\section{Residual and integral mesh cokernels}
\label{app:residual-mesh}

In this appendix we complete the periodic trace calculations required for the characteristic-two and integral branches of the main argument. Appendix~\ref{app:characteristic-two-calculations} establishes Proposition~\ref{prop:residual-correction-spaces} by bounding the positive-degree cokernels $\operatorname{coker}D_m(\C)$ ($m\geq2$) over a field of characteristic two, and Appendix~\ref{app:integral-corrections} lifts these calculations to $\mathcal O=\Z/4$ to prove Lemma~\ref{lem:integral-mesh-cokernels} (used in Section~\ref{sec:integral-deformations}). Throughout, we retain the periodic tree models and trace reductions of Appendix~\ref{app:periodic-mesh}.

\subsection{Bounds for the characteristic-two mesh cokernels}
\label{app:characteristic-two-calculations}

Throughout this subsection we assume $\charac k=2$ and prove Proposition~\ref{prop:residual-correction-spaces}. Recall from the proof of Proposition~\ref{prop:periodic-localization} that a row path runs from $\tau v$ to $v$ and indexes a row of the mesh derivative $D_m$. We show that on each periodic component of $\Gamma_A$, the direct sum of the mesh cokernels $\operatorname{coker}D_m(\C)$ over all degrees $m\geq2$ has dimension at most one, and when nonzero is spanned by the class of a single row path in $C_d(\tau v,v)$ for some vertex $v$.

As in Appendix~\ref{app:periodic-mesh}, all vertex deletions below retain $\tau$-invariant subsets of vertices; by~\eqref{eq:mesh-object-quotient}, the resulting object quotients are the mesh categories of the retained subquivers, allowing us to restrict nonzero paths from smaller models to rule out small periods under the brick hypothesis.

\subsubsection{Trees of type A}
\label{app:char2-line-models}

We show that type $A$ contributes no mesh cokernel. In the preprojective category $\Pi(A_M)$, number the vertices $1,\ldots,M$ from left to right and denote the rightward and leftward steps by $a_+$ and $a_-$. The interior mesh relations commute $a_+a_-$ with $a_-a_+$, while the endpoint backtracks at~$1$ and~$M$ vanish. Sorting the steps of any path starting at vertex $i$ shows that the paths
\begin{equation}
  a_+^p a_-^q\qquad(0\leq p\leq M-i,\ 0\leq q\leq i-1)
  \label{eq:a-rectangle}
\end{equation}
span the paths from $i$, because any step pushing $p$ past $M-i$ or $q$ past $i-1$ can be commuted to an endpoint backtrack and vanishes. To see that the paths~\eqref{eq:a-rectangle} are linearly independent, consider the representation of $\Pi(A_M)$ with basis vectors $v_{p,q}$ ($0\leq p\leq M-i$, $0\leq q\leq i-1$) placed at vertex $i+p-q$, on which $a_+$ and $a_-$ increment $p$ and $q$ respectively (and act by zero at the boundary $p=M-i$ or $q=i-1$). Since these increments commute and satisfy the endpoint relations, the paths~\eqref{eq:a-rectangle} map $v_{0,0}$ to the distinct basis vectors $v_{p,q}$. In particular, all paths of degree at least $M$ in $\Pi(A_M)$ vanish, and for each vertex $i$ there is a unique nonzero path of top degree $M-1$, which terminates at the reflected vertex $M+1-i$.

We now examine the three periodic models of type~$A$:
\begin{itemize}
\item \emph{Pure translation quotients $\Z A_M/\langle\tau^s\rangle$.} The positive ordinary trace of $\Pi(A_M)$ vanishes even in characteristic two: any closed path at $i$ sorts to $(a_+a_-)^q$, and cyclic rotation combined with the interior commutation relation transports $(a_+a_-)^q$ step by step to an endpoint, where it vanishes (without dividing by $q$). Lemma~\ref{lem:twisted-trace-surjection}(2) therefore gives $\operatorname{coker}D_{d-1}=0$ for all $d\geq3$ and every period $s\geq1$.

