%
\label{lem:idempotent-decomposition}
%   II.D35, II.T27, II.D45, II.T31
Let $f \in \mathcal{O}_\tau(\tau^3)$ be $\tau$-holomorphic.
Then:
\begin{enumerate}
    \item[\textup{(D1)}]
          \textbf{Channel holomorphy:}
          $f_+ = e_+ \cdot f$
          and $f_- = e_- \cdot f$
          are each $\tau$-holomorphic.
    \item[\textup{(D2)}]
          \textbf{Canonical splitting:}
          $f = f_+ + f_-$
          is the unique decomposition of~$f$
          into a sum of
          a $B$-channel holomorphic function
          and a $C$-channel holomorphic function.
    \item[\textup{(D3)}]
          \textbf{Completeness:}
          the canonical decomposition map
          $\Delta_\tau \colon
          \mathcal{O}_\tau(\tau^3) \to
          \mathcal{O}_\tau^{(+)}(\tau^3)
          \times \mathcal{O}_\tau^{(-)}(\tau^3)$
          is an isomorphism of $H_\tau$-modules.
\end{enumerate}

We prove each assertion in turn.

\smallskip
\noindent\textbf{Proof of (D1): Channel holomorphy.}
We must show that $f_+ = e_+ \cdot f$
is $\tau$-holomorphic.
By the Mutual Determination Theorem
(Theorem~\ref{thm:mutual-determination}, II.T27),
$\tau$-holomorphy of~$f$ is equivalent
to any of the five descriptions
(R), (S), (G), (C), (H).
We use the spectral description~(S),
which is the most transparent
for this argument.

By the finite spectral support theorem
(Theorem~\ref{thm:finite-spectral-support}, II.T31,
Chapter~\ref{ch:canonical-basis}),
$f$ has a canonical expansion
in the basis $\mathcal{B}_\tau$
(Definition~\ref{def:canonical-basis}, II.D45):
\[
    f
    \;=\;
    \sum_{(k,v,\sigma) \in S_f}
    c_{k,v}^{(\sigma)} \cdot E_{k,v}^{(\sigma)},
\]
where $S_f$ is a finite support set,
$c_{k,v}^{(\sigma)} \in H_\tau$
are the spectral coefficients,
and $E_{k,v}^{(\sigma)}$ are the cylinder generators
(Definition~\ref{def:cylinder-generator}, II.D46).
The index $\sigma \in \{B, C\}$
labels the bipolar channel.

The cylinder generators satisfy the channel rule:
\[
    e_+ \cdot E_{k,v}^{(B)} = E_{k,v}^{(B)},
    \quad
    e_+ \cdot E_{k,v}^{(C)} = 0,
    \qquad
    e_- \cdot E_{k,v}^{(B)} = 0,
    \quad
    e_- \cdot E_{k,v}^{(C)} = E_{k,v}^{(C)}.
\]
This follows from (I4):
$E_{k,v}^{(B)}$ lies in the $e_+$-eigenspace
and $E_{k,v}^{(C)}$ lies in the $e_-$-eigenspace
of the $\jj$-action.
Therefore:
\[
    f_+
    \;=\;
    e_+ \cdot f
    \;=\;
    \sum_{(k,v,\sigma) \in S_f}
    c_{k,v}^{(\sigma)} \cdot (e_+ \cdot E_{k,v}^{(\sigma)})
    \;=\;
    \sum_{(k,v,B) \in S_f}
    c_{k,v}^{(B)} \cdot E_{k,v}^{(B)}.
\]
The right side is a finite linear combination
of cylinder generators---all
from the $B$-channel---with
the \emph{same} spectral coefficients.
Since the support $S_f^{(+)} := \{(k,v,B) : (k,v,B) \in S_f\}$
is a subset of~$S_f$,
it is finite.
The resulting expansion
is a valid spectral tail
with finite support,
hence $f_+$ is $\tau$-holomorphic
by the converse direction
of the finite spectral support theorem.

The argument for $f_-$ is identical,
with $B$ replaced by $C$ throughout.

\smallskip
\noindent\textbf{Proof of (D2): Canonical splitting.}
The identity $f = f_+ + f_-$
follows from completeness (I3):
$f(x) = e_+ f(x) + e_- f(x)
= f_+(x) + f_-(x)$
for all $x \in \tau^3$.
By~(D1), both $f_+$ and $f_-$ are holomorphic.

Uniqueness: suppose $f = g + h$
where $g \in \mathcal{O}_\tau^{(+)}(\tau^3)$
and $h \in \mathcal{O}_\tau^{(-)}(\tau^3)$.
Then $g$ takes values in $e_+ \cdot H_\tau$,
so $e_- \cdot g = 0$;
similarly $e_+ \cdot h = 0$.
Apply $e_+$ to both sides:
$e_+ f = e_+ g + e_+ h = g + 0 = g$,
so $g = f_+$.
By the same argument, $h = f_-$.

\smallskip
\noindent\textbf{Proof of (D3): Isomorphism.}
The map $\Delta_\tau \colon f \mapsto (f_+, f_-)$
is a homomorphism of $H_\tau$-modules:
for $\lambda \in H_\tau$,
\[
    \Delta_\tau(\lambda \cdot f)
    = (e_+ \lambda f,\; e_- \lambda f)
    = (\lambda \cdot e_+ f,\; \lambda \cdot e_- f)
    = \lambda \cdot \Delta_\tau(f),
\]
where the second equality uses
the centrality of $e_\pm$ in~$H_\tau$
(they commute with all elements
because $H_\tau$ is commutative).
Additivity is immediate.

Injectivity:
if $\Delta_\tau(f) = (0, 0)$,
then $f_+ = f_- = 0$,
so $f = f_+ + f_- = 0$.

Surjectivity:
given $(g, h) \in
\mathcal{O}_\tau^{(+)}(\tau^3)
\times \mathcal{O}_\tau^{(-)}(\tau^3)$,
set $f := g + h$.
Then $f$ is holomorphic
(as a sum of holomorphic functions),
$f_+ = e_+ (g + h) = e_+ g + e_+ h = g + 0 = g$,
and similarly $f_- = h$.
So $\Delta_\tau(f) = (g, h)$.

Hence $\Delta_\tau$ is an isomorphism.