_i c_iw_i^h$. The exact conditional mean and its expansion are \[ \EE[f(x)\mid h] =\frac{\rho B(h)}{Z(h)+\rho S(h)} =\rho\frac{B(h)}{Z(h)}+O(\rho^2). \] There are finitely many profiles and $\min_h Z(h)>0$. The error is therefore uniform over all profiles for this fixed model. The functions $h\mapsto w^h$ with $w$ strictly positive span all functions on the finite profile set. Otherwise a nonzero linear combination of distinct monomials would vanish on the open positive orthant, that is, the open set of vectors with all coordinates positive. A nonzero polynomial cannot vanish on this open set. Choose the rows and coefficients so that $\sum_i c_i f_i w_i^h=Z(h)g(h)$ for all $h$. Given all right labels, the left clones are independent. In particular, the two distinct classes $u\in C,w\in D$ in \eqref{eq:deleted-raw} have conditional product mean \[ a^2\rho^2g(h_u)g(h_w)+O(\rho^3). \] The $O(\rho^3)$ error is uniform over the finitely many profiles, roots, and source pairs. The marginal background on right labels changes by $O(\rho)$ uniformly over its finite assignment set, since its weights are polynomials and its normalization is positive at $\rho=0$. Choose the rows and interpolation coefficients before varying $\rho$. For each fixed $\rho>0$, the preceding root-class limit gives the deleted-root inequality. Now divide its unnormalized form by $a^2\rho^2$ and let $\rho\downarrow0$. This proves \eqref{eq:profile-q} for positive backgrounds with inner activities $\mu_x\ell_x$ and outer activities $\mu_x$. \emph{Independent alphabets and zero entries.} To give inner and outer classes disjoint alphabets, assign each desired inner state activity $\eta m_x$ and multiplier $\eta^{-1}$; assign each outer state activity $m_x$ and multiplier zero. For each fixed $\eta>0$, first approach the zero multiplier from positive values, after taking the root-class and rare-row limits. Then let $\eta\downarrow0$. Inner states retain their desired effective activities, while their outer activities vanish; outer states have zero inner activity. This realizes arbitrary inner and outer activity vectors on disjoint alphabets. The unnormalized entries of $M$ are finite polynomials in the effective activity vectors and kernel entries. A final continuity argument therefore extends the inequality to all nonnegative finite models, including zero partition functions. The earlier error bounds are asserted only for each fixed positive model; no uniformity in these final boundary limits is needed. This proves the lemma. \end{proof} \begin{example}[A two-profile witness for the Fano plane]\label{ex:fano-profile} Let $(X,\mathcal B)$ be the Fano plane, $G$ its incidence bigraph, and $V_0$ a target line with distinct points $A,B$. Here \emph{lines} are the design's three-point blocks. The \emph{incidence kernel} is $K(x,J)=1$ if $x\in J$ and zero otherwise. Take $C=\{A\}$ and $D=X\setminus C$. Use inner target alphabet $V_0$, outer alphabet $X\setminus V_0$, right alphabet $\mathcal B\setminus\{V_0\}$, and the incidence kernel, with all activities one. Let $p_A$ record $b$ copies of each of the two lines in $N(A)\setminus\{V_0\}$, and define $p_B$ similarly. Two different inner source classes cannot both have profile $p_A$. Their left clones would all be at $A$, since the two target lines in that profile intersect only at $A$. Any nonroot source line through one class meets any nonroot line through the other at an outer source point. Their target lines must therefore contain both $A$ and a common point outside $V_0$, forcing them to coincide. All target lines in both profiles would be equal, contrary to the two distinct lines in $p_A$. The same argument applies to $p_B$. Thus the two diagonal entries are zero. The identity assignment at source root $V_0$ contributes to the off-diagonal entry, so the principal submatrix is \[ \begin{pmatrix}0&c\\c&0\end{pmatrix}, \qquad c>0. \] It has negative determinant. Since the Fano plane is flag-transitive, Lemma~\ref{lem:profile-psd} applies under a weak-norming assumption and excludes every $G(a,b)$. \end{example} \subsection{A two-profile minor in a design} We now choose the finite model that will test a design. The two profiles belong