\item \emph{Reflected odd trees $\Z A_{2m+1}/\langle\tau^s\rho\rangle$.} Here the brick hypothesis requires $s>m$: if $s\leq m$, retaining the central vertex and the first $s$ vertices of each arm yields the quotient $\Z A_{2s+1}/\langle\tau^s\rho\rangle$, in which the top-degree paths of length $2s$ in $\Pi(A_{2s+1})$ connect each vertex to its $\rho$-reflection while advancing the repetition coordinate by $s$, and hence project to nonzero radical endomorphisms at every vertex. With $s>m$ established, any row path from $\tau v$ to $v$ in $\Z A_{2m+1}/\langle\tau^s\rho\rangle$ has even path degree $d=2q\leq2m$ satisfying $s\mid(q-1)$ by Lemma~\ref{lem:twisted-trace-surjection}. Since $0\leq q-1<m<s$ for $q\geq1$, this divisibility gives $q=1$ ($d=2$), so no correction of path degree $d\geq3$ exists.

\item \emph{Odd quotients of even $A$ $\Z A_{2n}/\langle\eta^r\rangle$ ($r$ odd).} In the coordinates $(i,j)\in\{1,\ldots,n\}\times(\Z/r\Z)$ of Appendix~\ref{app:periodic-trees}, the arrows $\gamma_j\colon(1,j)\to(1,j+1)$, $a_{i,j}\colon(i,j)\to(i+1,j)$, and $b_{i,j}\colon(i+1,j)\to(i,j+2)$ have the property that any path starting and ending at the same level $i$ contains an equal number of $a$-steps and $b$-steps, so its total shift in the second coordinate $j$ equals its path degree $d$. If $r\leq2n-1$, retaining the first $(r+1)/2$ levels yields $\Z A_{r+1}/\langle\eta^r\rangle$; the top-degree paths of length $r$ in $\Pi(A_{r+1})$ connect each vertex to its reflection at the same level $i$ and shift $j$ by $r\equiv0\pmod r$, giving nonzero radical endomorphisms at every vertex and contradicting the brick hypothesis. Thus $r>2n-1$. Since any row path from $\tau(i,j)=(i,j-2)$ to $(i,j)$ starts and ends at level $i$, its path degree $d$ satisfies $d\equiv2\pmod r$; as $d\leq2n-1<r$, we must have $d=2$, ruling out any correction of degree $d\geq3$.
\end{itemize}

\subsubsection{The two short arms of type D}

For type $D$, we bound the possible degrees and periods of a nonzero mesh cokernel. Set $N=n-2$ for $\Delta=D_n$. By Lemma~\ref{lem:star-return-reduction}, the central corner $e_{x_0}\Pi(D_n)e_{x_0}$ is a quotient of
\[
  \Lambda^{D}_N=k\langle x,y\rangle/\bigl(x^2,y^2,(x+y)^N\bigr),
\]
where $x$ and $y$ represent the length-two closed paths along the two short arms. In word degree $j<N$, $\Lambda^{D}_N$ is spanned by the two alternating words $w_j$ and $w'_j$ of length $j$ starting with $x$ and $y$, respectively; in word degree $N$, the relation $(x+y)^N=0$ becomes $w_N+w'_N=0$. Multiplying $w_N+w'_N=0$ on the left or right by $x$ and $y$ shows that both alternating words of degree $N+1$ vanish, so $(\Lambda^{D}_N)_j=0$ for all $j>N$.

