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\title{%
  Bisection Residue: Self-Similarity Obstruction\\[4pt]
  \large and the Migration of Binary Structure}
\author{L\'eon Fernando Vlegels}
\date{\today}

\theoremstyle{definition}
\newtheorem{definition}{Definition}[section]

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}{Proposition}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{conjecture}{Conjecture}[section]

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\newtheorem{remark}{Remark}[section]

\begin{document}
\maketitle

\begin{abstract}
When a hypercube cannot be bisected into two self-similar copies,
what symmetry --- if any --- survives?
We show that the obstruction relocates the binary structure rather than
eliminating it: the number $2$ migrates from its role as a self-similarity
count to the order of a residual symmetry group.
The precise condition for $\mathbb{Z}_2$ to appear as this residue is
algebraic --- a \emph{coset involution criterion} involving the stabilizer
of the bisection (Theorem~\ref{thm:criterion}).
We exhibit a compact subset of $\mathbb{R}^2$, the \emph{chiral pinwheel},
in which the minimal residual bisection symmetry is $\mathbb{Z}_4$, not
$\mathbb{Z}_2$, demonstrating that the residue depends on the geometric
structure of the bisection and not merely on the obstruction.
\end{abstract}

\tableofcontents

% -----------------------------------------------------------------------
\section{Introduction}
% -----------------------------------------------------------------------

A \emph{rep-$k$ tile} is a shape admissible for dissection into $k$
mutually congruent pieces each similar to the original.
The theory of rep-tiles, introduced by Golomb~\cite{G} and developed
extensively since~\cite{SW}, includes the basic observation that a
hypercube $[0,1]^n$ is a rep-$k$ tile only for $k = m^n$ with $m$ a
positive integer (Proposition~\ref{prop:tiling} below).
In particular, hypercubes are not rep-$2$ tiles for $n \geq 2$.

This paper investigates what replaces self-similar bisection when it is
obstructed.
The replacement condition is weaker: rather than requiring each piece to
resemble the whole, we ask only that the two pieces resemble \emph{each
other}.
This shifts the governing structure from a self-similarity equation to a
group action, and raises the question: what is the minimal symmetry group
witnessing the resulting bisection?

We call this the \emph{bisection residue}.
Our main results are:

\begin{enumerate}
  \item[(A)] A necessary and sufficient algebraic condition for the
    bisection residue to be $\mathbb{Z}_2$ --- the \emph{coset involution
    criterion} (Theorem~\ref{thm:criterion}).
  \item[(B)] The quaternion group $Q_8$ provides a minimal algebraic
    example where the residue is $\mathbb{Z}_4$
    (Theorem~\ref{thm:Q8}).
  \item[(C)] Every compact subset of $\mathbb{R}^n$ admits at least one
    bisection with $\mathbb{Z}_2$ residue
    (Theorem~\ref{thm:halfspace}).
  \item[(D)] The chiral pinwheel --- a compact subset of $\mathbb{R}^2$
    with 4-fold rotational symmetry and chiral arms --- admits a natural
    bisection whose minimal residue is $\mathbb{Z}_4$, not $\mathbb{Z}_2$
    (Theorem~\ref{thm:pinwheel}).
\end{enumerate}

Result~(D) shows that $\mathbb{Z}_2$ is not universally forced by the
self-similarity obstruction: the residue depends on the \emph{type} of
bisection available, not merely on the obstruction itself.

Throughout, $G$ denotes an isometry group acting on a metric space $X$,
and $G_A = \{g \in G : g(A) = A\}$ denotes the stabilizer of a piece $A$.