to two target points $A,B$ on a root block $V_0$. Give both points a small activity $\varepsilon$. Then the identity assignment contributes to the off-diagonal entry at order $\varepsilon^{2a}$. If the corresponding diagonal entry has higher order, a $2$-by-$2$ principal minor is negative. The rest of the proof determines what a diagonal assignment of order $\varepsilon^{2a}$ would force. Let $\mathcal D=(X,\mathcal B)$ be a design with the parameters of Theorem~\ref{thm:designs}, in the notation of Definition~\ref{def:design}. Every point lies in \[ r=\frac{\lambda(v-1)}{k-1}>\lambda \] blocks. Its incidence bigraph $F$ is connected and biregular. It is twin-free: distinct points have only $\lambda0$. Let $p_A$ contain $b$ copies of each block in $N(A)\setminus\{V_0\}$, and define $p_B$ analogously. The profiles are distinct. If an inner source class has profile $p_A$, every one of its left clones is at $A$: after restoring the deleted root class with every clone labeled $V_0$, a \emph{virtual root block}, its target point is incident with every member of $N(A)$, and equal point degrees and twin-freeness force that point to be $A$. The same holds for $p_B$ and $B$. The assignment sending every point clone to its base point and every nonroot block clone to its base block, at source root $V_0$, contributes to the off-diagonal entry for the source pair $A,B$. Consequently, for some constant $c_0>0$ independent of $\varepsilon$, \begin{equation}\label{eq:minor-orders} M_{p_A,p_B}\ge c_0\varepsilon^{2a},\qquad M_{p_B,p_B}=O(\varepsilon^{2a}). \end{equation} If no $p_A,p_A$ assignment has degree exactly $2a$ in $\varepsilon$, then $M_{p_A,p_A}=O(\varepsilon^{2a+1})$. The determinant of this two-profile principal submatrix is then negative for small $\varepsilon$, contradicting Lemma~\ref{lem:profile-psd}. Call an assignment contributing at degree exactly $2a$ in $\varepsilon$ a \emph{minimal assignment}. We record what it would entail. At some source root $V$, there would be distinct classes $u\in V\cap C$, $w\in V\cap D$ all of whose clones are at $A$. All other point clones would avoid $A,B$. Selecting one actual target image from each point class gives a map $f:X\to X$ with \begin{equation}\label{eq:collapse-fibers} f^{-1}(A)=\{u,w\},\qquad f^{-1}(B)=\varnothing. \end{equation} Select actual target images of all block classes as well, and restore a virtual root row $V\mapsto V_0$. This gives an incidence-preserving map $h$ with point restriction $f$. More information comes from the clone counts on $N(u)$. Divide these counts by $b$ to form a matrix $P$ with rows $N(u)$ and columns $N(A)$; the virtual root row has its single entry at $V_0$. Every row and column sum is one because the prescribed profile contains exactly $b$ copies of every nonroot column. Thus $P$ is \emph{doubly stochastic}: its entries are nonnegative and each row and column sums to one. Complete adjacency between clone classes implies \begin{equation}\label{eq:transport-compatible} P_{E,J}>0\quad\Longrightarrow\quad f(x)\in J \quad\text{for every }x\in E. \end{equation} The selected image of each block in $N(u)$ can be chosen from its positive columns. \subsection{Closure lines and saturated transport}\label{sec:closure-lines} For distinct points $u,x$, put \begin{equation}\label{eq:closure-lines} S_x^u=N(u)\cap N(x),\qquad L(u,x)=\{z\in X:S_x^u\subseteq N(z)\}. \end{equation} Thus $S_x^u$ is the set of blocks through both points and $L(u,x)$ is their intersection, called the \emph{closure line}. Each $S_x^u$ has size $\lambda$. If $z,t$ are distinct members of $L(u,x)$, then the $\lambda$ defining blocks contain both, and therefore $N(z)\cap N(t)=S_x^u$. It follows that $L(z,t)=L(u,x)$. These \emph{closure lines} form a linear space: every pair belongs to a unique line, and distinct lines meet in at most one point. Any design block containing two points of a closure line contains the entire line. Call a line \emph{thick} if it has at least three points, and \emph{thin} if it has two. For $x\notin\{u,w\}$, \eqref{eq:collapse-fibers} gives $f(x)\ne A$. By \eqref{eq:transport-compatible}, the $\lambda$ rows in $S_x^u$ send all their mass to the $\lambda$ columns in $S_{f(x)}^A$. The latter columns have total capacity $\lambda$, so they are saturated and no other row can send mass to them. Hence \begin{equation}\label{eq:saturation} P_{E,J}>0\quad\Longrightarrow\quad \bigl(E\in S_x^u\ \Longleftrightarrow\ J\in S_{f(x)}^A\bigr). \end{equation} In particular, distinct source patterns $S_x^u,S_y^u$ have distinct target patterns $S_{f(x)}^A,S_{f(y)}^A$. If the latter sets were equal, their saturated preimages would be equal. \begin{lemma}[A collapse creates a thin flag]\label{lem:thin-flag} Any minimal assignment described in \eqref{eq:collapse-fibers}--\eqref{eq:transport-compatible} produces a target flag $(A,J)$ such that every closure line through $A$ contained in $J$ has size two. \end{lemma} \begin{proof} Choose a block $E$ containing $w$ but not $u$, using $r>\lambda$. Its $k-1$ other points lie on distinct closure lines through $u$. Otherwise $E$ would contain two points on such a line and hence the whole line, including $u$. Their patterns $S_x^u$ are thus distinct. By \eqref{eq:saturation}, their images lie on $k-1$ distinct closure lines through $A$. The block $J=h(E)$ contains $A=f(w)$ and all these $k-1$ images, which are distinct and different from $A$. They exhaust its $k$ points. If a closure line through $A$ in $J$ had a third point, that point would be one of these images and would duplicate an already represented closure line. This contradicts their distinctness. \end{proof} If the design has any thick line, a defining block and a point on that line give a flag whose block contains a thick line through its point. Flag transitivity forces this property at every flag, contrary to Lemma~\ref{lem:thin-flag}. Thus no minimal assignment exists. Choose any two distinct target points in a block and any partition placing them in opposite classes. The minor \eqref{eq:minor-orders} contradicts weak norming. We conclude: under our weak-norming assumption, every closure line has size two. Equivalently, for fixed $u$ the patterns $S_x^u$, $x\ne u$, are all distinct. \subsection{Thin closure lines force a point transposition} Two-point closure lines alone do not force a complete design. For example, complementing the seven blocks of the Fano plane gives a $2$-$(7,4,2)$ design. Two distinct complemented blocks intersect in two points, so every closure line has size two. The additional information comes from the collapse assignment forced by profile positivity. We now show that it produces a point transposition preserving all blocks. Suppose now that all closure lines have size two and that a minimal assignment exists. By \eqref{eq:saturation}, $f$ is injective outside $\{u,w\}$. Together with \eqref{eq:collapse-fibers}, this gives a bijection \begin{equation}\label{eq:outside-bijection} f:X\setminus\{u,w\}\longrightarrow X\setminus\{A,B\}. \end{equation} Distinct target blocks through $A$ are distinguished by their incidence with $X\setminus\{A,B\}$. Indeed, two equal-sized sets both containing $A$ cannot become identical after deleting $B$ unless they were identical already. By \eqref{eq:saturation}--\eqref{eq:outside-bijection}, all positive columns of any row of $P$ have the same such signature. Each row therefore has only one positive column. Since $P$ is doubly stochastic, it is a \emph{permutation matrix}, with one entry one in each row and column and all others zero. All right classes in $N(u)$ are monochromatic, meaning that every clone in a class has the same target label, and $h$ maps $N(u)$ bijectively to $N(A)$. \begin{lemma}\label{lem:design-transposition} For a minimal assignment in a design with only two-point closure lines, the transposition $(u\ w)$ preserves the entire block family. \end{lemma} \begin{proof} Change only $f(w)=A$ to $\pi(w)=B$. By \eqref{eq:outside-bijection}, the resulting map $\pi$ is a point permutation, with $\pi(u)=A$. There are $r-\lambda$ blocks through $u$ avoiding $w$. On such a block $E$, $f$ is injective, so the $k$ points of $f(E)$ exhaust the image block $h(E)$. Thus $h(E)=f(E)=\pi(E)$ is a block through $A$ avoiding $B$. The star bijection shows that these images exhaust all $r-\lambda$ target blocks through $A$ avoiding $B$. The remaining $\lambda$ blocks