By Lemma~\ref{lem:arm-trace-reduction}, the positive twisted trace of $\Pi(D_n)$ is a quotient of the corresponding twisted trace of $\Lambda^{D}_N$:
\begin{itemize}
\item For the \emph{ordinary trace} ($\theta=\id$), the degree-one classes $[x]$ and $[y]$ vanish by~\eqref{eq:arm-power-trace-zero}. Any alternating word of odd length $j\geq3$ begins and ends with the same letter, so cyclic rotation produces $x^2$ or $y^2$ and the class vanishes. For even $j$, cyclic rotation identifies $[w_j]=[w'_j]$. Hence the positive ordinary trace is at most one-dimensional in each path degree $d=2j$ with $j$ even and $2\leq j\leq N$, and vanishes in all other positive degrees.
\item For the \emph{short-arm interchange} ($\theta=\iota$ swapping $x$ and $y$), twisted cyclic rotation $[uv]=[v\iota(u)]$ turns an alternating word of even length $j$ into a word containing $x^2$ or $y^2$, while for odd $j$ it identifies $[w_j]=[w'_j]$. Hence the positive $\iota$-twisted trace is at most one-dimensional in each path degree $d=2j$ with $j$ odd and $1\leq j\leq N$, and vanishes in all other positive degrees.
\end{itemize}
In either case, Lemma~\ref{lem:twisted-trace-surjection} shows that a nonzero cokernel in path degree $d=2j\geq3$ can occur only when
\begin{equation}
  2\leq j\leq N\qquad\text{and}\qquad s\mid(j-1).
  \label{eq:d-possible-correction}
\end{equation}
(These degree and parity restrictions match Riedtmann's characteristic-two analysis of type~$D$ in~\cite[Section~2, p.~487, and Sections~3.3, 3.9]{Riedtmann1984}.)

For $D_4$ ($N=2$), we also check a $3$-cycle $\rho$ permuting the three arms $x,y,x+y$. Here $\Lambda^{D}_2$ is spanned by $1,x,y,xy=yx$ with $(\Lambda^{D}_2)_j=0$ for $j\geq3$. In word degree~$1$, twisted cyclic rotation under $\rho$ identifies $[x]=[y]=[x+y]$, whence $[x]=0$; in word degree~$2$, $[xy]=[y\rho(x)]=[y^2]=0$ (and for $\rho^2$, $[yx]=[x\rho^2(y)]=[x^2]=0$). The only possible ordinary trace class has path degree $d=4$ ($j=2$), for which $s\mid(j-1)$ gives $s=1$; then $l=(1-j)/s=-1$, so the induced twist $\theta=\rho^{-1}$ is nontrivial and no mesh cokernel class arises from a $3$-cycle on $D_4$.

To show that~\eqref{eq:d-possible-correction} admits at most one value of $j$ when $\theta\in\{\id,\iota\}$, we bound $N$ in terms of the period $s$ using the top-degree socle paths of $\Pi(D_n)$ (cf.\ \cite[Section~6.2]{Kaplan}).

\begin{lemma}
\label{lem:d-top-paths}
Let $\Pi(D_n)$ be the preprojective algebra of type $D_n$ over an arbitrary field, set $N=n-2$, and let $\nu\in\Aut(D_n)$ be the identity if $N$ is even and the short-arm interchange if $N$ is odd. For every vertex $x$ of $D_n$, there is a nonzero path of degree $2N$ in $\Pi(D_n)$ from $x$ to $\nu x$.
\end{lemma}
\begin{proof}
For $D_n$, the Coxeter number is $2n-2$, and the Nakayama permutation
fixes the long arm and interchanges the short arms exactly when $n$ is odd
\cite[Sections~4.1--4.2 and~4.4]{BBK}. Thus
$e_x\Pi(D_n)_{2n-4}e_{\nu x}$ is one-dimensional over any field.
Since $2n-4=2N$ and $n\equiv N\pmod2$, a nonzero path in this space
has the required degree and endpoints.
\end{proof}