% -----------------------------------------------------------------------
\section{The Self-Similarity Obstruction}
\label{sec:obstruction}
% -----------------------------------------------------------------------

We record the obstruction to self-similar bisection.
These results are known; we include them to fix notation and make the
paper self-contained.

\begin{proposition}[Tiling count; cf.\ {\cite{SW}}]
\label{prop:tiling}
If $[0,1]^n$ is partitioned into $m$ mutually congruent axis-aligned
$n$-cubes, then $m = k^n$ for some positive integer $k$.
\end{proposition}

\begin{proof}
Each cube has side length $s$; the intervals of length $s$ along each
coordinate axis partition $[0,1]$, forcing $s = 1/k$ for some
$k \in \mathbb{Z}^+$.
The total count is then $k^n$.
\end{proof}

\begin{corollary}
\label{cor:noRep2}
$[0,1]^n$ is not a rep-$2$ tile for $n \geq 2$.
\end{corollary}

\begin{proof}
$m = 2$ requires $k = 2^{1/n}$, which is not a positive integer for
$n \geq 2$, since $1^n < 2 < 2^n$.
\end{proof}

\begin{remark}
The same argument shows that $2$ is not of the form $k^n$ for any
$n \geq 2$, because it is not a perfect $n$-th power.
The following proposition reinforces this at the metric level.
\end{remark}

\begin{proposition}[Scale irrationality]
\label{prop:irrational}
For $n \geq 2$, $\;2^{1/n}$ is irrational.
\end{proposition}

\begin{proof}
Suppose $2^{1/n} = p/q$ in lowest terms.
Then $p^n = 2q^n$, so $p^n$ is even, hence $p$ is even.
Write $p = 2r$; then $2^{n-1}r^n = q^n$, so $q^n$ is even, hence $q$
is even --- contradicting $\gcd(p,q)=1$.
\end{proof}

The two propositions attack the same root from different directions:
the counting argument says $2$ is not in the sequence $\{k^n\}$; the
metric argument says the required scaling factor $2^{-1/n}$ cannot be
rational.

% -----------------------------------------------------------------------
\section{The Residue Criterion}
\label{sec:criterion}
% -----------------------------------------------------------------------

After ruling out self-similar bisection, we relax: seek any partition
$\{A,B\}$ of $X$ such that $A$ and $B$ are isometric to each other
(not to the whole).

\begin{definition}[Bisection residue]
\label{def:residue}
The \emph{bisection residue} of a partition $\{A,B\}$ of $X$ with
$A \cong B$ is the cyclic group $\mathbb{Z}_m$, where $m$ is the
minimal order of any isometry witnessing $A \cong B$.

Let $G$ act on $X$ by isometries and let $\sigma \in G$ satisfy
$\sigma(A) = B$.
Since $\sigma$ is bijective and $B = X\setminus A$, we have
$\sigma(B) = A$, hence $\sigma^2 \in G_A$.
The full set of witnessing isometries is the coset $\sigma G_A$; the
bisection has $\mathbb{Z}_2$ residue if and only if $\sigma G_A$ contains
an element of order~$2$.

When $X \subset \mathbb{R}^n$, two choices of $G$ are natural.
Taking $G = \mathrm{Isom}(X)$ gives the \emph{intrinsic residue}, which
depends only on the symmetries of $X$ itself.
Taking $G = E(n)$, the full Euclidean group, gives the \emph{ambient
residue}, which allows isometries of $\mathbb{R}^n$ that need not
preserve $X$ as a set.
Sections~\ref{sec:criterion}--\ref{sec:Q8} work intrinsically;
Sections~\ref{sec:geometric}--\ref{sec:pinwheel} work with ambient
isometries, where $E(n)$ supplies involutions independent of the
internal symmetry of $X$.
The two notions coincide when the minimal witness happens to lie in
$\mathrm{Isom}(X)$, and can differ otherwise ---
Theorem~\ref{thm:pinwheel} exhibits both in the same example.
\end{definition}

\begin{theorem}[Coset involution criterion]
\label{thm:criterion}
The coset $\sigma G_A$ contains an element of order~$2$ if and only if
there exists $h \in G_A$ such that
\begin{equation}
  \sigma h \sigma = h^{-1}.
  \label{eq:criterion}
\end{equation}
\end{theorem}

\begin{proof}
$(\Rightarrow)$:
Suppose $(\sigma h)^2 = e$ for some $h \in G_A$.
Then $\sigma h \sigma h = e$, so $\sigma h \sigma = h^{-1}$.

