\section{Periodic meshes and correction vanishing outside characteristic two}
\label{app:periodic-mesh}

In this appendix we prove the two results on periodic mesh categories used in Part~\ref{part:graded-form}: the Dynkin-tree covering description of periodic components (Proposition~\ref{prop:periodic-tree-model}, used in Section~\ref{sec:scalar-normalization}) and the vanishing of the mesh cokernels $\operatorname{coker}D_m(\C)$ outside characteristic two (Proposition~\ref{prop:mesh-correction-spaces}, which proves Theorems~A and~B when $\charac k\ne2$). The characteristic-two and modulo-four calculations in Appendix~\ref{app:residual-mesh} reuse the tree models and path identities established here.

\subsection{Finite trees and periodic mesh categories}
\label{app:periodic-trees}

We first prove Proposition~\ref{prop:periodic-tree-model}, stated in Subsection~\ref{subsec:mesh-walk-cover}, by lifting sectional paths to construct a repetition quiver cover. This proof also fixes the explicit Dynkin tree models used in our subsequent cokernel calculations. Both the repetition-quiver covering and its construction via sectional paths are due to Riedtmann~\cite[Sections~1.5--1.6 and~2.1]{Riedtmann1980}.

\begin{proof}[Proof of Proposition~\ref{prop:periodic-tree-model}]
We divide the proof into four steps.

\medskip
\noindent\textbf{Step 1: Sectional paths and a finite rooted tree.}
A path in a stable translation quiver $Q$ is called \emph{sectional} if it contains no consecutive pair of arrows of the form $(\sigma a)a$. Because every term in a mesh relation is of the form $(\sigma a)a$, a sectional path can never appear as a monomial in any contextual mesh relation $u r_v w$. Consequently, the indicator functional of a sectional path annihilates the mesh ideal, so every sectional path represents a nonzero morphism in the mesh category $C(Q,c)$. Since $C(Q,c)$ is finite-dimensional, there is an integer $N$ such that all paths of length at least $N$ vanish in $C(Q,c)$; hence every sectional path in $Q$ has length strictly less than $N$.

Fix a base vertex $x_0$ in the connected periodic quiver $Q$, and let $\Delta$ be the directed graph whose vertices are all sectional paths in $Q$ starting at $x_0$ (including the trivial path at $x_0$), with a directed edge $p\to q$ whenever $q$ is obtained by appending a single arrow to $p$. By definition, $\Delta$ is a finite tree rooted at the trivial path, oriented away from the root. Its repetition quiver $\Z\Delta$ has vertices $(p,j)$ for $p\in\Delta_0$ and $j\in\Z$, translation $\tau(p,j)=(p,j-1)$, and for each directed edge $p\to q$ in $\Delta$ (where $q=pa$ for an arrow $a$ of $Q$) a pair of arrows
\[
  (p,j)\longrightarrow(q,j),\qquad
  (q,j)\longrightarrow(p,j+1).
\]
We define a translation-quiver morphism $\pi\colon\Z\Delta\to Q$ on vertices by
\[
  \pi(p,j)=\tau^{-j}(\operatorname{target}p),
\]
and on the two repetition arrows associated to $q=pa$ by sending $(p,j)\to(q,j)$ to $\tau^{-j}a$ and $(q,j)\to(p,j+1)$ to $\tau^{-j}(\sigma^{-1}a)$.

At the root $p=x_0$, the children $q$ correspond to all outgoing arrows from $x_0$. At a nonroot path $p$ ending with an arrow $a$, the children $q$ correspond to all outgoing arrows from $\operatorname{target}p$ except $\sigma^{-1}a$, while the returning arrow $(p,j)\to(p',j+1)$ to the parent $p'$ of $p$ maps to $\tau^{-j}(\sigma^{-1}a)$. Thus $\pi$ restricts to a bijection on the outgoing arrows at every vertex of $\Z\Delta$, and $\sigma$ gives a bijection from the incoming arrows at $z$ to the
outgoing arrows at $\tau z$. Since $\pi$ commutes with $\tau$ and
$\sigma$, its bijectivity on outgoing arrows at $\tau z$ implies its
bijectivity on incoming arrows at $z$. In particular, the image of $\pi$ is a union of connected components, hence equals all of $Q$.

Because $\pi$ is a local isomorphism on incoming and outgoing stars, paths in $Q$ lift uniquely to $\Z\Delta$ once a starting vertex in $\Z\Delta$ is fixed. Pulling back the mesh coefficients of $Q$ along $\pi$ equips $\Z\Delta$ with a mesh category whose contextual mesh relations from a fixed starting vertex $(p,j)$ project bijectively onto those of $Q$ starting at $\pi(p,j)$. Taking the direct sum over all endpoints in $\Z\Delta$ therefore identifies the space of paths in $\Z\Delta$ starting at $(p,j)$, modulo lifted mesh relations, with the space of paths in $C(Q,c)$ starting at $\pi(p,j)$. In particular, all paths of length at least $N$ vanish in the lifted mesh category on $\Z\Delta$.

\medskip
\noindent\textbf{Step 2: The tree $\Delta$ is a Dynkin diagram.}
We next rescale the arrows of $\Z\Delta$ so that the lifted mesh relations take the standard $\pm1$ signs of the signed preprojective algebra $\Pi(\Delta)$. Along each tree edge $u\to v$ of $\Delta$, denote the repetition arrows by $a_j\colon(u,j)\to(v,j)$ and $b_j\colon(v,j)\to(u,j+1)$. The mesh at $(v,j)$ contains $b_{j-1}a_j$, and that at $(u,j)$
contains $a_{j-1}b_{j-1}$. Prescribing coefficients $+1$ and $-1$ for
these terms, respectively, imposes equations of the form $\xi_{a,j}\xi_{b,j-1}=\alpha_{v,j}$ and $\xi_{a,j-1}\xi_{b,j-1}=\beta_{u,j}$ on the arrow scales $\xi$. Here $\alpha_{v,j}$ and $\beta_{u,j}$ are the ratios of the prescribed signed coefficients to the corresponding original nonzero mesh coefficients. Taking their ratio determines $\xi_{a,j}$ recursively for all $j\in\Z$ from $\xi_{a,0}=1$, and then determines every $\xi_{b,j}$ (since the index $j$ runs over $\Z$, there is no periodicity obstruction, and distinct edges of $\Delta$ involve disjoint sets of arrows).

Recall that the \emph{signed preprojective algebra} $\Pi(\Delta)$ is the quotient of the path algebra of the doubled quiver $\overline\Delta$ (obtained by adding a reverse arrow $a^*\colon v\to u$ for each tree edge $a\colon u\to v$) by the quadratic relations
\[
  \sum_{a:x\to y}aa^*-\sum_{a:y\to x}a^*a=0\qquad(x\in\Delta_0).
\]
Forgetting the integer coordinate $j$ in $\Z\Delta$ identifies the rescaled mesh relations on $\Z\Delta$ with the defining relations of $\Pi(\Delta)$, and every path in $\overline\Delta$ starting at $u$ lifts uniquely to a path in $\Z\Delta$ starting at $(u,0)$. The same lifting applies to contextual relations, so for each $u,v$
the path quotient $\Pi(\Delta)(u,v)$ is the direct sum of the lifted
mesh spaces from $(u,0)$ to $(v,j)$ over all $j$. Consequently, all paths of length at least $N$ vanish in $\Pi(\Delta)$, so $\Pi(\Delta)$ is finite-dimensional.

