% main.tex -- Sequential Products on Euclidean Jordan Algebras:
%             Classification in Rank at Least Three and the Complex Qubit
%% VENUE: JPA (J. Phys. A) first, double-anonymous; JMP second. Standalone operational-algebra paper.
% ASSERT_CONVENTION: sequential_product=a.b (dot notation), standard/Lueders
%   product=a bullet b = Q_{sqrt a} b, jordan_product=a circ b, jordan_mult_op=T_x,
%   quadratic_rep=Q_x, axiom_source=arXiv:1803.11139 Definition 2 EXCLUSIVELY,
%   cited machinery=arXiv:1803.08453 (Wetering2018three) Section 5 + EJA appendix,
%   order-iso=>Jordan-iso via arXiv:1904.09278 (van Imhoff-Roelands) Cor 2.5.

\documentclass[12pt]{article}
\usepackage[margin=1in]{geometry}
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% Standard (Lueders) reference product a . b = Q_{sqrt a} b, written with \bullet
% to distinguish it from the unknown/general sequential product a . b (\seqp).
\newcommand{\lprod}[2]{#1 \mathbin{\bullet} #2}

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\ifanonymous\hypersetup{pdfauthor={}}\else\hypersetup{pdfauthor={Bryan Ehrlich}}\fi

\title{Sequential Products on Euclidean Jordan Algebras:\\
Classification in Rank at Least Three and the Complex Qubit}

\ifanonymous
  \author{}
\else
  \author{Bryan Ehrlich\\[4pt]
    \normalsize Independent researcher, Trumbull, Connecticut 06611, USA\\
    \normalsize ehrlich.bryan@gmail.com}
\fi
\date{}

\begin{document}

\maketitle

\begin{abstract}
A sequential product takes an ordered pair of effects $(a, b)$ to a new effect
$\seqp{a}{b}$, read as ``measure $a$, then $b$,'' subject to seven standard
axioms, one of them norm-continuity in the first argument.  We classify these
products on every finite-dimensional Euclidean Jordan algebra whose rank-two
summands are real or complex qubits.  The classification is typewise, because a
sequential product cannot couple distinct simple summands.  On real
$\MnKsa{n}{\mathbb{R}}$ ($n \geq 2$), quaternionic $\MnKsa{n}{\mathbb{H}}$
($n \geq 3$), exceptional $H_3(\mathbb{O})$, and rank-one summands, the
L\"uders product $\lprod{a}{b} = Q_{\sqrt a}\,b$ is forced; here $Q$ is the
quadratic representation, so on the matrix types this reads
$\sqrt{a}\,b\,\sqrt{a}$.  On each complex summand $\Mnsa{n}$ ($n \geq 3$) the
products are instead exactly Liu--Wu's twisted family
$a \circ_t b = a^{1/2+it}\,b\,a^{1/2-it}$, one for each real $t$.  Its members
agree with L\"uders on sharp first effects and on commuting pairs, and differ
from it by an outcome-conditioned unitary back-action, detectable
interferometrically.  The complex qubit $\Mnsa{2}$ carries a function space
$C(\mathbb{RP}^2, \mathbb{R})$ of products, in place of the single parameter
that survives from rank three on.  The remaining rank-two spin factors
are left open.  Adding trace-form symmetry to these axioms forces the
L\"uders product on every simple type of rank $\geq 3$, and we compute the
freedom such selectors remove.  That selector
is van de Wetering's inner-product symmetry, and the
Jordan form of Gudder--Latr\'emoli\`ere's trace duality.  The proof runs through van de
Wetering's comparison map and a frame-stabilizer argument on each type.
\end{abstract}


%% =====================================================================
%%  SECTION 1: INTRODUCTION
%% =====================================================================
\section{Introduction}
\label{sec:intro}

How far the operational axioms of \Cref{def:sp} determine a sequential product
is a question going back to Gudder's open-problems list for sequential effect
algebras~\cite{Gudder2005}.  Gudder and
Latr\'emoli\`ere~\cite{GudderLatremoliere2008} proved that five conditions, one
of which is a trace duality, characterize the L\"uders product
$\lprod{a}{b} = \sqrt{a}\,b\,\sqrt{a}$ on Hilbert-space effects.  Without the
trace duality, the other four conditions no longer determine the product.  Liu
and Wu~\cite[Rem.~1]{LiuWu2009} exhibited the
one-parameter twisted family
\begin{equation}\label{eq:twisted-intro}
  a \circ_t b = a^{1/2+it}\, b\, a^{1/2-it},
  \qquad t \in \mathbb{R}.
\end{equation}
Every member of this family satisfies all seven standard axioms and agrees with
the L\"uders product whenever the first argument is a projection.  The algebraic
clauses are checked in Liu and Wu's own proof and in
Shen--Wu~\cite[Thm.~3.1, Ex.~3.1]{ShenWu2009}; norm-continuity \ref{ax:S2}
follows from continuity of the functional calculus.

Van de Wetering~\cite{vandeWetering2019b} showed every finite-dimensional
sequential-product space is order-isomorphic to a Euclidean Jordan algebra;
Euclidean Jordan algebras are also the arena of the operational
reconstruction program for quantum theory~\cite{BarnumGraydonWilce2020}.
The question is then how much freedom the axioms leave on a simple one.

The three closest results stop short of answering it.  Liu--Wu~\cite{LiuWu2009}
establish the \emph{existence} of the constant-twist
family~\eqref{eq:twisted-intro}.
Shen--Wu~\cite{ShenWu2009} classify the products of a functional-calculus
ansatz, under axioms not including~\ref{ax:S2} and permitting arbitrary frame
dependence at rank two.  Van de Wetering~\cite{Wetering2018three}, on the
complex Hilbert-space type and for simple invertible first arguments, reduces
the residual comparison automorphism to conjugation by a bicommutant unitary
$u_a$.  His own uniqueness theorems reach every type, but each needs an
added selector on the product: invariance, inner-product symmetry, or
invertibility preservation.  None of the three results classifies the
admissible dependence $a \mapsto u_a$ under the bare axioms, without an ansatz
or an added hypothesis on the product, and none settles the
residual freedom across the four simple Euclidean Jordan types of rank at
least three
(\Cref{sec:priorart}).  \Cref{mthm:master} supplies that classification,
and \Cref{cor:qubit-classification} identifies the sharp
rank-two failure in the complex series as $C(\mathbb{RP}^2, \mathbb{R})$.
The freedom the seven axioms leave in a finite-dimensional effect update is
therefore at most an outcome-conditioned unitary back-action on each simple type
of rank at least three.  Where that back-action survives, it shifts sequential
measurement probabilities in an explicit single-qubit interferometer
(\Cref{rem:twist-detection}).

\begin{maintheorem}[Sequential-product classification, rank $\geq 3$]
\label{mthm:master}
  Let $J$ be a finite-dimensional simple Euclidean Jordan algebra of
  rank $n \geq 3$, and let $(\cdot)$ be any sequential product on its
  effects---any binary operation
  $\eff{J} \times \eff{J} \to \eff{J}$ satisfying the seven axioms
  \ref{ax:S1}--\ref{ax:S7} of \Cref{def:sp}, including first-variable
  norm-continuity \ref{ax:S2}.  Then:
  \begin{itemize}[nosep]
    \item if $J$ is of real type $\MnKsa{n}{\mathbb{R}}$, quaternionic
      type $\MnKsa{n}{\mathbb{H}}$, or exceptional type
      $H_3(\mathbb{O})$, the product is the L\"uders product
      $\seqp{a}{b} = Q_{\sqrt a}\,b$;
    \item if $J$ is of complex type $H_n(\mathbb{C}) = \Mnsa{n}$, there
      is a unique $t \in \mathbb{R}$ with
      $\seqp{a}{b} = a^{1/2+it}\,b\,a^{1/2-it}$ for all effects $a, b$
      (with $0^{1/2\pm it} := 0$).
  \end{itemize}
  The complex-type parameter $t$ is stated for the fixed labelled algebra
  $\Mnsa{n}$, on which it is unique (\Cref{rem:moduli}); quotienting by all
  Jordan automorphisms further identifies $t$ with $-t$ through the transpose.
  Beyond the standing structural hypotheses (finite dimension, simplicity,
  rank $\geq 3$), the seven axioms are the only constraints imposed on the
  product: Peirce-block-diagonality of
  the update $a \mapsto \seqp{a}{(\cdot)}$ is derived, not assumed through any
  pair-locality, decomposition-independence, or functional-calculus
  template (\Cref{sec:priorart}).
\end{maintheorem}

The threshold is sharp in the complex series: \Cref{thm:qubit-boundary}
exhibits a continuous, frame-dependent product on $\Mnsa{2}$ satisfying all seven
axioms and conjugate to no constant-twist member
of~\eqref{eq:twisted-intro}---a continuous specialization of Shen--Wu's
construction~\cite{ShenWu2009}, here verified against the full axiom list
including \ref{ax:S2}.  (The real qubit is already rigid by
\Cref{prop:real}, and the complex qubit is classified completely: its
norm-continuous \ref{ax:S1}--\ref{ax:S7} products are parametrized bijectively
by $C(\mathbb{RP}^2, \mathbb{R})$, one continuous twist value per unordered
frame (\Cref{cor:qubit-classification}).  The
rank-two spin factors $\mathbb{R} \oplus \mathbb{R}^k$ beyond the real and
complex qubits, including $H_2(\mathbb{H})$ and $H_2(\mathbb{O})$, remain
unsettled.  Sharpness of the rank-three threshold is therefore witnessed within
the complex series, not by a classification of every rank-two type.)  The mechanism
at the threshold is frame overlap: globalizing the per-frame
parameter compares two
overlapping frames on the cross-coherence space of a shared third atom
(\Cref{lem:orientation,lem:adjacent}), and a qubit frame has no third atom to
share, so the
per-frame parameter never globalizes.  The four
coordinate types exhaust the simple algebras of rank $\geq 3$, so
\Cref{mthm:master} is a complete typewise classification there.  A general
sequential-product space is a direct sum of simple algebras
(\Cref{thm:vdw1}), and every product decomposes over that sum
(\Cref{prop:central})---distinct summands cannot couple---so the simple
rank-$\geq 3$ case is the whole content apart from the rank-$\leq 2$
summands.  The one rank-one simple algebra $J = \mathbb{R}$ is
immediate: \Cref{lem:homog}(i) gives a positive linear $L_a : \mathbb{R} \to
\mathbb{R}$ and \Cref{lem:aone} forces $L_a(\id) = a$, so $\seqp{a}{b} = ab$ is
L\"uders.  Within finite-dimensional Euclidean Jordan algebras under
\ref{ax:S1}--\ref{ax:S7}, only one classification problem then remains: the
other rank-two spin factors.  Assembling the pieces gives the following
classification.

\begin{maintheorem}[Finite-dimensional classification with resolved rank-two
  summands]
\label{mthm:omnibus}
  Let $J = \bigoplus_{\alpha=1}^{m}J_\alpha$ be a finite-dimensional Euclidean
  Jordan algebra, decomposed into simple summands, in which every rank-two
  summand is isomorphic to the real qubit $\MnKsa{2}{\mathbb{R}}$ or the
  complex qubit $\Mnsa{2}$, and let $(\cdot)$ satisfy
  \ref{ax:S1}--\ref{ax:S7}.  Fix an isomorphism of each summand onto its model
  algebra in the Jordan--von Neumann--Wigner list, and read $J_\alpha$ as that
  model.  Then $(\cdot)$ decomposes over the summands,
  $\seqp{a}{b} = \sum_\alpha \seqp{a_\alpha}{b_\alpha}$, and on each summand
  it is:
  \begin{itemize}[nosep]
    \item the L\"uders product, if $J_\alpha$ is $\mathbb{R}$,
      $\MnKsa{k}{\mathbb{R}}$ ($k \geq 2$), $\MnKsa{k}{\mathbb{H}}$
      ($k \geq 3$), or $H_3(\mathbb{O})$;
    \item $\seqp{a}{b} = a^{1/2+i\,t_\alpha}\,b\,a^{1/2-i\,t_\alpha}$ for a unique
      $t_\alpha \in \mathbb{R}$, if $J_\alpha = H_k(\mathbb{C})$ with
      $k \geq 3$;
    \item the frame-dependent product of \Cref{prop:n2-sufficiency} for a
      unique continuous $t_\alpha : \mathbb{RP}^2 \to \mathbb{R}$, if
      $J_\alpha = \Mnsa{2}$.
  \end{itemize}
  Conversely every independent choice of these parameters defines such a
  product, so relative to those identifications there is a bijection of sets
  \[
    \mathcal{M}_{\mathrm{SP}}(J)
    \;\cong\;
    \prod_{\alpha \,:\; J_\alpha \textnormal{ complex of rank}\,{\geq}\,3}
      \mathbb{R}
    \;\;\times\!\!\!
    \prod_{\alpha \,:\; J_\alpha \,=\, \Mnsa{2}}
      C(\mathbb{RP}^2, \mathbb{R}),
  \]
  every other summand contributing a singleton.  Here
  $\mathcal{M}_{\mathrm{SP}}(J)$ denotes the set of norm-continuous
  \ref{ax:S1}--\ref{ax:S7} sequential products on the fixed (labelled) effect
  algebra of $J$ (\Cref{rem:moduli}).
\end{maintheorem}

\begin{proof}
The decomposition, and the componentwise assembly giving the converse, are the
two halves of \Cref{prop:central}; the rank-$\geq 3$ summands are
\Cref{mthm:master}; the
rank-one summand $\mathbb{R}$ is the computation just given; the real qubit is
\Cref{prop:real} at $n = 2$; the complex qubit is
\Cref{cor:qubit-classification}.  Uniqueness of each parameter on its labelled
summand is \Cref{rem:moduli} in the first case and the injectivity half of
\Cref{cor:qubit-classification} in the second.  The converse also needs each
listed value to be attained, so that the displayed sets are exact rather than
upper bounds.  The L\"uders product is a norm-continuous \ref{ax:S1}--\ref{ax:S7}
product on every summand (\Cref{lem:simple-bridge}(i), with \ref{ax:S2} from
continuity of the functional calculus), so the singleton factors are nonempty.
Every $t_\alpha \in \mathbb{R}$ is attained by \Cref{lem:twist-sufficiency},
and every continuous $t_\alpha$ on $\mathbb{RP}^2$ by
\Cref{prop:n2-sufficiency}.
\end{proof}

\paragraph{The complex dial as unitary back-action.}
The parameter $t$ has an operational reading.  Write $s(a)$ for the
support projection and
$U^{t}_a := (\id - s(a)) + \sum_{\lambda > 0}\lambda^{it}p_\lambda$ for the
unitary from the \emph{support} logarithm of $a$, so that $U^t_a = a^{it}$ when
$a$ is invertible.  Then $a^{1/2+it} = \sqrt{a}\,U^t_a$ and
\begin{equation}\label{eq:backaction-intro}
  a \circ_t b
  = \sqrt{a}\,\bigl(U^t_a\, b\, (U^t_a)^{*}\bigr)\,\sqrt{a}
  = \Ad_{U^t_a}\!\bigl(\lprod{a}{b}\bigr):
\end{equation}
on effects (Heisenberg picture, $\Ad_U(x) := UxU^{*}$) the complex-type
product is a L\"uders update followed by conjugation by the
outcome-conditioned unitary $U^t_a$.  Dually, the unnormalized state operation is
$\rho \mapsto (U^t_a)^{*}(\sqrt{a}\,\rho\,\sqrt{a})\,U^t_a$: the L\"uders
state followed by the opposite unitary $a^{-it}$.  The resulting operator has
trace $\operatorname{tr}(a\rho)$, and conditioned on the outcome the normalized
posterior divides by $\operatorname{tr}(a\rho)$.  \Cref{rem:twist-detection}
gives the interferometric signature and the invisibility to sharp first
effects.  Throughout, the object classified is this effect-level product
$(a, b) \mapsto \seqp{a}{b}$, not a quantum instrument; the
sequential-product/instrument boundary is discussed in \Cref{sec:priorart}.

\paragraph{Proof inputs and provenance.}
Van de Wetering's uniqueness section~\cite[\S 5]{Wetering2018three} supplies,
for each invertible effect $a$, the comparison map
$\Theta_a := Q_{\sqrt a}^{-1}L_a$, a unital linear order isomorphism fixing the \emph{effects} that
operator-commute with $a$; extending that fixing to the full operator
commutant is a span argument given here (\Cref{sec:machinery}).  Two
ingredients of the proof are, to our knowledge, not in the prior literature.
The first is the verification that this comparison machinery applies
without a normality hypothesis in the finite-dimensional EJA setting (through
finite spectrality, \Cref{lem:simple-bridge}).  The second is the
frame-stabilizer argument (\Cref{sec:common,sec:branches}), which eliminates the
parameter on the real, quaternionic, and exceptional types and leaves one scalar
direction on the complex torus.  The upgrade of $\Theta_a$ to a Jordan automorphism is
itself imported (the van Imhoff--Roelands order-isomorphism
theorem~\cite[Cor.~2.5]{vanImhoffRoelands2019}); invoking it in place of the
Hilbert-space bicommutant-unitary reduction allows the argument to reach the
exceptional type.  \Cref{sec:priorart} positions the result against prior
art, in particular against van de Wetering's Hilbert-space reduction of the
comparison automorphism to a bicommutant unitary.  For the complex type in finite dimension and rank $\geq 3$,
\Cref{mthm:master} pins that residual dependence $a \mapsto \Ad_{u_a}$ to
$\Ad_{a^{it}}$ for one global $t$.  The Supplementary
Material collects the full input ledger.


%% =====================================================================
%%  SECTION 2: PRELIMINARIES
%% =====================================================================
\section{Preliminaries}
\label{sec:prelim}

\subsection{Euclidean Jordan algebras and the Peirce decomposition}
\label{sec:eja}

A \emph{Euclidean Jordan algebra} (EJA) is a finite-dimensional real
commutative (not necessarily associative) algebra $(J, \circ)$ with unit $e$,
equipped with an associative inner product
$\langle x, y\rangle = \operatorname{tr}(x \circ y)$, satisfying the
Jordan identity $x \circ (x^2 \circ y) = x^2 \circ (x \circ y)$.  Here
$\operatorname{tr}$ is the Jordan trace, the sum of eigenvalues in any spectral
resolution; on the matrix types it is the ordinary, unnormalized matrix trace.
The interior
$\Omega = \operatorname{int}(J^+)$ of the cone $J^+$ of squares is a
symmetric cone.  We follow Faraut and
Kor\'anyi~\cite{FarautKoranyi1994}.  The \emph{effect space} is
$\eff{J} = \{a : 0 \leq a \leq e\}$.  For $x \in J$ the multiplication
operator is $T_x(y) = x \circ y$ and the quadratic representation is
$Q_x = 2T_x^2 - T_{x^2}$; for invertible $a \in \Omega$, $Q_{\sqrt a}$ is
a cone automorphism~\cite[Prop.~III.2.2]{FarautKoranyi1994}.  A
\emph{Jordan frame} is a complete system of orthogonal primitive
idempotents $F = \{p_1, \ldots, p_n\}$, $\sum_i p_i = e$; $n$ is the rank
of $J$.  Two elements \emph{operator commute} if $[T_x, T_y] = 0$,
equivalently if they are simultaneously diagonalizable in a common Jordan
frame.

