\section{Mesh cokernels and deformation of module categories}
\label{sec:positive-deformations}

We retain the graded category $\C=\gr_\JJ\ind(\modu A)$ and its standard
form $A_0$. We describe the mesh cokernels in characteristic two and
illustrate positive filtered deformations by a nonstandard algebra. We then prove
the extension bounds needed to lift graded $A_0$-modules and morphisms to
first-order deformations of $A_0$. These lifts will relate the Hochschild cohomology
classes represented by degree-raising two-cocycles to the mesh cokernels in Section~\ref{sec:weights-and-parameters}.

\subsection{Mesh cokernels in characteristic two}

By Proposition~\ref{prop:periodic-localization}, the cokernel of the mesh
derivative $D_m(\C)$ splits as the direct sum of the cokernels on the
periodic components of $\Gamma_A$. In
Appendix~\ref{app:characteristic-two-calculations} we evaluate these
cokernels on the tree models of Proposition~\ref{prop:periodic-tree-model}:
every type-$A$ periodic component has zero cokernel, while each remaining
periodic component has at most a one-dimensional cokernel concentrated in a
single degree. For a scalar mesh category $C=C(Q,c)$, we call a periodic
component of $Q$ containing at least one arrow \emph{residual} if
$\bigoplus_{m\geq2}\operatorname{coker}D_m(C)$ has a nonzero class supported
on that component, and we let $I$ index the residual components.

\begin{proposition}
\label{prop:residual-correction-spaces}
Let $C=C(Q,c)$ be a finite-dimensional scalar mesh category on a finite
translation quiver $Q$ with at most one arrow between each ordered pair of
vertices. Assume that $\charac k=2$ and that $C$ satisfies the brick
hypothesis of Subsection~\ref{subsec:brick-hypothesis}.
\begin{enumerate}[label={\upshape(\arabic*)}]
\item The $i$th residual component ($i\in I$) is isomorphic to
$\Z\Delta/\langle\tau^{s_i}\rho\rangle$, where $\Delta$ is a Dynkin tree of
type $D$ or $E$ and $s_i\geq1$, as in
Proposition~\ref{prop:periodic-tree-model}(2).
\item For each $i\in I$, there exist a vertex $v_i$ of the $i$th residual
component and a path $p_i:\tau v_i\to v_i$ of length $d_i=2s_i+2$ such that,
viewing $p_i$ as an element of $V_{d_i}(C)$ supported in the summand $C_{d_i}(\tau v_i,v_i)$
and writing $[p_i]$ for its class in $\operatorname{coker}D_{d_i-1}(C)$, we
have
\[
 \bigoplus_{m\geq2}\operatorname{coker}D_m(C)
                      =\bigoplus_{i\in I}k[p_i].
\]
In particular, $d_i=2s_i+2$ is the unique path degree in which the $i$th
residual component has a nonzero cokernel class.
\end{enumerate}
\end{proposition}

By Lemma~\ref{lem:deletion-brick}, Proposition~\ref{prop:residual-correction-spaces}
applies directly to $C=\C$. In Section~\ref{sec:weights-and-parameters} we
will show that the mesh errors arising from positive filtered deformations of
$A_0$ embed into $\bigoplus_{i\in I}k[p_i]$ via the comparison map of
Proposition~\ref{prop:hochschild-mesh-restriction}.

\subsection{Positive deformations and a nonstandard example}

The preceding cokernels measure errors in mesh relations. To compare
these errors with changes in the algebra itself, we now study filtered
algebras whose leading multiplication is that of $A_0$. We first give an
example in which changing a zero product preserves a multiplicative basis;
we will then lift modules to such deformations to obtain the comparison
with mesh cokernels in Proposition~\ref{prop:hochschild-mesh-restriction}.

\begin{definition}
\label{def:positive-deformation}
A \emph{positive filtered deformation} of a finite-dimensional nonnegatively
graded algebra $\Lambda=\bigoplus_{d\geq0}\Lambda_d$ is an algebra $B$
equipped with a finite decreasing multiplicative filtration $F^\bullet B$
and a specified graded algebra isomorphism $\gr_F B\cong\Lambda$.
\end{definition}

In particular, the categorical radical filtration
$F^n A(x,y)=\JJ^n(P_y,P_x)$ of Section~\ref{sec:projective-rays} makes the
original algebra $A$ a positive filtered deformation of its standard form
$A_0$.