$(\Leftarrow)$:
Suppose $\sigma h \sigma = h^{-1}$ for some $h \in G_A$.
Then
\[
  (\sigma h)^2
  = \sigma h \cdot \sigma h
  = (\sigma h \sigma)\cdot h
  = h^{-1}\cdot h
  = e.
\]
So $\sigma h \in \sigma G_A$ is an involution.
\end{proof}

\begin{remark}
Condition~\eqref{eq:criterion} says that $\sigma$ \emph{inverts} $h$ by
conjugation --- equivalently, $h$ and $h^{-1}$ are conjugate in $G$ via
$\sigma$.
This is a condition on the stabilizer element $h$, not on $\sigma$
itself.
The $Q_8$ example (Theorem~\ref{thm:Q8}) illustrates why the condition
can fail globally: the unique involution $-1$ of $Q_8$ is central, so
it lies in every stabilizer $G_A$ but can never be inverted by
conjugation (since $j(-1)j = -j^2 = 1 \neq (-1)^{-1} = -1$), and no
element $h \in H$ satisfies $jhj = h^{-1}$.
\end{remark}

\begin{corollary}[Trivial stabilizer forces $\mathbb{Z}_2$]
\label{cor:trivialStab}
If $G_A = \{e\}$, then $\sigma^2 \in G_A$ forces $\sigma^2 = e$.
The unique witnessing isometry $\sigma$ is itself an involution;
$\mathbb{Z}_2$ residue holds necessarily.
\end{corollary}

\begin{corollary}[Self-witnessing involutions]
\label{cor:selfWitness}
If $\sigma$ has order $2$, then taking $h = e$ gives
$\sigma e \sigma = \sigma^2 = e = e^{-1}$.
The criterion is satisfied; $\mathbb{Z}_2$ residue holds.
\end{corollary}

% -----------------------------------------------------------------------
\section{The Minimal Algebraic Counterexample}
\label{sec:Q8}
% -----------------------------------------------------------------------

Without a constraint on the stabilizer, $\mathbb{Z}_2$ residue can fail.
The quaternion group $Q_8$ provides the minimal example.

\begin{theorem}[$Q_8$ bisection has no $\mathbb{Z}_2$ witness]
\label{thm:Q8}
Let $G = Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$ act on itself by left
multiplication, $H = \langle i \rangle = \{1,-1,i,-i\}$, $\sigma = j$.
The partition $\{A = H,\; B = jH\}$ satisfies $A \cong B$ via $\sigma=j$,
which has order~$4$.
No involution witnesses $A \cong B$.
\end{theorem}

\begin{proof}
The unique element of order~$2$ in $Q_8$ is $-1$.
We verify that criterion~\eqref{eq:criterion} fails for every
$h \in H$:

\medskip
\begin{center}
\begin{tabular}{cccc}
\toprule
$h$ & $\sigma h\sigma = jhj$ & $h^{-1}$ & Equal? \\
\midrule
$1$  & $j\cdot 1\cdot j = j^2 = -1$ & $1$  & No \\
$-1$ & $j(-1)j = -j^2 = 1$          & $-1$ & No \\
$i$  & $j\cdot i\cdot j = (ji)j = (-k)j = i$ & $-i$ & No \\
$-i$ & $j(-i)j = -(jij) = -i$       & $i$  & No \\
\bottomrule
\end{tabular}
\end{center}
\medskip

The criterion fails for every $h \in H$, so $jH = \{j,-j,k,-k\}$
contains no involution.
The minimal-order witnessing isometry has order~$4$; the residue is
$\mathbb{Z}_4$.
\end{proof}

\begin{remark}
The failure requires three simultaneous conditions:
(i) the unique involution $-1$ is central, so it normalizes every
subgroup;
(ii) every subgroup of $Q_8$ contains $-1$ as the unique element of
order~$2$;
(iii) therefore $-1 \in A$ for every coset bisection, forcing $-1$ to
\emph{stabilize} $A$ rather than swap it with $B$.
If any condition fails --- if the group has non-central involutions, or
if the unique involution does not lie in every subgroup ---
$\mathbb{Z}_2$ residue is restored.
\end{remark}

% -----------------------------------------------------------------------
\section{The Geometric Program}
\label{sec:geometric}
% -----------------------------------------------------------------------