The Peirce decomposition relative to $F$ is
\begin{equation}\label{eq:peirce}
  J = \bigoplus_i \mathbb{R}p_i \;\oplus\; \bigoplus_{i<j} V_{ij},
  \qquad
  V_{ij} = \{x : p_i \circ x = \tfrac12 x = p_j \circ x\},
\end{equation}
with $\dim V_{ij} = d$ independent of $i < j$ when $J$ is simple, the Peirce
multiplicity, equal to
$1, 2, 4, 8$ for $\MnKsa{n}{\mathbb{R}}$, $H_n(\mathbb{C})$,
$\MnKsa{n}{\mathbb{H}}$, $H_3(\mathbb{O})$
respectively~\cite{FarautKoranyi1994}.  (We write $H_n(\mathbb{K})$ and
$\MnKsa{n}{\mathbb{K}}$ interchangeably for the Hermitian $n \times n$
matrices over $\mathbb{K}$ with the symmetrized product.)  For a single idempotent $q$,
$J = J_2(q) \oplus J_1(q) \oplus J_0(q)$ with
$J_k(q) = \{x : q \circ x = \tfrac{k}{2}x\}$; here $J_2(q)$ and $J_0(q)$
are EJAs with units $q$ and $e - q$, and their elements operator commute
across the pair.  All Jordan frames of a simple
EJA are conjugate under $\Aut(J)$~\cite[Thm.~IV.2.5]{FarautKoranyi1994}.
By the Jordan spectral theorem every $a \in J$ has a spectral resolution
$a = \sum_i \lambda_i p_i$ into a Jordan frame; every effect is therefore
\emph{simple} (a finite real combination of orthogonal idempotents; the term
applies to elements, not to $J$).

The standard (L\"uders) product on $J$ is
$\lprod{a}{b} = Q_{\sqrt a}\,b$; on $\Mnsa{n}$ it is
$\sqrt{a}\,b\,\sqrt{a}$.  Three conventions stand throughout.  First, $e$,
$\id$ and $I$ all
denote the unit.  Second, for an effect with
spectral resolution $a = \sum_k \lambda_k p_k$, the \emph{support projection} is
$\lceil a\rceil := \sum_{\lambda_k > 0} p_k$ (also written $s(a)$), equal to
$e$ exactly when $a$ is invertible.  Third, every functional-calculus expression
takes the value $0$ at eigenvalue $0$, so in particular
$0^{1/2 \pm it} := 0$.


\subsection{Sequential products}
\label{sec:sp-axioms}

Following Gudder--Greechie~\cite{GudderGreechie2002} and van de
Wetering~\cite{vandeWetering2019b}:

\begin{definition}[Sequential product space,
  {\cite[Def.~2]{vandeWetering2019b}}]
\label{def:sp}
  Let $(V, \leq, \id)$ be an order unit space with a binary operation
  $(\cdot) : \eff{V} \times \eff{V} \to \eff{V}$.  Write
  $a \compatible b$ when $\seqp{a}{b} = \seqp{b}{a}$.  Then $(V, \cdot)$
  is a \emph{sequential product space} if:
  \begin{enumerate}[label=\textup{(S\arabic*)},leftmargin=3em]
    \item\label{ax:S1} \textbf{Additivity:}
      $\seqp{a}{(b+c)} = \seqp{a}{b} + \seqp{a}{c}$ whenever
      $b + c \leq \id$.
    \item\label{ax:S2} \textbf{Continuity:}
      $a \mapsto \seqp{a}{b}$ is continuous in the order unit norm.
    \item\label{ax:S3} \textbf{Unitality:}
      $\seqp{\id}{a} = a$ for all $a$.
    \item\label{ax:S4} \textbf{Symmetry of orthogonality:}
      $\seqp{a}{b} = 0 \Rightarrow \seqp{b}{a} = 0$.
    \item\label{ax:S5} \textbf{Associativity of compatible effects:}
      $a \compatible b \Rightarrow
      \seqp{a}{(\seqp{b}{c})} = \seqp{(\seqp{a}{b})}{c}$.
    \item\label{ax:S6} \textbf{Additivity of compatibility:}
      $a \compatible b \Rightarrow a \compatible (\id - b)$; if also
      $a \compatible c$ then $a \compatible (b + c)$ when
      $b + c \leq \id$.
    \item\label{ax:S7} \textbf{Multiplicativity of compatibility:}
      $a \compatible b$ and $a \compatible c \Rightarrow
      a \compatible \seqp{b}{c}$.
  \end{enumerate}
\end{definition}

These are van de Wetering's Definition~2~\cite{vandeWetering2019b}
restated, under lightly reworded clause names, with the domain riders made explicit;
\ref{ax:S2} is his norm-continuity in the first argument, which he adds to the
sequential-effect-algebra list and flags as new.  His structure
theorem places the classification on Euclidean Jordan algebras.

\begin{theorem}[{\cite[Thm.~1]{vandeWetering2019b}}]\label{thm:vdw1}
  A finite-dimensional sequential product space is order-isomorphic to a
  Euclidean Jordan algebra.
\end{theorem}

Thus any finite-dimensional order unit space carrying an operation satisfying
\ref{ax:S1}--\ref{ax:S7} is (order-isomorphic to) a direct sum of simple
Euclidean Jordan algebras,\footnote{The identification of \Cref{thm:vdw1} is
unital---van de Wetering's Jordan product takes the order unit as its Jordan
unit---so the unit-relative axioms \ref{ax:S1}--\ref{ax:S7} and the effect
interval transport to each summand, as \Cref{prop:central} requires.} over
which the product decomposes with no coupling between summands
(\Cref{prop:central}; the scope consequences are drawn in
\Cref{sec:intro}).  The
Jordan--von Neumann--Wigner classification~\cite{JordanVonNeumannWigner1934}
lists the simple types; those of rank $\geq 3$ are
$\MnKsa{n}{\mathbb{R}}$, $H_n(\mathbb{C})$, $\MnKsa{n}{\mathbb{H}}$, and
$H_3(\mathbb{O})$.

\begin{lemma}[Effects span; linear extension]\label{lem:span}
In an order unit space, $\eff{V}$ contains the closed ball of radius
$\tfrac12$ about $\tfrac12\id$, hence spans $V$; two linear maps agreeing
on $\eff{V}$ are equal.  The same holds for the effect interval $[0, q]$
of the Peirce subalgebra $J_2(q)$, which spans $J_2(q)$.
\end{lemma}

\begin{proof}
If $\|v\| \leq \tfrac12$ then $0 \leq \tfrac12\id + v \leq \id$; agreement
on this ball forces agreement on $V$ by linearity and scaling.  Applied
to the order unit space $(J_2(q), q)$ this gives the last claim.
\end{proof}


%% =====================================================================
%%  SECTION 3: CITED MACHINERY -- THE COMPARISON MAP
%% =====================================================================
\section{The Comparison Map}
\label{sec:machinery}

Fix a finite-dimensional Euclidean Jordan algebra $J$.  Let $(\cdot)$ be a norm-continuous sequential
product on $\eff{J}$, with left action $L_a(b) = \seqp{a}{b}$, and let
$\lprod{a}{b} = Q_{\sqrt a}b$ be the standard reference product, with left
action $L^{0}_a = Q_{\sqrt a}$.

\subsection{Simplicity and order preservation}
\label{sec:simple}

\begin{lemma}[Positive linear extension and first-variable homogeneity]
\label{lem:homog}
For each effect $a$: \textup{(i)} $L_a : b \mapsto \seqp{a}{b}$ is additive
by \ref{ax:S1} and order bounded, hence extends uniquely to a positive
linear map $L_a : J \to J$ (\Cref{lem:span}); \textup{(ii)}
$(\lambda a)\cdot b = \lambda\,(\seqp{a}{b})$ for all $\lambda \in [0,1]$
and effects $b$.
\end{lemma}

\begin{proof}
(i) $L_a$ is additive by \ref{ax:S1} and maps $[0,\id]$ into $[0,\id]$, so
it is order bounded.  For $x \geq 0$ with $x/n \leq \id$ set
$\widetilde{L}_a(x) := n\,L_a(x/n)$; additivity makes this independent of $n$
and additive on $J^+$, order boundedness gives real homogeneity, and
extension by differences yields a unique positive linear map $L_a : J \to J$
on $J = \operatorname{span}\eff{J}$ (\Cref{lem:span}).
(ii) First-variable homogeneity is van de Wetering's linearity property
(his Proposition~3.9~\cite{Wetering2018three}).  His proof passes from rational
to real scalars through an infimum.  That passage is available because his
$\sigma$-SEA definition provides both the infimum of a descending sequence and
two conditions tying it to the product,
$b \cdot \bigwedge a_i = \bigwedge (b \cdot a_i)$ and
$b \compatible a_i \ \forall i \Rightarrow b \compatible \bigwedge a_i$; his
proof invokes the second.  Here
\ref{ax:S2} supplies that passage in its place, as he notes in
\cite[Prop.~3.9, footnote~2]{Wetering2018three}.  The same statement is proved
directly from
\ref{ax:S1}--\ref{ax:S7}, with \ref{ax:S2} in exactly that role, in
\cite[Prop.~2(3)]{vandeWetering2019b}.
\end{proof}

\begin{lemma}[Positive-cone extension]\label{lem:cone-ext}
Each product extends from effect first arguments to arbitrary first
arguments $v$ in the positive cone $J^{+}$ by
\[
  \seqp{v}{b} := \mu\,\bigl(\seqp{(v/\mu)}{b}\bigr),
  \qquad \mu \geq \|v\|,\ \mu > 0,
\]
for effects $b$: the value is independent of the chosen normalization
$\mu$, the extension is positively homogeneous in the first argument, and
it agrees with the original operation on $\eff{J}$.
\end{lemma}

\begin{proof}
If $0 < \mu \leq \mu'$ with $\mu \geq \|v\|$, then $v/\mu$ and $v/\mu'$ are
effects ($0 \leq v$ and $\|v\| \leq \mu$ give $v \leq \mu e$ in an order
unit space) and $v/\mu' = (\mu/\mu')(v/\mu)$ with $\mu/\mu' \in (0,1]$, so
\Cref{lem:homog}(ii) gives
$\mu'\,(\seqp{(v/\mu')}{b}) = \mu'\,(\mu/\mu')\,(\seqp{(v/\mu)}{b})
= \mu\,(\seqp{(v/\mu)}{b})$: any two admissible normalizations agree.
Taking $\mu = 1$ for a nonzero $v \in \eff{J}$ recovers the original
operation (and $\seqp{0}{b} = 0$ by \ref{ax:S1} and \ref{ax:S4}, as in
\Cref{lem:aone}, consistently with every $\mu$); positive homogeneity
$\seqp{(\lambda v)}{b} = \lambda\,(\seqp{v}{b})$ for $\lambda > 0$ follows
by normalizing $\lambda v$ with $\lambda\mu$.  For the standard product the
extension is $Q_{\sqrt v}$ itself, by the homogeneity
$Q_{\sqrt{\lambda v}} = \lambda\,Q_{\sqrt v}$.
\end{proof}

\begin{lemma}[Finite-spectral bridge]\label{lem:simple-bridge}
Throughout, a \emph{simple effect} means an effect in van de Wetering's
class $E_0$~\cite{Wetering2018three}: one with a finite spectral
decomposition $a = \sum_k \lambda_k p_k$ into orthogonal sharp effects (not
to be confused with a simple EJA).  Sharpness is order-theoretic---$p$ is
sharp when $p \wedge (e - p) = 0$~\cite[Def.~3.14]{Wetering2018three}---and
under the standard product it is equivalent to
$\lprod{p}{p} = p$~\cite[Prop.~3.15]{Wetering2018three}, i.e.\ $p^{2} = p$:
the sharp effects of an EJA are exactly its idempotents.  On a
finite-dimensional EJA $J$:
\begin{enumerate}[label=\textup{(\roman*)},nosep]
  \item the standard product makes $\eff{J}$ a convex $\sigma$-sequential
    effect algebra, hence a sequential effect space
    (SES)~\cite[Thm.~A.6]{Wetering2018three};
  \item every effect is simple ($E = E_0$), so the simplicity hypothesis
    of Propositions~5.2, 5.5, 5.7 and~4.20 of~\cite{Wetering2018three}
    is met by every effect, for the unknown product $(\cdot)$ as well
    (their remaining hypotheses are verified at each use);
    normality of $(\cdot)$ is never assumed;
  \item every invertible effect is order preserving for both products
    (Proposition~4.20 of~\cite{Wetering2018three} gives every simple
    effect a pseudo-inverse, and Proposition~4.19 there that a
    pseudo-inverse renders the effect order preserving)\footnote{A
    \emph{pseudo-inverse} of a simple effect is its spectral inverse on
    its support.  The \emph{order preserving} property
    (Definition~4.17 of~\cite{Wetering2018three}) has two clauses: the
    induced left multiplication $L_a$ carries the effect interval onto the
    order interval below $a$ (lower-interval surjectivity), and it reflects
    the order relative to the support of $a$; for a full-support invertible
    effect the second clause is order reflection, so $L_a$ is an order
    embedding onto its image.  See \cite[Def.~4.17,
    Props.~4.19--4.20]{Wetering2018three}.};
  \item the standard product is \emph{invariant}: for every unital order
    isomorphism $\Phi$ of $J$, $\Phi(\lprod{a}{b}) = \lprod{\Phi a}{\Phi b}$
    (it is defined from the Jordan product, and unital order isomorphisms
    of an EJA are Jordan isomorphisms).
\end{enumerate}
\end{lemma}

\begin{proof}
(i) is van de Wetering's Theorem~A.6~\cite{Wetering2018three}, stated there
for an arbitrary Euclidean Jordan algebra.
(ii) holds by the Jordan spectral theorem: every element is a finite real
combination of orthogonal idempotents.  Consequently the simple-effect
clauses cover every effect, and the normality hypothesis needed only to
pass to non-simple effects is vacuous.  (iii): the pseudo-inverse of a nonzero simple effect has norm $\geq 1$ and,
unless the effect is sharp, lies outside the effect interval.  The
positive-cone extension of \Cref{lem:cone-ext} applies both products to it
in the first argument, and the linear extension of \Cref{lem:homog}(i) in the
second, independently of the normalization.
Proposition~4.20 of~\cite{Wetering2018three} supplies a
pseudo-inverse for every simple effect from its finite spectral
representation $a = \sum_k \lambda_k p_k$ inside the commutative subalgebra
of orthogonal sharp effects: explicitly the spectral inverse on the support,
$a^{-} = \sum_{\lambda_k > 0}\lambda_k^{-1}p_k$, with
$\seqp{a}{a^{-}} = \seqp{a^{-}}{a} = \lceil a\rceil$ for either
product.  Proposition~4.19 there
turns that pseudo-inverse into order preservation, again for either product.
Neither proposition is the inverse-law hypothesis of his Theorem~5.19.
(iv) is van de Wetering's remark that the Jordan-defined standard product
is invariant~\cite{Wetering2018three}.
\end{proof}

\begin{lemma}[Automatic normality]\label{lem:normality}
On a finite-dimensional order-unit space $V$, any operation $(\cdot)$ on effects
satisfying \ref{ax:S1} and \ref{ax:S2} is \emph{normal} in van de Wetering's
sense~\cite[Defs.~2.3 and~5.1]{Wetering2018three}: for every descending sequence
of effects $b_k \downarrow b$ with order-infimum $b = \bigwedge_k b_k$ and every
effect $a$, \textup{(i)}~$\seqp{a}{b_k} \downarrow \seqp{a}{b}$, and
\textup{(ii)}~if $a \compatible b_k$ for all $k$ then $a \compatible b$.
\end{lemma}

\begin{proof}
By \Cref{lem:homog}(i)---whose proof uses only the order-unit structure and
\ref{ax:S1}, and so applies verbatim with $V$ in place of $J$---the map
$L_a : x \mapsto \seqp{a}{x}$ is positive linear on $V$, hence
norm-continuous in finite dimension.  A descending sequence of effects $b_k$
lies in the compact interval $[0, b_1]$, so it has convergent subsequences.
By closedness of the cone every subsequential limit is a lower bound for
$\{b_k\}$ and dominates every other lower bound, hence equals the
order-infimum $\bigwedge_k b_k = b$.  The whole sequence therefore converges:
$b_k \to b$ in norm.  \textup{(i)}~Positivity of $L_a$ makes $\seqp{a}{b_k}$
descending, and continuity gives $\seqp{a}{b_k} \to \seqp{a}{b}$ in norm, so
$\bigwedge_k \seqp{a}{b_k} = \seqp{a}{b}$.  \textup{(ii)}~If
$\seqp{a}{b_k} = \seqp{b_k}{a}$ for all $k$, continuity of $L_a$ sends the
left side to $\seqp{a}{b}$ and \ref{ax:S2} (first-variable continuity,
$b_k \to b$) the right side to $\seqp{b}{a}$, so $a \compatible b$.  The
argument uses only the finite-dimensional order structure with \ref{ax:S1}
and \ref{ax:S2}, so it is prior to and independent of the Jordan
identification.
\end{proof}

\begin{remark}[The role of \ref{ax:S2}]
\label{rem:S2-scope}
In finite dimension \ref{ax:S1} and \ref{ax:S2} already make the unknown
product \emph{normal} (\Cref{lem:normality})---derived, never assumed---while
\Cref{lem:simple-bridge}(ii) independently covers every effect by the
simple-effect clauses with no normality extension invoked;
\Cref{rem:import-hyps} states both routes and discharges every imported
hypothesis on the second, self-contained one.  Either way the sole
normality-type input is van de Wetering's scalar-homogeneity step
(\Cref{lem:homog}(ii)), where \ref{ax:S2} supplies it.
\end{remark}