The following classical pair shows why a multiplicative basis can persist
in a nontrivial deformation.

\begin{example}[Riedtmann's example; {\cite[Sections~2.1, 2.3 and~7.6]{BongartzSurvey}}]
\label{ex:nonstandard-pair}
Let $\charac k=2$, and let $Q$ be the quiver with arrows
$a:1\to2$, $b:2\to1$, and a loop $q$ at $2$. Consider the family of algebras
\[
 B_t=kQ/(ba-q^2,\ ab-taqb,\ q^4)\qquad(t\in k),
\]
in which $B_0$ and $B_1$ form the classical representation-finite standard
and nonstandard pair. For every $t\in k$, the defining relations imply
$aq^2=taq^3$ and $q^2b=tq^3b$; since $q^4=0$, multiplying once more by $q$
gives $aq^3=q^3b=0$ and hence $aq^2=q^2b=0$. Thus $B_t$ is spanned by the ten
elements
\[
 e_1,e_2,\ q,q^2,q^3,\ a,aq,\ b,qb,\ w=aqb,
\]
whose nonzero products, apart from multiplication by the primitive
idempotents and powers of $q$, are
\[
 a q=aq,\quad q b=qb,\quad ba=q^2,\quad
 b(aq)=(qb)a=q^3,\quad a(qb)=(aq)b=w,\quad ab=tw.
\]
Direct inspection shows that this multiplication table is associative on the
ten formal basis vectors and satisfies the defining relations of $B_t$;
acting with this formal algebra on the sum of the two idempotents $e_1,e_2$
confirms that the ten elements are linearly independent in $B_t$. Assigning
degree one to $a,b,q$ makes $B_t$ a positive filtered deformation of $B_0$,
and at both $t=0$ and $t=1$ the shared basis is normed and multiplicative.

Although the product $ab=tw$ depends on $t$, its value lies in
$a(\rad B_t(2,2))b=kw$, so the ray product of $a$ and $b$ remains zero for
every $t\in k$ by Definition~\ref{def:ray-product}. The ray category of $B_t$
is therefore independent of $t$, and its linearization is $B_0$.

To see directly that $B_1\not\cong B_0$, observe that any isomorphism would
preserve the two primitive idempotents up to conjugation because the diagonal
corners have dimensions two and four. The images in $B_1$ of the generators
of $B_0$ would then have the form
\[
 a'=u a+v aq,\qquad b'=s b+zqb,\qquad
 q'=c q+d q^2+h q^3,\qquad usc\ne0.
\]
Since $(q')^2=c^2q^2$ in characteristic two, the relation $b'a'=(q')^2$
requires $su=c^2$ and $sv+zu=0$, which implies
\[
 a'b'=(us+uz+vs)w=usw\ne0,
\]
contradicting $ab=0$ in $B_0$. Thus modifying the zero product $ab$ changes
the isomorphism class of the algebra while preserving its ray category. The
linear coefficient of $t$ in the multiplication of $B_t$ is the Hochschild
two-cocycle sending $(a,b)$ to $w$ and all other basis pairs to zero;
Proposition~\ref{prop:hochschild-mesh-restriction} will pair each such
cohomology class with a mesh cokernel class.
\end{example}

\subsection{Degrees of extensions between normalized modules}
\label{subsec:ext-degrees}

To relate deformations of $A_0$ to the mesh cokernel classes $[p_i]$, we
lift its graded indecomposable modules and their morphisms to first-order
deformations. The obstruction to lifting a module action lies in
$\Ext^2$; once the actions are lifted, the obstruction to lifting a
morphism lies in $\Ext^1$. Lemma~\ref{lem:square-zero-selected-lifts}
gives the relative bar-complex calculation proving these assertions,
following the deformation theory of Gerstenhaber and Lowen
\cite{Gerstenhaber,Lowen}. We first establish the required degree bounds.