Returning to $\Z D_n/\langle\tau^s\rho\rangle$ with $\rho\in\{\id,\iota\}$, the brick hypothesis requires $N<2s$: otherwise, deleting the outer $N-2s$ vertices of the long arm leaves $\Z D_{2s+2}/\langle\tau^s\rho\rangle$, in which Lemma~\ref{lem:d-top-paths} (with $N'=2s$ even) provides nonzero paths of degree $4s$ fixing every vertex of $D_{2s+2}$ and advancing the repetition coordinate by $2s$, which project to nonzero radical endomorphisms at every vertex for either choice of $\rho$. Combining $N<2s$ with~\eqref{eq:d-possible-correction} yields $1\leq j-1\leq N-1<2s-1$ and $s\mid(j-1)$, which leaves only $j=s+1$ and $d=2s+2$ (occurring only when $s+1\leq N<2s$, with $s$ odd for the trivial twist $\theta=\id$ and $s$ even for $\theta=\iota$).

\subsubsection{Exceptional upper bounds}

For $\Delta=E_7,E_8$, we retain the presentation $\Lambda^{E}_m=k\langle x,y\rangle/(x^m,y^3,(x+y)^2)$ from Lemma~\ref{lem:e78-traces} for $m=4,5$, respectively, and use the analogous
presentation for $E_6$ with $m=3$ from Lemma~\ref{lem:star-return-reduction}.
For $E_6$, the variables $x,y$ represent the two long-arm loops, which
are interchanged by reflection; the short-arm loop is $-x-y$. By \cite[Section~4.1]{BBK}, $\Pi(\Delta)$ vanishes in path degrees
greater than $h-2$, where $h=12,18,30$ for $E_6,E_7,E_8$,
respectively. Consequently, in its central corner the images of words
in $x,y$ vanish above word degrees $5,8,14$, respectively. We apply
these bounds to the trace of $\Pi(\Delta)$, which is a quotient of the
trace spaces in~\eqref{eq:exceptional-trace-spaces}.

Applying Lemma~\ref{lem:e78-traces}(1) in characteristic two with $a=yx^3$ and $b=yx^4$, the relations $2=0$ and $3[ab]=0$ simplify the block identities to
\[
  [a^2W]=0\ (W\ne1),\qquad [abW]=[baW]\ (W\ne1),\qquad [ab]=0,\qquad [b^q]=0\ (q\geq2).
\]
In the trace of $\Pi(\Delta)$, any word in $a,b$ of length at least~$3$ containing at least two $a$'s vanishes by commuting the $a$'s to be adjacent, while a word with a single $a$ reduces to $[ab^r]$, which vanishes for $r=1$ (as $[ab]=0$) and for $r\geq3$ (as $\deg(ab^r)=4+5r\geq19>14$). Among the single-$y$ words, $3[yx^2]=0$ and $5[yx^4]=0$ still give $[yx^2]=[yx^4]=0$ in characteristic two, leaving only the following spanning representatives for the positive ordinary trace:
\[
  \begin{array}{c|c|c}
  \text{Dynkin type}&\text{spanning trace representatives}&\text{path degrees }d\\ \hline
  E_6&[yx]&4\\
  E_7&[yx],\ [yx^3],\ [(yx^3)^2]&4,\ 8,\ 16\\
  E_8&[yx],\ [yx^3],\ [(yx^3)^2],\ [(yx^3)(yx^4)^2]&4,\ 8,\ 16,\ 28
  \end{array}
\]
Each listed path degree is at most one-dimensional, and all other positive ordinary trace degrees vanish.

