[step:Identify the canonical relations of the two half-wave branches]
For the half-wave operators $E_+(t)$ and $E_-(t)$, the relevant phase functions are, microlocally away from $\eta=0$,
\begin{align*}
\phi_+ : \mathbb{R}_t \times \mathbb{R}_x^n \times \mathbb{R}_y^n \times (\mathbb{R}_\eta^n \setminus \{0\}) \to \mathbb{R}
\end{align*}
defined by
\begin{align*}
\phi_+(t,x,y,\eta) := (x-y)\cdot \eta + t|\eta|,
\end{align*}
and
\begin{align*}
\phi_- : \mathbb{R}_t \times \mathbb{R}_x^n \times \mathbb{R}_y^n \times (\mathbb{R}_\eta^n \setminus \{0\}) \to \mathbb{R}
\end{align*}
defined by
\begin{align*}
\phi_-(t,x,y,\eta) := (x-y)\cdot \eta - t|\eta|.
\end{align*}
The associated canonical relations in
\begin{align*}
T^*(\mathbb{R}_t \times \mathbb{R}_x^n) \times T^*\mathbb{R}_y^n
\end{align*}
are the graphs obtained from
\begin{align*}
\partial_\eta \phi_\pm(t,x,y,\eta) = 0.
\end{align*}
For $\phi_+$ this critical equation is
\begin{align*}
x-y+t\frac{\eta}{|\eta|}=0,
\end{align*}
and for $\phi_-$ it is
\begin{align*}
x-y-t\frac{\eta}{|\eta|}=0.
\end{align*}
The output covectors are computed from the phase derivatives:
\begin{align*}
\tau = \partial_t\phi_\pm = \pm |\eta|, \qquad \xi = \partial_x\phi_\pm = \eta.
\end{align*}
Therefore the two branches send an input covector $(y,\eta)$, $\eta \neq 0$, to covectors satisfying
\begin{align*}
(t,x,\tau,\xi) = \left(t, y \mp t\frac{\eta}{|\eta|}, \pm|\eta|,\eta\right),
\end{align*}
with the upper sign corresponding to $\phi_+$ and the lower sign corresponding to $\phi_-$.
Define the canonical graphs
\begin{align*}
\Lambda_\pm \subset T^*(\mathbb{R}_t \times \mathbb{R}_x^n) \setminus 0 \times T^*\mathbb{R}_y^n \setminus 0
\end{align*}
by declaring that $((t,x,\tau,\xi),(y,\eta)) \in \Lambda_\pm$ exactly when $\eta \neq 0$ and
\begin{align*}
(t,x,\tau,\xi) = \left(t, y \mp t\frac{\eta}{|\eta|}, \pm|\eta|,\eta\right).
\end{align*}
Thus $\Lambda_\pm(A)$ denotes the image of a conic set $A \subset T^*\mathbb{R}^n \setminus 0$ under this graph relation.
The Hamilton vector field of
\begin{align*}
p(t,x,\tau,\xi)=\tau^2-|\xi|^2
\end{align*}
is
\begin{align*}
H_p = 2\tau\,\partial_t - 2\sum_{j=1}^n \xi_j\,\partial_{x_j}.
\end{align*}
Since $p$ is independent of $t$ and $x$, $\tau$ and $\xi$ are constant along its integral curves. The curves just described are exactly reparametrizations of the integral curves of $H_p$ through
\begin{align*}
(0,y,\pm|\eta|,\eta).
\end{align*}
They are null because
\begin{align*}
p(0,y,\pm|\eta|,\eta)=|\eta|^2-|\eta|^2=0.
\end{align*}
[/step]