For graded right
$\Lambda$-modules $M$ and $N$, a homomorphism or extension has
\emph{internal degree} $t\in\Z$ if it is represented on a graded projective
resolution of $M$ by a cochain that shifts degrees by $+t$. Using the shift
convention $(N\langle a\rangle)_q=N_{q-a}$ (so that a degree-zero map
$M\langle a\rangle\to N\langle b\rangle$ has underlying internal degree
$a-b$), we have
\[
 \Ext^n_\Lambda(M,N)_t
   =\Ext^n_{\mathrm{gr}\text{-}\Lambda}(M,N\langle-t\rangle),
\]
where $\Ext_{\mathrm{gr}\text{-}\Lambda}$ is computed in the category of
graded right $\Lambda$-modules with degree-zero morphisms.

For $\Lambda=A_0$, enumerate the objects of $\C$ as $X_1,\ldots,X_h$, and
let $M_1,\ldots,M_h$ be the corresponding graded indecomposable right
$A_0$-modules obtained by projective evaluation, graded by
$(M_i)_d e_x=\C_d(P_x,X_i)$. We refer to $M_1,\ldots,M_h$ as the
\emph{degree-normalized} indecomposable $A_0$-modules; note that the
indecomposable projective $A_0$-modules $e_x A_0$ carry their natural
gradings under this normalization. Full faithfulness of projective
evaluation (Proposition~\ref{prop:projective-evaluation}(1)) immediately
gives
\begin{equation}
 \Hom_{A_0}(M_i,M_j)_r=0\quad(r<0),\qquad
 \Hom_{A_0}(M_i,M_j)_0=
 \begin{cases}k\cdot\id_{M_i},&\text{if }i=j,\\0,&\text{if }i\ne j.\end{cases}
 \label{eq:normalized-hom-degrees}
\end{equation}

The bound below includes internal degree zero. This case is used to lift
the modules over $\Z/4$ in Proposition~\ref{prop:fixed-integral-companion};
positive degrees suffice for filtered deformations over $k$.

\begin{lemma}
\label{lem:positive-ext-vanishing}
Let $A_0$ be the standard form of Construction~\ref{con:projective-algebra},
and let $M_1,\ldots,M_h$ be the degree-normalized indecomposable
$A_0$-modules defined above.
\begin{enumerate}[label={\upshape(\arabic*)}]
\item Every finite-dimensional graded right $A_0$-module is isomorphic to a
direct sum of degree shifts of the modules $M_i$.
\item For all $i,j$ and all $n\geq1$,
\[
                 \Ext^n_{A_0}(M_i,M_j)_t=0\qquad\text{for all }t>-2n.
\]
\end{enumerate}
\end{lemma}
\begin{proof}
Let $X$ be a finite-dimensional graded right $A_0$-module that is
indecomposable as a graded module. Decomposing $X$ as an ungraded
$A_0$-module into copies of the indecomposable modules $M_i$, let
$p_a:X\to M_{i_a}$ and $j_a:M_{i_a}\to X$ be the projection and inclusion
maps of the summands, so that $1_X=\sum_a p_a j_a$ in path order. Passing to
the homogeneous component of degree zero gives
\[
                   1_X=\sum_{a,r}(p_a)_r(j_a)_{-r}
\]
in the local ring $\End_{\mathrm{gr}\text{-}A_0}(X)$. Some single term
$(p_a)_r(j_a)_{-r}$ is therefore invertible, exhibiting $X$ as a graded
direct summand of $M_{i_a}\langle-r\rangle$. Since $M_{i_a}$ is already
indecomposable as an ungraded module, $X\cong M_{i_a}\langle-r\rangle$,
proving (1).

