% CALIBRATED_SELECTION_E.tex
\section{Calibrated Minimizers and Selection Principles}\label{sec:calibrated-selection}

This section isolates the precise sense in which one may \emph{restrict} attention to calibrated (locally optimal) minimizers,
while keeping clear what this does \emph{not} imply for arbitrary one-sided mean-minimizers.
The main point is that calibration is an \emph{edgewise} (one-step) equality condition,
so calibrated minimizers live entirely on the tight subgraph, whereas general minimizers are only asymptotically tight.

\subsection{Calibrated minimizers}

Fix a calibrated subaction $u:V\to\mathbb{R}$ in the sense of Definition~\ref{def:calibrated-subaction},
and write $c_u$ for the associated reduced costs \eqref{eq:reduced-cost-u}.

\begin{definition}[Calibrated bi-infinite minimizer]\label{def:calibrated-minimizer}
A bi-infinite admissible edge sequence
$\omega=(\dots,e_{-1},e_0,e_1,\dots)\in E^{\mathbb{Z}}$ is a \emph{calibrated minimizer for $u$}
if consecutive edges match and
\begin{equation}\label{eq:calibrated-equality}
 c_u(e_t)=0\quad\text{for all }t\in\mathbb{Z}.
\end{equation}
Equivalently, $\omega$ is a bi-infinite path in the tight subgraph $G^{(0)}=(V^{(0)},E^{(0)})$ defined by $u$.
\end{definition}

\begin{lemma}[Calibrated minimizers are supported on tight edges]\label{lem:calibrated-implies-tight}
If $\omega$ is a calibrated minimizer for $u$, then every edge of $\omega$ lies in $E^{(0)}=\{e\in E: c_u(e)=0\}$.
In particular, every calibrated minimizer is minimizing with asymptotic mean cost $\lambda_*$.
\end{lemma}

\begin{proof}
The first claim is immediate from \eqref{eq:calibrated-equality}.
For the mean cost, sum the identity $c(e_t)-\lambda_*+u(v_t)-u(v_{t+1})=0$ along any finite window,
exactly as in Lemma~\ref{lem:tight-path-mean}, and divide by the window length.
\end{proof}

\subsection{Existence on the Aubry--tight core}

The Aubry--tight core $\mathcal{A}$ from Definition~\ref{def:aubry-core} is designed so that tight cycles exist through every vertex of $\mathcal{A}$.
This yields calibrated minimizers without any recurrence hypothesis.

\begin{proposition}[Periodic calibrated minimizers on the core]\label{prop:periodic-calibrated}
For every vertex $v\in\mathcal{A}$ there exists a periodic bi-infinite calibrated minimizer $\omega\in\Sigma_{\mathcal{A}}$
that visits $v$.
Consequently, $\Sigma_{\mathcal{A}}$ is nonempty whenever $\mathcal{A}\neq\varnothing$.
\end{proposition}

\begin{proof}
By Definition~\ref{def:aubry-core}, the vertex $v$ lies on a directed tight cycle $\gamma$ in $G^{(0)}$.
Let $w$ be the finite edge word given by $\gamma$ starting and ending at $v$.
Repeating $w$ periodically in both time directions produces a bi-infinite admissible edge sequence $\omega\in (E^{(0)})^{\mathbb{Z}}$.
Then $c_u(e_t)=0$ for all $t$, so $\omega$ is calibrated by Definition~\ref{def:calibrated-minimizer}.
\end{proof}

\subsection{Minimizing measures are tight}

The previous subsection constructs explicit calibrated minimizers on $\mathcal{A}$.
For applications, it is also useful to know that every \emph{invariant} minimizer is supported on the tight shift.

\begin{proposition}[Invariant minimizers live on the tight shift]\label{prop:minimizing-measure-tight}
Let $\mu$ be a $\sigma$-invariant probability measure on the two-sided edge shift of $G$.
If $\mu$ minimizes the average cost, i.e.
\[\int c\,d\mu = \lambda_*,\]
then $\mu$ is supported on the tight edge shift $(E^{(0)})^{\mathbb{Z}}$.
In particular, $c_u(e)=0$ for $\mu$-almost every edge occurrence.
\end{proposition}

