%
\label{thm:spectral-decomposition}
Let $\mathbb{L} = H_\tau = \hat{\mathbb{Z}}_\tau[j]$
be the algebraic lemniscate
(Theorem~\ref{thm:algebraic-lemniscate}).
For every element $x \in \mathbb{L}$,
there exist unique elements
$\alpha, \beta \in \hat{\mathbb{Z}}_\tau$
such that:
\[
    \boxed{%
    x = \alpha \cdot e_+ + \beta \cdot e_-
    = \chi_+(x) \cdot e_+ + \chi_-(x) \cdot e_-.}
\]
The coefficients are:
\begin{align*}
    \alpha &= \chi_+(x) = a + b, \\
    \beta  &= \chi_-(x) = a - b,
\end{align*}
where $x = a + bj$ with $a, b \in \hat{\mathbb{Z}}_\tau$.
This decomposition is:
\begin{enumerate}
    \item \textbf{Canonical}:
          it depends only on the algebraic structure of $\mathbb{L}$,
          not on any choice of basis or coordinate system.
    \item \textbf{Orthogonal}:
          the $e_+$ and $e_-$ components do not interact
          ($e_+ \cdot e_- = 0$).
    \item \textbf{Complete}:
          $e_+ + e_- = 1$, so no information is lost.
    \item \textbf{Multiplicative}:
          $\chi_\pm(xy) = \chi_\pm(x) \cdot \chi_\pm(y)$
          for all $x, y \in \mathbb{L}$.
\end{enumerate}

\textbf{Existence.}
Let $x = a + bj \in \hat{\mathbb{Z}}_\tau[j]$.
By Proposition~\ref{prop:idempotent-decomposition}
(Chapter~\ref{ch:split-complex-scalars}),
\[
    x = (a + b) e_+ + (a - b) e_-.
\]
Setting $\alpha = a + b$ and $\beta = a - b$
gives the decomposition.
These are precisely the character evaluations:
$\chi_+(x) = a + b$ and $\chi_-(x) = a - b$
(Definition~\ref{def:lemniscate-characters}).

\textbf{Uniqueness.}
Suppose $x = \alpha' e_+ + \beta' e_-$
for some $\alpha', \beta' \in \hat{\mathbb{Z}}_\tau$.
Multiplying both sides by $e_+$:
\[
    x \cdot e_+ = \alpha' e_+^2 + \beta' e_- e_+
    = \alpha' e_+ + 0
    = \alpha' e_+.
\]
But also $x \cdot e_+ = (a + b) e_+$ from the computation
in Proposition~\ref{prop:idempotent-decomposition}.
Since $e_+ \neq 0$,
and $\hat{\mathbb{Z}}_\tau$ acts faithfully
on each idempotent sector,
we conclude $\alpha' = a + b = \alpha$.
Similarly, $\beta' = a - b = \beta$.

\textbf{Multiplicativity.}
Let $x = a + bj$ and $y = c + dj$.
Then $xy = (ac + bd) + (ad + bc)j$, and:
\begin{align*}
    \chi_+(xy)
    &= (ac + bd) + (ad + bc) \\
    &= (a + b)(c + d) \\
    &= \chi_+(x) \cdot \chi_+(y).
\end{align*}
The computation for $\chi_-$ is analogous:
$\chi_-(xy) = (a - b)(c - d) = \chi_-(x) \cdot \chi_-(y)$.

\textbf{Additivity} follows similarly:
$\chi_\pm(x + y) = \chi_\pm(x) + \chi_\pm(y)$.
Thus $\chi_+$ and $\chi_-$ are ring homomorphisms,
confirming their character property.