We prove (2) first for $n=1$. The vanishing is trivial if $M_i$ is
projective. If $X_i$ is nonprojective, Proposition~\ref{prop:graded-realization}(1)
and Lemma~\ref{lem:mesh-simple-tops} provide an exact sequence of
representable functors
\[
 0\longrightarrow\C(-,\tau X_i)\longrightarrow
 \bigoplus_{Y\to X_i}\C(-,Y)\longrightarrow\C(-,X_i)
\]
whose cokernel is the simple functor $\C_0(-,X_i)$. Restricting these
functors to the projective objects $P_x$ annihilates the simple cokernel
(since $X_i$ is nonprojective) and yields a short exact sequence of graded
$A_0$-modules with degree-one maps. By Corollary~\ref{cor:same-ar-quiver},
this is the almost split sequence ending at $M_i$. Inserting degree shifts
so that its maps become homogeneous of degree zero gives
\[
 0\longrightarrow M_{\tau i}\langle2\rangle\longrightarrow
 \bigoplus_{Y\to X_i}M_Y\langle1\rangle\longrightarrow M_i
 \longrightarrow0.
\]
Now suppose for a contradiction that
$0\to M_j\langle-t\rangle\to E\xrightarrow{v}M_i\to0$ is a nonsplit graded
extension with $t>-2$. It remains nonsplit as an ungraded sequence (since any
ungraded section would have a degree-zero component that is a graded
section). By the lifting property of the right almost split map, $v$ factors
through $\bigoplus_{Y\to X_i}M_Y\langle1\rangle\to M_i$, and taking the
degree-zero component produces a graded factorization. The induced graded
map on kernels $M_j\langle-t\rangle\to M_{\tau i}\langle2\rangle$ has
underlying internal degree $-t-2<0$ and therefore vanishes by
\eqref{eq:normalized-hom-degrees}. Consequently, the factorization descends
to a section of the almost split sequence ending at $M_i$, a contradiction.
Thus $\Ext^1_{A_0}(M_i,M_j)_t=0$ for all $t>-2$.

For $n\geq2$, observe that in a minimal graded projective presentation of a
nonprojective module $M_i$, every projective summand $P\langle a\rangle$ of
the projective cover has $a\geq1$ because its nonzero map to $M_i$ lies in
positive degree; similarly, every indecomposable summand $Z\langle b\rangle$
of the first syzygy $\Omega^1 M_i$ maps into $P\langle a\rangle$ via a
radical morphism of positive degree, so $b-a\geq1$ and hence $b\geq2$. By
induction on $r$, every indecomposable summand of the $r$th syzygy
$\Omega^r M_i$ has the form $Z\langle b\rangle$ with $b\geq 2r$. Dimension
shifting then gives
\[
 \Ext^n_{A_0}(M_i,M_j)_t
 =\Ext^1_{A_0}(\Omega^{n-1}M_i,M_j)_t=0
 \qquad\text{for all }t>-2n,
\]
because each indecomposable summand $Z\langle b\rangle$ of
$\Omega^{n-1}M_i$ has $b\geq 2(n-1)$ and thus contributes
$\Ext^1_{A_0}(Z,M_j)_{t+b}=0$ since $t+b>-2$.
\end{proof}

\subsection{Lifting actions and morphisms}

Before lifting modules to a positive filtered deformation $B$, we choose a
semisimple subalgebra of $B$ lifting $B/F^1B$ and split its filtration
as bimodules over that subalgebra.

For an algebra $\Lambda$ over a commutative ring $R$ and a unital subalgebra
$S\subseteq\Lambda$, the space of \emph{normalized $S$-relative Hochschild
$n$-cochains} ($n\geq1$) is
\begin{equation}
 C^n_S(\Lambda,\Lambda)
   =\Hom_{S\text{-}S}\bigl((\Lambda/S)^{\otimes_S n},\Lambda\bigr),
 \label{eq:normalized-relative-cochains}
\end{equation}
with $C^0_S(\Lambda,\Lambda)=\{a\in\Lambda:as=sa\text{ for all }s\in S\}$
in degree zero and the standard Hochschild differential (in particular,
$(df)(a,b)=af(b)-f(ab)+f(a)b$ for a $1$-cochain $f$). Viewed as an
$R$-multilinear map on $\Lambda^n$, a normalized $S$-relative $n$-cochain is
$S$-bilinear in the outer variables, $S$-balanced across each adjacent pair
of inputs, and vanishes whenever any input belongs to $S$. We will use these
cochains over fields and over $\Z/4$.
For a graded algebra, an $n$-cochain has internal degree $q$ if it sends
inputs of degrees $d_1,\ldots,d_n$ to degree $d_1+\cdots+d_n+q$.
The differential preserves internal degree; we write
$\HH^2(\Lambda)_{>0}=\bigoplus_{q>0}\HH^2(\Lambda)_q$ for its positive part.