A connected finite acyclic quiver with at least two vertices has a finite-dimensional preprojective algebra if and only if its underlying graph is Dynkin~\cite[Section~4.1, p.~345]{BBK}; since a one-vertex tree is already of type $A_1$, it follows that $\Delta$ is a Dynkin diagram.

\medskip
\noindent\textbf{Step 3: The fibres of $\pi$ are deck-group orbits.}
Write $R=\Z\Delta$ and let $o=(x_0,0)$ be the root at coordinate~$0$. Given any vertex $z\in R$ with $\pi(z)=\pi(o)$, we construct a translation-quiver endomorphism $H_z\colon R\to R$ over $\pi$ as follows. On the initial slice $\Delta\times\{0\}$, we set $H_z(o)=z$ and proceed outward along the tree: once $H_z(p,0)$ is defined and $p\to q$ is a tree edge, we lift the arrow $\pi(p,0)\to\pi(q,0)$ starting at $H_z(p,0)$ and define $H_z(q,0)$ to be its target (since every nonroot vertex of $\Delta$ has a unique parent, no conflicting choices arise). We then extend $H_z$ to all of $R$ by $\tau$-equivariance, setting $H_z(p,j)=\tau^{-j}H_z(p,0)$ on vertices and sending the repetition arrows at level~$j$ of a tree edge whose lift at level~$0$ is $b$ to $\tau^{-j}b$ and $\tau^{-j}(\sigma^{-1}b)$. These image arrows have the endpoints prescribed by $H_z$ and satisfy $\pi H_z=\pi$. Because both incoming and outgoing stars at every vertex project bijectively onto the corresponding stars in $Q$, the map $H_z$ is locally bijective, so its image is open and closed in the connected quiver $R$ and hence equals $R$. Moreover, any translation-quiver morphism $R\to R$ over $\pi$ is uniquely determined by the image of a single vertex via unique lifting of undirected walks. Choosing $z'\in R$ with $H_z(z')=o$, we have $\pi(z')=\pi(o)$, and the compositions $H_z\circ H_{z'}$ and $H_{z'}\circ H_z$ fix $o$ and $z'$ respectively, so $H_z$ is an automorphism of $R$ over $\pi$.

Consequently, the deck group $\Theta$ of translation-quiver automorphisms of $R$ over $\pi$ acts freely on $R$ and transitively on the fibre $\pi^{-1}(\pi(o))$. For any two vertices $u,v\in R$ with $\pi(u)=\pi(v)$, projecting an undirected walk from $o$ to $u$ down to $Q$ and lifting its reverse from $v$ produces a vertex $z\in\pi^{-1}(\pi(o))$ such that $H_z(u)=v$. Thus $\Theta$ acts transitively on every fibre of $\pi$, giving $R/\Theta\cong Q$.

\medskip
\noindent\textbf{Step 4: The deck group $\Theta$ is cyclic.}
Every translation-commuting automorphism of $\Z\Delta$ has the form
\[
  (v,j)\longmapsto(\rho(v),j+n_v),
\]
where $\rho$ is an undirected graph automorphism of $\Delta$ and the integer shifts $n_v\in\Z$ satisfy $n_v=n_u$ along a directed edge $u\to v$ if $\rho$ preserves its orientation, and $n_v=n_u+1$ if $\rho$ reverses its orientation. Thus the automorphism is completely determined by $\rho$ and its shift at a single vertex, and changing the orientation of $\Delta$ simply shifts the repetition coordinates of the vertices of $\Z\Delta$.

When $\Delta$ is of type $D_n$, $E_6$, $E_7$, $E_8$, or $A_{2n+1}$, every graph automorphism of $\Delta$ fixes a canonical root vertex (the trivalent vertex or the central vertex of $A_{2n+1}$). Orienting $\Delta$ away from this root makes every $\rho\in\Aut(\Delta)$ orientation-preserving, so $n_v$ is constant across $v\in\Delta_0$ and the full group of translation-commuting automorphisms of $\Z\Delta$ is $\Z\times\Aut(\Delta)$ (with $\Z$ generated by $\tau$). Since $\Theta$ acts freely on vertices and any element in the kernel of the projection $\Theta\to\Z$ fixes the chosen graph root at coordinate zero, the projection $\Theta\to\Z$ is injective; since $Q=\Z\Delta/\Theta$ is finite, its image is a nontrivial subgroup $s\Z$ ($s\geq1$). Hence $\Theta=\langle\tau^s\rho\rangle$ for some $s\geq1$ and $\rho\in\Aut(\Delta)$, proving the first case of Proposition~\ref{prop:periodic-tree-model}(2).

When $\Delta=A_{2n}$, label the two halves of the tree by $L_i$ and $R_i$ ($1\leq i\leq n$) from the center outward, orient the central edge as $L_1\to R_1$, and orient both halves outward ($L_i\to L_{i+1}$ and $R_i\to R_{i+1}$). The translation-quiver automorphism
\[
  \eta(L_i,j)=(R_i,j),\qquad \eta(R_i,j)=(L_i,j+1)
\]
satisfies $\eta^2=\tau^{-1}$. A graph automorphism of $A_{2n}$ is either
the identity or the reflection exchanging its halves. In the first case,
the edge-shift equations give the same shift at every vertex, so the
translation-quiver automorphism is a power of $\tau$. In the second case,
composing with $\eta^{-1}$ fixes every graph vertex and reduces to the
first case. Hence every translation-commuting automorphism is a power
of $\eta$. Thus $\Theta=\langle\eta^r\rangle$ for some $r\geq1$ (where even $r$ gives a pure translation quotient and odd $r$ twists translation by the reflection of $A_{2n}$), completing the proof of Proposition~\ref{prop:periodic-tree-model}.
\end{proof}

For our later calculations on the odd quotient $\Z A_{2n}/\langle\eta^r\rangle$ ($r$ odd), it is convenient to identify $L_{i,j}$ with $(i,2j)$ and $R_{i,j}$ with $(i,2j+1)$. The quotient quiver then has vertices $(i,j)$ with $1\leq i\leq n$ and $j\in\Z/r\Z$, translation $\tau(i,j)=(i,j-2)$, and arrows
\[
  \gamma_j\colon(1,j)\to(1,j+1),\qquad
  a_{i,j}\colon(i,j)\to(i+1,j),\qquad
  b_{i,j}\colon(i+1,j)\to(i,j+2).
\]

\subsection{Mesh corrections outside characteristic two}
\label{app:mesh-calculations}