through both $u,w$ map bijectively to the target blocks through both $A,B$. Their image point sets have $k-1$ points and avoid $B$, so in that case $h(E)=f(E)\cup\{B\}=\pi(E)$ as well. For a block $E'$ through $w$ avoiding $u$, $f$ is again injective on its point set. Therefore $h(E')=f(E')=\pi(u\ w)(E')$. These $r-\lambda$ images are distinct and exhaust the same target family through $A$ avoiding $B$. Pairing the two image families shows that for every block $E$ through $u$ avoiding $w$, the set $(u\ w)(E)$ is a block through $w$ avoiding $u$. The transposition exchanges these two families, and fixes every block containing both points or neither. It preserves all blocks. \end{proof} The corresponding involution of the incidence bigraph is a genuine atomic fold. Its nonfixed subgraph consists of two disjoint stars, centered at $u,w$, whose leaves are the blocks containing exactly their respective center. The two stars are exchanged. The following argument needs only the design automorphism, but this observation explains its connection with the main theorem. \subsection{A binary partition forces the complete design} Let $T\le\Aut(\mathcal D)$ be generated by all point transpositions preserving the design. Suppose $T$ has at least two point orbits. Choose one orbit $C$ and put $D=X\setminus C$. Since $\lambda\ge1$, there is a target block $V_0$ containing $A\in C$ and $B\in D$. Apply the colored profile matrix with this partition and these target profiles. Any minimal $p_A,p_A$ assignment must collapse a source pair $u\in C,w\in D$. By Lemma~\ref{lem:design-transposition}, $(u\ w)\in T$, contrary to their different $T$-orbits. Thus no minimal assignment exists, and the two-profile determinant from \eqref{eq:minor-orders} is negative. Weak norming forces $T$ to be transitive. Form a graph on $X$ whose edges are the pairs whose transpositions preserve the design. Its components are exactly the $T$-orbits. Transpositions on the edges of a connected graph generate the full symmetric group: swaps along a path generate a transposition between its endpoints. Therefore $T=\Sym(X)$. The nonempty block family is invariant under $\Sym(X)$ and consists of $k$-sets, so it must be all of $\binom{X}{k}$. \begin{proof}[Proof of Theorem~\ref{thm:designs}] If $F(a,b)$ cut-percolates, it is weakly norming by \cite{CL}. The preceding argument then forces $\mathcal B=\binom{X}{k}$. Conversely, suppose $\mathcal B=\binom{X}{k}$. Take $W=\Sym(v)$ with its standard type-$A_{v-1}$ simple system. The stabilizers of $\{1\}$ and $\{1,\ldots,k\}$ are standard parabolic subgroups, and their orbit pairs are exactly incidences between singletons and $k$-subsets. Thus $F$ is reflection. As $F(a,b)=F\otimes K_{a,b}$, every indicated blowup is reflection. The known reflection-to-cut-percolation implication of \cite{CL} shows that this blowup cut-percolates and is weakly norming. The preceding analytic argument shows that weak norming forces the design to be complete. Thus completeness implies reflection, reflection implies cut-percolation, cut-percolation implies weak norming, and weak norming implies completeness. This proves all the claimed equivalences independently of Theorem~\ref{thm:classification}. \end{proof} \Needspace{8\baselineskip} \begin{corollary}\label{cor:symmetric-design} For a nontrivial simple symmetric $2$-$(v,k,\lambda)$ design, its incidence bigraph and every uniform twin blowup are cut-percolating exactly when $k=v-1$, equivalently $\lambda=k-1$. The same criterion holds for weak norming. These incidence bigraphs are the \emph{crown graphs}: $K_{v,v}$ with a perfect matching deleted, that is, with $v$ disjoint edges covering all vertices removed. \end{corollary} \begin{proof} The complete design has $\binom vk$ blocks. Under $2\le k\lambda$. On the point subspace $\mathbf1_X^{\perp}$ the matrix $NN^{\mathsf T}$ is the positive scalar $r-\lambda$. On the constants its eigenvalue is $rk$. Put $\theta$ for the smaller root of \[ (r-\theta)(k-\theta)=r-\lambda. \] It lies strictly between zero and $\min\{r,k\}$. The Fiedler eigenspace, the lowest nonconstant Laplacian eigenspace from Section~\ref{sec:spectral-setup}, is exactly \[ \left\{\left(u,\frac{N^{\mathsf