\begin{proof}
By Definition~\ref{def:calibrated-subaction}, the reduced cost $c_u(e)\ge 0$ for all edges.
For a bi-infinite path $\omega$ with vertex sequence $(v_t)_{t\in\mathbb{Z}}$ we have
\[
 c_u(e_t)=c(e_t)-\lambda_*+u(v_t)-u(v_{t+1}).
\]
Integrating with respect to a $\sigma$-invariant measure $\mu$ yields
\[
\int c_u\,d\mu = \int (c-\lambda_*)\,d\mu + \int (u\circ\mathrm{tail}-u\circ\mathrm{head})\,d\mu.
\]
The last integral vanishes by $\sigma$-invariance (it is the integral of a coboundary), hence
\[
\int c_u\,d\mu = \int c\,d\mu - \lambda_*.
\]
If $\int c\,d\mu=\lambda_*$, then $\int c_u\,d\mu=0$.
Since $c_u\ge 0$, it follows that $c_u=0$ $\mu$-almost surely.
Equivalently, $\mu$ is supported on the set of paths using only edges in $E^{(0)}$.
\end{proof}

\begin{corollary}[Calibrated representatives exist in minimizing supports]\label{cor:calibrated-in-support}
If $\mu$ is a $\sigma$-invariant minimizing measure, then $\mu$-almost every $\omega$ is a calibrated minimizer for $u$.
In particular, the support of any minimizing measure is contained in the boundary subshift $\Sigma_{\mathcal{A}}$.
\end{corollary}

\begin{proof}
The first statement is immediate from Proposition~\ref{prop:minimizing-measure-tight}.
For the second, note that any bi-infinite tight path must lie in an SCC of $G^{(0)}$ that contains a cycle.
By Definition~\ref{def:aubry-core}, the union of such SCCs is exactly the Aubry--tight core $\mathcal{A}$,
so every bi-infinite tight path lies in $\Sigma_{\mathcal{A}}$.
\end{proof}

\subsection{What calibration does and does not buy for one-sided minimizers}

The results above justify a common proof strategy:
when proving statements about the \emph{symbolic carrier} of extremality (the boundary subshift and its SCCs),
one may work entirely on calibrated minimizers.
However, one must not confuse this with a statement about a \emph{given} one-sided minimizing trajectory.

\begin{proposition}[Calibrated selection from a one-sided minimizer]\label{prop:selection-from-onesided}
Let $\pi=(e_0,e_1,\dots)$ be a one-sided minimizing path.
Then there exists a bi-infinite calibrated minimizer $\omega$ and a sequence of times $n_k\to\infty$ such that
$\sigma^{n_k}\pi\to \omega$ in the product topology on $E^{\mathbb{Z}}$.
Equivalently, the $\omega$-limit set of $\pi$ contains a calibrated minimizer.
\end{proposition}

\begin{proof}
By Proposition~\ref{prop:asymptotic-tight}, the set $T=\{t\ge 0: e_t\in E^{(0)}\}$ has asymptotic density $1$.
Fix an integer $m\ge 1$.
Because $T$ has density $1$, there exist arbitrarily large times $n$ such that the window
$\{n-m,\dots,n+m\}$ is contained in $T$.
Choose $n(m)$ with this property.
The shift space $E^{\mathbb{Z}}$ is compact, so by a diagonal subsequence argument there exists a sequence $n_k\to\infty$
with $\sigma^{n_k}\pi$ converging to some bi-infinite admissible limit path $\omega$.
By construction of the windows, for each fixed coordinate $t\in\mathbb{Z}$, the edge at position $t$ in $\sigma^{n_k}\pi$
belongs to $E^{(0)}$ for all sufficiently large $k$; hence the limiting edge $e_t(\omega)$ also lies in $E^{(0)}$.
Therefore $\omega$ uses only tight edges and is a calibrated minimizer by Definition~\ref{def:calibrated-minimizer}.
\end{proof}

\begin{remark}[Scope and limitations]\label{rem:calibrated-limitations}
Proposition~\ref{prop:selection-from-onesided} is a \emph{selection} statement:
from a minimizing trajectory one can extract calibrated minimizers as limit points.
It does not imply that the original trajectory is eventually tight, trapped in a single tight SCC, or recurrent.
The only trajectory-level facts used later are
\begin{itemize}
\item density-$1$ tightness (Proposition~\ref{prop:asymptotic-tight}), and
\item positive-lower-density selection of at least one tight SCC (Proposition~\ref{prop:tight-dichotomy}).
\end{itemize}
Any stronger recurrence input (bounded gaps of a certified Doeblin word, or a syndetic modulus from an automaton) is treated as a separate layer.
\end{remark}