\subsection{The compatibility bridge}
\label{sec:bridge}

\begin{proposition}[Compatibility bridge;
  {\cite[Props.~A.1 and~A.3]{Wetering2018three}}]
\label{prop:bridge}
For effects $a, b$ of a Euclidean Jordan algebra, with the standard
product $\lprod{a}{b} = Q_{\sqrt a}b$,
\[
  \lprod{a}{b} = \lprod{b}{a}
  \iff [Q_a, Q_b] = 0
  \iff [T_a, T_b] = 0,
\]
i.e.\ standard-product compatibility is exactly Jordan operator
commutation.  In particular simultaneously diagonalizable effects are
compatible for the standard product.
\end{proposition}

The two equivalences are van de Wetering's Propositions~A.1
and~A.3~\cite{Wetering2018three}; the second
($[Q_a,Q_b] = 0 \iff [T_a,T_b] = 0$) is the one he notes he could not find
in the literature and which is treated in~\cite{Wetering2019commutativity}.
Combined with Proposition~5.2 of~\cite{Wetering2018three} (for simple
$a, b$, standard-product compatibility transfers the product value and
the compatibility to the unknown product), simultaneously diagonalizable
effects are compatible for the unknown product and their two products
agree.

\begin{lemma}[The unit as second argument]\label{lem:aone}
$\seqp{a}{\id} = a$ for every effect $a$.
\end{lemma}

\begin{proof}
Using \ref{ax:S1}, \ref{ax:S3}, \ref{ax:S4}, \ref{ax:S6}: by \ref{ax:S1}
with $b = c = 0$, $\seqp{a}{0} = 0$; by \ref{ax:S4}, $\seqp{0}{a} = 0$; so
$a \compatible 0$; by \ref{ax:S6}, $a \compatible 0^{\perp} = \id$; and by
\ref{ax:S3} with compatibility, $\seqp{a}{\id} = \seqp{\id}{a} = a$.
\end{proof}

The reduction to a simple algebra is itself a consequence of the axioms.

\begin{proposition}[Central decomposition]\label{prop:central}
Let $J = \bigoplus_{\alpha=1}^m J_\alpha$ be any finite-dimensional Euclidean
Jordan algebra with central summand units $e_\alpha$, and let $(\cdot)$ be a
sequential product satisfying \ref{ax:S1}--\ref{ax:S7}.  For an effect $a$
write $a_\alpha := e_\alpha \circ a \in \eff{J_\alpha}$.  Then
\[
  \seqp{a}{b} = \sum_{\alpha=1}^m \seqp{a_\alpha}{b_\alpha},
  \qquad \seqp{a_\alpha}{b_\alpha} \in J_\alpha,
\]
and each restriction is a sequential product satisfying
\ref{ax:S1}--\ref{ax:S7} on the order-unit space $(J_\alpha, \leq, e_\alpha)$.
A general finite-dimensional sequential product is thus the direct sum of its
summand products; distinct simple summands cannot couple.  Conversely, given
any sequential products $(\cdot)_\alpha$ satisfying \ref{ax:S1}--\ref{ax:S7}
on the summands, $\seqp{a}{b} := \sum_\alpha (a_\alpha \cdot_\alpha b_\alpha)$
satisfies \ref{ax:S1}--\ref{ax:S7} on $J$ and restricts to them.
\end{proposition}

\begin{proof}
The bridge (\Cref{prop:bridge}) and the Proposition~5.2 transfer of value and
compatibility hold on any finite-dimensional EJA, using only that the effects
involved are simple (Jordan spectral theorem), not that $J$ is simple; the
comparison map $\Theta_a$ is not used.  Each $e_\alpha$ is central, hence
operator-commutes with every effect, so $a$ and $e_\alpha$ are compatible for
the unknown product and $\seqp{a}{e_\alpha} = \lprod{a}{e_\alpha} =
Q_{\sqrt a}e_\alpha = a_\alpha$; likewise $\seqp{e_\alpha}{b_\alpha} =
Q_{\sqrt{e_\alpha}}b_\alpha = b_\alpha$ for $b_\alpha \leq e_\alpha$.  By
\ref{ax:S5} with $a \compatible e_\alpha$,
\[
  \seqp{a}{b_\alpha} = \seqp{a}{(\seqp{e_\alpha}{b_\alpha})}
  = \seqp{(\seqp{a}{e_\alpha})}{b_\alpha} = \seqp{a_\alpha}{b_\alpha},
\]
and monotonicity of the positive map $L_a$ (\Cref{lem:homog}) gives
$0 \leq \seqp{a}{b_\alpha} \leq \seqp{a}{e_\alpha} = a_\alpha \leq e_\alpha$,
so it lies in $[0, e_\alpha] \subseteq J_\alpha$.  Summing over $\alpha$ by \ref{ax:S1}
gives the formula.  Block-diagonality makes compatibility componentwise: for
$x, y \in \eff{J_\alpha}$ their compatibility in $J_\alpha$ coincides with
their global compatibility, and the component formula carries the global
complement $\id - y$ to the summand complement $e_\alpha - y$ inside
$J_\alpha$.  Explicitly, $\id - y = (e_\alpha - y) + \sum_{\beta \neq \alpha}
e_\beta$, and since $x \in \eff{J_\alpha}$ the component formula gives
$\seqp{x}{(\id - y)} = \seqp{x}{(e_\alpha - y)}$ and
$\seqp{(\id - y)}{x} = \seqp{(e_\alpha - y)}{x}$, so global compatibility of
$x$ with $\id - y$ is compatibility with the summand complement $e_\alpha - y$.
Hence the complement clause of \ref{ax:S6} restricts to each summand; its
additivity clause and \ref{ax:S1}--\ref{ax:S5}, \ref{ax:S7} restrict directly,
all on $(J_\alpha, \leq, e_\alpha)$, whose order-unit norm is the restriction
of the direct-sum norm.  For the converse, every axiom is verified coordinate
by coordinate.  Effects, the order, and complements are componentwise, giving
\ref{ax:S1}, \ref{ax:S3}, \ref{ax:S4}.  Compatibility $a \compatible b$ holds
in $J$ exactly
when $a_\alpha \compatible b_\alpha$ for every $\alpha$, so
\ref{ax:S5}--\ref{ax:S7} reduce to the summand clauses.  Finally \ref{ax:S2} holds
because the direct-sum order-unit norm is $\max_\alpha\|\cdot\|_\alpha$, so
norm convergence is convergence in every coordinate.
\end{proof}

The classification over $J$ is therefore exactly the product over $\alpha$ of
the summand classifications.

\subsection{The comparison map is a Jordan automorphism}
\label{sec:theta}

The comparison-map construction needs one hypothesis at every invertible
effect: order preservation \emph{for the unknown product as well as the
standard one}.  \Cref{lem:simple-bridge}(iii) records this; because every
argument below depends on that hypothesis, we display the transfer step
rather than leave it to be reassembled from the imported propositions.

\begin{proposition}[Pseudo-inverses transfer to the unknown product]
\label{prop:pseudo-transfer}
Let $a$ be an invertible effect of a finite-dimensional EJA $J$, with spectral
inverse $a^{-1} \in J^{+}$.  Then
\[
  \seqp{a}{a^{-1}} = \seqp{a^{-1}}{a} = \id
\]
for the unknown product as well as the standard one, so $a^{-1}$ is a
pseudo-inverse of $a$ for both; consequently $a$ is order preserving for both,
and the hypotheses of Proposition~5.3 of~\cite{Wetering2018three} hold at every
invertible effect.
\end{proposition}

\begin{proof}
Fix $\mu \geq \|a^{-1}\|$, so that $a^{-1}/\mu$ is an effect.  Being a
polynomial in $a$, it is diagonal in a spectral frame of $a$, hence operator
commutes with $a$ and so is compatible with it for the standard product
(\Cref{prop:bridge}).  Both are effects of a finite-dimensional EJA, hence
simple (\Cref{lem:simple-bridge}(ii)), so Proposition~5.2
of~\cite{Wetering2018three} applies to the pair in either order: they are
compatible for the unknown product too, and there the product takes its
standard value.  Undo the normalization on each side.  In the second argument
$L_a$ is positive linear (\Cref{lem:homog}(i)), so
\[
  \seqp{a}{a^{-1}} = \mu\,\seqp{a}{(a^{-1}/\mu)}
  = \mu\,\lprod{a}{(a^{-1}/\mu)} = \lprod{a}{a^{-1}}
  = Q_{\sqrt a}\,a^{-1} = \id .
\]
In the first argument the positive-cone extension is positively homogeneous and
independent of the normalization (\Cref{lem:cone-ext}), so likewise
$\seqp{a^{-1}}{a} = \mu\,\seqp{(a^{-1}/\mu)}{a}
= \mu\,\lprod{(a^{-1}/\mu)}{a} = \lprod{a^{-1}}{a}
= Q_{\sqrt{a^{-1}}}\,a = \id$.  Thus $a^{-1}$ is a pseudo-inverse of $a$ for
both products, and Proposition~4.19 of~\cite{Wetering2018three} makes $a$ order
preserving for both.  Note the order of dependence: the transfer uses only
compatibility for the \emph{standard} product together with simplicity of the
two effects, so nothing here presupposes the comparison map or any conclusion
drawn from it.
\end{proof}

\begin{proposition}[Comparison map; {\cite[\S 5]{Wetering2018three}} +
  {\cite[Cor.~2.5]{vanImhoffRoelands2019}}]
\label{prop:theta}
For every invertible effect $a$ of $J$ there is a Jordan automorphism
$\Theta_a$ of $J$ with
\[
  \seqp{a}{b} = \lprod{a}{\Theta_a(b)} = Q_{\sqrt a}\Theta_a(b)
  \qquad \text{for all } b,
\]
equivalently $L_a = Q_{\sqrt a}\,\Theta_a$.  Moreover $\Theta_a$ fixes
every $b$ that operator-commutes with $a$; and for operator-commuting invertible
$a, a'$, $\Theta_{a\bullet a'} = \Theta_a \Theta_{a'} = \Theta_{a'}\Theta_a$.
\end{proposition}

\begin{proof}
Proposition~5.3 of~\cite{Wetering2018three} constructs a
\emph{unital linear order isomorphism} $\Theta_a$ of $J$ with
$L_a = Q_{\sqrt a}\Theta_a$ (its multiplication-operator form; $Q_{\sqrt a}$
is invertible since $a$ is invertible).  Linearity and the order-iso
property come with Proposition~5.3's construction (its $\Theta_a$ arises by the
universal property of the quotient of his Proposition~4.22, whose maps are
positive and linear); unitality is
$\Theta_a(\id) = Q_{\sqrt a}^{-1}(\seqp{a}{\id}) = Q_{\sqrt a}^{-1}(a) = \id$,
using $\seqp{a}{\id} = a$ (\Cref{lem:aone}) and $Q_{\sqrt a}(\id) = a$.
Since $J$ is a finite-dimensional JB-algebra, the unital linear order
isomorphism $\Theta_a$ is a Jordan automorphism
(van Imhoff--Roelands~\cite[Cor.~2.5]{vanImhoffRoelands2019}; classically
Alfsen--Shultz~\cite{AlfsenShultz2003}, cited for exactly this by van de
Wetering~\cite{Wetering2018three}).  Thus $\Theta_a \in \Aut(J)$.

The fixing property is Proposition~5.5 of~\cite{Wetering2018three}: for
$b$ compatible with $a$ under the standard product (which by
\Cref{prop:bridge} is operator commutation) and
$b \leq \lceil a\rceil = e$ ($a$ being invertible), $\Theta_a(b) = b$.  Proposition~5.5 needs compatibility under the
\emph{reference} product only; that $a$ and $b$ are then also compatible
for the unknown product is a conclusion (Proposition~5.2
of~\cite{Wetering2018three}), not a second hypothesis.  This fixing extends
from the effects $b \leq e$ to the whole operator commutant of $a$ by
linearity: the commutant is an order-unit subspace, linearly spanned by its
unit-interval effects, on each of which $\Theta_a$ is the identity.  The
cocycle law is Proposition~5.7
of~\cite{Wetering2018three}, whose hypothesis is invariance of the
\emph{reference} product (\Cref{lem:simple-bridge}(iv)); we impose no
invariance on the unknown product.
\end{proof}

\Cref{prop:theta} reduces the classification to a single Jordan
automorphism: every type-dependent step below turns on a property of
$\Theta_a$.  The proposition's inputs are all imported; the
provenance split is stated in \Cref{sec:intro} and the imported hypotheses
are discharged in \Cref{rem:import-hyps}.

\begin{remark}[Hypotheses of the imported propositions, discharged in place]
\label{rem:import-hyps}
The comparison propositions invoked above (his 5.2, 5.3, 5.5, 5.7) have
their simplicity and normality hypotheses discharged here on two independent
grounds: the
unknown product is \emph{normal} (\Cref{lem:normality}), as is the standard
one (\Cref{lem:simple-bridge}(i)), so van de Wetering's normality-based
conclusions---which ask normality of both products---apply to every effect
directly; and,
independently, every effect of a finite-dimensional EJA is simple
($E = E_0$, \Cref{lem:simple-bridge}(ii)), so the normality-free
simple-effect clauses apply to every effect as well---\emph{not} through
simplicity of the ambient EJA.  (Of the remaining rows, those for
Propositions~4.19--4.20 and Corollary~2.5 are normality-free; the
row for his Proposition~3.9 carries the sole normality-type input, supplied in its
place by \ref{ax:S2}---see \Cref{rem:S2-scope}.)  The table discharges every
hypothesis on the second, self-contained route.  We write $(\cdot)$ for the unknown product and
$\lprod{a}{b} = Q_{\sqrt a}\,b$ for the standard reference product; the
comparison automorphism acts on the reference argument,
$L_a = Q_{\sqrt a}\,\Theta_a$, as in~\cite{Wetering2018three}.

\begin{center}
\footnotesize
\begin{tabular}{>{\raggedright\arraybackslash}p{0.34\textwidth}>{\raggedright\arraybackslash}p{0.27\textwidth}>{\raggedright\arraybackslash}p{0.27\textwidth}}
Imported result & Hypotheses & Discharged by \\
\hline
Prop.~3.9 of~\cite{Wetering2018three}: first-variable scalar homogeneity &
a sequential effect space; $\sigma$-normality, or first-variable norm-continuity
in its place &
\ref{ax:S2} (\Cref{rem:S2-scope}); the same statement is proved from
\ref{ax:S1}--\ref{ax:S7} directly in
\cite[Prop.~2(3)]{vandeWetering2019b} \\[2pt]
Props.~4.19--4.20~\cite{Wetering2018three}: simple effects have
pseudo-inverses, and a pseudo-inverse makes the effect order preserving &
$a$ a simple effect & $E = E_0$ (\Cref{lem:simple-bridge}(ii));
pseudo-inverse reached by \Cref{lem:cone-ext} \\[2pt]
Prop.~5.2~\cite{Wetering2018three}: transfer of value and compatibility &
$a, b$ simple effects in a sequential effect space, compatible for the
\emph{reference} product &
\Cref{lem:simple-bridge}(i)--(ii); reference compatibility = operator
commutation (\Cref{prop:bridge}), checked at each use \\[2pt]
Prop.~5.3~\cite{Wetering2018three}: the comparison map $\Theta_a$ &
$a$ order preserving \emph{for both products} & \Cref{lem:simple-bridge}(iii) \\[2pt]
Prop.~5.5~\cite{Wetering2018three}: commutant fixing &
$a, b$ simple; $b \leq \lceil a\rceil$; reference-product compatibility &
$E = E_0$; $a$ invertible, so $\lceil a\rceil = e$; \Cref{prop:bridge} \\[2pt]
Prop.~5.7~\cite{Wetering2018three}: cocycle law &
commuting invertible simple effects; \emph{reference} product invariant
(his Def.~5.6) &
$E = E_0$; \Cref{lem:simple-bridge}(iv); operator commutation verified at each use \\[2pt]
Cor.~2.5~\cite{vanImhoffRoelands2019}: unital linear order isomorphism
$\Rightarrow$ Jordan automorphism & a unital linear bijection of a
JB-algebra & $J$ a finite-dimensional EJA, hence a JB-algebra \\
\hline
\end{tabular}
\end{center}

Two standing clarifications.  None of the cited propositions assumes the
\emph{ambient} algebra is simple---only that the effects involved are; this
licenses \Cref{prop:central} inside a non-simple direct sum.  And
Proposition~5.3's extension to all of $J$ for invertible $a$ uses only that
pseudo-inverses are product-independent on simple effects.  (Two further
clarifications---unknown-product compatibility as Proposition~5.2's
\emph{conclusion}, and no invariance imposed on the unknown product---are
recorded in the proof of \Cref{prop:theta}.)
\end{remark}

\begin{remark}[No Hilbert-space unitary is needed]\label{rem:no-unitary}
Van de Wetering's Lemma~5.8 further reduces $\Theta_a$, for simple invertible
$a$, to $\Ad_{u_a}$ for a bicommutant unitary $u_a$, using that all
automorphisms of $B(H)$ are inner.  That step is Hilbert-space-specific and has no
octonionic analogue.  (The proof of his Theorem~5.11 states that the
Theorem~5.9 argument carries over to $B(H)\sa$ for real, complex,
quaternionic or octonionic $H$.  The three associative types are indeed
covered; the Albert algebra is not the self-adjoint part of any associative
algebra, and $\Aut(H_3(\mathbb{O})) = F_4$ is not a group of inner
automorphisms, so the bicommutant unitary is unavailable there.
\Cref{thm:albert} below reaches the exceptional type by a different route.)
We do not use the step: the frame-stabilizer argument
below needs only that $\Theta_a$ is a Jordan automorphism, which
\Cref{prop:theta} supplies on every type.
\end{remark}


%% =====================================================================
%%  SECTION 4: THE COMMON REDUCTION
%% =====================================================================
\section{The Common Reduction}
\label{sec:common}

Fix a simple EJA $J$ of rank $n \geq 2$, a Jordan frame
$F = \{p_1, \ldots, p_n\}$, and for $r \in (-\infty, 0]^n$ the invertible
effect $a(r) = \sum_i e^{r_i}p_i$ diagonal in $F$.  Write
$\Theta_r := \Theta_{a(r)}$.  The lemmas of this section hold at every
rank; the rank hypothesis $n \geq 3$ enters only in the typewise branches
of \Cref{sec:branches}.

