\section{A local obstruction that survives twins}\label{app:local-obstruction}
The profile inequality in Lemma~\ref{lem:profile-psd} also lifts a
local distance obstruction to arbitrary uniform twin blowups.
The unblown weighted-root method of \cite[Theorem~3]{KMPW}
uses the same principle: weight vertices by their distance from a
root, and compare the leading terms of a two-by-two minor.
Here the profiles retain enough information to choose two
neighborhood bijections even when the source vertices have twins.

Write $\operatorname{dist}_F(x,y)$ for graph distance, the minimum
number of edges in a path from $x$ to $y$. A rooted homomorphism
$\phi:F\to F$ fixing $v$ is \emph{distance-preserving from $v$}
if $\operatorname{dist}_F(v,\phi(z))=\operatorname{dist}_F(v,z)$
for every vertex $z$.

\begin{proposition}\label{prop:local-twin-obstruction}
Let $F=(A,B;E)$ be a connected $C_4$-free biregular bigraph,
with left degree at least three, right degree at least two,
and $\Aut(F)$ transitive on $B$. Fix $v\in B$ and $u_1\in N_F(v)$.
Suppose that for no distinct $x,y\in N_F(v)$ is there a
distance-preserving rooted homomorphism $\phi:F\to F$ such that
\[
 \phi(v)=v,\qquad \phi(x)=\phi(y)=u_1,
\]
and such that both maps
$\phi:N_F(x)\to N_F(u_1)$ and
$\phi:N_F(y)\to N_F(u_1)$ are bijections.
Then $F(a,b)$ is not weakly norming for any $a,b\ge1$.
\end{proposition}

\begin{proof}
The degree and $C_4$-freeness assumptions imply twin-freeness.
Choose $u_2\in N_F(v)\setminus\{u_1\}$, which exists
by the right-degree assumption. Suppose that $F(a,b)$ is weakly
norming. First we obtain a
profile matrix at one fixed source root, with no source coloring.
Average the matrices of Lemma~\ref{lem:profile-psd} over all
partitions $A=C\sqcup D$, assigning each vertex independently
to either side with probability $1/2$. Every ordered pair of
distinct inner classes appears with the same positive factor.
The average is positive semidefinite. Source automorphisms
taking one root to another preserve the uncolored matrix,
including the inner and outer activity vectors, by the same
relabeling argument as in that lemma. Dividing out the common
root count and the positive averaging factor shows that the
following matrix at source root $v$ is positive semidefinite:
\[
 J_{p,q}=\sum_{\substack{x,y\in N_F(v)\\x\ne y}}
       \text{unnormalized weight of }(h_x=p,\ h_y=q).
\]
It is symmetric because the source pairs are ordered.

Delete all right clones above $v$. Use the inner left target
alphabet $N_F(v)$, the outer left alphabet $A\setminus N_F(v)$,
and the right alphabet $B\setminus\{v\}$, with the incidence
kernel of $F$. Give every target state $z$ activity
$\gamma^{\operatorname{dist}_F(v,z)}$, where $\gamma>0$.
Lemma~\ref{lem:profile-psd} allows these separate alphabets
and zero kernel entries.

Every inner source class is monochromatic in a positive-weight
assignment. Indeed, two different labels in $N_F(v)$ have
no common right neighbor other than $v$, by $C_4$-freeness;
that label has been deleted, and a surviving neighboring right
clone exists.

For $u\in N_F(v)$, let $p_u$ be the profile containing $b$
copies of each label in $N_F(u)\setminus\{v\}$.
These profiles are distinct. They contain at least two
different labels, whose unique common left neighbor is $u$,
so a class with profile $p_u$ has every left clone at $u$.
Moreover, this profile permits a bijective choice of one label
from each remaining neighboring right base class. To see this,
form a bipartite graph between these base classes and the
labels in $N_F(u)\setminus\{v\}$, joining a class to every
label appearing on one of its clones. Each class has $b$
clones, and each target label occurs $b$ times in the profile.
For any set of source classes, its $b$ times as many slots fit
only into the labels adjacent to it, each with capacity $b$.
Every set of classes thus has at least as many adjacent labels.
This is Hall's matching condition, and it gives the desired
bijection.

Reinsert a virtual source root mapped to $v$. Every target
label assigned to a clone above $z$ is at distance at most
$\operatorname{dist}_F(v,z)$: lift a shortest source path to
the clones and apply the assignment to it. Consequently the
total exponent of $\gamma$ is at most
\[
 D=a\sum_{z\in A}\operatorname{dist}_F(v,z)
       +b\sum_{z\in B\setminus\{v\}}\operatorname{dist}_F(v,z).
\]
Equality holds exactly when every clone preserves distance.
The identity assignment attains this exponent.
Let $L_{ij}$ be the coefficient of $\gamma^D$ in
$J_{p_{u_i},p_{u_j}}$. The identity gives $L_{12}>0$.
We claim that $L_{11}=0$.

A contribution to $L_{11}$ would give two distinct inner
classes $x,y$, both assigned $u_1$, with the prescribed
profiles. Their remaining right neighborhoods are disjoint,
since a shared class would give a four-cycle through $v$.
We can therefore make the two matching choices above
independently, and choose one arbitrary clone label at
every other base vertex. The complete edges between clone
classes make the selected map a homomorphism of $F$.
Reinserting $v\mapsto v$ gives bijections on both marked
neighborhoods, and maximality of the exponent $D$ makes
the map distance-preserving. This contradicts the hypothesis.