We prove Proposition~\ref{prop:mesh-correction-spaces} for a
finite-dimensional scalar mesh category $C=C(Q,c)$ satisfying the brick
hypothesis, with $\charac k\ne2$. By
Proposition~\ref{prop:periodic-localization}, $\operatorname{coker}D_m(C)$
is the direct sum of the cokernels on the periodic components of $Q$.
A periodic component without arrows contributes zero, since its
positive-degree morphism spaces vanish. By Proposition~\ref{prop:periodic-tree-model}, each nontrivial periodic component is modeled on a Dynkin repetition quotient $\Z\Delta/\Theta$. We first normalize the mesh coefficients on these periodic models (diagonal rescalings of arrows and mesh relations conjugate $D_m$ by invertible diagonal matrices, preserving both $\operatorname{coker}D_m$ and the brick hypothesis). We then show that $\operatorname{coker}D_{d-1}$ is a quotient of a \emph{twisted trace space} of the signed preprojective algebra $\Pi(\Delta)$, and compute these twisted trace spaces for each Dynkin type. Note that our normalization, the twisted-trace surjection, and the nonvanishing results in Lemmas~\ref{lem:d4-four-return} and~\ref{lem:e7-sixteen-return} hold in arbitrary characteristic; we will point out explicitly where the invertibility of~$2$ is used.

\subsubsection{Scalar coefficients in the periodic models}

Consider first a rooted periodic model $Q(\Delta,s,\rho)=\Z\Delta/\langle\tau^s\rho\rangle$, where $\Delta$ is oriented away from a root fixed by $\rho$. The orbit graph $\Delta/\langle\rho\rangle$ is again a finite rooted tree. For each $\rho$-orbit $v$ of vertices in $\Delta$, let $m_v=s|v|$ denote the period of the corresponding $\tau$-cycle in $Q(\Delta,s,\rho)$. Along any parent-child edge orbit $u\to v$, the $\rho$-stabilizer of a child vertex fixes its unique parent in the tree, so $|u|$ divides $|v|$ and hence $m_u\mid m_v$.

To compare the rooted models with the signed preprojective algebra, we
rescale their arrows and mesh relations using the following lemma.

\begin{lemma}
\label{lem:periodic-scalar-gauges}
Any system of nonzero mesh coefficients $c$ in $C(Q,c)$ can be normalized as follows:
\begin{enumerate}[label={\upshape(\arabic*)}]
\item On $Q=\Z\Delta/\langle\tau^s\rho\rangle$, where $\Delta$ is a finite tree oriented away from a root fixed by $\rho$, the coefficients $c$ can be rescaled to the standard $\pm1$ coefficients of the signed preprojective algebra $\Pi(\Delta)$.
\item On the odd quotient $\Z A_{2n}/\langle\eta^r\rangle$ ($r$ odd), written in the coordinates at the end of Appendix~\ref{app:periodic-trees}, the coefficients $c$ can be rescaled so that every length-two mesh path has coefficient $+1$.
\end{enumerate}
\end{lemma}
\begin{proof}
For~(1), label the arrows in $Q$ lying over a parent-child edge orbit $u\to v$ by $j\in\Z/m_v\Z$ as
\[
  a_j\colon(u,j\bmod m_u)\to(v,j),\qquad
  b_j\colon(v,j)\to(u,j+1\bmod m_u).
\]
Let $\xi_{a,j},\xi_{b,j}\in k^*$ be arrow scales and let $\zeta_{v,j}\in k^*$ scale the mesh relation at $(v,j)$. Transforming the given mesh coefficients into the target $\pm1$ coefficients requires solving equations of the form
\[
  \xi_{a,j}\xi_{b,j-1}=t_{v,j}\zeta_{v,j},\qquad
  \xi_{a,j-1}\xi_{b,j-1}=t_{u,j}\zeta_{u,j\bmod m_u},
\]
for prescribed nonzero scalars $t_{v,j},t_{u,j}\in k^*$ (when $|\text{orbit }v|>|\text{orbit }u|$, several child indices $j$ share the same parent mesh $\zeta_{u,j\bmod m_u}$). Dividing the two equations gives a first-order recurrence for $\xi_{a,j}$ around the cycle $j\in\Z/m_v\Z$, which has a solution if and only if the product of the ratios around the cycle equals~$1$:
\[
  \prod_{j\in\Z/m_v\Z}\zeta_{v,j}=c_{uv}\left(\prod_{j\in\Z/m_u\Z}\zeta_{u,j}\right)^{m_v/m_u},
\]
where $c_{uv}=\prod_{j}(t_{u,j}/t_{v,j})\in k^*$. Because the orbit graph $\Delta/\langle\rho\rangle$ is a rooted tree, we may set $\zeta_{\text{root},j}=1$ at the root and inductively choose $\zeta_{v,j}\in k^*$ down the tree so that the cycle-product condition holds on every edge orbit, after which the recurrences determine all arrow scales $\xi_{a,j},\xi_{b,j}\in k^*$.

For~(2), on $\Z A_{2n}/\langle\eta^r\rangle$ ($r$ odd), we set the scales of the central arrows $\gamma_j$ equal to~$1$ and choose the level-$1$ mesh scales $\zeta_{1,j}$ so that each $\gamma_{j-2}\gamma_{j-1}$ has coefficient $+1$. Along each arm edge between level $i$ and level $i+1$, the ratio recurrence shifts the index $j\in\Z/r\Z$ by~$2$; because $r$ is odd, shifting by~$2$ generates the entire cycle $\Z/r\Z$, so the compatibility condition simply prescribes the product $\prod_{j\in\Z/r\Z}\zeta_{i+1,j}$ in terms of $\prod_{j\in\Z/r\Z}\zeta_{i,j}$ and can be solved inductively for $i=1,\ldots,n-1$.
\end{proof}

\subsubsection{Rotation and twisted traces}

Let $\Pi=\Pi(\Delta)$ be the signed preprojective algebra of a rooted tree $\Delta$. Given an orientation-preserving automorphism $\theta\in\Aut(\Delta)$, we define the \emph{$\theta$-twisted trace space} of $\Pi$ to be the graded vector space quotient
\begin{equation}
  \operatorname{Tr}_\theta(\Pi)=\left(\bigoplus_{x\in\Delta_0}\Pi(x,\theta x)\right)
  \Big/\Span\bigl\{uv-v\theta(u):u\in\Pi(x,y),\ v\in\Pi(y,\theta x)\bigr\}.
  \label{eq:twisted-trace-space}
\end{equation}
In $\operatorname{Tr}_\theta(\Pi)$, the relation $[uv]=[v\theta(u)]$ allows one to cut a path $uv$ from $x$ to $\theta x$ and move its prefix $u$ to the end after applying $\theta$; in particular, $[w]=[\theta(w)]$ for every $w\in\Pi(x,\theta x)$. When $\theta=\id$, $\operatorname{Tr}_{\id}(\Pi)=\Pi/[\Pi,\Pi]$ is the ordinary cyclic trace space of $\Pi$.

