\section{Maxwell reduction and Floquet conventions}
\subsection{Physical assumptions}
Assume a spatially homogeneous, scalar, instantaneous constitutive law
$D=\varepsilon(t)E$ and $B=\mu H$, with constant $\mu>0$ and measurable
$T$-periodic permittivity satisfying
\begin{equation}\label{eq:epsbounds}
0<\varepsilon_-\leq\varepsilon(t)\leq\varepsilon_+<\infty.
\end{equation}
There are no imposed free-current impulses at switching times. This is an
ideal nondispersive model; the external modulation can supply energy, so
``lossless constitutive parameters'' does not imply conservation of wave
energy under modulation.

For a transverse plane wave proportional to $e^{ikx}$, write its electric
induction and magnetic induction amplitudes as $d(t)$ and $b(t)$. With one
consistent polarization convention, Maxwell's equations are
\begin{equation}
\dot d=-\frac{ik}{\mu}b,\qquad
\dot b=-\frac{ik}{\varepsilon(t)}d.
\end{equation}
Since $\mu$ is constant, differentiation of the first equation gives
\begin{equation}\label{eq:maxwellscalar}
\ddot d+\frac{k^2}{\mu\varepsilon(t)}d=0.
\end{equation}
At a temporal jump, $d$ and $b$ remain continuous; therefore $d$ and $\dot d$
are the continuous state variables in this model. Differentiating $E$ instead
would introduce coefficient derivatives and is not the reduction used here.

Fix $k\neq0$ and suppose $\varepsilon_-<\varepsilon_+$. Define
\begin{equation}\label{eq:scaling}
\omega_-:=\frac{|k|}{\sqrt{\mu\varepsilon_+}},\quad
R:=\sqrt{\frac{\varepsilon_+}{\varepsilon_-}}>1,\quad
s=\omega_-t,\quad S=\omega_-T,\quad
u(s):=\frac{\varepsilon_+}{\varepsilon(s/\omega_-)}.
\end{equation}
Then $1\leq u\leq R^2$ and the problem is
\begin{equation}\label{eq:oscillator}
y''+u(s)y=0,\qquad
x'=A(u)x,\qquad
A(u)=\begin{pmatrix}0&1\\-u&0\end{pmatrix},\qquad
x=(y,y')^{\mathsf T}.
\end{equation}
Primes in \eqref{eq:oscillator} denote $s$-derivatives.

\subsection{Measurable coefficients, monodromy, and gaps}
For bounded measurable $u$, the integral equation
$X(s)=\id+\int_0^s A(u(v))X(v)\dd v$ has a unique absolutely continuous
fundamental matrix. Picard iteration converges uniformly on finite time
intervals: its $j$th increment is bounded by $(Cs)^j/j!$ when
$\|A\|\leq C$. Uniqueness follows by iterating the same integral estimate
for the difference of two solutions. The determinant is one, since its
almost-everywhere derivative is $\tr A\det X=0$ and $\det X(0)=1$.

Let $M_u=X(S)$ and set
\begin{equation}\label{eq:objective}
L(u):=\log\rho(M_u),\qquad
D=\tfrac12\tr M_u.
\end{equation}
For a real determinant-one $2\times2$ matrix, its multipliers are the roots
of $z^2-2Dz+1=0$. Consequently,
\begin{equation}\label{eq:discriminantgrowth}
L(u)=\begin{cases}\arcosh|D|,&|D|>1,\\0,&|D|\leq1.\end{cases}
\end{equation}
The physical amplitude growth exponent is $L(u)/T$. A momentum gap is a
connected component of the open set of real $k$ for which $|D(k)|>1$.
At $|D|=1$, Jordan growth can be polynomial, but its exponential rate is
zero. No branch choice of a complex logarithm is needed for $L$.

For a constant segment with $u=w^2$ and duration $h$, the state transfer is
\begin{equation}\label{eq:segment}
E(w,h)=\begin{pmatrix}
\cos(wh)&\sin(wh)/w\\-w\sin(wh)&\cos(wh)
\end{pmatrix}.
\end{equation}
Chronological segments multiply with the latest factor on the left.


\paragraph{Zero momentum and the analytic inverse convention.}
The forward normalization explicitly assumes $k\ne0$. At $k=0$, the physical
Maxwell $(d,b)$ evolution is the identity. The formally extended scalar state
$(y,\dot y)$ has a shear, because its relation to $(d,b)$ becomes singular
at zero momentum. Both have spectral radius one. The inverse jets below
are derivatives of the analytic scalar transfer family at $k=0$; they do
not assert two independently controllable physical zero-momentum amplitudes.
This distinguishes the two conventions more precisely than the v1.0 audit.