\label{thm:prime-polarity-scaling}
Let $\pi$ be an automorphic representation with conductor $q$. Choose level $n$ such that $q \mid n$. Then:
\begin{enumerate}
\item The Label$_n$ decomposition~\eqref{eq:ch27-euler-decomposition} is well-defined and independent of $n$ for $n \gg 0$.
\item The spectral trichotomy (III.T14) applies to each sector: $H_L = H_B \oplus H_C \oplus H_X$.
\item The zeros of $L(s, \pi)$ decompose into $B$-sector, $C$-sector, and $X$-sector zeros.
\item GRH for $\pi$ is equivalent to spectral purity in each sector: all zeros of $L_B$, $L_C$, $L_X$ on $\Re(s) = \frac{1}{2}$.
\end{enumerate}

(1) The Label$_n$ classifier stabilizes for $n$ divisible by the conductor $q$ (III.T13). For $n' > n$ with $n \mid n'$, the sets of $B$-, $C$-, $X$-primes refine but preserve asymptotic density.

(2) The Hilbert space $H_L$ decomposes spectrally by the bipolar characters $\chi_+$ and $\chi_-$ (Book II, Central Theorem). The crossing point eigenspace $H_C$ is orthogonal to $H_B = H_+ \oplus H_-$ by the spectral trichotomy (III.T14). The exceptional sector $H_X$ is finite-dimensional.

(3) By~\eqref{eq:ch27-euler-decomposition}, zeros of $L(s, \pi)$ are unions of zeros of $L_B$, $L_C$, $L_X$.

(4) Each sector's $L$-function corresponds to a spectral determinant on the respective subspace. Spectral reality on each subspace implies all zeros real, hence on $\Re(s) = \frac{1}{2}$ by functional equation symmetry.