\begin{lemma}
\label{lem:twisted-trace-surjection}
Let $Q=Q(\Delta,s,\rho)=\Z\Delta/\langle\tau^s\rho\rangle$ be a rooted periodic mesh model equipped with the signed preprojective mesh coefficients of $\Pi=\Pi(\Delta)$, and let $D_{d-1}$ be the mesh derivative~\eqref{eq:mesh-derivative} in path degree $d\geq3$.
\begin{enumerate}[label={\upshape(\arabic*)}]
\item The cokernel $\operatorname{coker}D_{d-1}$ vanishes unless $d$ is even and $s$ divides $d/2-1$.
\item When $d$ is even and $s\mid(d/2-1)$, set $l=(1-d/2)/s$ and $\theta=\rho^l$. Then there is a canonical surjection
\[
  \operatorname{Tr}_\theta(\Pi)_d\twoheadrightarrow\operatorname{coker}D_{d-1}.
\]
\end{enumerate}
\end{lemma}
\begin{proof}
A path of length $d$ in $Q$ from $\tau(x,j)=(x,j-1)$ to $(x,j)$ lifts to a path in $\Z\Delta$ from $(x,j-1)$ to $(\rho^l x,j-sl)$ for some integer $l\in\Z$. Because $x$ and $\rho^l x$ lie at the same distance from the root of $\Delta$, the lifted path takes equal numbers of steps away from the root (which preserve the second coordinate) and toward the root (which increment the second coordinate by~$1$). Hence $d$ must be even and $(j-1)+d/2=j-sl$, which gives $d/2-1=-sl$; this proves~(1) and shows that the lifted path projects in $\Pi(\Delta)$ to a path of length $d$ from $x$ to $\theta x=\rho^l x$.

In the signed mesh presentation on $Q$, let $a\colon u\to v$ be an arrow. By the column description in the proof of Proposition~\ref{prop:periodic-localization}, modifying its mate $\sigma a\colon\tau v\to u$ by a parallel path $w$ of length $d-1$ adds $c[wa]+c'[(\tau a)w]$ to $\im D_{d-1}$ in the rows at $v$ and $u$, respectively, where $c$ and $c'$ are the two mesh coefficients on the tree edge underlying $a$ (here $wa\colon\tau v\to v$ and $(\tau a)w\colon\tau u\to u$ both have the required row endpoints). Because $c$ and $c'$ have opposite signs $\{+1,-1\}$, this identifies $[wa]=[(\tau a)w]$ in $\operatorname{coker}D_{d-1}$. Iterating this single-arrow rotation $d$ times along a row path $p$ of length $d$ gives $[p]=[\tau p]$, so shifting the repetition coordinate of a lift in $\Z\Delta$ does not change its class in $\operatorname{coker}D_{d-1}$ (and the deck relation $\tau^s\rho=\id$ in $Q$ likewise identifies lifts differing by $\tau^s\rho$).

When we rotate a prefix of a lifted path to the back, transporting that prefix across the full lift from $(x,j-1)$ to $(\rho^l x,j-sl)$ acts on its tree vertices and arrows by $\theta=\rho^l$. After forgetting repetition coordinates, cutting a path at an intermediate vertex $y$ and rotating the prefix $u\colon x\to y$ past the suffix $v\colon y\to\theta x$ therefore replaces $uv$ with $v\theta(u)$, where $\theta(u)\colon\theta x\to\theta y$. Thus the single-arrow rotations induce the relation $[uv]=[v\theta(u)]$ in $\operatorname{coker}D_{d-1}$. Since contextual mesh relations in $\Pi(\Delta)$ lift to contextual mesh relations in $Q$, the resulting projection $\operatorname{Tr}_\theta(\Pi)_d\to\operatorname{coker}D_{d-1}$ is well defined and surjective.
\end{proof}

To compute $\operatorname{Tr}_\theta(\Pi)$ for star-shaped trees $\Delta$, we first express closed paths at the central vertex $o$ in terms of the length-two loops along its arms, following~\cite[Sections~11.3 and~13.1]{Schedler} and incorporating the arm-permuting twist $\theta$.

\begin{lemma}
\label{lem:star-return-reduction}
Let $\Delta$ be a star tree rooted at its central vertex $o$, with arms of lengths $L_1,\ldots,L_t$, and for each $i=1,\ldots,t$ let $x_i\in\Pi(o,o)_2$ denote the length-two closed path that steps out along the $i$-th arm and immediately returns to $o$. Then the corner algebra $\Pi(o,o)$ is generated by $x_1,\ldots,x_t$ and is a quotient of
\begin{equation}
  k\langle x_1,\ldots,x_t\rangle\big/\Bigl(\sum_{i=1}^t x_i,\ x_1^{L_1+1},\ldots,x_t^{L_t+1}\Bigr).
  \label{eq:star-corner-presentation}
\end{equation}
\end{lemma}
\begin{proof}
Label the vertices along the $i$-th arm as $v_0=o,v_1,\ldots,v_{L_i}$. At any internal vertex $v_{j-1}$ ($j\geq2$), the mesh relation replaces the outward backtrack $v_{j-1}\to v_j\to v_{j-1}$ (up to sign) by the inward backtrack $v_{j-1}\to v_{j-2}\to v_{j-1}$, strictly decreasing the sum of the distances from $o$ along the path. Iterating this replacement reduces every closed path at $o$ to a polynomial in $x_1,\ldots,x_t$. The mesh relation at $o$ gives $\sum_{i=1}^t x_i=0$, and, writing $a_j\colon v_{j-1}\to v_j$ and $b_j\colon v_j\to v_{j-1}$, successive mesh replacements give
\[
 x_i^q=\pm a_1\cdots a_qb_q\cdots b_1\qquad(1\leq q\leq L_i).
\]
For $q=L_i+1$, the same replacement reaches the factor $b_{L_i}a_{L_i}=0$ at the leaf, so $x_i^{L_i+1}=0$.
\end{proof}

The branching-vertex return algebras in types $D$ and $E$ also occur
in the deformed preprojective algebras of Bia\l kowski, Erdmann and
Skowro\'nski \cite[Lemma~3.1]{BES}. Their construction perturbs the
relation at that vertex, subject to admissibility conditions; for the
corrected condition in type $E$, see Bia\l kowski \cite{BialkowskiAlgorithm}.
Our calculation instead concerns periodic mesh errors and allows diagram
twists, so we do not identify these errors with admissible preprojective
deformations. We next reduce twisted traces to the same return paths.

\begin{lemma}
\label{lem:arm-trace-reduction}
Let $\Delta$ be a star tree with center $o$, and let $\theta\in\Aut(\Delta)$ be an automorphism fixing $o$ and permuting the arms. In every positive degree, $\operatorname{Tr}_\theta(\Pi)_+$ is spanned by the classes of positive-degree words in the arm loops $x_1,\ldots,x_t$ at $o$. Moreover, if the $i$-th arm is fixed pointwise by $\theta$, then
\begin{equation}
  [x_i^q]=0\quad\text{in }\operatorname{Tr}_\theta(\Pi)\qquad\text{for all }q\geq1.
  \label{eq:arm-power-trace-zero}
\end{equation}
\end{lemma}
\begin{proof}
If a path $w$ from $v$ to $\theta v$ passes through the center $o$, writing $w=fg$ with $f\colon v\to o$ and $g\colon o\to\theta v$ gives $[fg]=[g\theta(f)]$ in $\operatorname{Tr}_\theta(\Pi)$, which is a closed path at $\theta o=o$ and hence a linear combination of words in $x_1,\ldots,x_t$ by Lemma~\ref{lem:star-return-reduction}.