For the nontrivial graph automorphism $\theta$ of $E_6$, which swaps
$x$ and $y$, the relations
\[
 y^2=yx+xy+x^2,\qquad yxy=xyx,\qquad
 yx^2y=xyx^2+x^2yx
\]
follow from~\eqref{eq:e8-word-identities} after imposing $x^3=0$.
They reduce every word in $\Lambda^{E}_3$ to a combination of $x^i$
and $x^iyx^j$ with $0\leq i,j\leq2$. Its spanning words in positive
word degrees are therefore:
\[
  \begin{array}{c|l}
  \text{word degree}& \text{spanning words in }\Lambda^{E}_3\\ \hline
  1&x,\ y\\
  2&x^2,\ xy,\ yx\\
  3&\xi_1=x^2y,\ \xi_2=xyx,\ \xi_3=yx^2\\
  4&\xi_4=x^2yx,\ \xi_5=xyx^2\\
  5&\xi_6=x^2yx^2
  \end{array}
\]
The $\theta$-twisted trace relations $[xw]=[wy]$ and $[yw]=[wx]$ identify the words in degree~$1$ ($[x]=[y]$) and in degree~$2$ ($[x^2]=[xy]=[yx]$, using $y^2=xy+yx+x^2$). In degrees~$3$ and~$4$, using the algebra expansions $xy^2=x^2y+xyx=\xi_1+\xi_2$, $x^2y^2=x^2yx=\xi_4$, and $y^2x^2=xyx^2=\xi_5$ (since $x^3=0$), the following five twisted-trace relations show that all spanning classes vanish:
\begin{align*}
  w=x^2\text{ in }[xw]=[wy]&\implies [\xi_1]=[x^2y]=[x^3]=0,\\
  w=x^2\text{ in }[yw]=[wx]&\implies [\xi_3]=[yx^2]=[x^3]=0,\\
  w=xy\text{ in }[xw]=[wy]&\implies [\xi_1]=[xy^2]=[\xi_1]+[\xi_2]\implies [\xi_2]=0,\\
  w=x^2y\text{ in }[xw]=[wy]&\implies 0=[x^3y]=[x^2y^2]=[\xi_4],\\
  w=yx^2\text{ in }[yw]=[wx]&\implies [\xi_5]=[y^2x^2]=[yx^3]=0.
\end{align*}
Degree~$5$ is spanned by $\xi_6$ alone, and the fixed short arm contributes no new classes. Thus the positive $\theta$-twisted trace of $\Pi(E_6)$ is at most one-dimensional in path degrees $d=2,4,10$ and vanishes in all other positive degrees.

Applying Lemma~\ref{lem:twisted-trace-surjection} to these candidate degrees:
\begin{itemize}
\item For an ordinary trace class ($\theta=\id$) in path degree $d\in\{4,8,16,28\}$, the period $s$ must divide $d/2-1\in\{1,3,7,13\}$. (In the reflected $E_6$ model $\Z E_6/\langle\tau^s\rho\rangle$, Lemma~\ref{lem:twisted-trace-surjection}(2) requires $(d/2-1)/s$ to be even for the untwisted trace, since $\theta=\rho^{-(d/2-1)/s}$ and $\rho$ has order two; for the sole $E_6$ candidate $d=4$, $s\mid1$ gives $s=1$, for which $(d/2-1)/s=1$ is odd and hence twisted.)
\item For a $\theta$-twisted trace class on $E_6$ in path degree $d\in\{4,10\}$, Lemma~\ref{lem:twisted-trace-surjection}(2) requires $s\mid(d/2-1)$ with $(d/2-1)/s$ odd, which gives $s=1$ for $d=4$ and $s=4$ for $d=10$.
\end{itemize}
Finally, period $s=1$ is excluded in every case by deleting the vertices outside the central $D_4$ subquiver and applying Lemma~\ref{lem:d4-four-return} (noting that the degree-$4$ closed paths in $\Pi(D_4)$ advance the repetition coordinate by~$2$ and therefore also close in the reflected quotient $\Z D_4/\langle\tau\rho\rangle$ since $\rho^2=\id$). Consequently, the only remaining possibilities for a positive-degree mesh cokernel class of degree $d\geq3$ in types $E_6,E_7,E_8$ are:
\begin{equation}
  \label{eq:exceptional-correction-bounds}
  \begin{array}{c|c|c}
  \text{Dynkin type and action}&\text{period }s&\text{path degree }d\\ \hline
  E_6\text{ (with reflection }\rho\text{)}&4&10\\
  E_7&3,\ 7&8,\ 16\text{ respectively}\\
  E_8&3,\ 7,\ 13&8,\ 16,\ 28\text{ respectively}
  \end{array}
\end{equation}
In every case, each periodic component has at most one nonzero mesh cokernel $\operatorname{coker}D_{d-1}$, occurring in path degree $d=2s+2$ and spanned by the class of a single row path $p_i\colon\tau v_i\to v_i$ with $v_i$ lying over the central vertex of the tree. Taking $I$ to index those periodic components of $\Gamma_A$ whose mesh cokernel is nonzero and choosing a spanning row path $p_i$ on each completes the proof of Proposition~\ref{prop:residual-correction-spaces}.