\begin{lemma}
\label{lem:split-positive-deformation}
Let $k$ be a field, let $\Lambda=\bigoplus_{d\geq0}\Lambda_d$ be a
finite-dimensional nonnegatively graded $k$-algebra whose degree-zero part
$S=\Lambda_0$ is split semisimple, and let $B$ be a positive filtered
deformation of $\Lambda$. Then $S$ lifts to a unital subalgebra of $B$, and
the filtration $F^\bullet B$ splits as a filtration of $S$-bimodules. Under
the resulting filtered $S$-bimodule identification $B\cong\Lambda$ (inducing
the given identification $\gr_F B=\Lambda$), the multiplication of $B$ takes
the form $\mu_0+\sum_{q>0}\mu_q$, where $\mu_0$ is the multiplication of
$\Lambda$ and each $\mu_q\in C^2_S(\Lambda,\Lambda)$ is a normalized
$S$-relative $2$-cochain raising internal degree by $q$.
\end{lemma}
\begin{proof}
Since $\mathfrak n=F^1B$ is a nilpotent ideal with $B/\mathfrak n=S$, we can
lift the diagonal matrix units of $S$ to orthogonal idempotents $p_i\in B$
within each simple block. Lifting the matrix units $E_{i1}$ and
$E_{1i}$ of a given block to elements $u_i\in p_iBp_1$ and $v_i\in p_1Bp_i$
and replacing $v_i$ by $(v_i u_i)^{-1}v_i$, we obtain $v_i u_i=p_1$; then
$p_i-u_i v_i$ is a nilpotent idempotent in $p_iBp_i$ and therefore vanishes,
so the products $u_i v_j$ form a complete system of matrix units defining a
unital subalgebra $S\subseteq B$.

Because $S\otimes_k S^{\op}$ is semisimple, the filtration $F^\bullet B$
splits as a filtration of $S$-bimodules with this lifted copy of $S$ in degree
zero. Under the resulting filtered $S$-bimodule identification $B\cong\Lambda$,
the degree-zero component of the multiplication of $B$ is $\mu_0$, and each
higher-degree component $\mu_q$ is $S$-bilinear and vanishes whenever either
input belongs to $S$, so $\mu_q\in C^2_S(\Lambda,\Lambda)_q$.
\end{proof}

We now lift modules and morphisms over $R[\varepsilon]/(\varepsilon^2)$.
These first-order lifts are used in Sections~\ref{sec:weights-and-parameters}
and~\ref{sec:integral-deformations}. For modules $M,N$ that are free over
$R$, the normalized relative bar complex has cochains
\[
 C^n_S(M,N)=\Hom_S(M\otimes_S(\Lambda/S)^{\otimes_S n},N).
\]
Here $\Hom_S$ means right $S$-linear maps. When $S=R^n$ and the vertex
components of $M$ and $\Lambda/S$ are free over $R$, the bar resolution
has projective terms and computes $\Ext^n_\Lambda(M,N)$.