\begin{lemma}[Frame fixing; block-diagonality is a conclusion]
\label{lem:frame-fix}
$\Theta_r$ fixes each $p_i$ and the diagonal $\bigoplus_i \mathbb{R}p_i$
pointwise, preserves each Peirce block $V_{ij}$ (the common $\tfrac12$-eigenspace
of $T_{p_i}$ and $T_{p_j}$), and lies in the identity
component $\Stab(F)^{\circ}$ of the frame stabilizer
$\Stab(F) = \{g \in \Aut(J) : gp_i = p_i\}$.  Consequently
$L_{a(r)} = Q_{\sqrt{a(r)}}\,\Theta_r$ is Peirce-block-diagonal.
\end{lemma}

\begin{proof}
Each $p_i$ is diagonal in $F$, hence operator-commutes with $a(r)$; by
\Cref{prop:theta} (with \Cref{prop:bridge}), $\Theta_r(p_i) = p_i$, and by
linearity $\Theta_r = \mathrm{id}$ on $\bigoplus_i \mathbb{R}p_i$.  A
Jordan automorphism fixing every $p_i$ preserves each $V_{ij}$: for
$x \in V_{ij}$, $p_i \circ \Theta_r(x) = \Theta_r(p_i \circ x) =
\tfrac12\Theta_r(x)$ and likewise for $p_j$, so
$\Theta_r(x) \in V_{ij}$.  Thus $\Theta_r \in \Stab(F)$.  For continuity of
$r \mapsto \Theta_r$, fix an effect basis $b_1, \ldots, b_m$ of $J$
(\Cref{lem:span}).  \ref{ax:S2} makes each $a \mapsto L_a(b_k)$ norm-continuous,
so the matrix coefficients of $L_a$ in this basis are continuous, giving
operator-norm continuity of $a \mapsto L_a$ in finite dimension.  Since
$a \mapsto Q_{\sqrt a}^{-1} = Q_{a^{-1/2}}$ is continuous on invertibles as well,
$r \mapsto \Theta_r = Q_{a(r)^{-1/2}}L_{a(r)}$ is continuous.  With the path
$t \mapsto \Theta_{tr}$ from $\Theta_0 = \mathrm{id}$ this places
$\Theta_r \in \Stab(F)^{\circ}$;
$\Stab(F) = \bigcap_i\{g \in \Aut(J) : gp_i = p_i\}$ is closed in the
compact Lie group $\Aut(J)$, hence a closed Lie subgroup.
Block-diagonality of $L_{a(r)}$ follows from
$L_{a(r)} = Q_{\sqrt{a(r)}}\Theta_r$, both factors block-diagonal.
\end{proof}

\begin{lemma}[Coalescence]\label{lem:coalescence}
If $r_i = r_j$, then $\Theta_r$ fixes $V_{ij}$ pointwise.  More generally,
if an \emph{invertible} effect $a$ has the form $a = \lambda q + a_0$ with
$q = p_i + p_j$, $\lambda > 0$, and $a_0 \in J_0(q)$ invertible in
$J_0(q)$---that is, if $a$ is scalar on the Peirce $2$-space of $q$ and has no
Peirce $1$-part---then $\Theta_a|_{J_2(q)} = \mathrm{id}$.
\end{lemma}

\begin{proof}
Let $q = p_i + p_j$ and $a = \lambda q + a_0$, $a_0 \in J_0(q)$.  Take an
effect $b$ with $0 \leq b \leq q$, so $b \in J_2(q)$.  Spectrally
decompose $b = \sum_\alpha \mu_\alpha e_\alpha$
($\sum_\alpha e_\alpha = q$) inside $J_2(q)$ and
$a_0 = \sum_\beta \nu_\beta f_\beta$ ($\sum_\beta f_\beta = e - q$) inside
$J_0(q)$.  Every $e_\alpha$ is orthogonal to every $f_\beta$, so the
combined family is a complete orthogonal system of idempotents (refining to a
Jordan frame of $J$), and
$a = \sum_\alpha \lambda e_\alpha + \sum_\beta \nu_\beta f_\beta$ is
diagonal in it; hence $a$ and $b$ are simultaneously diagonalizable and
operator commute.  By \Cref{prop:bridge} they are compatible for the
standard product, and since $a$ is invertible, $\lceil a\rceil = e \geq b$,
so \Cref{prop:theta} gives $\Theta_a(b) = b$.  Effects supported by $q$
span $J_2(q)$ (\Cref{lem:span}), so $\Theta_a|_{J_2(q)} = \mathrm{id}$; in
particular $\Theta_a$ fixes $V_{ij} \subset J_2(q)$.  The first statement
is the case $a = a(r)$ with $r_i = r_j$.  The argument is spectral Peirce
calculus, valid for every coordinate type including the exceptional one.
\end{proof}

\begin{lemma}[Diagonal homomorphism and hyperplane factorization]
\label{lem:homomorphism}
The map $r \mapsto \Theta_r$ is a continuous homomorphism
$((-\infty,0]^n, +) \to \Stab(F)^{\circ}$; it extends to a continuous
homomorphism $\chi : (\mathbb{R}^n, +) \to \Stab(F)^{\circ}$ with a
real-linear differential $d\chi : \mathbb{R}^n \to \mathfrak{stab}(F)$,
and the block representation $\rho_{ij} : \mathfrak{stab}(F) \to
\mathfrak{so}(V_{ij})$---the differential of the group action
$\Stab(F)^{\circ} \to SO(V_{ij})$, at which Lie-algebra level all
faithfulness claims below are made---satisfies
\[
  \rho_{ij}(d\chi(r)) = (r_i - r_j)\, T_{ij}
  \qquad \text{for a single } T_{ij} \in \mathfrak{so}(V_{ij}).
\]
\end{lemma}

\begin{proof}
For operator-commuting diagonal $a(r), a(r')$ the standard product satisfies
$a(r) \bullet a(r') = a(r + r')$, so Proposition~5.7 of~\cite{Wetering2018three}
(\Cref{prop:theta}) gives $\Theta_{r+r'} = \Theta_r \Theta_{r'}$; the image
is abelian.  Extend to $\mathbb{R}^n = S - S$, $S = (-\infty,0]^n$, by
$\chi(s - t) := \Theta_s \Theta_t^{-1}$, well defined by commutativity
($s - t = s' - t' \Rightarrow s + t' = s' + t \Rightarrow
\Theta_s\Theta_{t'} = \Theta_{s'}\Theta_t \Rightarrow
\Theta_s\Theta_t^{-1} = \Theta_{s'}\Theta_{t'}^{-1}$, the last step because the
image is abelian) and continuous via the canonical
continuous decomposition $x = s(x) - t(x)$ with
$s_i(x) = \min(x_i, 0)$, $t_i(x) = \min(-x_i, 0)$.  A continuous homomorphism
between finite-dimensional Lie groups is smooth, so
$d\chi : \mathbb{R}^n \to \mathfrak{stab}(F)$ is real-linear.  A
continuous homomorphism out of the connected group $\mathbb{R}^n$ is moreover
determined by its
differential, so recovering $d\chi$ recovers $\chi$ and each $\Theta_r$
(used in the branches to pass from $d\chi = 0$ to $\Theta_r = \mathrm{id}$).
For $h \in S$ with $h_i = h_j$, \Cref{lem:coalescence} gives
$\chi(sh)|_{V_{ij}} = \mathrm{id}$ for all $s \geq 0$; right-differentiating at
$s = 0$ yields $\rho_{ij}(d\chi(h)) = 0$.  Differences of such $h$ span the
hyperplane $\{r_i = r_j\}$: any $v$ with $v_i = v_j$ is $v = h - h'$ with
$h = v - t(1,\ldots,1)$, $h' = -t(1,\ldots,1)$, $t \geq \max(0, \max_k v_k)$,
both in the orthant and both satisfying the equality.  The real-linear map
$r \mapsto \rho_{ij}(d\chi(r))$ therefore vanishes on that hyperplane and is a multiple
of $(r_i - r_j)$: $\rho_{ij}(d\chi(r)) = (r_i - r_j)T_{ij}$ with
$T_{ij} \in \mathfrak{so}(V_{ij})$ (the stabilizer lies in $\Aut(J)$, hence
preserves the trace form and acts orthogonally on each block, as in
\Cref{prop:stabilizers}).  At the level of the differential, no global logarithm
and no $2\pi$ ambiguity enter.
\end{proof}

Throughout, $\Stab(F)$ is the \emph{pointwise} frame stabilizer (fixing
each $p_i$), not the larger setwise stabilizer that also permutes the
atoms.

\begin{proposition}[Frame-stabilizer data]\label{prop:stabilizers}
Let $J$ be a simple EJA of rank $n$ of one of the four types listed below---these
exhaust the simple EJAs of rank $\geq 3$, though not those of rank
two---and $\Stab(F)$ the pointwise stabilizer of a Jordan frame.  As $\Stab(F) \subset \Aut(J)$ preserves the trace form
$\langle x,y\rangle = \operatorname{tr}(x\circ y)$, it acts on each Peirce
block $V_{ij}$ through $O(V_{ij})$, and $\mathfrak{stab}(F)$ through
$\mathfrak{so}(V_{ij})$.  The identity component $\Stab(F)^{\circ}$ and the
induced block action are, by type:
\begin{center}
\begin{tabular}{lccl}
  $J$ & $d = \dim V_{ij}$ & $\Stab(F)^{\circ}$ & $\mathfrak{stab}(F)$ on $V_{ij}$ \\
  \hline
  $\MnKsa{n}{\mathbb{R}}$ & $1$ & $\{1\}$ & $0$ \\
  $H_n(\mathbb{C})$ & $2$ & $U(1)^n/U(1) \cong T^{n-1}$ & $z \mapsto i(\theta_i - \theta_j)z$ \\
  $\MnKsa{n}{\mathbb{H}}$ & $4$ & $Sp(1)^n/\{\pm1\}$ & $x \mapsto \xi_i x - x\xi_j$, $\xi_\bullet \in \operatorname{Im}\mathbb{H}$ \\
  $H_3(\mathbb{O})$ & $8$ & $\Spin(8)$ & triality: the $8_v, 8_s, 8_c$ differentials
\end{tabular}
\end{center}
\end{proposition}

\begin{proof}
Trace-form invariance of $\Aut(J)$ is standard~\cite[Ch.~III]{FarautKoranyi1994},
giving the orthogonality statement.  For the matrix types the identity
component of $\Aut(\MnKsa{n}{\mathbb{K}})$ acts by $X \mapsto UXU^{*}$.
Fixing every $p_i = E_{ii}$ forces $U = \operatorname{diag}(u_i)$ with
$u_i \in \mathbb{K}$, $|u_i| = 1$, so the induced action on
$V_{ij} \cong \mathbb{K}$ is $x \mapsto u_i x \bar u_j$: real
($u_i = \pm1$, discrete), complex ($U(1)^n/U(1)$, with differential
$z \mapsto i(\theta_i - \theta_j)z$), and quaternionic
($Sp(1)^n/\{\pm1\}$, differential $x \mapsto \xi_i x - x\xi_j$; the quotient
by the discrete central $\{\pm1\}$ does not change the Lie algebra
$(\operatorname{Im}\mathbb{H})^n$).  For
$H_3(\mathbb{O})$ the pointwise stabilizer is $\Spin(8)$ acting on
$V_{12}$, $V_{13}$, $V_{23}$ by the three inequivalent triality representations
$8_v, 8_s, 8_c$, in some order
(Yokota~\cite[Thms.~2.7.1,~1.16.2]{Yokota2009}; see
also~\cite{SpringerVeldkamp2000}).  The data
are otherwise classical~\cite{FarautKoranyi1994, Loos1977}.
\end{proof}

The real stabilizer is discrete, so its identity component is trivial;
for the complex type the transpose lies outside the connected torus and
plays no role.  The remaining branches turn on different
representation-theoretic mechanisms.  In the quaternionic type the two-slot
map $(\xi_i, \xi_j) \mapsto (x \mapsto \xi_i x - x\xi_j)$ is injective, so it
separates the two atoms of a coherence block---even though for $n \ge 3$ an
individual $\rho_{ij}$ is \emph{not} faithful there, the spectator
coordinates lying in its kernel.  In the exceptional type each block differential is a
nontrivial representation of the simple Lie algebra $\mathfrak{spin}(8)$,
hence faithful, so a single block already detects the Lie-algebra element.
Only for $\mathbb{C}$ does the action retain a scalar direction (it sees
$\theta_i, \theta_j$ only through their difference), and there too, for
$n \geq 3$, $\rho_{ij}$ is not individually faithful.  Hence at rank $\geq 3$
only the complex type survives.


%% =====================================================================
%%  SECTION 5: THE TYPEWISE BRANCHES
%% =====================================================================
\section{The Typewise Branches}
\label{sec:branches}

We run the frame-stabilizer reduction of \Cref{sec:common} type by type; by
\Cref{prop:stabilizers} the stabilizer block data collapse the residual freedom, at
rank $\geq 3$, on every type but the complex, which retains one scalar direction.

\begin{proposition}[Real type]\label{prop:real}
On $\MnKsa{n}{\mathbb{R}}$, $n \geq 2$: $\Theta_r = \mathrm{id}$ for all
$r$, so $\seqp{a}{b} = Q_{\sqrt a}b$.  In particular the real qubit
($n = 2$) is rigid.
\end{proposition}

\begin{proof}
The pointwise frame stabilizer $\{\pm1\}^n/\{\pm1\}$ is discrete at every
rank $n \geq 2$, so $\Stab(F)^{\circ} = \{1\}$; the continuous homomorphism
$\chi$ from connected $\mathbb{R}^n$ into it is trivial, giving
$\Theta_r = \mathrm{id}$ and $L_{a(r)} = Q_{\sqrt{a(r)}}$ for every
invertible $a(r)$.  Every invertible effect is diagonal in its own frame,
so this holds for all invertible effects, and for all effects by
\Cref{prop:singular}.
\end{proof}

\begin{theorem}[Quaternionic type]\label{thm:quaternionic}
On $\MnKsa{n}{\mathbb{H}}$, $n \geq 3$: $\Theta_r = \mathrm{id}$ for all
$r$, so $\seqp{a}{b} = Q_{\sqrt a}b$.
\end{theorem}

\begin{proof}
Here $\mathfrak{stab}(F) = (\operatorname{Im}\mathbb{H})^n$ and, by
\Cref{lem:homomorphism}, $d\chi(r) = (\xi_1(r), \ldots, \xi_n(r))$ with
$\xi_i(r) = \sum_l r_l a_{il}$, $a_{il} \in \operatorname{Im}\mathbb{H}$,
acting on $V_{ij} \cong \mathbb{H}$ by $x \mapsto \xi_i(r)x - x\xi_j(r)$,
pinned to $(r_i - r_j)T_{ij}$:
\begin{equation}\label{eq:H-dagger}
  \xi_i(r)\,x - x\,\xi_j(r) = (r_i - r_j)\,T_{ij}(x)
  \qquad \forall x, \forall r.
\end{equation}
Fix $i \neq l$; since $n \geq 3$ choose $j \notin \{i, l\}$.  Matching the
coefficient of $r_l$ in~\eqref{eq:H-dagger} for block $(i,j)$: the right
side is $\propto (r_i - r_j)$ with $l \neq i, j$, so its $r_l$-coefficient
is $0$, giving $a_{il}x = x a_{jl}$ for all $x$; setting $x = 1$,
$a_{il} = a_{jl}$, then $a_{il}x = xa_{il}$ forces
$a_{il} \in Z(\mathbb{H}) \cap \operatorname{Im}\mathbb{H} = \{0\}$.  So
$a_{il} = 0$ for $i \neq l$.  Matching $r_i$ gives $T_{ij}(x) = a_{ii}x$
and $r_j$ gives $T_{ij}(x) = x a_{jj}$, so $a_{ii}x = xa_{jj}$ for all
$x$, forcing $a_{ii} = a_{jj} \in Z(\mathbb{H}) \cap \operatorname{Im}\mathbb{H}
= \{0\}$, whence $T_{ij} = 0$ and $d\chi = 0$.  Thus $\Theta_r = \mathrm{id}$
for every frame and $\seqp{a}{b} = Q_{\sqrt a}b$ on invertibles, hence
(\Cref{prop:singular}) everywhere.
\end{proof}

\begin{theorem}[Exceptional type]\label{thm:albert}
On $H_3(\mathbb{O})$: $\Theta_r = \mathrm{id}$ for all $r$, so
$\seqp{a}{b} = Q_{\sqrt a}b$.
\end{theorem}

\begin{proof}
The pointwise frame stabilizer is
$\{\alpha \in F_4 = \Aut(H_3(\mathbb{O})) : \alpha E_i = E_i,\ i = 1,2,3\} \cong \Spin(8)$
(Yokota~\cite[Thm.~2.7.1]{Yokota2009}; see
also~\cite{SpringerVeldkamp2000}), realized as the triality triple
$\{(a_1,a_2,a_3) \in SO(8)^3 : (a_1x)(a_2y) = a_3(xy)\}$
(Yokota~\cite[Thm.~1.16.2]{Yokota2009}).  Its three $SO(8)$ factors act on
$V_{12}$, $V_{13}$, $V_{23}$ as the three inequivalent $8$-dimensional
representations $8_v, 8_s, 8_c$ of $\mathfrak{spin}(8)$~\cite[\S 2.4,
4.2]{Baez2002}, in some order; the labelling is permuted by the outer
automorphisms, and only nontriviality
is used below.  By \Cref{lem:homomorphism},
$d\chi(r) = \sum_{k=1}^3 r_k \eta_k$ with $\eta_k \in \mathfrak{spin}(8)$
and $\rho_{ij}(d\chi(r)) = (r_i - r_j)T_{ij}$, the $\rho_{12}, \rho_{13},
\rho_{23}$ being these three triality representations.  Each is a
nontrivial representation of the simple Lie algebra $\mathfrak{spin}(8)$,
hence \emph{faithful}---unlike the individual quaternionic and complex block
representations at rank $\geq 3$, and unlike the three \emph{group}
representations $8_v,
8_s, 8_c$, each of which has a two-element central kernel; the argument uses
only the differential.  For block $V_{12}$, the coefficient of $r_3$ gives
$\rho_{12}(\eta_3) = 0$ (coalescence, RHS $\propto r_1 - r_2$), so
$\eta_3 = 0$; symmetrically $\eta_2 = \eta_1 = 0$.  Thus $d\chi \equiv 0$,
$\Theta_r = \mathrm{id}$.  Rank is exactly $3$, so each frame is directly a
three-index configuration and no globalization is needed; all frames are
conjugate~\cite[Thm.~IV.2.5]{FarautKoranyi1994}.  Working with $\Spin(8)$
abstractly sidesteps octonionic non-associativity.  Hence
$\seqp{a}{b} = Q_{\sqrt a}b$ on invertibles, and (\Cref{prop:singular})
everywhere.
\end{proof}