\subsection{Correction spaces over the integers modulo four}
\label{app:integral-corrections}

We now lift the preceding calculations to $\mathcal O=\Z/4$ for the integral mesh category $E_4$ constructed in Proposition~\ref{prop:fixed-integral-companion}.


\begin{proof}[Proof of Lemma~\ref{lem:integral-mesh-cokernels}]
Set $\mathcal Q_m=\operatorname{coker}D_m(E_4)$. By Proposition~\ref{prop:fixed-integral-companion}, reducing the source, target, and matrix of $D_m(E_4)$ modulo~$2$ recovers $D_m(\mathcal C^{(2)})$, so right exactness of the tensor product gives degree- and weight-preserving identifications
\[
  \mathcal Q_m/2\mathcal Q_m\cong\operatorname{coker}D_m(\mathcal C^{(2)}),\qquad
  \operatorname{coker}D_m(\mathcal C^{(2)})\otimes_{\F_2}k\cong\operatorname{coker}D_m(\C).
\]
By Proposition~\ref{prop:residual-correction-spaces}, the paths $p_i$ of path degree $d_i=m+1$ span $\operatorname{coker}D_m(\C)$ over $k$ and hence already form an $\F_2$-basis of $\operatorname{coker}D_m(\mathcal C^{(2)})$. Lifting these paths to $E_4$ and letting $J_m\subseteq \mathcal Q_m$ denote the $\mathcal O$-submodule they generate, Nakayama's lemma (or directly $\mathcal Q_m=J_m+2\mathcal Q_m=J_m+2(J_m+2\mathcal Q_m)=J_m$ since $4\mathcal Q_m=0$) shows that $\mathcal Q_m=J_m$, and in particular $\mathcal Q_m=0$ in all degrees and weights with no such $p_i$. Since the mod-$2$ reductions of these paths are linearly independent over $\F_2$, it suffices to show that twice each $p_i$ lies in $\operatorname{im}D_m(E_4)$, i.e., vanishes in $\mathcal Q_m$.

First, any row path that leaves its periodic component passes through a nonperiodic vertex and therefore lies in a finite triangular incidence chain as in the proof of Proposition~\ref{prop:periodic-localization}; because the mesh coefficients of $E_4$ are units in $\mathcal O$, eliminating from an exposed end expresses any such row path as a derivative over $\mathcal O$. Consequently, all terms in a contextual mesh relation that leave the periodic component vanish in $\mathcal Q_m$. Likewise, the scalar normalizations of Lemma~\ref{lem:periodic-scalar-gauges}(1) hold over $\mathcal O$, since they divide only by unit mesh coefficients and rescale arrows and rows by units in $\mathcal O$. After this normalization to the signed preprojective coefficients $\pm1$, each arrow-correction column on a retained periodic component has opposite unit coefficients $\pm\alpha$ in its two row incidences, giving $\alpha[wa]-\alpha[(\tau a)w]=0$ and hence $[wa]=[(\tau a)w]$ in $\mathcal Q_m$. Thus the class of a path $p_i$ in $\mathcal Q_m$ factors through the positive $\theta$-twisted trace of the integral signed preprojective category $\Pi_{\mathcal O}(\Delta)$.