If instead $w$ never visits $o$, then $w$ is supported entirely on a single arm of $\Delta$. Because $w$ goes from $v$ to $\theta v$, that arm must be fixed pointwise by $\theta$, so $w$ is a closed path and twisted rotation on that arm is ordinary cyclic rotation. Rotating $w$ so that it begins at a vertex $v_j$ of minimum depth along $w$, we see that its two incident steps at $v_j$ form an inward-pointing turn $v_{j+1}\to v_j\to v_{j+1}$; applying the mesh relation at $v_{j+1}$ replaces this turn by $v_{j+1}\to v_{j+2}\to v_{j+1}$ (or by~$0$ at the leaf), strictly increasing the sum of the vertex depths along the cyclic word. Since the depth sum of a word of fixed length is bounded above, repeating this step eventually pushes the word off the leaf and shows that $[w]=0$. Applying the exact same outward-pushing argument to the loop power $x_i^q$ on any $\theta$-fixed arm yields~\eqref{eq:arm-power-trace-zero}.
\end{proof}

\subsubsection{Types A and D}

Assume now that $\charac k\ne2$. We apply Lemma~\ref{lem:twisted-trace-surjection} and Lemma~\ref{lem:arm-trace-reduction} to rule out all nonzero classes in $\operatorname{coker}D_{d-1}$ ($d\geq3$) for the rooted models of types $A$ and $D$:
\begin{itemize}
\item \emph{Type $A$ with trivial twist ($\theta=\id$).} This covers all pure translation quotients of type $A$, including $\Z A_{2n}/\langle\eta^r\rangle$ for even $r$ (since $\eta^r=\tau^{-r/2}$). Rooting the path $A_n$ at one of its endpoints makes it a star with a single arm, so $\Pi(o,o)$ is generated by $x_1$ subject to $x_1=0$ by Lemma~\ref{lem:star-return-reduction}; hence $\operatorname{Tr}_{\id}(\Pi)_+=0$.
\item \emph{Type $A_{2n+1}$ with reflection ($\theta=\rho\ne\id$).} Rooting $A_{2n+1}$ at its central vertex gives two arms of length $n$ with loops $x_1=x$ and $x_2=-x$ exchanged by $\theta(x)=-x$. For every $q\geq1$, twisted rotation of one factor $x$ gives $[x^q]=[x^{q-1}\theta(x)]=-[x^q]$, so $2[x^q]=0$ and hence $[x^q]=0$ because $\charac k\ne2$.
\item \emph{Type $D_n$ with $\theta=\id$ or the short-arm transposition.} For $D_4$, any transposition exchanges two arms and fixes the third;
choose the exchanged arms as those defining $x$ and $y$ below. Let $x$ and $y$ be the loops along the two short arms of length~$1$, and let $z=-x-y$ be the loop along the long arm of length $n-3$. Since $x^2=y^2=0$, the corner $\Pi(o,o)$ is spanned by alternating words in $x$ and $y$.
  \begin{itemize}
  \item When $\theta=\id$, any alternating word of odd length $2q+1\geq3$ has the same letters at its two ends and therefore rotates to a word containing $x^2=0$ or $y^2=0$, while $[x]=[y]=0$ by~\eqref{eq:arm-power-trace-zero}. In even word length $2q$, cyclic rotation identifies $[(xy)^q]=[(yx)^q]$, while~\eqref{eq:arm-power-trace-zero} applied to the fixed long arm gives $0=[z^{2q}]=[(x+y)^{2q}]=[(xy)^q]+[(yx)^q]=2[(xy)^q]$. Since $\charac k\ne2$, we obtain $[(xy)^q]=0$.
  \item When $\theta$ interchanges $x$ and $y$, any alternating word of even length $2q$ twisted-rotates to a word containing $x^2=0$ or $y^2=0$ (for instance, $[(xy)^q]=[(yx)^{q-1}y\theta(x)]=[(yx)^{q-1}y^2]=0$). In odd word length $2q+1$, the two alternating words $w=x(yx)^q$ and $w'=y(xy)^q$ satisfy $[w]=[\theta(w)]=[w']$, while the fixed long arm gives $0=[(-z)^{2q+1}]=[(x+y)^{2q+1}]=[w]+[w']=2[w]$, so again $[w]=0$.
  \end{itemize}
\item \emph{Type $D_4$ with a three-cycle $\theta$.} When $\Delta=D_4$, all three arms have length~$1$, so $x^2=y^2=(x+y)^2=0$. Hence $yx=-xy$, and every word of word degree~$3$ (path degree~$6$) vanishes. In word degree~$2$ (path degree~$4$), label the arms so that $\theta(x)=y$. Then
$[xy]=[y\theta(x)]=[y^2]=0$, and the algebra relation $yx=-xy$ gives
$[yx]=0$ in the same $\theta$-twisted trace space. Thus $\operatorname{Tr}_\theta(\Pi)_d=0$ for all $d\geq3$.
\end{itemize}

\subsubsection{The odd quotients of even A}

Consider next the odd quotient $Q=\Z A_{2n}/\langle\eta^r\rangle$ ($r$ odd), normalized by Lemma~\ref{lem:periodic-scalar-gauges}(2) so that every mesh path of length~$2$ has coefficient $+1$ in the coordinates $(i,j)$ ($1\leq i\leq n$, $j\in\Z/r\Z$) of Appendix~\ref{app:periodic-trees}. Any path from $\tau(i,j)=(i,j-2)$ to $(i,j)$ takes equal numbers of outward steps $a$ (which change $j$ by~$0$) and inward steps $b$ (which change $j$ by $+2$); since each $\gamma$ step changes $j$ by $+1$, the total change in the second coordinate $j\in\Z/r\Z$ equals the path length $d$. Thus a path of length $d$ from $\tau(i,j)$ to $(i,j)$ can exist only when $d\equiv2\pmod r$.

Because all mesh coefficients equal $+1$, the column of $D_{d-1}$ corresponding to any arrow $u$ imposes the relation $[wu]+[(\tau u)w]=0$ in $\operatorname{coker}D_{d-1}$, so rotating a path by one arrow negates its class in $\operatorname{coker}D_{d-1}$ while preserving its cyclic sequence of levels $i\in\{1,\ldots,n\}$. Given any path of length $d$ from $\tau v$ to $v$ that reaches a maximum level $i+1>1$, the step into level $i+1$ and the step out of it must be $a_i b_i$ (as there are no level-preserving arrows above level~$1$). Rotating the path if necessary so that $a_i b_i$ does not straddle the cut, we may apply the mesh relation at level $i$ to replace $a_i b_i$ by $-b_{i-1}a_{i-1}$ when $i>1$ (or by $-\gamma\gamma$ when $i=1$), strictly decreasing the sum of the levels along the path. Repeating this reduction expresses every class in $\operatorname{coker}D_{d-1}$ as a linear combination of the classes $w_j=[\gamma_{j-2}\gamma_{j-1}\cdots\gamma_{j+d-3}]$ of pure level-$1$ paths of length $d$ ending at $(1,j)$ ($j\in\Z/r\Z$).