To globalize the complex per-frame parameter across overlapping frames
(\Cref{lem:adjacent}), we need a generator on the shared cross-coherence space
that does not depend on how the shared rank-two block splits into atoms; the
next lemma constructs it.

\begin{lemma}[Splitting-independent complex structure on a cross-coherence space]
\label{lem:orientation}
Let $J = \Mnsa{n}$, $n \geq 3$, let $q$ be a rank-two projection and $p_k$ an
atom with $q \circ p_k = 0$, and let
$X = \{x : q \circ x = \tfrac12 x = p_k \circ x\}$ be the cross-coherence
space, identified with the self-adjoint part of
$qM_n(\mathbb{C})p_k \oplus p_kM_n(\mathbb{C})q$: each $x \in X$ is
$x = z + z^{*}$ with corner $z = qxp_k$.  Define the real-linear map
\[
  \mathcal{J}_{q,k}(x) = i z - i\,(z^{*}), \qquad z = qxp_k .
\]
Then:
\begin{enumerate}[label=\textup{(\roman*)},nosep]
  \item $\mathcal{J}_{q,k}$ maps $X$ to $X$ with
    $\mathcal{J}_{q,k}^{2} = -\id_X$, and depends only on $q$, $p_k$, and the
    fixed ambient complex scalar $i$ of the labelled matrix algebra---\emph{not}
    on any splitting $q = p_1 + p_2$ into atoms.
  \item Every inner automorphism $\Ad_u$ ($u$ unitary) with $uqu^{*} = q$ and
    $up_ku^{*} = p_k$ commutes with $\mathcal{J}_{q,k}$ on $X$.
  \item Let $\{p_l\}_{l \geq 3}$ be pairwise-orthogonal atoms with
    $q + \sum_{l \geq 3} p_l = e$ and $p_k$ among them, and
    $a = \lambda q + \sum_{l \geq 3}\lambda_l p_l$ with
    $\lambda, \lambda_l \in (0,1)$.  Then for every $t \in \mathbb{R}$,
    \[
      \Ad_{a^{it}}\big|_X
        = \exp\!\bigl(t\,(\log\lambda - \log\lambda_k)\,\mathcal{J}_{q,k}\bigr),
    \]
    a rotation of $X$ in the orientation fixed by $\mathcal{J}_{q,k}$.
\end{enumerate}
\end{lemma}

\begin{proof}
The corner $z = qxp_k$ is uniquely determined by $x$ since $q, p_k$ are fixed.
As $(i z^{*})^{*} = -i z$, one has $(iz - iz^{*})^{*} = -iz^{*} + iz$, so
$\mathcal{J}_{q,k}(x)$ is self-adjoint and lies in
$qM_n(\mathbb{C})p_k \oplus p_kM_n(\mathbb{C})q$, i.e. in $X$; its corner is
$iz$.  Applying $\mathcal{J}_{q,k}$ again,
$\mathcal{J}_{q,k}^{2}(x) = i(iz) - i(iz)^{*} = -z - z^{*} = -x$, giving (i);
only $q, p_k, i$ enter the definition.  For (ii), $u q u^{*} = q$ and
$u p_k u^{*} = p_k$ give
$q(uxu^{*})p_k = u\,(qxp_k)\,u^{*} = uzu^{*}$, so
$\mathcal{J}_{q,k}(\Ad_u x) = i\,uzu^{*} - i\,(uzu^{*})^{*}
= u(iz - iz^{*})u^{*} = \Ad_u(\mathcal{J}_{q,k} x)$.  For (iii), $a$ is
invertible, so $a^{it} = e^{it\log\lambda}q + \sum_{l\geq3}
e^{it\log\lambda_l}p_l$ is unitary; since $z = qzp_k$,
$\Ad_{a^{it}}(x)$ has corner $a^{it} z (a^{it})^{*}
= e^{it\log\lambda}\, z\, e^{-it\log\lambda_k}
= e^{i\varphi} z$ with $\varphi = t(\log\lambda - \log\lambda_k)$.  On $X$,
multiplying the corner by $e^{i\varphi} = \cos\varphi + i\sin\varphi$ is
$\cos\varphi\,\id_X + \sin\varphi\,\mathcal{J}_{q,k}
= \exp(\varphi\,\mathcal{J}_{q,k})$, using $\mathcal{J}_{q,k}^{2} = -\id_X$.
\end{proof}

\begin{proposition}[One parameter per frame]\label{prop:per-frame}
Let $J = H_n(\mathbb{C}) = \Mnsa{n}$, $n \geq 3$, and let
$F = \{p_1, \ldots, p_n\}$ be a Jordan frame.  There is a single
$t_F \in \mathbb{R}$, depending only on the unordered frame, with
\[
  \Theta_r = \Ad_{a(r)^{it_F}},
  \qquad\text{i.e.}\quad
  \seqp{a(r)}{b} = a(r)^{1/2+it_F}\,b\,a(r)^{1/2-it_F}
  \text{ for all effects } b,
\]
for every $r \in (-\infty, 0]^n$.
\end{proposition}

\begin{proof}
Here $\mathfrak{stab}(F)$ is the torus
$\{i\,\operatorname{diag}(\theta_1,\ldots,\theta_n)\}/(\mathbb{R}\,i\,\id)$
and $\rho_{ij}(d\chi(r)) = i(\theta_i(r) - \theta_j(r))$ on
$V_{ij} \cong \mathbb{C}$.  Only the differences $\theta_i - \theta_j$ are
gauge-invariant (the torus Lie algebra is
$\mathbb{R}^n/\mathbb{R}(1,\ldots,1)$), so work with the block weights
$\ell_{ij}(r) := \theta_i(r) - \theta_j(r)$ directly, avoiding any choice of
representative.  \Cref{lem:homomorphism} gives
$\ell_{ij}(r) = (r_i - r_j)\,t_{ij}$ for reals $t_{ij}$, and the weights obey
the cocycle $\ell_{ik} = \ell_{ij} + \ell_{jk}$.  Hence for distinct
$i, j, k$ (available since $n \geq 3$),
\[
  t_{ik}(r_i - r_k) = t_{ij}(r_i - r_j) + t_{jk}(r_j - r_k)
  \quad\text{for all } r;
\]
comparing the coefficients of $r_i$ and of $r_k$ gives $t_{ik} = t_{ij}$ and
$t_{ik} = t_{jk}$.  So $t_{ij} =: t_F$ is one constant per frame, and
$\rho_{ij}(d\chi(r)) = i\,t_F(r_i - r_j)$ on every block.  This constant is
independent of the labelling of the atoms of $F$: reversing an ordered pair
sends $\ell_{ij} \mapsto \ell_{ji} = -\ell_{ij}$ and
$r_i - r_j \mapsto r_j - r_i = -(r_i - r_j)$, leaving the coefficient $t_F$ in
$\ell_{ij}(r) = t_F(r_i - r_j)$ unchanged; since permutations are generated by
transpositions, $t_F$ depends only on the unordered frame.  So
$\Theta_r$ acts on $V_{ij}$ by the rotation $e^{\,i\,t_F(r_i - r_j)}$, i.e.
$\Theta_r = \Ad_{a(r)^{it_F}}$ (with $a(r)^{it_F} = \sum_k e^{it_F r_k}p_k$,
whose conjugation rotates the $(i,j)$ corner by $e^{i t_F(r_i-r_j)}$), and
$L_{a(r)} = Q_{\sqrt{a(r)}}\Ad_{a(r)^{it_F}}$, i.e.
$\seqp{a(r)}{b} = a(r)^{1/2+it_F}\,b\,a(r)^{1/2-it_F}$.
\end{proof}

\begin{lemma}[Adjacent-frame constancy]\label{lem:adjacent}
Let $n \geq 3$, and let $F$ and $F'$ be adjacent Jordan frames of
$\Mnsa{n}$: frames differing by a rotation inside a rank-two block
$q = p_1 + p_2 = p'_1 + p'_2$ and sharing the atoms $p_3, \ldots, p_n$.
Then $t_F = t_{F'}$.
\end{lemma}

\begin{proof}
Fix $k \geq 3$ and let
$X = \{x : q \circ x = \tfrac12 x = p_k \circ x\}$ be the cross-coherence
space, so $X = V_{1k} \oplus V_{2k} = V'_{1k} \oplus V'_{2k}$ in the two
frames.  Take $a = \lambda q + \sum_{l \geq 3}\lambda_l p_l$ with
$\lambda, \lambda_l \in (0,1)$ distinct.  By \Cref{lem:coalescence} $\Theta_a$
fixes $J_2(q)$ \emph{pointwise}, and \Cref{prop:per-frame} gives
$\Theta_a = \Ad_{a^{it_F}}$ with $a^{it_F}$ commuting with $q$ and $p_k$; so
\Cref{lem:orientation}(iii) yields
\[
  \Theta_a\big|_X
    = \exp\!\bigl(t_F\,(\log\lambda - \log\lambda_k)\,\mathcal{J}_{q,k}\bigr),
\]
the two subblocks $V_{1k}, V_{2k}$ carrying the same phase because $a$ has the
repeated eigenvalue $\lambda$ on $\range(q)$.  The oriented generator
$\mathcal{J}_{q,k}$ depends only on $q$, $p_k$, and the ambient $i$
(\Cref{lem:orientation}(i)), \emph{not} on the splitting $q = p_1 + p_2$; so
the same operator $\Theta_a$, computed in $F'$, acts on $X$ as
$\exp\!\bigl(t_{F'}(\log\lambda - \log\lambda_k)\,\mathcal{J}_{q,k}\bigr)$
against the \emph{same} generator.  The comparison is licensed by two further facts.  First,
$\Theta_a \in \Stab(F)^{\circ}$ is \emph{inner}---the identity component of
$\Aut(\Mnsa{n})$ acts by $X \mapsto UXU^{*}$ (\Cref{prop:stabilizers})---hence
$\mathbb{C}$-linear on $X$ and commuting with $\mathcal{J}_{q,k}$
(\Cref{lem:orientation}(ii)); the orientation-reversing transpose, which would
reverse $\mathcal{J}_{q,k}$, is excluded.  Second, the identification is
independent of the unitary matrix units aligning a general frame with the
standard one: such a change conjugates $\mathcal{J}_{q,k}$ by a unitary fixing
$q$ and $p_k$, under which it is invariant (\Cref{lem:orientation}(ii)).
Writing $s = \log\lambda - \log\lambda_k$ and letting $\lambda$ vary over an
open interval, $s$ ranges over an interval and
$\exp(t_F s\,\mathcal{J}_{q,k}) = \exp(t_{F'} s\,\mathcal{J}_{q,k})$ for all
such $s$.  Differentiating at an interior point and cancelling the common
invertible exponential gives
$t_F\,\mathcal{J}_{q,k} = t_{F'}\,\mathcal{J}_{q,k}$; since
$\mathcal{J}_{q,k}^{2} = -\id_X$ (so $\mathcal{J}_{q,k} \neq 0$), this forces
$t_F = t_{F'}$ \emph{exactly} (not $t_F = -t_{F'}$), with no $2\pi$
ambiguity.
\end{proof}

\begin{theorem}[Complex type]\label{thm:complex}
On $H_n(\mathbb{C}) = \Mnsa{n}$, $n \geq 3$: there is a single
$t \in \mathbb{R}$ with $\seqp{a}{b} = a^{1/2+it}b\,a^{1/2-it}$ for all
effects $a, b$.
\end{theorem}

\begin{proof}
Each frame carries one parameter (\Cref{prop:per-frame}), adjacent frames
carry the same one (\Cref{lem:adjacent}), and any two frames are joined by
finitely many adjacent moves (\Cref{lem:frame-connectivity}), so a single
global $t$ results.  Every invertible effect is diagonal in some frame, so
$\seqp{a}{b} = a^{1/2+it}b\,a^{1/2-it}$ for all invertible $a$; density
(\Cref{prop:singular}, both sides norm-continuous) extends it to all
effects.
\end{proof}

\begin{lemma}[Frame connectivity]\label{lem:frame-connectivity}
For $n \geq 2$, the graph on Jordan frames of $\Mnsa{n}$ with
$F \sim F'$ iff they share $n - 2$ atoms (differ only inside a rank-two
projection $P$) is connected.
\end{lemma}

\begin{proof}
Frames are orthonormal bases up to phase and ordering.  Let $p = ww^{*}$ be an
atom of the target frame and write $w = \sum_i c_iv_i$ in the current
orthonormal basis $\{v_i\}$ (so the current frame is $\{v_iv_i^{*}\}$).  If
$m > 1$ of the coefficients are nonzero, choose $i \neq j$ with
$c_i, c_j \neq 0$, put $\rho = (|c_i|^2 + |c_j|^2)^{1/2} > 0$ and
$u = \rho^{-1}(c_iv_i + c_jv_j)$, and complete $u$ to an orthonormal pair
$\{u, u^{\perp}\}$ of $\operatorname{span}\{v_i, v_j\}$.  Replacing
$\{v_i, v_j\}$ by $\{u, u^{\perp}\}$ changes exactly two atoms of the
frame, one edge of the graph.  In the new basis $w$ has component
$\rho \neq 0$ along $u$ and $0$ along $u^{\perp}$: its remaining coordinates
are orthogonal to $u^{\perp} \in \operatorname{span}\{v_i, v_j\}$, and
$\langle u^{\perp}, w\rangle = \rho\,\langle u^{\perp}, u\rangle = 0$.  So
the number of nonzero coordinates of $w$ drops by one.  After $\leq n - 1$
edges $p$ is an atom of the current frame; all subsequent moves may be taken
inside $p^{\perp}$, and induction on $n$ installs the remaining target
atoms, the base case $n = 2$ being a single move.
\end{proof}

\begin{lemma}[Singular extension]\label{prop:singular}
In each of \Cref{prop:real} and \Cref{thm:quaternionic,thm:albert,thm:complex}, the
stated formula extends from invertible to all effects by
\ref{ax:S2}-continuity.
\end{lemma}

\begin{proof}
For any effect $a$, $a_\varepsilon = (1-\varepsilon)a + \varepsilon e$ is
invertible for $\varepsilon \in (0,1]$ and $a_\varepsilon \to a$;
\ref{ax:S2} gives $L_{a_\varepsilon}(b) \to L_a(b)$, and the Jordan
functional calculus makes $Q_{\sqrt{a_\varepsilon}}b$ (resp.
$a_\varepsilon^{1/2\pm it}$, with value $0$ at eigenvalue $0$) continuous,
so the invertible-case equality passes to the limit.
\end{proof}

\begin{proof}[Proof of \Cref{mthm:master}]
Combine \Cref{prop:real} (real), \Cref{thm:quaternionic} (quaternionic),
\Cref{thm:albert} (exceptional), and \Cref{thm:complex} (complex), with
\Cref{prop:singular}.  The four coordinate types exhaust the simple
Euclidean Jordan algebras of rank $\geq 3$.  For the real, quaternionic,
and exceptional types the per-frame conclusion $\Theta_r = \mathrm{id}$ is
already global, because each invertible effect is diagonal in its own frame, so
no frame-connectivity step is needed.  Only the complex type, whose
per-frame parameter $t_F$ survives, requires the frame-overlap
globalization of \Cref{thm:complex}.  The three typewise kill
mechanisms---stabilizer discreteness (real), two-slot separation
(quaternionic), and block-representation faithfulness (exceptional)---are
isolated in the discussion following \Cref{prop:stabilizers}.
\end{proof}

\begin{lemma}[Sufficiency of the twisted products]\label{lem:twist-sufficiency}
For every $n \geq 2$ and every $t \in \mathbb{R}$, the twisted product
$\circ_t$ of~\eqref{eq:twisted-intro} is a norm-continuous \ref{ax:S1}--\ref{ax:S7} sequential
product on the effects of $H_n(\mathbb{C})$.
\end{lemma}
\begin{proof}
Set $g(\lambda) = \lambda^{1/2+it}$ with $g(0) := 0$, so that
$\lvert g(\lambda)\rvert^2 = \lambda$ and $g$ is multiplicative on
$[0,\infty)$: $g(\lambda\mu) = g(\lambda)g(\mu)$.  Then
$\seqp{a}{b} = a^{1/2+it}\,b\,a^{1/2-it} = f_a(a)\,b\,\overline{f_a}(a)$ with
$f_a = g|_{\mathrm{sp}(a)}$ is the functional-calculus product of this
character family.  Shen--Wu's construction
criterion~\cite[Thm.~3.1]{ShenWu2009} requires of the family $\{f_a\}$
exactly two conditions: \textup{(a)} $\lvert f_a(\lambda)\rvert =
\sqrt{\lambda}$ on the spectrum of $a$, and \textup{(b)} for commuting
$a, b$, $f_a(a)f_b(b) = \zeta\,f_{ab}(ab)$ for some unimodular scalar
$\zeta$.  Here \textup{(a)} is $\lvert g(\lambda)\rvert^{2} = \lambda$, and
\textup{(b)} holds with phase $\zeta \equiv 1$: commuting positive matrices
are jointly diagonalizable, and entrywise $g(\alpha\beta) =
g(\alpha)g(\beta)$ (with $g(0) = 0$ covering zero eigenvalues), so
$g(a)g(b) = g(ab)$ exactly.  Hence the algebraic axioms
\ref{ax:S1} and \ref{ax:S3}--\ref{ax:S7} hold~\cite[Thm.~3.1,
Ex.~3.1]{ShenWu2009}.  For first-variable norm-continuity \ref{ax:S2}: $g$ is
continuous on $[0,\infty)$ (since $\lvert g(\lambda)\rvert = \lambda^{1/2} \to
0$), so $a \mapsto a^{1/2+it}$ is norm-continuous on the positive cone by
continuity of the functional calculus, and therefore so is
$a \mapsto \seqp{a}{b}$ for each fixed $b$.  All seven axioms hold, for every
real $t$.
\end{proof}

\begin{remark}[The labelled classification space of products]\label{rem:moduli}
For a fixed finite-dimensional EJA $J$, let
$\mathcal{M}_{\mathrm{SP}}(J)$ denote the set of norm-continuous \ref{ax:S1}--\ref{ax:S7}
sequential products on its fixed (labelled) effect algebra---a parameter set
\emph{before} any quotient by Jordan automorphisms and with no topology
imposed, so the identifications below are bijections of sets.
\Cref{mthm:master} computes it at rank $n \geq 3$:
\[
  \mathcal{M}_{\mathrm{SP}}(J) = \{\bullet\}
  \ \text{ for } \MnKsa{n}{\mathbb{R}},\ \MnKsa{n}{\mathbb{H}},\ H_3(\mathbb{O}),
  \qquad
  \mathcal{M}_{\mathrm{SP}}(H_n(\mathbb{C})) \cong \mathbb{R},\ \ t \mapsto \circ_t,
\]
with $\{\bullet\}$ the L\"uders singleton and $\circ_t$ the twisted
product~\eqref{eq:twisted-intro}.  For the bijection $t \mapsto \circ_t$,
well-definedness
is \Cref{lem:twist-sufficiency}, surjectivity is \Cref{mthm:master}, and
injectivity follows from \Cref{rem:twist-detection}, whose $2 \times 2$
interferometric computation embeds in a coherence corner of
$H_n(\mathbb{C})$ and there separates every real $t$.  Peirce exchange
covariance (\Cref{cor:selectors}) selects the point $t = 0$; quotienting by
all Jordan automorphisms identifies $t$ with $-t$ through the transpose.  At
rank two the complex qubit instead carries
$\mathcal{M}_{\mathrm{SP}}(\Mnsa{2}) \cong C(\mathbb{RP}^2, \mathbb{R})$
(\Cref{cor:qubit-classification}), into which the constant twists embed.
\end{remark}


%% =====================================================================
%%  SECTION 6: THE RANK-TWO BOUNDARY
%% =====================================================================
\section{The Rank-Two Boundary}
\label{sec:qubit}

At rank two, the preceding argument does not impose cross-frame
constancy.  Throughout this section, write rank-one projections of
$\Mnsa{2}$ as $P_n = \tfrac12(\id + n\cdot\sigma)$ for $n \in S^2$
($\sigma$ the Pauli vector; here $n$ is the Bloch vector, not the rank), so
$P_{-n} = \id - P_n$.  The unordered frames
$\{P_n, P_{-n}\}$ are the points $[n]$ of $S^2/\{\pm1\} = \mathbb{RP}^2$,
and $\mathrm{fr}\,a \in \mathbb{RP}^2$ denotes the frame of a nonscalar
effect $a$.