\begin{lemma}
\label{lem:square-zero-selected-lifts}
Let $R$ be a field or $\Z/4$, let $\Lambda$ be a nonnegatively graded
$R$-algebra whose underlying $R$-module is finitely generated and free,
with $\Lambda_0=S=R^n$, and let
$M_1,\ldots,M_h$ be finitely generated graded right $\Lambda$-modules that
are free over $R$. Suppose that each graded module $\Hom_\Lambda(M_i,M_j)$ is
free over $R$ and concentrated in nonnegative degrees, and that
\[
 \Ext^2_\Lambda(M_i,M_i)_t=0,\qquad
 \Ext^1_\Lambda(M_i,M_j)_t=0\qquad\text{for all }i,j\text{ and all }t>0.
\]
Let $\mu\in C^2_S(\Lambda,\Lambda)$ be a normalized $S$-relative Hochschild
$2$-cocycle of internal degree $q>0$, and form the graded first-order
deformation
$\Lambda_\mu=(\Lambda\otimes_R R[\varepsilon]/(\varepsilon^2),m+\varepsilon\mu)$
where $m$ denotes the multiplication of $\Lambda$ and
$\deg\varepsilon=-q$. Then:
\begin{enumerate}[label={\upshape(\arabic*)}]
\item Each $M_i$ lifts to a graded right $\Lambda_\mu$-module $M_i^\mu$ free
over $R[\varepsilon]/(\varepsilon^2)$ (taking $M_i^\mu=e_x\Lambda_\mu$ when
$M_i=e_x\Lambda$), uniquely up to a degree-zero $\Lambda_\mu$-module
isomorphism reducing to the identity modulo $\varepsilon$.
\item Every homogeneous $\Lambda$-linear map $f:M_i\to M_j$ lifts to a
homogeneous $\Lambda_\mu$-linear map $M_i^\mu\to M_j^\mu$ of the same degree,
and for any homogeneous $R$-basis of $\Hom_\Lambda(M_i,M_j)$, any chosen
homogeneous lifts form an $R[\varepsilon]/(\varepsilon^2)$-basis of
$\Hom_{\Lambda_\mu}(M_i^\mu,M_j^\mu)$.
\item If $\Lambda$ and the $M_i$ carry a compatible grading by an abelian
group $L$ and $\mu$ is homogeneous of $L$-degree $\chi$, then assigning
$\varepsilon$ the $L$-degree $-\chi$ allows all module and morphism lifts to
be chosen homogeneous for the $L$-grading as well.
\end{enumerate}
\end{lemma}
\begin{proof}
Because the vertex components of $\Lambda$ and $M_i$ are direct summands of
finitely generated free $R$-modules, they are projective over $R$, so the
normalized $S$-relative bar resolution of $M_i$ consists of graded projective
$\Lambda$-modules and computes $\Ext^\bullet_\Lambda(M_i,M_j)$ over $R$ (including
when $R=\Z/4$). Writing a candidate lifted action on
$M_i\otimes_R R[\varepsilon]/(\varepsilon^2)$ as
$m\cdot a=ma+\varepsilon\nu_i(m,a)$ with $\nu_i\in C^1_S(M_i,M_i)_q$, the
associativity condition modulo $\varepsilon^2$ is
\[
 \nu_i(ma,b)-\nu_i(m,ab)+\nu_i(m,a)b=m\mu(a,b).
\]
Since $d\mu=0$, the right-hand side is a relative bar $2$-cocycle of internal
degree $q>0$, so the vanishing of $\Ext^2_\Lambda(M_i,M_i)_q$ guarantees the
existence of $\nu_i$. The difference of any two solutions is a relative bar
$1$-cocycle of degree $q$, which by $\Ext^1_\Lambda(M_i,M_i)_q=0$ is the
coboundary of a degree-$q$ endomorphism $h$ of $M_i$, so the two lifts are
identified by the automorphism $1+\varepsilon h$. This proves (1).

Similarly, a homogeneous map $f:M_i\to M_j$ of degree $r\geq0$ lifts to a
$\Lambda_\mu$-linear map $f+\varepsilon g$ if and only if
\[
 g(ma)-g(m)a=\nu_j(f(m),a)-f(\nu_i(m,a)).
\]
The right-hand side is a relative bar $1$-cocycle of internal degree
$r+q>0$, so the vanishing of $\Ext^1_\Lambda(M_i,M_j)_{r+q}$ provides the correction $g$.
Let $F_1,\ldots,F_s$ be homogeneous lifts of a homogeneous $R$-basis
$f_1,\ldots,f_s$ of $\Hom_\Lambda(M_i,M_j)$. Given any lifted morphism
$\Phi\in\Hom_{\Lambda_\mu}(M_i^\mu,M_j^\mu)$, its reduction modulo
$\varepsilon$ is a $\Lambda$-linear map $\sum_l a_lf_l$, so the difference
$\Phi-\sum_l a_lF_l$ has the form $\varepsilon H$ for an $R$-linear map
$H:M_i\to M_j$ that is $\Lambda$-linear because $\varepsilon^2=0$. Expanding
$H=\sum_l b_lf_l$ and using $\varepsilon F_l=\varepsilon f_l$ gives
$\Phi=\sum_l(a_l+\varepsilon b_l)F_l$, so $F_1,\ldots,F_s$ span
$\Hom_{\Lambda_\mu}(M_i^\mu,M_j^\mu)$. Moreover, if
$\sum_l(a_l+\varepsilon b_l)F_l=0$, reducing modulo $\varepsilon$ yields
$a_l=0$ for all $l$, and comparing the $\varepsilon$-coefficients of the
remaining equality $\varepsilon\sum_l b_lf_l=0$ between free
$R[\varepsilon]/(\varepsilon^2)$-modules yields $b_l=0$ for all $l$. This
proves (2), and projecting any bar-complex solution onto its homogeneous
component of the required $L$-degree gives (3).
\end{proof}