We now turn to compact subsets of $\mathbb{R}^n$, where the ambient
Euclidean group $E(n)$ provides additional resources.

\begin{theorem}[Halfspace bisections have $\mathbb{Z}_2$ residue]
\label{thm:halfspace}
Let $X \subset \mathbb{R}^n$ be compact.
Let $H$ be a hyperplane and $\{A,B\} = \{X\cap H^+,\; X\cap H^-\}$.
The reflection $\tau_H$ across $H$ restricts to an isometric bijection
$A\to B$ and has order~$2$; the bisection has $\mathbb{Z}_2$ residue.
\end{theorem}

\begin{proof}
$\tau_H:\mathbb{R}^n\to\mathbb{R}^n$ is an isometry with
$\tau_H^2 = \mathrm{id}$.
It maps $H^+$ to $H^-$, hence restricts to an isometric bijection
$A = X\cap H^+ \to X\cap H^- = B$.
\end{proof}

\begin{corollary}[Existence of $\mathbb{Z}_2$ bisection]
\label{cor:existence}
For any compact $X\subset\mathbb{R}^n$, there exists at least one
bisection with $\mathbb{Z}_2$ residue.
\end{corollary}

\begin{proof}
Since $X$ is compact, there exists a hyperplane $H$ with
$X\cap H^+\neq\emptyset$ and $X\cap H^-\neq\emptyset$.
Apply Theorem~\ref{thm:halfspace}.
\end{proof}

Theorem~\ref{thm:halfspace} uses an \emph{ambient} isometry of
$\mathbb{R}^n$ --- one that need not preserve $X$ as a set.
This is why it succeeds where $Q_8$ fails: the ambient Euclidean space
supplies involutions regardless of the intrinsic symmetry of $X$.

% -----------------------------------------------------------------------
\section{The Chiral Pinwheel: A Geometric $\mathbb{Z}_4$ Counterexample}
\label{sec:pinwheel}
% -----------------------------------------------------------------------

Theorem~\ref{thm:halfspace} guarantees $\mathbb{Z}_2$ for halfspace
bisections.
We now exhibit a compact set $X\subset\mathbb{R}^2$ admitting a
\emph{non-halfspace} bisection whose minimal residue is $\mathbb{Z}_4$.

\medskip
\noindent\textbf{Construction.}
Let $\alpha\subset\{(x,y):x>0,\,y>0\}$ be a compact connected set
satisfying:
\begin{enumerate}
  \item[(C1)] $\alpha$ is \emph{chiral}: $\mathrm{Isom}(\alpha)=\{e\}$.
  \item[(C2)] $\alpha$ lies strictly in the open northeast quadrant,
    away from the axes.
\end{enumerate}
Such $\alpha$ exists explicitly: take $\alpha$ to be the filled triangle
with vertices $(1,2)$, $(3,1)$, $(4,4)$.
Its three side lengths $\sqrt{5}$, $\sqrt{10}$, $\sqrt{13}$ are all
distinct, so $\alpha$ is scalene and admits no reflective symmetry;
combined with having no rotational symmetry, $\mathrm{Isom}(\alpha)=\{e\}$,
confirming~(C1).
All vertices lie in the open northeast quadrant, confirming~(C2).

Let $R$ denote counterclockwise rotation by $90^\circ$ about the origin.
Define
\[
  X = \alpha \cup R(\alpha) \cup R^2(\alpha) \cup R^3(\alpha),
\]
with arms $a_0=\alpha$, $a_1=R(\alpha)$, $a_2=R^2(\alpha)$,
$a_3=R^3(\alpha)$, and bisection
\[
  A = a_0\cup a_2, \qquad B = a_1\cup a_3.
\]

\begin{proposition}
\label{prop:pinwheelStructure}
\begin{enumerate}
  \item[(a)] $R(A)=B$, so $A\cong B$ via $R$, which has order~$4$.
  \item[(b)] $\mathrm{Isom}(X)=\{e,R,R^2,R^3\}\cong\mathbb{Z}_4$.
  \item[(c)] $G_A=\{e,R^2\}$.
\end{enumerate}
\end{proposition}

\begin{proof}
(a) $R(a_0)=a_1$ and $R(a_2)=a_3$, so $R(A)=B$.