\paragraph{Ordered frames and their orientation.}
A point $n \in S^2$ is the same datum as an \emph{ordered} frame
$(P_n, P_{-n})$, the antipodal map $n \mapsto -n$ reversing the order.  Write
$W_n := \{x : P_n \circ x = \tfrac12 x = P_{-n} \circ x\} \cong \mathbb{R}^2$
for the Peirce coherence block~\eqref{eq:peirce} of that frame; inside the
traceless self-adjoint $3$-space $Y \cong \mathbb{R}^3$ it is the plane
$n^{\perp}$.  Orient $W_n$ by
\begin{equation}\label{eq:J-ordered}
  \mathcal{J}_n(x) := i\,P_n x P_{-n} - i\,P_{-n} x P_n .
\end{equation}
Writing $x \in W_n$ in the $\{P_n, P_{-n}\}$ eigenbasis as the off-diagonal
$\begin{psmallmatrix}0 & z\\ \bar z & 0\end{psmallmatrix}$ gives
$\mathcal{J}_n(x) = \begin{psmallmatrix}0 & iz\\ -i\bar z & 0\end{psmallmatrix}$,
so $\mathcal{J}_n$ maps $W_n$ to $W_n$ with $\mathcal{J}_n^{2} = -\id_{W_n}$:
the orientation is anchored to the ambient $i$ on the coherence corner and
invokes no right-hand rule on $n$.  Since $n \mapsto -n$ swaps
$P_n \leftrightarrow P_{-n}$,
\begin{equation}\label{eq:J-odd}
  \mathcal{J}_{-n} = -\mathcal{J}_n :
\end{equation}
reversing the order of a frame reverses the orientation of its block.
Formula~\eqref{eq:J-ordered} applied on all of $Y$ annihilates the diagonal
$\mathbb{R}n$ (there $P_n x P_{-n} = 0$), so $\mathcal{J}_n$ extends to $Y$
with $\mathcal{J}_n^{2} = -\Pi_{W_n}$, minus the orthogonal projection of $Y$
onto $W_n$, so that $\mathcal{J}_n^{2}$ has trace $-2$ on $Y$.  Since $P_n$
depends continuously,
indeed real-analytically, on $n \in S^2$, so does $\mathcal{J}_n$ as a
real-linear operator on $Y$.

\begin{proposition}[Necessity at rank two]\label{prop:n2-necessity}
Let $(\cdot)$ be a norm-continuous \ref{ax:S1}--\ref{ax:S7} sequential product
on $\Mnsa{2}$.  For each ordered frame $(P_n, P_{-n})$, $n \in S^2$, there is
a single real $\widetilde t(n)$ such that every invertible effect
$a = \lambda_+P_n + \lambda_-P_{-n}$ diagonal in that frame has comparison map
\begin{equation}\label{eq:n2-block}
  \Theta_a\big|_{W_n}
  = \exp\!\bigl(\ell\;\widetilde t(n)\;\mathcal{J}_n\bigr),
  \qquad \ell := \log(\lambda_+/\lambda_-),
\end{equation}
the log-ratio being taken \emph{in the given order} and allowed either sign:
one real parameter per ordered frame, independent of which effect diagonal in
that frame is used.  No cross-frame constancy follows; that $\widetilde t$
depends only on the unordered frame, and so descends to $\mathbb{RP}^2$, is
\Cref{lem:n2-descent}, proved below.  Singular effects are covered only by
continuity (by the $\epsilon$-regularization argument of \Cref{prop:singular},
carried out for this rank-two case in the proof of
\Cref{cor:qubit-classification}), the ratio $\lambda_+/\lambda_-$ being
undefined at a zero eigenvalue.
\end{proposition}

\begin{proof}
By \Cref{prop:theta} (valid at any rank), $\Theta_a$ is a Jordan
automorphism fixing $P_n$ and $P_{-n}$.  By \Cref{lem:frame-fix} it lies in the
identity component of the frame stabilizer, so it acts on the single block
$W_n$ as a rotation in $\mathrm{SO}(2)$, not a reflection: in the orientation
\eqref{eq:J-ordered}, as $\exp(\theta\,\mathcal{J}_n)$ for a real angle
$\theta$.  For diagonal
$a(r) = e^{r_1}P_n + e^{r_2}P_{-n}$ the standard product satisfies
$a(r)\bullet a(r') = a(r+r')$, so Proposition~5.7
of~\cite{Wetering2018three} (\Cref{prop:theta}) makes
$r \mapsto \Theta_{a(r)}|_{W_n}$ a continuous homomorphism on the negative orthant,
extended to a homomorphism $\mathbb{R}^2 \to \mathrm{SO}(2)$ by the
cancellation extension of \Cref{lem:homomorphism} (valid at any rank).
By coalescence (\Cref{lem:coalescence}, valid at any rank) it is trivial
when $r_1 = r_2$.  Since $\mathbb{R}^2$ is simply connected, a continuous
homomorphism $\mathbb{R}^2 \to \mathrm{SO}(2)$ lifts through the universal
cover $\mathbb{R} \to \mathrm{SO}(2)$ to a continuous homomorphism
$\mathbb{R}^2 \to \mathbb{R}$, i.e.\ to a linear functional.  One killing the
diagonal is a multiple of $r_1 - r_2$, so the homomorphism is
$r \mapsto \exp\bigl((r_1 - r_2)\,\widetilde t(n)\,\mathcal{J}_n\bigr)$ for a
single real $\widetilde t(n)$.  Here $r_1 - r_2$ is exactly the ordered
log-ratio $\ell$, which gives~\eqref{eq:n2-block} for either sign of $\ell$.
No magnitude ordering of the two eigenvalues is imposed anywhere in this
argument: the parameter is attached to the ordered frame, and the sign of
$\ell$ is carried by the formula.  Distinct frames of
$\Mnsa{2}$ share no atom and the complement of a rank-one $P$ is rank one,
so there is no third projection to compare frames.  The frame-overlap step
of \Cref{lem:adjacent}, which acts on the cross-coherence space of a shared
spectator atom, does not apply.
\end{proof}

In fact the rank-two products are completely classified.  We prove the
constructive half first: every continuous frame function yields a product
(\Cref{prop:n2-sufficiency}), with an explicit frame-dependent family
witnessing that the freedom is real (\Cref{thm:qubit-boundary}).  The
converse then closes: the frame parameter of \Cref{prop:n2-necessity} is
bounded (\Cref{lem:n2-bounded}), continuous (\Cref{lem:n2-continuity}), and
descends to $\mathbb{RP}^2$ (\Cref{lem:n2-descent}), so the norm-continuous \ref{ax:S1}--\ref{ax:S7} products
on $\Mnsa{2}$ are exactly $C(\mathbb{RP}^2, \mathbb{R})$
(\Cref{cor:qubit-classification}).

\begin{remark}[Rank-two selection consequences]\label{rem:n2-selection}
Two selector consequences follow for the two rank-two qubits without any
exhaustive classification.  First, \Cref{prop:real} applies at rank two: the real
qubit's unique norm-continuous \ref{ax:S1}--\ref{ax:S7} product is the L\"uders product, whose
coherence-block action $\sqrt{xy}\,\mathrm{Id}$ is Peirce exchange
covariant.  Second, for $\Mnsa{2}$: by \Cref{prop:n2-necessity} and
\Cref{lem:frame-fix} the block action of an invertible nonscalar
$a = \lambda_+P_n + \lambda_-P_{-n}$ is
$L_a|_{W_n} = Q_{\sqrt a}\,\Theta_a|_{W_n} =
\sqrt{\lambda_+\lambda_-}\,\exp(\ell\,\widetilde t(n)\,\mathcal{J}_n)$ with
$\ell = \log(\lambda_+/\lambda_-)$, the scalar factor being
exchange-symmetric.  Here Peirce exchange covariance (\Cref{sec:selectors}) is the
fixed-orientation condition: for the fixed ordered frame $(P_n, P_{-n})$,
exchanging the two eigenvalues---which replaces $\ell$ by $-\ell$---leaves
unchanged the induced endomorphism of the \emph{same} coherence space $W_n$,
in its fixed ambient orientation $\mathcal{J}_n$.  (Relabeling the frame instead
would also send $\mathcal{J}_n$ to $\mathcal{J}_{-n} = -\mathcal{J}_n$
by~\eqref{eq:J-odd} and make the two sign changes cancel; the condition
constrains the block action at fixed orientation.)  It forces
\[
  \exp(\ell\,\widetilde t(n)\,\mathcal{J}_n)
  = \exp(-\ell\,\widetilde t(n)\,\mathcal{J}_n)
\]
for $\ell$ over an interval, hence $\widetilde t(n) = 0$ for every frame; with scalar
effects L\"uders and singular effects covered by \ref{ax:S2}-continuity,
every norm-continuous \ref{ax:S1}--\ref{ax:S7} product on $\Mnsa{2}$ satisfying Peirce
exchange covariance is the L\"uders product, the products that fail it being
left to the full classification of \Cref{cor:qubit-classification}.
\end{remark}

\begin{proposition}[Sufficiency: every continuous frame function is a product]
\label{prop:n2-sufficiency}
For every continuous $t : \mathbb{RP}^2 \to \mathbb{R}$, the operation
\[
  \seqp{a}{b} := a^{1/2+i\,t_a}\,b\,a^{1/2-i\,t_a},
  \qquad t_a := t(\mathrm{fr}\,a) \text{ for nonscalar } a,
  \quad t_a := 0 \text{ for scalar } a
\]
is a norm-continuous \ref{ax:S1}--\ref{ax:S7} sequential product on $\Mnsa{2}$; we write
$\circ_t$ for it, extending the constant-twist notation
of~\eqref{eq:twisted-intro}.
\end{proposition}

The algebraic content of this family is Shen--Wu's
Example~3.2~\cite{ShenWu2009}, which assigns an arbitrary real parameter to
each unordered frame.  This paper adds \ref{ax:S2} (Step~3), the axiom
that on the necessity side trades their arbitrary frame parameter for a
continuous one, together with a self-contained verification of all seven
axioms.

\begin{proof}
Throughout, $F_a := a^{1/2 + i\,t_a}$, a normal matrix with
$F_aF_a^{*} = a$, so that uniformly $\seqp{a}{b} = F_a b F_a^{*}$ (for
scalar $a$, $F_a = \sqrt a$).  The steps use only that $t_a$ is a real
number attached to the unordered frame of $a$, bounded and continuous in
the frame, and they are self-contained.  The displayed algebraic
identities, with the twist value carried as a free real
parameter, are additionally corroborated in exact symbolic arithmetic by
\texttt{verify\_n2.py} (labels V1--V8; V9, particular to the family of
\Cref{thm:qubit-boundary}, appears in the same script; see the Supplementary
Material).

\emph{Step 0.}  For scalar $a = \lambda\id$, $\seqp{a}{b} = \lambda b$
(V2); a nonscalar effect has two distinct eigenvalues and a unique
unordered frame $\{P, \id - P\}$, so $t_a$ is well defined.

\emph{Step 1 (block form).}  For nonscalar
$a = \lambda_1 P + \lambda_2 P^{\perp}$, $t = t_a$,
\[
  \seqp{a}{b} = \lambda_1 PbP + \lambda_2 P^{\perp}bP^{\perp}
  + \sqrt{\lambda_1\lambda_2}\bigl(e^{it\log(\lambda_1/\lambda_2)}PbP^{\perp}
  + e^{-it\log(\lambda_1/\lambda_2)}P^{\perp}bP\bigr)
\]
(V1), the L\"uders value for $\lambda_2 = 0$ (V3).  Cone-preservation:
$b \mapsto F_abF_a^{*}$ is positive
and $b \leq \id$ gives $F_abF_a^{*} \leq a \leq \id$.

\emph{Step 2 (S1, S3, S4).}  \ref{ax:S1}: $b \mapsto F_abF_a^{*}$ is linear.
\ref{ax:S3}: $\id$ is scalar with $\lambda = 1$ (Step 0).  \ref{ax:S4}:
suppose $\seqp{a}{b} = (F_a\sqrt{b})(F_a\sqrt{b})^{*} = 0$, so
$F_a\sqrt{b} = 0$.  With the partial isometry
$u = \sum_{\lambda_j > 0}\lambda_j^{-i\,t_a}p_j$ one has $\sqrt{a} = uF_a$,
so $\sqrt{a}\sqrt{b} = u\,F_a\sqrt{b} = 0$; taking adjoints,
$\sqrt{b}\sqrt{a} = 0$.  Writing $F_b = v\sqrt{b}$ with
$v = \sum_{\mu_j > 0}\mu_j^{\,it_b}q_j$ for the spectral data
$b = \sum_j \mu_j q_j$ (note the opposite sign of exponent from $u$), we get $F_b\sqrt{a} = v\,\sqrt{b}\sqrt{a} = 0$, whence
$\seqp{b}{a} = (F_b\sqrt{a})(F_b\sqrt{a})^{*} = 0$ (instances: V8).

\emph{Step 3 (S2: continuity of $a \mapsto \seqp{a}{b}$).}  At a nonscalar
$a_0$: the eigenvalues are continuous, the frame
$P(a) = (a - \lambda_2\id)/(\lambda_1 - \lambda_2)$ and with it
$\mathrm{fr}\,a \in \mathbb{RP}^2$ are continuous on the nonscalar locus,
so $t_a = t(\mathrm{fr}\,a)$ is continuous there.  The map
$(z, s) \mapsto z^{1/2+is}$ (with
value $0$ at $z = 0$) is jointly continuous on $[0,1] \times \mathbb{R}$,
since its modulus is $\sqrt{z}$.  Hence $a \mapsto F_a$, and with it
$a \mapsto \seqp{a}{b}$, is continuous at $a_0$, including rank-deficient
$a_0$.  At a scalar $a_0 = \lambda_0\id$: for nonscalar $a$ with eigenvalues
$\lambda_1, \lambda_2$, frame $P$, twist value $t_a$,
expanding $b$ into its four $P$-blocks (each blocking map a contraction),
\[
  \|\seqp{a}{b} - \lambda_0 b\|
  \leq \Bigl(|\lambda_1 - \lambda_0| + |\lambda_2 - \lambda_0|
  + 2\,\bigl|\sqrt{\lambda_1\lambda_2}\,e^{i\theta} - \lambda_0\bigr|\Bigr)\|b\|,
  \quad \theta = t_a\log(\tfrac{\lambda_1}{\lambda_2}),
\]
the cross coefficient read as its actual value $0$ when $\lambda_2 = 0$.  If
$\lambda_0 > 0$, then eventually $\lambda_2 > 0$ along $a \to \lambda_0\id$
and $\lambda_1, \lambda_2 \to \lambda_0$, while
$|t_a| \leq \|t\|_{\infty} < \infty$ ($t$ is continuous on the compact
$\mathbb{RP}^2$), so $\theta \to 0$ and every term
vanishes.  If $\lambda_0 = 0$: since $b \leq \id$,
$0 \leq \seqp{a}{b} = F_abF_a^{*} \leq F_aF_a^{*} = a$, so
$\|\seqp{a}{b}\| \leq \|a\| = \max(\lambda_1,\lambda_2) \to 0$ (using no bound on $t$).
Either way $\seqp{a}{b} \to \lambda_0 b$.