The leading minor is consequently strictly negative:
\[
                       L_{11}L_{22}-L_{12}^2=-L_{12}^2<0.
\]
It is the coefficient of $\gamma^{2D}$ in the corresponding
minor of $J$. For sufficiently large finite $\gamma$, that
minor is negative, contradicting positive semidefiniteness.
The finite boundary model with zero incidence entries is
already permitted by Lemma~\ref{lem:profile-psd}.
\end{proof}

\begin{corollary}\label{cor:binary-not-norming}
The binary-coset bigraph $H_4$ is not weakly norming.
The same holds for every uniform twin blowup of $H_4$.
\end{corollary}
\begin{figure}[htbp]
\centering
\input{figures/coset-corner-drawing}
\caption{The local distance obstruction for $H_4$.
Points are filled and coset vertices are hollow. Choose
$x\in U_i$, $y\in U_j$ so that $z=x+y$ is at root distance
four. A putative collapse of both root cosets to $U_k$,
bijective on their point neighborhoods, allows a choice with
$x'=\phi(x)$ and $y'=\phi(y)$ having different ordinary
coordinate $b_k$ values. Any point collinear with both must
lie in $U_k$, since all other directions preserve $b_k$.
Thus $z'=\phi(z)$ has root distance at most two. The target
diagram shows membership in this one coset; target points
may coincide. Corollary~\ref{cor:binary-not-norming} makes
this obstruction analytic and retains it under twins.}
\label{fig:coset-corner}
\end{figure}
\begin{proof}
Apply Proposition~\ref{prop:local-twin-obstruction} to $H_4$
with the parts reversed: the left vertices are cosets and
the right vertices are points. Both degrees are four,
the graph is $C_4$-free, and translations are transitive on
points. Take the root point to be $0$.
Suppose a distance-preserving homomorphism $\phi$ maps two distinct
root cosets $U_i,U_j$ to the same root coset $U_k$, bijectively
on both point neighborhoods.

The three nonzero points of each source coset map bijectively
to $U_k\setminus\{0\}$. Choose $x\in U_i\setminus\{0\}$
and $y\in U_j\setminus\{0\}$ with images $a_k$ and $b_k$,
respectively. The point $z=x+y$ is collinear with both
$x$ and $y$ and has root distance four. To see the latter,
it suffices to show that $z$ lies in no $U_\ell$.
We have $z\notin U_i\cup U_j$ because $U_i\cap U_j=\{0\}$.
If either $x$ or $y$ has a nonzero ordinary coordinate,
that coordinate forces direction $i$ or $j$, already excluded.
Otherwise $z=a_i+a_j$, which is none of the $a_\ell$:
the only nonempty dependence among the four protected
directions is $a_1+a_2+a_3+a_4=0$.

Put $x'=\phi(x)$, $y'=\phi(y)$, and $z'=\phi(z)$.
The point $z'$ shares a coset with both $x'$ and $y'$.
If $z'\notin U_k$, neither of these cosets has direction $k$.
Every other direction preserves coordinate $b_k$, forcing
$b_k(z')=b_k(x')=b_k(y')$, a contradiction. Hence
$z'\in U_k$, so its distance from $0$ is at most two,
contrary to distance preservation. The proposition excludes
all uniform twin multiplicities.
\end{proof}

\begin{corollary}\label{cor:two-kernel-coset-witness}
For $H_4$, decorated H\"older already fails with two strictly
positive kernels: one on edges in a chosen coset direction
and one on all remaining edges. These two edge classes are
preserved by every oriented fold.
\end{corollary}
\begin{proof}
Work with the reversed parts as above. Partition the left
coset vertices into direction one and all other directions,
and choose the target profiles of $U_1,U_2$ at root $0$.
Translations preserve this source partition and act transitively
on roots, so the colored profile matrices at every root agree.
The two-profile proof of
Proposition~\ref{prop:local-twin-obstruction} now applies to
this colored matrix directly. The identity still contributes
to its off-diagonal entry, and the distance-preserving collapse
is still impossible for every pair of distinct source cosets.
Thus its leading minor is negative.

Choose a profile function giving a strictly negative quadratic
form. The rare-row interpolation and successive finite
approximations in Lemma~\ref{lem:profile-psd} then give a
finite model with strictly positive common kernel $W$ and a real
edge observable $F$ for which
$\Cov(X,Y)<0$. Here $X,Y$ are the sums of $F$ over the two
edge classes, as in Lemma~\ref{lem:cross-covariance}.
Strict negativity persists when zero entries are replaced by
sufficiently small positive entries; all parameters can be
chosen finite.

Let $Z(s,t)$ be the partition function with $We^{sF}$ on
the chosen direction and $We^{tF}$ on the others, and put
$\alpha=1/4$.
For the logarithmic H\"older gap
\[
 D(s,t)=\alpha\log Z(s,s)+(1-\alpha)\log Z(t,t)-\log Z(s,t)
\]
edge transitivity gives $\nabla D(0,0)=0$. Also $D(s,s)=0$
and $D_{st}(0,0)=-\Cov(X,Y)$. Consequently
\[
 \left.\frac{d^2}{d\varepsilon^2}D(\varepsilon,-\varepsilon)
 \right|_{\varepsilon=0}
                =4\Cov(X,Y)<0.
\]
For sufficiently small $\varepsilon>0$ the two positive
kernels $We^{\varepsilon F}$ and $We^{-\varepsilon F}$
therefore violate decorated H\"older on these two edge classes.
Coset direction is preserved by every atom, hence by every
oriented fold, proving the last assertion.
\end{proof}