Since $d\equiv2\pmod r$, the path of length $d-1$ formed by $d-1$ consecutive $\gamma$-arrows starting at $(1,j-1)$ has the same source and target as $\gamma_{j-1}\colon(1,j-1)\to(1,j)$. Evaluating $D_{d-1}$ on this arrow correction in the two root meshes containing $\gamma_{j-2}\gamma_{j-1}$ and $\gamma_{j-1}\gamma_j$ yields
\[
  w_{j-1}+w_j=0\qquad(j\in\Z/r\Z).
\]
Because $r$ is odd, going all the way around the $r$-cycle gives $w_j=(-1)^r w_j=-w_j$, so $2w_j=0$ and hence $w_j=0$ whenever $\charac k\ne2$ (and for $n=1$ all paths of length~$d\geq2$ vanish by the root mesh relation $\gamma_{j-2}\gamma_{j-1}=0$).

\subsubsection{Type $E_6$ and the small-period exclusion}

For $\Delta=E_6$, the three arms have lengths $1,2,2$. Let $x$ be the loop along the short arm and $y,z$ the loops along the two arms of length~$2$, so that $x+y+z=0$ and $x^2=y^3=z^3=0$. Assuming $\charac k\ne2$, set $v=y-z=x+2y$, so that $2y=v-x$ and $2z=-v-x$. Using $x^2=0$, the two cubic relations expand as
\[
 0=(v-x)^3=v^3-(v^2x+vxv+xv^2)+xvx,
\]
\[
 0=(-v-x)^3=-v^3-(v^2x+vxv+xv^2)-xvx.
\]
Adding and subtracting these equations and dividing by $2$ gives
\begin{equation}
  x^2=0,\qquad v^3=-xvx,\qquad v^2x+vxv+xv^2=0.
  \label{eq:e6-corner-relations}
\end{equation}
The relations~\eqref{eq:e6-corner-relations} give the following spanning
words in word degree $q=d/2\geq1$:
\[
\begin{array}{c|l}
\text{Word degree }q & \text{Spanning words} \\\hline
1 & x,\ v \\
2 & xv,\ vx,\ v^2 \\
3 & xvx,\ xv^2,\ vxv \\
4 & \psi_4=xvxv,\ \psi_4'=vxv^2 \\
5 & \psi_5=xvxv^2 \\
\geq6 & 0.
\end{array}
\]
To verify that these words span, we start from $1$ and multiply on the right by $x$ and $v$ degree by degree. In word degrees at most~$3$, every product outside the listed words reduces immediately via $x^2=0$, $v^2x=-vxv-xv^2$, and $v^3=-xvx$. Passing to word degree~$4$, multiplying~\eqref{eq:e6-corner-relations} on the left or right by $x$ gives $xv^2x=-\psi_4$, $vxvx=\psi_4$, and $xv^3=v^3x=0$. Finally, in word degrees~$5$ and~$6$, right multiplication on $\psi_4$, $\psi_4'$, and $\psi_5$ yields $\psi_4x=x(vxvx)=0$, $\psi_4v=\psi_5$, $\psi_4'x=-v\psi_4=-\psi_5$, $\psi_4'v=vxv^3=0$, $\psi_5v=xvxv^3=0$, and $\psi_5x=xvx(v^2x)=-xvx(vxv+xv^2)=-\psi_4xv=0$.

When $\theta=\rho$ is the nontrivial reflection of $E_6$, $\theta$ fixes $x$ and negates $v$. In $\operatorname{Tr}_\rho(\Pi)$, every word containing an odd number of $v$'s satisfies $[w]=[\theta(w)]=-[w]$ and therefore vanishes since $\charac k\ne2$. Among words with an even number of $v$'s, $[x]=0$ by~\eqref{eq:arm-power-trace-zero}, $[v^2]=[v\theta(v)]=-[v^2]=0$, the last relation of~\eqref{eq:e6-corner-relations} gives $0=[v^2x]+[vxv]+[xv^2]=[xv^2]-[xv^2]+[xv^2]=[xv^2]$ in word degree~$3$, and $[\psi_4]=[(xv)\theta(xv)]=-[\psi_4]=0$ in word degree~$4$. Thus $\operatorname{Tr}_\rho(\Pi)_+=0$.

When $\theta=\id$ (ordinary trace), all pure arm powers vanish by~\eqref{eq:arm-power-trace-zero}. Expanding $0=4[y^2]=[(v-x)^2]$ and $0=4[z^2]=[(v+x)^2]$ gives $[xv]=[vx]=[v^2]=0$ in word degree~$2$, since $2$ is invertible. In word degree~$3$, $[xvx]=[x^2v]=0$ and $[vxv]=[xv^2]$, while the last relation of~\eqref{eq:e6-corner-relations} gives $3[xv^2]=0$. In word degree~$4$, $[\psi_4]=-[xv^2x]=-[x^2v^2]=0$ and $[\psi_4']=[xv^3]=0$, and in word degree~$5$, $[\psi_5]=-[\psi_4'x]=-[x\psi_4']=-[\psi_5]=0$. Consequently, the only possible nonzero class in $\operatorname{Tr}_{\id}(\Pi)_+$ occurs in word degree~$3$ (path degree $d=6$) when $\charac k=3$, where it is spanned by $[xv^2]$.

To show that even this degree-$6$ class in characteristic~$3$ cannot occur in a periodic component satisfying the brick hypothesis, we use the following characteristic-free lemma, which will also be needed for $E_7$ and $E_8$.

\begin{lemma}
\label{lem:d4-four-return}
Over every field $k$, every vertex of $\Pi(D_4)$ admits a nonzero closed path of path degree~$4$. Consequently, the pure translation mesh quotients $\Z D_4/\langle\tau^s\rangle$ of periods $s=1$ and $s=2$ contain no bricks.
\end{lemma}
\begin{proof}
Orient the three arms of $D_4$ outward from the center $0$ to the leaves $1,2,3$ with arrows $a_i\colon0\to i$ and $b_i\colon i\to0$, and set $x_i=a_ib_i$. The defining relations of $\Pi(D_4)$ are $b_ia_i=0$ ($i=1,2,3$) and $x_1+x_2+x_3=0$. At the center~$0$, the paths of length~$4$ are the six products $x_ix_j$ ($i\ne j$), subject only to the relations $x_i(x_1+x_2+x_3)=0$ and $(x_1+x_2+x_3)x_j=0$; in the order $(12,13,21,23,31,32)$, their classes in $\Pi(D_4)_4$ are $(z,-z,-z,z,z,-z)$ with $z=x_1x_2\ne0$. Likewise, at each leaf $i$, the two length-$4$ closed paths $b_ia_jb_ja_i$ ($j\ne i$) are subject only to the single relation $\sum_{j\ne i}b_ia_jb_ja_i=0$, so each is nonzero in $\Pi(D_4)_4$.