\emph{Step 4 (compatibility equals commutation).}  We show
$\seqp{a}{b} = \seqp{b}{a} \iff ab = ba$.  ($\Leftarrow$)  If $a = \lambda\id$
is scalar, $\seqp{a}{b} = \lambda b$ (Step 0) and
$\seqp{b}{a} = F_b(\lambda\id)F_b^{*} = \lambda F_bF_b^{*} = \lambda b$;
symmetrically if $b$ is scalar.  Otherwise $a, b$ share their unique frame,
$F_a$ is a functional-calculus function of $a$, so $[F_a, b] = 0$ and
$\seqp{a}{b} = F_aF_a^{*}b = ab = ba = \seqp{b}{a}$.  ($\Rightarrow$)  Suppose
$ab \neq ba$; then $a, b$ are both nonscalar.  Since
$F_a^{*}aF_a = aF_a^{*}F_a = a^2$ (functional calculus) and the trace is
cyclic, $\operatorname{tr}((\seqp{a}{b})a) = \operatorname{tr}(F_abF_a^{*}a)
= \operatorname{tr}(bF_a^{*}aF_a) = \operatorname{tr}(a^2b)$.  Computing the
other pairing in $b$'s eigenbasis, with $b = \mu_1 Q + \mu_2 Q^{\perp}$
($\mu_1 > \mu_2 \geq 0$), $\hat a$ the matrix of $a$ there, and
$\theta_b = t_b\log(\mu_1/\mu_2)$,
\[
  \operatorname{tr}(a^2b) - \operatorname{tr}((\seqp{b}{a})a)
  = \bigl(\mu_1 + \mu_2 - 2\sqrt{\mu_1\mu_2}\cos\theta_b\bigr)\,|\hat a_{12}|^2
\]
(and $= \mu_1|\hat a_{12}|^2$ if $\mu_2 = 0$).  Here
$\mu_1 + \mu_2 - 2\sqrt{\mu_1\mu_2}\cos\theta_b =
(\sqrt{\mu_1} - \sqrt{\mu_2})^2 + 2\sqrt{\mu_1\mu_2}(1 - \cos\theta_b) > 0$
for $\mu_1 \neq \mu_2$, and $\hat a_{12} \neq 0$ exactly when $ab \neq ba$,
so the two pairings differ and $\seqp{a}{b} \neq \seqp{b}{a}$.  (The
displayed trace identities are corroborated symbolically as V4.)

\emph{Step 5 (S6, S7).}  Compatibility is commutation (Step 4);
$[a,b] = 0$ gives $[a, \id - b] = [a, b+c] = 0$ and, as $F_b$ is a
function of $b$, $[a, \seqp{b}{c}] = F_b[a,c]F_b^{*} = 0$.

\emph{Step 6 (S5; the phase cocycle).}  Let $a \compatible b$, so $ab = ba$
and $\seqp{a}{b} = ab$ (Step 4).  We claim $F_aF_b = \zeta(a,b)F_{ab}$ with
$|\zeta(a,b)| = 1$, from which \ref{ax:S5} follows:
$\seqp{a}{(\seqp{b}{c})} = F_aF_b\,c\,F_b^{*}F_a^{*} = F_{ab}\,c\,F_{ab}^{*}
= \seqp{(ab)}{c} = \seqp{(\seqp{a}{b})}{c}$.  Since $a, b$ commute they are
jointly diagonalizable, with eigenvalue pairs $(\alpha_k, \beta_k)$.  If
$a, b, ab$ are all nonscalar they share one frame, hence one twist value
$t$, and entrywise $\alpha_k^{1/2+it}\beta_k^{1/2+it} =
(\alpha_k\beta_k)^{1/2+it}$ (exact power laws for nonnegative reals, with
$0 = 0$ at a zero eigenvalue), so $\zeta = 1$.  If both are scalar,
$\zeta = 1$.  If $b = \beta\id$ is scalar ($\beta > 0$) and $a$ nonscalar,
then $\beta a$ is nonscalar with the same frame, $t_{\beta a} = t_a
=: s$ (the frames coincide), and $F_aF_b = \sqrt{\beta}\,F_a =
\beta^{-is}F_{\beta a}$, so
$\zeta = \beta^{-is}$; if $\beta = 0$ both sides vanish, and symmetrically
for $a$ scalar.  If $a, b$ are nonscalar but $ab = \gamma\id$ is scalar with
$\gamma > 0$, then $\log(\alpha_1/\alpha_2) = -\log(\beta_1/\beta_2)$ and
$F_aF_b = \sqrt{\gamma}\,e^{it\log\gamma}\id = e^{it\log\gamma}F_{ab}$, so
$\zeta = e^{it\log\gamma}$; if $\gamma = 0$ both sides vanish.  In every
case $\zeta$ may be taken with $|\zeta| = 1$.  (The case verification is corroborated symbolically as
V5--V6.)
\end{proof}

\begin{theorem}[The rank-two boundary: rigidity genuinely fails]
\label{thm:qubit-boundary}
Let $V = \Mnsa{2}$.  Fix a rank-one projection $R$ and, for a nonscalar
effect $a$ with spectral frame $F = \{P, \id - P\}$, set
\begin{equation}\label{eq:tau-def}
  \tau(a) := \bigl(2\operatorname{tr}(PR) - 1\bigr)^{2} \in [0,1],
\end{equation}
a well-defined function of the frame; $\tau(a) := 0$ for scalar $a$.
Define
\begin{equation}\label{eq:qubit-family}
  \seqp{a}{b} := a^{1/2 + i\tau(a)}\; b\; a^{1/2 - i\tau(a)},
  \qquad 0^{z} := 0.
\end{equation}
Then:
\begin{enumerate}[label=(\roman*)]
\item the operation~\eqref{eq:qubit-family} maps effects to effects; on the
  ordered frame $(P_n, P_{-n})$ of a nonscalar effect
  $a = \lambda_+P_n + \lambda_-P_{-n}$ its coherence action on $W_n$ is the
  rotation
  \[
    \sqrt{\lambda_+\lambda_-}\;
    \exp\!\bigl(\ell\,\tau(a)\,\mathcal{J}_n\bigr),
    \qquad \ell = \log(\lambda_+/\lambda_-)
  \]
  (for $\lambda_- > 0$; at $\lambda_- = 0$ the prefactor
  $\sqrt{\lambda_+\lambda_-}$ vanishes and, by $0^{z} := 0$, the coherence
  action is the L\"uders value $0$);
\item it satisfies all seven axioms \ref{ax:S1}--\ref{ax:S7};
\item it is genuinely frame-dependent: there is no pair $(\Phi, t)$,
  consisting of a Jordan automorphism $\Phi$ of $V$ and a constant
  $t \in \mathbb{R}$, such that
  $\seqp{a}{b} = \Phi^{-1}\bigl(\Phi(a) \circ_t \Phi(b)\bigr)$ for all
  effects.  In particular the complex conclusion of \Cref{mthm:master}
  fails at rank two.
\end{enumerate}
\end{theorem}

\begin{proof}
\emph{$\tau$ is a continuous frame function.}  Replacing $P$ by $\id - P$
negates $2\operatorname{tr}(PR) - 1$, which the square in~\eqref{eq:tau-def}
undoes, so $\tau$ is a well-defined function of the unordered frame.  As a
function of $[n] \in \mathbb{RP}^2$ it is continuous, being a polynomial in
$\operatorname{tr}(P_nR)$ with $P_n$ continuous in $n$ and the value even
under $n \mapsto -n$.  For scalar $a$ both~\eqref{eq:qubit-family} and the
operation of \Cref{prop:n2-sufficiency} give the L\"uders value, so the
operation~\eqref{eq:qubit-family} \emph{is} the operation of
\Cref{prop:n2-sufficiency} at the frame function $\tau$.  (ii) follows from
\Cref{prop:n2-sufficiency}, and Step~1 there gives (i): the block form, the
coherence rotation with twist value $\tau(a)$, the L\"uders value at
$\lambda_- = 0$ (V3), and effect preservation, on taking
$P = P_n$ and $(\lambda_1, \lambda_2) = (\lambda_+, \lambda_-)$ there.

\emph{Frame dependence (iii).}  $\tau = 1$ on the frame of $R$ and
$0$ on any frame with $\operatorname{tr}(PR) = \tfrac12$ (V9).  The unital
Jordan automorphisms of $\Mnsa{2}$ are $X \mapsto UXU^{*}$ and
$X \mapsto UX^{\mathsf{T}}U^{*}$ ($U \in SU(2)$),\footnote{A unital Jordan
automorphism $\Phi$ of $\Mnsa{2}$ fixes the trace, and since
$(x\cdot\sigma)^{2} = |x|^{2}I$ for the Pauli vector $\sigma$ it preserves the
Euclidean norm on the traceless part, so it acts there as some
$O \in \mathrm{O}(3)$.  The adjoint action of $SU(2)$ realizes all of $SO(3)$ and
the transpose realizes the reflection reversing the $\sigma_y$-axis, giving the
two stated forms.} acting on the Bloch ball as
$SO(3)$ and its orientation-reversing coset, i.e.\ as $\mathrm{O}(3)$; the
conjugations preserve the block orientation $\mathcal{J}_n$ (hence $t$) and the
transpose reverses it (hence $t \mapsto -t$).  So any Jordan automorphism
$\Phi$ and constant $t$ transport $\circ_t$ to a product whose twist is
$\pm t$ on every frame, constant in magnitude; since $\tau$ takes both
values $0$ and $1$, no $(\Phi, t)$ reproduces~\eqref{eq:qubit-family}.
\end{proof}

\begin{lemma}[The frame parameter is bounded]\label{lem:n2-bounded}
For any norm-continuous \ref{ax:S1}--\ref{ax:S7} product on $\Mnsa{2}$,
$M := \sup_{n \in S^2}|\widetilde t(n)| < \infty$, with $\widetilde t$ the
ordered-frame parameter of \Cref{prop:n2-necessity}.
\end{lemma}

\begin{proof}
Norms here are the operator norms on the finite-dimensional space of
endomorphisms of $\Mnsa{2}$ induced by the order unit norm; all norms on that
space are equivalent, so which one is used affects no statement below.  By
\ref{ax:S2} and finite dimension $a \mapsto L_a$ is operator-norm
continuous on invertible effects, and $a \mapsto Q_{a^{-1/2}}$ is continuous,
so $a \mapsto \Theta_a = Q_{a^{-1/2}}L_a$ is operator-norm continuous (as in
\Cref{lem:frame-fix}).  If $|\widetilde t(n_k)| \to \infty$ along some
sequence $n_k \in S^2$, fix $\lambda \in (0,1)$, put
$s_k = \pi/|\widetilde t(n_k)| \to 0$ and
$a_k = \lambda e^{s_k/2}P_{n_k} + \lambda e^{-s_k/2}P_{-n_k}$ (an effect for all large $k$: since
$s_k \to 0$, $\lambda e^{s_k/2} \to \lambda < 1$, so the finitely many
terms with $\lambda e^{s_k/2} > 1$ are discarded, and the contradiction
uses only the tail).  Then $a_k = \lambda\id + O(s_k) \to
\lambda\id$ uniformly in the frame (the perturbation is $O(s_k)$ and
$\|P_{n_k}\| = 1$), so $\Theta_{a_k} \to \Theta_{\lambda\id} = \id$.  (Here
$\Theta_{\lambda\id} = \id$ because $\lambda\id$ operator-commutes with every
effect, so \Cref{prop:theta} fixes the whole algebra.)  But $a_k$ has ordered
log-ratio $s_k$ in the frame $n_k$, so by~\eqref{eq:n2-block} its rotation
angle on $W_{n_k}$ is $\widetilde t(n_k)s_k = \pm\pi$, giving
$\Theta_{a_k}|_{W_{n_k}} = -\id$.  Any norm-one $x_k \in W_{n_k}$ then has
$(\Theta_{a_k} - \id)\,x_k = -2x_k$, so $\|\Theta_{a_k} - \id\| \geq 2$ even
though the plane $W_{n_k}$ varies with $k$, a contradiction.  Hence
$\sup_{n}|\widetilde t(n)| < \infty$.
\end{proof}

\begin{lemma}[Continuity of the ordered-frame parameter]
\label{lem:n2-continuity}
The ordered-frame parameter $\widetilde t$ of \Cref{prop:n2-necessity} is
continuous on $S^2$.
\end{lemma}

\begin{proof}
Fix $\lambda \in (0,1)$ and $s > 0$ with $sM < \pi$ ($M$
finite by \Cref{lem:n2-bounded}) and $\lambda e^{s/2} \leq 1$, and set
$a_n(s) = \lambda e^{s/2}P_n + \lambda e^{-s/2}P_{-n}$, an effect with ordered
log-ratio $s$, continuous in $n$.  Work on the \emph{fixed} traceless
self-adjoint $3$-space $Y$.  By \Cref{prop:n2-necessity} and \Cref{lem:frame-fix},
$\Theta_{a_n(s)}|_Y = \id_{\mathbb{R}n} \oplus (\text{rotation of } W_n) \in SO(3)$
with rotation angle $s\,\widetilde t(n)$, $|s\,\widetilde t(n)| \leq sM < \pi$.
It therefore has no eigenvalue $-1$, and its principal logarithm is defined and is a
continuous $\mathfrak{so}(3)$-valued function of $n$,
\[
  K_n := \operatorname{Log}\!\bigl(\Theta_{a_n(s)}|_Y\bigr) = s\,\widetilde t(n)\,\mathcal{J}_n
\]
(both sides annihilate $\mathbb{R}n$ and act as the angle-$s\widetilde t(n)$
generator on $W_n$).  Recover the scalar by the explicit formula
\[
  \widetilde t(n) = -\frac{1}{2s}\,\operatorname{tr}_{Y}\!\bigl(\mathcal{J}_n K_n\bigr),
\]
valid because $\operatorname{tr}_{Y}(\mathcal{J}_n K_n)
= s\,\widetilde t(n)\,\operatorname{tr}_{Y}(\mathcal{J}_n^2)
= s\,\widetilde t(n)\cdot(-2)$, since $\mathcal{J}_n^2 = -\Pi_{W_n}$ has trace
$-2$ on $Y$ (killing the $n$-axis).  As $n \mapsto \mathcal{J}_n$ and
$n \mapsto K_n$ are continuous, $\widetilde t$ is continuous on $S^2$; the trace
is now taken over the fixed space $Y$, not over a moving $2$-plane.
\end{proof}

\begin{lemma}[Evenness and descent to $\mathbb{RP}^2$]\label{lem:n2-descent}
$\widetilde t(-n) = \widetilde t(n)$ for all $n \in S^2$: reversing the order
of a frame reverses the orientation of its block~\eqref{eq:J-odd} and negates
the ordered log-ratio, and those two sign changes cancel
in~\eqref{eq:n2-block}.  The frame parameter therefore descends to a single
continuous, bounded function
$t : S^2/\{\pm1\} = \mathbb{RP}^2 \to \mathbb{R}$ of the unordered
frame---a genuine $\mathbb{RP}^2$ function, with no residual line-bundle
twist.
\end{lemma}

\begin{proof}
With $a_n(s) = \lambda e^{s/2}P_n + \lambda e^{-s/2}P_{-n}$ as in
\Cref{lem:n2-continuity} (the same fixed $s$, so that $sM < \pi$; both
$a_n(\pm s)$ are effects because $\lambda e^{s/2} \leq 1$), the swap
$P_n \leftrightarrow P_{-n}$ gives the single effect two descriptions,
$a_{-n}(s) = a_n(-s)$, hence one comparison map
$\Theta_{a_{-n}(s)} = \Theta_{a_n(-s)}$, and $W_{-n} = W_n$.  Read off each
description by~\eqref{eq:n2-block}: in the ordered frame $-n$ the ordered
log-ratio is $+s$, and in the ordered frame $n$ it is $-s$, so
\[
  \exp\!\bigl(s\,\widetilde t(-n)\,\mathcal{J}_{-n}\bigr)
  = \exp\!\bigl(-s\,\widetilde t(-n)\,\mathcal{J}_n\bigr)
  \qquad\text{must equal}\qquad
  \exp\!\bigl(-s\,\widetilde t(n)\,\mathcal{J}_n\bigr),
\]
the first equality by the oddness~\eqref{eq:J-odd} of the orientation.  Both
exponents have angle in $(-\pi,\pi)$ against the same generator
$\mathcal{J}_n$, so equality of the two rotations forces
$\widetilde t(-n) = \widetilde t(n)$.  No eigenvalue-magnitude
comparison enters: \eqref{eq:n2-block} holds at either sign of the ordered
log-ratio, so the two readings of the same effect are
comparable.  Boundedness is \Cref{lem:n2-bounded}, and continuity passes to
the quotient.
\end{proof}

\begin{corollary}[Rank-two classification]\label{cor:qubit-classification}
The map $t \mapsto \circ_t$ of \Cref{prop:n2-sufficiency} is a bijection from
$C(\mathbb{RP}^2, \mathbb{R})$ onto the norm-continuous \ref{ax:S1}--\ref{ax:S7} sequential
products on $\Mnsa{2}$, an explicit bijection of sets for the fixed labelled
complex matrix algebra $\Mnsa{2}$:
\[
  \mathcal{M}_{\mathrm{SP}}(\Mnsa{2}) \;\cong\; C(\mathbb{RP}^2, \mathbb{R}).
\]
\end{corollary}

\begin{proof}
\Cref{prop:n2-sufficiency} gives the map.  For the inverse, any product's
frame parameter is bounded and continuous, and it descends to
$\mathbb{RP}^2$ (\Cref{prop:n2-necessity};
\Cref{lem:n2-bounded,lem:n2-continuity,lem:n2-descent}).
The identity $L_a = Q_{\sqrt a}\,\Theta_a$ from \Cref{prop:theta}, the
diagonal fixing from \Cref{lem:frame-fix}, and the coherence action from
\Cref{prop:n2-necessity} together show that the product equals $\circ_t$ on
invertible nonscalar effects.  The two sides are matched by taking
$P = P_n$ and $(\lambda_1, \lambda_2) = (\lambda_+, \lambda_-)$, so that
$\log(\lambda_1/\lambda_2) = \ell$: Step~1 of \Cref{prop:n2-sufficiency} puts
$\sqrt{\lambda_1\lambda_2}\,e^{\,i\,t\log(\lambda_1/\lambda_2)}$ on the
$PbP^{\perp}$ corner, while $L_a = Q_{\sqrt a}\Theta_a$
with~\eqref{eq:n2-block} puts
$\sqrt{\lambda_+\lambda_-}\,e^{\,i\,\ell\,\widetilde t(n)}$ there---the same
orientation $\mathcal{J}_n$ and the same sign of $\ell$ on both sides, so
$t = \widetilde t$.  Since both sides are \ref{ax:S2}-continuous
(\Cref{prop:n2-sufficiency}) and the invertible nonscalar effects are dense,
they agree on all effects.
Injectivity: if $t \neq t'$ they differ at some frame $F = [n]$; choose
$s > 0$ small enough that $0 < s\,|t(F) - t'(F)| < 2\pi$, and take the
nonscalar effect $a$ diagonal in $F$ with eigenvalues in $(0,1)$ and ordered
log-ratio $\ell = s$ in the ordering $(P_n, P_{-n})$.  Then the coherence
rotations $\exp(s\,t(F)\,\mathcal{J}_n)$ and $\exp(s\,t'(F)\,\mathcal{J}_n)$
are distinct---their angles differ by less than a full turn---so they differ
on any $b$ with nonzero $P_nbP_{-n}$ corner, and $\circ_t \neq \circ_{t'}$.
The target is the function space $C(\mathbb{RP}^2, \mathbb{R})$, not
a twisted-section space: the orientation datum cancels in the frame parameter
(\Cref{lem:n2-descent}), so $\widetilde t$ is a function on $\mathbb{RP}^2$ with
no residual bundle obstruction.
\end{proof}