T}u}{k-\theta}\right): u\in\mathbf1_X^{\perp}\right\}. \] Indeed, the two nonconstant eigenvalue branches come from that quadratic equation; its larger root exceeds $\max\{r,k\}$. The remaining eigenvalues are $0,r+k$ on the constant vectors and possibly $k$ on $\ker N$. Restriction to the point coordinates thus identifies the Fiedler representation of every graph automorphism with its ordinary permutation action on $\mathbf1_X^{\perp}$. That action is faithful, meaning that only the identity acts as the identity: fixing every point fixes every block because the design is simple. Lemma~\ref{lem:atomic} says an atom acts as a reflection or the identity on this space. It cannot act as the identity by faithfulness. An involutive point permutation has one negative direction on $\mathbf1_X^{\perp}$ for each disjoint point transposition in its cycle decomposition. Hence an atom must transpose exactly two points. Conversely, if $(x\ y)$ preserves the design, the moved vertices of the induced incidence automorphism are $x,y$ and the blocks containing exactly one of them. The moved subgraph consists of two exchanged stars, centered at $x,y$. Both stars have $r-\lambda>0$ leaves. Thus this automorphism is an atomic fold. All permutations within each exchangeability class are generated by these transpositions, giving the formula. Corollary~\ref{cor:all-twins}, whose group formula needs no cut-percolation assumption, shows that every fold of $G(a,b)$ preserves the union of left clones above each $X_i$. No word can merge points from different classes, so its left image has at least $c$ vertices. In the base graph, orient transposition folds within $X_i$ toward one selected point to collapse that class. Lift these folds while retaining clone indices; then collapse the remaining clones by twin transpositions. This attains exactly one image vertex per class, hence $c$. \end{proof} \begin{corollary}\label{cor:design-left-exact} For every such design and every $a,b\ge1$, the following are equivalent: left cut-percolation of $G(a,b)$; transitivity of its fold group on the point part; completeness of the design. \end{corollary} \begin{proof} The first two properties both mean $c=1$. Then all point transpositions preserve the design, so the full symmetric group preserves it. This group is transitive on the $k$-subsets; the nonempty block family must contain them all. Conversely, a complete design has every transposition, so $c=1$. \end{proof} \begin{corollary}\label{cor:design-no-atoms} If $\lambda=1$ and $k\ge3$, the base incidence bigraph has no nonidentity fold. The same holds for every incomplete symmetric two-design, where symmetric means $|\mathcal B|=v$. \end{corollary} \begin{proof} In a Steiner two-design, meaning $\lambda=1$, a point transposition $(x\ y)$ would take a block containing $x$ but not $y$ to a different block sharing $k-1\ge2$ points. Those common points would occur together twice, contradicting $\lambda=1$. For a symmetric design, $r=k$ and $N$ is square and invertible. The induced point and block permutation matrices are conjugate by $N$, so a point transposition would move exactly two blocks. They differ only in exchanging $x,y$ and thus intersect in $k-1$ points. Distinct blocks in a symmetric design meet in $\lambda$ points, so $\lambda=k-1$. The parameter identity $k(k-1)=\lambda(v-1)$ gives $k=v-1$, the complete design. Apply Theorem~\ref{thm:design-fold-group} in both cases. \end{proof} \subsection{An automorphism question for nested designs} \label{sec:nested-designs} The exchangeability classes also recover a prescribed partition without knowing it in advance. Repeating this operation recovers an entire nested chain. Amarra, Devillers, and Praeger \cite[Problem~2]{ADP} asked whether the full automorphism group of their flag-transitive two-designs is the iterated wreath product used in their construction. They proved this for chains of length two and three. We give the equality for all lengths. The reconstruction itself applies to a larger class of set families and needs no two-design assumption. \paragraph{Recovering a nested partition.} Let $s\ge1$, and let $e_1,\ldots,e_s$ and $t_1,\ldots,t_s$ be integers with \[ 2\le t_i