(b) Any isometry of $X$ permutes the four arms, so $\mathrm{Isom}(X)$ is
finite.
Since $X$ is bounded, no non-trivial translation preserves $X$.
Glide reflections have infinite order and are therefore absent from any
finite isometry group.
Since each arm is a rotation of $\alpha$ and $\alpha$ is chiral, no
reflection preserves $X$.
The only remaining isometries are rotations; since $X$ has exactly
4-fold rotational symmetry, $\mathrm{Isom}(X)\cong\mathbb{Z}_4$.

(c) $R^2(a_0)=a_2$ and $R^2(a_2)=a_0$, so $R^2(A)=A$.
Neither $R$ nor $R^3$ fixes $A$.
\end{proof}

\begin{theorem}[Chiral pinwheel --- minimal residue is $\mathbb{Z}_4$]
\label{thm:pinwheel}
For the chiral pinwheel $X$ with bisection $\{A,B\}$ as above:
\begin{enumerate}
  \item[(i)] \textup{(Intrinsic.)} No element of $\mathrm{Isom}(X)$ of
    order~$2$ witnesses $A\cong B$.
  \item[(ii)] \textup{(Ambient.)} No element of $E(2)$ of order~$2$
    witnesses $A\cong B$.
\end{enumerate}
In particular, both the intrinsic and ambient residues of $\{A,B\}$
are~$\mathbb{Z}_4$.
\end{theorem}

\begin{proof}
\textit{(i)}
The only involution in $\mathrm{Isom}(X)\cong\mathbb{Z}_4$ is $R^2$.
By Proposition~\ref{prop:pinwheelStructure}(c), $R^2(A)=A$, so $R^2$
does not witness $A\cong B$.

\textit{(ii)}
Elements of $E(2)$ of order~$2$ are reflections across lines and
$180^\circ$ rotations about points.

\medskip
\noindent\textit{Reflections.}
Suppose $\rho_\ell(A)=B$.
Then $\rho_\ell(a_0)$ must equal one of the arms in $B$ as a subset of
$\mathbb{R}^2$ (by disjointness and connectivity of the arms).

If $\rho_\ell(a_0) = a_1 = R(a_0)$, then $\rho_\ell\circ R^{-1}$ maps
$a_0$ to $a_0$ isometrically, so $\rho_\ell\circ R^{-1}\in
\mathrm{Isom}(a_0)=\{e\}$ by~(C1), giving $\rho_\ell = R$.
But $\rho_\ell$ is orientation-reversing and $R$ is
orientation-preserving --- contradiction.

If $\rho_\ell(a_0) = a_3 = R^3(a_0)$, then $\rho_\ell\circ R^{-3}\in
\mathrm{Isom}(a_0)=\{e\}$, so $\rho_\ell = R^3$.
Again $\rho_\ell$ is orientation-reversing and $R^3$ is
orientation-preserving --- contradiction.

In both cases we reach a contradiction; no reflection maps $A$ to $B$.

\medskip
\noindent\textit{$180^\circ$ rotations.}
Let $\rho_c$ denote the $180^\circ$ rotation about $c\in\mathbb{R}^2$.
Suppose $\rho_c(A)=B$; then $\rho_c(a_0)\in\{a_1,a_3\}$.

If $\rho_c(a_0) = a_1 = R(a_0)$, then $\rho_c\circ R^{-1}\in
\mathrm{Isom}(a_0)=\{e\}$ by~(C1), so $\rho_c = R$.
But $R$ has order $4\neq 2$, contradicting $\rho_c$ being a
$180^\circ$ rotation.

If $\rho_c(a_0) = a_3 = R^3(a_0)$, then $\rho_c\circ R^{-3}\in
\mathrm{Isom}(a_0)=\{e\}$, so $\rho_c = R^3$, again order
$4\neq 2$.
Contradiction.