By the path-space direct-sum identification in Step~2 of the proof of
Proposition~\ref{prop:periodic-tree-model}, these nonzero paths have
nonzero lifts to $\Z D_4$. Passing to a translation quotient again
identifies its path space with a direct sum over lifted endpoints, so
their images remain nonzero. In $\Z D_4$, each of these length-$4$ closed paths increases the repetition coordinate by~$2$, and therefore projects in $\Z D_4/\langle\tau^s\rangle$ for $s\in\{1,2\}$ to a nonzero radical endomorphism at every vertex, ruling out the existence of any brick.
\end{proof}

Returning to $E_6$ in path degree $d=6$, Lemma~\ref{lem:twisted-trace-surjection}(1) requires $s\mid(6/2-1)=2$, so $s\in\{1,2\}$. Deleting the outer vertices of the two long arms of $E_6$ at every repetition coordinate leaves a translation-invariant subquiver isomorphic to a $D_4$ quotient of period $s$, whose mesh category is the object quotient of the $E_6$ mesh category by~\eqref{eq:mesh-object-quotient}. The brick hypothesis requires every nonempty object quotient to contain a
brick, so a quotient with nonzero radical endomorphisms at every object
is impossible. For a pure translation quotient ($s=1$ or $2$), Lemma~\ref{lem:d4-four-return} shows that this $D_4$ quotient has no bricks; for the reflected quotient $\Z E_6/\langle\tau^s\rho\rangle$, $s=2$ gives an odd twist $l=-1$ where $\operatorname{Tr}_\rho(\Pi)_6=0$, while $s=1$ gives a $D_4$ quotient with $(\tau\rho)^2=\tau^2$, on which the degree-$4$ closed paths of Lemma~\ref{lem:d4-four-return} again close up to form nonzero radical endomorphisms at every vertex. Thus no periodic $E_6$ component satisfying the brick hypothesis has a nonzero mesh cokernel class.

\subsubsection{The remaining exceptional traces}

We treat $\Delta=E_7$ and $\Delta=E_8$ simultaneously. Let $x,y,z$ denote the length-two loops at the center $o$ along the long, medium, and short arms, respectively (note that $x$ now denotes the long-arm loop, whereas in $E_6$ it denoted the short-arm loop). By Lemma~\ref{lem:star-return-reduction}, their relations in $\Pi(o,o)$ are
\[
  x+y+z=0,\qquad x^m=y^3=z^2=0,\qquad
  \text{where }m=\begin{cases}4 & \text{for }E_7,\\ 5 & \text{for }E_8.\end{cases}
\]
Eliminating $z=-x-y$, we see from Lemmas~\ref{lem:star-return-reduction} and~\ref{lem:arm-trace-reduction} that the positive-degree part of the ordinary trace space $\operatorname{Tr}_{\id}(\Pi)_+$ is a quotient of the positive-degree part of $\mathsf{Q}^{E}_m$, where for any commutative coefficient ring $R$ we define the graded $R$-algebra $\Lambda^{E}_m$ and its $R$-module trace quotient $\mathsf{Q}^{E}_m$ by
\begin{equation}
  \begin{split}
  \Lambda^{E}_m &= R\langle x,y\rangle/\bigl(x^m,y^3,(x+y)^2\bigr),\\
  \mathsf{Q}^{E}_m &= \Lambda^{E}_m\big/\bigl([\Lambda^{E}_m,\Lambda^{E}_m]+\Span_R\{x^q,y^q,(x+y)^q:q\geq1\}\bigr),
  \end{split}
  \label{eq:exceptional-trace-spaces}
\end{equation}
with word degree in $x,y$ equal to half the path degree in $\Pi(\Delta)$. We first compute a spanning set and torsion relations for $(\mathsf{Q}^{E}_5)_+$ over $\Z$; these bounds follow the integral trace calculation in~\cite[Theorem~13.1.1]{Schedler} and will be used both here (for $\charac k\ne2$) and in Appendix~\ref{app:residual-mesh} (for $\charac k=2$ and $\mathcal O=\Z/4$).

\begin{lemma}
\label{lem:e78-traces}
Let $\Lambda^{E}_m$ and $\mathsf{Q}^{E}_m$ be defined as in~\eqref{eq:exceptional-trace-spaces} over an arbitrary commutative ring, and write $[w]$ for the class of a word $w$ in $\mathsf{Q}^{E}_m$.
\begin{enumerate}[label={\upshape(\arabic*)}]
\item Over any commutative ring, set $a=yx^3$, $b=yx^4$, and $t_j=[yx^j]$. The positive-degree part $(\mathsf{Q}^{E}_5)_+$ is spanned by the elements
\[
  t_1,\ t_2,\ t_3,\ t_4,\qquad [a^2],\qquad [ab^r]\quad(r\geq1),
\]
which satisfy the torsion relations
\begin{equation}
  \begin{gathered}
  2t_1=2t_3=0,\qquad 3t_2=0,\qquad 5t_4=0,\\
  2[a^2]=0,\qquad 3[ab]=0,\qquad 2[ab^r]=0\quad(r\geq2).
  \end{gathered}
  \label{eq:integral-exceptional-trace-relations}
\end{equation}
Moreover, for every nonempty word $W$ in the blocks $a$ and $b$,
\begin{equation}
  [a^2W]=0,\qquad [abW]=-[baW],\qquad [b^q]=0\quad(q\geq2),
  \label{eq:integral-exceptional-block-relations}
\end{equation}
and $\mathsf{Q}^{E}_4$ is the quotient of $\mathsf{Q}^{E}_5$ obtained by setting $x^4=0$.
\item If $2$ is invertible in the coefficient ring, then $(\mathsf{Q}^{E}_5)_+$ is spanned by
\[
  [yx^2],\qquad [yx^4],\qquad [(yx^3)(yx^4)],
\]
which are annihilated by $3$, $5$, and $3$, respectively, and $(\mathsf{Q}^{E}_4)_+$ is spanned by $[yx^2]$, annihilated by $3$.
\end{enumerate}
\end{lemma}
\begin{proof}
\textit{(1)} In $\Lambda^{E}_5$ over $\Z$, expanding $(x+y)^2=0$ and $y^3=0$ yields the four identities
\begin{equation}
\begin{aligned}
  y^2 &= -yx-xy-x^2,\\
  yxy &= xyx+x^3,\\
  yx^2y &= -yx^3-xyx^2-x^2yx-x^3y-2x^4,\\
  0 = F &:= yx^3yx+yx^4y+xyx^3y-x^2yx^3-x^3yx^2.
\end{aligned}
  \label{eq:e8-word-identities}
\end{equation}
(Indeed, substituting the first identity into $y^2\cdot y=0$ gives the second; expanding $y^2(xy)=(y^2x)y$ gives the third; and equating $(yxy)x^2y=xyx^3y+x^5y$ with $yx(yx^2y)=-yx^5+x^2yx^3+x^3yx^2+x^6-yx^3yx-yx^4y$ using $x^5=0$ gives the fourth.)