\begin{remark}[Attribution]\label{rem:qubit-attribution}
The non-L\"uders freedom is Shen--Wu's~\cite[Thm.~3.1, Ex.~3.2]{ShenWu2009}
(their frame parameter arbitrary and not required continuous);
\Cref{thm:qubit-boundary} specializes it to a continuous example verified
against the full seven-axiom list, and the new content is the
necessity half of \Cref{cor:qubit-classification}---see the Liu--Wu/Shen--Wu
paragraph of \Cref{sec:priorart}.  At rank two compatible nonscalar effects
share their unique frame, so \ref{ax:S5}--\ref{ax:S7} never compare distinct
nonscalar frames.
\end{remark}

\begin{remark}[Operational detectability of the twist]
\label{rem:twist-detection}
The parameter $t$ in the complex family
$a\circ_t b = a^{1/2+it}\,b\,a^{1/2-it}$ has an interferometric signature
on a single qubit.  Take the unsharp first effect
$a_r=\operatorname{diag}(e^{-r},1)$, the sharp second effect
$b=\lvert+\rangle\langle+\rvert$, and prepare
$\rho_\phi=\tfrac12\bigl(\begin{smallmatrix}1 & e^{-i\phi}\\[1pt] e^{i\phi}&1\end{smallmatrix}\bigr)$.
The ordered joint probability that $a_r$ then $b$ both return ``yes'' is
\[
  p_t(r,\phi)=\operatorname{tr}\!\bigl(\rho_\phi\,(a_r\circ_t b)\bigr)
  =\frac{1+e^{-r}}{4}+\frac{e^{-r/2}}{2}\,\cos(\phi-tr).
\]
Scanning the state phase isolates two quadratures,
$p_t(r,0)\propto\cos(tr)$ and $p_t(r,\tfrac\pi2)\propto\sin(tr)$ above the
$t$-independent pedestal $\tfrac14(1+e^{-r})$; a single unsharp $r>0$ fixes
$t$ modulo $2\pi/r$, and two values with irrational ratio fix
$t\in\mathbb R$ in the noiseless model.  Unsharpness is essential: for a
projection $a$ one has $a^{1/2\pm it}=a$, hence $a\circ_t b=aba$ and the
$t$-dependence vanishes.%
\footnote{This is a tomography calculation for the Liu--Wu family, not new
structure.  Liu and Wu~\cite{LiuWu2009} introduce the one-parameter product
$A^{1/2+ti}BA^{1/2-ti}$, $t\in\mathbb R$ (their Remark~1), and compute its
explicit qubit form (their Corollary~4.4, $a, b > 0$ branch, with the $t$
restored in their Remark~1): for
$A=\operatorname{diag}(a^2,b^2)$ and
$B=\bigl(\begin{smallmatrix}x&y\\ \bar y&z\end{smallmatrix}\bigr)$ the
off-diagonal entry carries the phase $\theta=(\ln a^2-\ln b^2)t$, exactly
the phase $e^{-itr}$ behind the $\cos(\phi-tr)$ above; they already read
$t$ as a physical ``time
parameter.''  Shen and Wu~\cite{ShenWu2009} generalise the construction but
state no state-dependent detection formula.}
\end{remark}


%% =====================================================================
%%  SECTION 7: SELECTORS
%% =====================================================================
\section{Selectors: Prior-Art Corollaries}
\label{sec:selectors}

For the complex type the master theorem leaves the one-parameter family;
several classical side conditions select $t = 0$.  The last two are van de
Wetering's or Gudder--Latr\'emoli\`ere's, in greater generality; we record
them as corollaries to locate where selection happens, not as new
characterizations.  Write $E(x,y)$ for the action of $\seqp{a}{(\cdot)}$ on a
coherence block $V_{ij}$ when $a$ has eigenvalues $x, y$ on the two atoms
$p_i, p_j$; by \Cref{thm:complex} it is
$E(x,y) = \sqrt{xy}\,\exp(t\log(x/y)\,\mathcal{J})$ with $\mathcal{J}$ the
block's complex structure.  Here \emph{Peirce exchange covariance} means
$E(x,y) = E(y,x)$ with the
block's complex structure $\mathcal{J}$ held \emph{fixed} (the fixed-orientation
condition of \Cref{rem:n2-selection}, not a relabelling of $p_i, p_j$).

\begin{corollary}[Selectors that fix $t = 0$]\label{cor:selectors}
On $H_n(\mathbb{C})$, $n \geq 3$, the complex-type product of
\Cref{mthm:master} is the L\"uders product ($t = 0$) if any of the
following holds: (i)~Peirce exchange covariance,
$E(x,y) = E(y,x)$ on every coherence block; (ii)~trace-form symmetry,
$\langle \seqp{a}{b}, c\rangle = \langle b, \seqp{a}{c}\rangle$;
(iii)~covariance under every unital order automorphism (on $\Mnsa{n}$ the
transpose suffices).
\end{corollary}

\begin{proof}
Write the block action as $\sqrt{xy}\,\exp(t\log(x/y)\,\mathcal{J})$.
(i) $E(x,y) = E(y,x)$ for all $x,y$ iff $t = 0$.
(ii) Trace-orthogonality of
Peirce spaces makes symmetry equivalent to self-adjointness of the block
rotation $\exp(t\log(x/y)\,\mathcal{J})$ for all admissible $x,y$.  A real rotation is
self-adjoint exactly at angles in $\pi\mathbb{Z}$, but $t\log(x/y)$ ranges over
an interval as $x,y$ vary, so this forces $t = 0$.
(iii) The transpose acts on
each block by $\mathcal{J} \mapsto -\mathcal{J}$, forcing $t = -t$.
\end{proof}

Attribution of the selectors, and the weakenings that fail to select, are
recorded with the prior art in \Cref{sec:priorart}.


%% =====================================================================
%%  SECTION 8: PRIOR ART AND NOVELTY
%% =====================================================================
\section{Prior Art, Novelty, and Provenance}
\label{sec:priorart}

\paragraph{Gudder--Latr\'emoli\`ere (2008).}
They characterize the L\"uders product on Hilbert-space effects from five
conditions~\cite{GudderLatremoliere2008}, one a trace-duality condition
$\operatorname{tr}((\seqp{a}{b})c) = \operatorname{tr}(b(\seqp{a}{c}))$,
the Hilbert-space form of selector~(ii).
The twisted products satisfy Gudder--Latr\'emoli\`ere's other four conditions,
including their weak associativity $\seqp{A}{(\seqp{A}{B})} = \seqp{A^2}{B}$; this is
distinct from the SEA compatible-associativity axiom \ref{ax:S5}, which the twists
also satisfy (Liu--Wu, Shen--Wu).  \Cref{mthm:master} classifies without the trace-duality
condition.

\paragraph{Liu--Wu, Shen--Wu.}
The twisted family~\eqref{eq:twisted-intro} is Liu--Wu's~\cite[Thm.~4.3,
Cor.~4.4, Rem.~1]{LiuWu2009}, an existence result answering a
uniqueness question from Gudder's open-problems list~\cite{Gudder2005} in the
negative: they construct the family and
read $t$ as a physical time parameter, but do not classify the
\ref{ax:S1}--\ref{ax:S7} products.  Their Theorem~4.3 and Corollary~4.4 are
constructions, and their rank-two form result (their Theorem~3.2) forces a
single phase on the coherence corner for each fixed first argument while
leaving the dependence of that phase on the argument unconstrained.  None of
these shows that every such product lies in the constant-$t$ family, and at
rank two it does not: the products there are the whole function space
$C(\mathbb{RP}^2, \mathbb{R})$ (\Cref{cor:qubit-classification}), inside which
the constant-$t$ family is a single line.  Shen--Wu~\cite[Thm.~3.1,
Exs.~3.1--3.2]{ShenWu2009} classify the functional-calculus ansatz under
their SEA axioms, which do not include \ref{ax:S2} (Example~3.1 the
constant-$t$ family, Example~3.2 arbitrary frame dependence).  The
constant-$t$ Liu--Wu family is first-argument norm-continuous, whereas
Shen--Wu's rank-two frame-dependent construction permits arbitrary frame choices
and need not satisfy \ref{ax:S2}.  Their forced values are ansatz-free on all
commuting pairs (their Theorem~2.2), not merely for sharp first arguments.
Their Theorem~3.1 characterizes which functional-calculus families are
products but does not show every product has that form; \Cref{mthm:master} and
\Cref{cor:qubit-classification} obtain the classification with no
functional-calculus assumption, so that \emph{every} norm-continuous \ref{ax:S1}--\ref{ax:S7}
product is of that form.  \Cref{thm:qubit-boundary}'s continuous example specializes their
construction; for the interferometric read-out and two-value identifiability
of $t$ (\Cref{rem:twist-detection}) we are aware of no counterpart in either
paper.

\paragraph{Van de Wetering; the residual-automorphism question.}
\Cref{thm:vdw1} is his structure theorem.  His
Sections~4--5~\cite{Wetering2018three} supply the comparison machinery
(Propositions~4.20, 4.22, 5.2, 5.3, 5.5, 5.7) and three uniqueness theorems,
each adding one selector.  His EJA appendix supplies the compatibility
bridge (\Cref{prop:bridge}).  On the complex, Hilbert-space type he reduces
the residual comparison automorphism to $\Ad_{u_a}$ for a bicommutant unitary
$u_a$ (his Lemma~5.8, for simple invertible $a$; \Cref{rem:no-unitary}), but does not classify the
resulting dependence $a \mapsto \Ad_{u_a}$ under the bare axioms, and the
reduction has no exceptional-type analogue.  For the complex type, in finite
dimension and rank at least three,
\Cref{mthm:master} gives this comparison automorphism the form $\Ad_{a^{it}}$
with one global constant; equivalently, the residual unitary is $a^{it}$ up to
a scalar phase, invisible under conjugation.  At rank two the
residual is instead a full continuous phase field, $\mathcal{M}_{\mathrm{SP}}(\Mnsa{2})$
in bijection with $C(\mathbb{RP}^2, \mathbb{R})$
(\Cref{cor:qubit-classification}).  Within the complex matrix family only
the infinite-dimensional case remains open.  We know of no prior statement of
the typewise classification.  Separately, van de Wetering's thesis records
that his general update rule~(1.1), the fully general instrument form, is
not compatible with the sequential axioms, \emph{why} being ``still not
entirely clear'' (Remark~2.2.7~\cite{vandeWeteringThesis2021}).  That remark
concerns the sequential-product/instrument boundary, not the
residual-automorphism classification settled here.  (The same remark attributes
to Liu--Wu~\cite{LiuWu2009} the claim that every
$B(H)\sa$ update rule satisfying the axioms has the form
$b \mapsto u_a\sqrt{a}\,b\,\sqrt{a}\,u_a^{*}$ with $u_a$ a unitary commuting
with $a$.  That fixes the \emph{shape} of the update, and in finite dimension it
is his Lemma~5.8 above; it is the dependence $a \mapsto u_a$, not the
shape, that \Cref{mthm:master} determines.)  In a complementary
direction, Westerbaan, Westerbaan, and van de Wetering classify the normal
sequential effect algebras themselves into three structural
types~\cite{WesterbaanWesterbaanVdW2020}.  The present problem instead fixes the
effect algebra, a finite-dimensional Euclidean Jordan algebra, and
classifies the norm-continuous \ref{ax:S1}--\ref{ax:S7} products it carries,
with normality derived (\Cref{lem:normality}) rather than assumed.

\paragraph{Selector attributions; weakenings that fail.}
Conditions (ii), (iii) of \Cref{cor:selectors} are van de Wetering's in
strictly greater generality: inner-product symmetry selects the standard
product on any EJA (his Theorem~5.14), and covariance under all unital order
isomorphisms with norm-continuity does too (his Theorems~5.9/5.11).
\Cref{rem:no-unitary} discusses the exceptional case in the proof of his
Theorem~5.11.  On $B(H)$
the trace-duality
route is Gudder--Latr\'emoli\`ere's~\cite{GudderLatremoliere2008}.  Three
weakenings fail to select, with the twisted family as witness.
Transition-probability
symmetry does not select, because the twisted products have symmetric atomic
transition probabilities $\operatorname{tr}(pq)$.  Identity-component covariance
does not, because the
$\mathrm{SO}(2)$ rotations commute with $t\mathcal{J}$; only the orientation-reversing
transpose kills $t$.  Closure of the invertibles does not, because the twisted
product of invertibles is again invertible.  Van de Wetering's strictly stronger
\emph{invertibility preservation}~\cite[Def.~5.16]{Wetering2018three}, which
adds the inverse law $(\seqp{a}{b})^{-1} = \seqp{a^{-1}}{b^{-1}}$, does select
given norm-continuity~\cite[Thm.~5.19]{Wetering2018three}.

\paragraph{Ledger.}
The full input ledger of imported results and the
detailed catalogue of nearest prior results are collected in the
Supplementary Material.  That catalogue covers van de Wetering's own
Section~5 and uniqueness theorems, the purity/dagger selectors
of~\cite{WesterbaanWesterbaanVdW2018pure}, and the Westerbaan--van de
Wetering reconstruction~\cite{WesterbaanVdW2021}, the last of which explicitly
records the remaining assert-map uniqueness gap.
Two recent lines of work are adjacent but distinct.
Jen\v{c}ov\'a and Pulmannov\'a study spectrality in convex sequential effect
algebras~\cite{JencovaPulmannova2023}, characterizing it, under further
conditions, by monotone $\sigma$-completeness of the maximal commutative
subalgebras.  The finite spectrality that this paper needs comes for free in
finite dimension (\Cref{lem:simple-bridge}).  Gudder studies the
\emph{operation-relative} sequential product $a[\mathcal{I}]b$, whose value
depends on the operation $\mathcal{I}$ chosen to measure
$a$~\cite{Gudder2023}.  The products classified here are the effect-level ones
satisfying \ref{ax:S1}--\ref{ax:S7}, and on the complex type each is
realized by a canonical single-Kraus operation,
$\rho \mapsto (U^t_a)^{*}\sqrt{a}\,\rho\,\sqrt{a}\,U^t_a$
of~\eqref{eq:backaction-intro}.

\paragraph{Relation to the pair-local ansatz.}
A restrictive alternative posits that the update is pair-local and
decomposition-independent, acting by the compressions on the diagonal Peirce
blocks and blockwise on the coherence spaces.  \Cref{mthm:master} assumes no
such thing: Peirce-block-diagonality and the single block generator are
conclusions (\Cref{lem:frame-fix,lem:homomorphism}) drawn from the comparison
map.  Among \ref{ax:S1}--\ref{ax:S7} products the pair-local ansatz therefore
names a subclass already covered, not an input to the proof.

\paragraph{Niestegge.}
The symmetric transition probabilities built into the Euclidean-Jordan
(self-dual) setting of \Cref{mthm:master} are a restriction, not a
convention: there exist rank-two generalized spin factors with \emph{asymmetric}
transition probabilities---the non-Hilbertian examples over
$\ell^{p}$, $1 < p < \infty$, $p \neq 2$, of
Niestegge~\cite[Thm.~7.1, \S 10]{Niestegge2024}.  These lie outside the
present setting; we cite them only for that boundary fact.


%% =====================================================================
%%  SECTION 9: CLOSING
%% =====================================================================
\section{Closing Remarks and Open Questions}
\label{sec:closing}

Two questions remain.  First, the higher spin factors: \Cref{prop:real} settles
the real qubit and \Cref{sec:qubit} classifies the complex one, but the other
simple rank-two Jordan algebras---the spin factors $\mathbb{R} \oplus
\mathbb{R}^k$ beyond those two, including $H_2(\mathbb{H})$ and
$H_2(\mathbb{O})$---are untreated.  Second, infinite
dimensions: the argument uses finite-dimensional spectral resolutions and Lie
theory on a finite-dimensional stabilizer, and extending to type~I factors or
JBW-algebras would require new continuity arguments.


%% =====================================================================
%%  ACKNOWLEDGMENTS
%% =====================================================================
\section*{Acknowledgments}

\ifanonymous\else
The author thanks John van de Wetering for making his sequential product
theorems publicly available and clearly stated.

\fi
Large-language-model tools were used in the preparation of this work, as
follows.  Claude models (Anthropic; Opus~4.8 and Fable~5) were used for
literature search, exploratory mathematical conversations, manuscript
drafting, and assistance with the accompanying Lean~4 proof artifact and SymPy
scripts.  A GPT model (OpenAI; \texttt{gpt-5.6-sol}) was used for
pre-submission mathematical consistency checking and critical evaluation of
draft arguments.
No AI system is an author.  All definitions, theorem statements, proofs, and
attributions are the author's; all cited sources were independently inspected,
and all mathematical statements, proofs, attributions, and computational
outputs were verified by the author.  No reference was included on the basis
of AI-generated output alone.  The author assumes full responsibility for the
accuracy, originality, and integrity of the work.

\section*{Data and code availability}

The Lean~4 proof artifact and the SymPy script \texttt{verify\_n2.py} that
corroborate the paper's computations
\ifanonymous
accompany this submission as supplementary material, in an anonymized archive
containing the complete Lake project, the \texttt{lake-manifest.json} revision
pin, the script, a reproduction guide, and a theorem-to-file map.
On acceptance the same tree is deposited in a permanent public repository.
\else
% CAMERA-READY: this commit predates the SequentialProductsArtifact rename. Re-point
% at a deposit of the RENAMED tree before the non-anonymous build, or the supplement
% and the repository will name different module paths.
are deposited in a permanent public repository at the immutable commit
\href{https://github.com/ehrlich-b/research/tree/b7db3e8b6c7b225a53480a8b7c395a75ed543317/twist-normal-form-lean}{\texttt{ehrlich-b/research@b7db3e8}}
(directory \texttt{twist-normal-form-lean/}).
\fi
A Zenodo archival DOI will be
minted from the archived source before publication.  The Lean development is a
self-contained Lake project (Mathlib v4.28.0).  Which statements are
machine-checked, which are carried as cited interface hypotheses, and which are
not formalized at all are recorded in the archive's theorem-to-file map and in
the accompanying Supplementary Material, together with the axiom-closure audit
and the SymPy identity list.  No other data were generated or analysed in this
work.


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%%  BIBLIOGRAPHY
%% =====================================================================
\bibliographystyle{alpha}
\bibliography{refs}

\end{document}