Since no order-$2$ element of $E(2)$ witnesses $A\cong B$, and $R$
(order~$4$) does, the minimal residue is $\mathbb{Z}_4$.
\end{proof}

\begin{remark}
The proof of part~(ii) uses condition~(C1) essentially: if $\alpha$ had
nontrivial isometries, the composition $\rho_c\circ R^{-1}$ could be a
nontrivial element of $\mathrm{Isom}(\alpha)$ rather than the identity,
and the contradiction would not follow.
Chirality is not merely a convenience --- it is the property that forces
the argument through.
\end{remark}

% -----------------------------------------------------------------------
\section{Discussion}
\label{sec:discussion}
% -----------------------------------------------------------------------

\subsection{The three-level structure}

The results naturally separate into three regimes, summarised in
Table~\ref{tab:residues}.

\begin{center}
\begin{longtable}{lll}
\caption{Obstruction type and bisection residue by setting}
\label{tab:residues}\\
\toprule
Setting & Obstruction type & Minimal bisection residue \\
\midrule
\endfirsthead
\toprule
Setting & Obstruction type & Minimal bisection residue \\
\midrule
\endhead
$[0,1]^n$, $n\geq 2$
  & Dimensional arithmetic (rep-tile)
  & $\mathbb{Z}_2$ (halfspace bisection) \\
$Q_8$ coset bisection
  & Algebraic (central involution in every subgroup)
  & $\mathbb{Z}_4$ \\
Chiral pinwheel
  & Geometric (chirality $+$ 4-fold rotation)
  & $\mathbb{Z}_4$ \\
\bottomrule
\end{longtable}
\end{center}

The common algebraic mechanism in the two $\mathbb{Z}_4$ cases is
Theorem~\ref{thm:criterion}: the witnessing isometry $\sigma$ fails to
invert any element of the stabilizer $G_A$ by conjugation.

\subsection{Migration of binary structure}

In the self-similar setting, the number $2$ appears as the \emph{count}
of pieces.
In the residue setting, $2$ appears as the \emph{order} of the minimal
witnessing symmetry.
The self-similarity obstruction relocates $2$ between categorical
roles --- from cardinality to group order --- rather than eliminating it.
The chiral pinwheel shows that this relocation is not automatic: the
value that the binary structure migrates \emph{to} depends on the
geometry of the available bisections.

\subsection{Open problem}

\begin{conjecture}[Order characterisation of bisection residue]
\label{conj:order}
Let $X\subset\mathbb{R}^n$ be compact and $\{A,B\}$ a bisection of $X$
with $A\cong B$.
Define
\[
  k(A,B)
  := \min\bigl\{k\in\mathbb{Z}^+
     : \exists\,\varphi\in E(n)\text{ with }
       \mathrm{ord}(\varphi)=2k
       \text{ and }\varphi(A)=B\bigr\}.
\]
Then the bisection residue of $\{A,B\}$ is $\mathbb{Z}_{2k(A,B)}$.
\end{conjecture}

Theorems~\ref{thm:halfspace} and~\ref{thm:pinwheel} confirm this
conjecture in the two cases studied: $k(A,B)=1$ for every halfspace
bisection (ambient reflection witnesses, giving $\mathbb{Z}_2$ residue),
and $k(A,B)=2$ for the chiral pinwheel bisection ($R$ witnesses with
order~$4$, no order-$2$ element does, giving $\mathbb{Z}_4$ residue).
The general case --- whether the order of the minimal ambient witness
always determines the residue --- remains open.

% -----------------------------------------------------------------------
\begin{thebibliography}{9}

\bibitem{G}
S.~W.~Golomb,
``Replicating figures in the plane,''
\textit{Mathematical Gazette} \textbf{48} (1964), 403--412.

\bibitem{SW}
S.~L.~Snover, C.~Waiveris, J.~K.~Williams,
``Rep-tiling for triangles,''
\textit{Discrete Mathematics} \textbf{91} (1991), 193--200.

\bibitem{K}
J.~Kyn\v{c}l and Z.~Pat\'akov\'a,
``On the nonexistence of $k$-reptile simplices
in $\mathbb{R}^3$ and $\mathbb{R}^4$,''
\textit{Electronic Journal of Combinatorics}
\textbf{24}(3) (2017), P3.1.

\end{thebibliography}

\end{document}