Moreover, the star and arm reductions of Lemmas~\ref{lem:star-return-reduction} and~\ref{lem:arm-trace-reduction} hold verbatim over $\mathcal O$, because every replacement step uses only coefficients $\pm1$:
\begin{itemize}
\item Replacing an outward backtrack $a_j b_j$ along an arm by $\pm b_{j-1}a_{j-1}$ decreases the total arm depth until any closed path at the center is a word in the length-two arm loops, whose signed sum at the center vanishes.
\item Along an arm of length $L$, replacing $b_j a_j$ at internal vertices by $\pm a_{j+1}b_{j+1}$ rewrites $(a_1 b_1)^q$ as $\pm a_1\cdots a_q b_q\cdots b_1$ for $q\leq L$ and gives $(a_1 b_1)^{L+1}=0$ by the leaf relation $b_L a_L=0$.
\item In the twisted trace along a $\theta$-fixed arm, rotating a closed word to a vertex of minimal depth and replacing the inward backtrack by the outward backtrack strictly increases the depth sum until the word reaches the leaf and vanishes, showing that positive powers of a fixed-arm loop have zero trace over $\mathcal O$.
\end{itemize}

We now check that $2$ annihilates the trace classes of the paths $p_i$ in each Dynkin type that can contribute to $I$:
\begin{itemize}
\item \emph{Type $D_n$.} In the central corner quotient over $\mathcal O$, the short-arm relations $x^2=y^2=0$ leave the alternating words $w_j$ and $w'_j$ of word degree $j=s+1$. For the trivial twist $\theta=\id$ ($s$ odd, $j$ even), cyclic rotation gives $[w_j]=[w'_j]$; for the short-arm interchange $\theta=\iota$ ($s$ even, $j$ odd), twisted cyclic rotation $[uv]=[v\iota(u)]$ likewise gives $[w_j]=[w'_j]$. Since the long-arm loop equals $-(x+y)$ in the central mesh relation and its $j$-th power has zero trace along the fixed long arm, we obtain in both cases
\[
  0=[(-x-y)^j]=(-1)^j([w_j]+[w'_j])=\pm2[w_j],
\]
so $2[w_j]=0$ over $\mathcal O$.

\item \emph{Reflected type $E_6$ ($s=4$, $d=10$, word degree~$5$).} In the central corner $\Lambda^{E}_3$ over $\mathcal O$, the relations $x^3=y^3=(x+y)^2=0$ give
\[
  y^2=-yx-xy-x^2,\qquad yxy=xyx,\qquad yx^2y=-xyx^2-x^2yx,
\]
where the second identity follows from $0=y^3=(-yx-xy-x^2)y=-yxy-xy^2-x^2y=-yxy-x(-yx-xy-x^2)-x^2y=-yxy+xyx+x^3=-yxy+xyx$ and the third follows by comparing $(yxy)y=(xyx)y$ with $yx(y^2)=yx(-yx-xy-x^2)$. These three identities eliminate any pair of $y$'s separated by $0$, $1$, or $2$ factors of $x$, so $\Lambda^{E}_3$ is spanned over $\mathcal O$ by $x^i$ and $x^iyx^j$ ($0\leq i,j\leq2$), and word degree~$5$ is spanned by $\xi_6=x^2yx^2$. Applying the $\theta$-twisted trace relation $[xw]=[wy]$ ($\theta(x)=y$) together with the third identity and $x^3=0$ yields
\[
  [\xi_6]=[x(xyx^2)]=[(xyx^2)y]=[x(yx^2y)]=[-x^2yx^2-x^3yx]=-[\xi_6],
\]
so $2[\xi_6]=0$ over $\mathcal O$.

\item \emph{Types $E_7$ and $E_8$.} Lemma~\ref{lem:e78-traces}(1) holds over any commutative ring, and in particular over $\mathcal O=\Z/4$. Every spanning class in the positive ordinary trace is annihilated by $2$, $3$, or $5$; since $3$ and $5$ are units in $\Z/4$, the entire positive trace is annihilated by~$2$.
\end{itemize}
Thus $2\mathcal Q_m=0$ for all $m\geq2$, completing the proof.
\end{proof}