Any cyclic word in $x$ and $y$ can be described by the lengths of the $x$-gaps between consecutive occurrences of $y$. Whenever a gap has length $0$, $1$, or $2$, applying the first, second, or third identity in~\eqref{eq:e8-word-identities} rewrites the word in terms of words with strictly fewer occurrences of $y$. Because $x^5=0$, iterating this reduction expresses every class in $(\mathsf{Q}^{E}_5)_+$ as a linear combination of words with at most one $y$ (namely $t_j=[yx^j]$ for $1\leq j\leq4$, as $[x^q]=0$) and cyclic words in the two blocks $a=yx^3$ and $b=yx^4$.

Set $c=yx^2$. Note first that any word containing $c=yx^2$ flanked on both sides by $a$ or $b$ (so that the adjacent $x$-gaps have lengths $p,q\geq3$) has trace zero: rotating it to the form $yx^2yx^qUx^p$ and expanding $yx^2y$ via the third identity in~\eqref{eq:e8-word-identities} produces five terms, three of which contain $x^5=0$ and the other two of which begin with $x^2$ or $x^3$ and therefore contain $x^{p+2}=0$ or $x^{p+3}=0$ after cyclic rotation.

Using the relation $F=0$ from~\eqref{eq:e8-word-identities}, we now establish~\eqref{eq:integral-exceptional-block-relations}:
\begin{itemize}
\item For any nonempty word $W$ in $\{a,b\}$, taking the trace of $Fx^2W=0$ gives
\[
  [a^2W]+[bcW]+[xacW]-[x^3bW]=0.
\]
Here $[bcW]=0$ and $[xacW]=0$ by the preceding paragraph (or because rotating the initial $x$ to the end of $W$ produces $x^5=0$), and rotating $x^3$ to the end of $[x^3bW]$ likewise produces $x^6=0$ or $x^7=0$; hence $[a^2W]=0$. When $W$ is empty, $0=[Fx^2]=2[a^2]+[bc]$, and multiplying the third identity of~\eqref{eq:e8-word-identities} by $x^4$ gives $[bc]=[yx^2yx^4]=0$, so $2[a^2]=0$.
\item Taking the trace of $Fx^3W=0$ gives
\[
  [abW]+[baW]+[xaaW]=0.
\]
When $W$ is nonempty, rotating the initial $x$ of $[xaaW]$ to the end either creates $x^5=0$ (if $W$ ends in $b$) or gives $[a^2W'b]=0$ (if $W=W'a$); thus $[abW]=-[baW]$. When $W$ is empty, all three terms equal $[ab]$, giving $3[ab]=0$.
\item Taking the trace of $Fx^4b^r=0$ ($r\geq0$) gives $[b^{r+2}]+[xab^{r+1}]=0$; rotating the initial $x$ of the second term to the end creates $x^5=0$, so $[b^q]=0$ for all $q\geq2$.
\end{itemize}
From~\eqref{eq:integral-exceptional-block-relations}, in any word in $\{a,b\}$ of block length at least~$3$ containing at least two $a$'s, anticommuting $a$ and $b$ brings two $a$'s adjacent, so the trace vanishes by $[a^2W]=0$; in a word with a single $a$ and $r\geq2$ copies of $b$, anticommuting $a$ all the way through $b^r$ and cyclically rotating back gives $[ab^r]=(-1)^r[b^ra]=(-1)^r[ab^r]$ and also $[ab\cdot b^{r-1}]=-[bab^{r-1}]=-[ab^r]$, so $2[ab^r]=0$.

Finally, for $t_j=[yx^j]$ ($1\leq j\leq4$), taking the trace of the first identity in~\eqref{eq:e8-word-identities} gives $2t_1=0$; comparing the trace of the second identity with that of $y^2x$ gives $3t_2=0$; comparing the trace of the third identity with that of $y^2x^2$ gives $2t_3=0$; and taking the trace of the third identity multiplied by $x$ gives $t_4=-4t_4$, or $5t_4=0$. This proves~(1), and~(2) follows immediately on inverting~$2$.
\end{proof}

To exclude the remaining periods for $E_7$ and $E_8$ in
characteristics~$3$ and~$5$, we use the top degree of a Dynkin
preprojective algebra. Over any field, if $h$ is the Coxeter number and
$\nu$ the Nakayama permutation, then
$e_x\Pi(\Delta)_{h-2}e_{\nu x}$ is one-dimensional for every vertex
$x$; see Brenner, Butler and King
\cite[Sections~4.1--4.2 and~4.4]{BBK}. In particular, some path of
length $h-2$ from $x$ to $\nu x$ is nonzero.

\begin{lemma}
\label{lem:e7-sixteen-return}
Over every field $k$, every vertex of $\Pi(E_7)$ admits a nonzero closed path of path degree~$16$.
\end{lemma}
\begin{proof}
For $E_7$, the Coxeter number is $18$ and the Nakayama permutation is
the identity \cite[Section~4.4]{BBK}. The stated top-degree description
therefore gives a nonzero closed path of degree $18-2=16$ at every vertex.
\end{proof}

We can now complete the proof of Proposition~\ref{prop:mesh-correction-spaces} for $E_7$ and $E_8$ outside characteristic two. By Lemma~\ref{lem:e78-traces}(2), the only possible nonzero classes in $(\mathsf{Q}^{E}_4)_+$ and $(\mathsf{Q}^{E}_5)_+$ when $2$ is invertible are $[yx^2]$ (word degree~$3$, path degree $d=6$, in characteristic~$3$), $[yx^4]$ (word degree~$5$, path degree $d=10$, for $E_8$ in characteristic~$5$), and $[(yx^3)(yx^4)]$ (word degree~$9$, path degree $d=18$, for $E_8$ in characteristic~$3$). By Lemma~\ref{lem:twisted-trace-surjection}(1), path degrees $d=6,10,18$ require the period $s$ to divide $d/2-1=2,4,8$, respectively.

However, periods $s=1$ and $s=2$ are ruled out by deleting the outer vertices to pass to $\Z D_4/\langle\tau^s\rangle$ and applying Lemma~\ref{lem:d4-four-return}. For periods $s=4$ and $s=8$, the degree-$16$ closed paths at every vertex of $\Pi(E_7)$ from Lemma~\ref{lem:e7-sixteen-return} advance the repetition coordinate by~$8$, and therefore project to nonzero radical endomorphisms at every vertex of $\Z E_7/\langle\tau^s\rangle$ (which is also an object quotient of $\Z E_8/\langle\tau^s\rangle$ obtained by deleting the terminal vertex of the long arm). Thus no periodic component of type $E_7$ or $E_8$ satisfying the brick hypothesis can have period dividing~$8$. Since $E_7$ and $E_8$ have no nontrivial graph automorphisms, this completes the proof of Proposition~\ref{prop:mesh-correction-spaces}.