Agent-assistance disclosure: this work was prepared with AI-agent assistance (drafting, code, experiments) under human direction. The authors, writing as Third Projection Collective, take responsibility for its content. Open proofs are marked [OPEN] in the text.
\begin{document}\maketitle
\begin{abstract}
A recent proposal conjectures that a failure of gluing for verified results localizes a lying coalition''~\cite{fut}. We make part of this precise. (1) For certified values over quorum covers, the sheaf condition holds iff $n\ge 3f+1$. This is a restatement of Byzantine quorum intersection~\cite{mr98}; we checked it exhaustively for $n\le 10$. (2) When local results agree only up to a renaming, so that overlaps carry $\mathbb Z/2$ transition witnesses, every set-level (0-truncated) check accepts. The stack-level check rejects exactly when the observed witnesses are not a coboundary. A lying witness set is undetectable iff it is a cut of the overlap graph. (3) Localization is limited. In simulations (200 random overlap graphs per setting), the minimum-weight explanation recovered the exact liars in 36.5\% to 70.5\% of single-liar runs and in 0\% to 20.5\% of multi-liar runs. A general localization theorem, and a full protocol model with signatures and timing, remain \emph{open}.
\end{abstract}
\section{Setting}
There are $n$ nodes, of which at most $f$ are Byzantine, and quorums are the sets of size $n-f$. A \emph{certificate} for slot $s$ is a value signed by every member of some quorum, and honest nodes sign at most one value per slot. Let $\mathcal C(U)$ be the set of values certified by quorums contained in $U$.
\begin{proposition}\label{p:sheaf}
Two certificates for the same slot always agree iff any two quorums share an honest node. For $|q|=n-f$ this holds iff $n\ge 3f+1$. In that case, $\mathcal C$ restricted to the cover by quorums satisfies separation and gluing for each slot.
\end{proposition}
\begin{proof}
We have $|q\cap q'|\ge n-2f$, with equality attained, so the intersection is guaranteed to contain an honest node iff $n-2f\ge f+1$. If $n\le 3f$, place $f$ Byzantine nodes in an intersection of size at most $f$; they sign both values, which gives two conflicting local sections that do not glue. Our script \texttt{sim.py} checks the minimum intersection over all pairs of quorums for all $2\le n\le 10$ and $1\le f\le n/2$ (25 cases), and the criterion agrees in every case.
\end{proof}
This is the classical quorum intersection condition~\cite{mr98,cl02}, in sheaf language as in~\cite{ab11}. We claim no novelty for it.
\section{Renaming witnesses and a 1-stack} \begin{definition} Let $G=(V,E)$ be the overlap graph of regions. Each region holds a local result that is well defined up to a $\mathbb Z/2$ renaming (for instance a choice of orientation or labelling), and each overlap $e$ carries a witness $x_e\in\mathbb Z/2$ produced by one node. Honest witnesses are the true transitions, which we normalize to $0$; a liar may report $1$. A global section is a potential $p:V\to\mathbb Z/2$ with $p(u)+p(v)=x_{uv}$ for every edge. \end{definition} \begin{proposition}\label{p:stack} (i) The 0-truncation, which passes to renaming classes, forgets every witness, so the set-level gluing check always succeeds. (ii) A global section exists iff $x$ is a coboundary, that is, iff $x$ vanishes on the cycle space $H_1(G;\mathbb Z/2)$. (iii) For a liar set $L$ with $x=\mathbf 1_L$, the lie is undetected iff $\mathbf 1_L$ is a cut. In particular a single liar is detected iff its edge lies on a cycle. \end{proposition} \begin{proof} (i) Renaming classes are singletons, so every family of classes glues. (ii) This is standard: integrate $x$ along a spanning tree, and the result is consistent iff $x$ sums to $0$ around each fundamental cycle. (iii) Coboundaries of $\mathbb Z/2$ potentials are exactly the cuts $\delta S$, and an edge lies in some cut $\delta S$ with $\delta S=\{e\}$ iff it is a bridge. \end{proof} So the stack sees what the sheaf of renaming classes cannot. The syndrome is the restriction of $x$ to $H_1$, so at most $\beta_1(G)$ bits are available to locate liars.
\section{How well does the syndrome localize liars?} Decoding the syndrome means choosing a minimum-weight $y$ with $x+y$ a coboundary. This is the decoding problem of the graph's cycle code. \texttt{sim.py} draws a random spanning tree plus extra edges (so $\beta_1$ is 3 or 6), picks random liars, and decodes by brute force (seed 0, 200 graphs per row): \begin{center}\begin{tabular}{rrrrr} $|V|$&$\beta_1$&liars&detected&exact unique recovery\\\hline 6&3&1&187&73\\6&3&2&176&0\\8&6&1&195&141\\8&6&2&195&41\\8&6&3&199&0 \end{tabular}\end{center} Detection is near-universal, but exact localization is rare once there is more than one liar. The site's conjecture is therefore false as a general claim: failure to glue localizes a coalition only up to the coset $L+\{\text{cuts}\}$. That this coset is the exact limit of localization is Proposition~\ref{p:stack}(iii), which we prove; the table only quantifies how often the minimum-weight member of the coset is the true $L$. \todo{OPEN: conditions on $G$ (e.g.\ girth, expansion) that guarantee unique recovery for $|L|\le t$, using cycle-code distance; a full protocol model with signatures, asynchrony and timing; non-abelian renaming groups and $H^1$ with non-abelian coefficients.} \section*{Disclosure}Drafted with AI-agent assistance; the authors (Third Projection Collective) take responsibility for the content. All numbers come from \texttt{code/sim.py} (\texttt{results.json}). \bibliographystyle{eptcs}\bibliography{refs} \end{document}