On the simplicity of the three-particle amplitude in holomorphic configuration I am reading Clifford Cheung's 2017 TASI Lectures on Scattering Amplitudes. In section 3, "Bootstrapping Amplitudes", the procedure for bootstrapping the three-particle amplitude for massless particles is described. Momentum conservation implies severe restrictions on the spinor helicity variables. Specifically, the amplitude is found to have support on only two possible kinematic configurations, either of,
\begin{align}
[12] = [23] = [31] = 0 \implies \tilde{\lambda}_1 \propto \tilde{\lambda}_2 \propto \tilde{\lambda}_3, \tag{3.2}
\end{align}
or
\begin{align}
\langle 12\rangle  = \langle 23\rangle = \langle 31\rangle = 0 \implies \lambda_1 \propto \lambda_2 \propto \lambda_3. \tag{3.3}
\end{align}
Then the following claim is made, on page 13,

Both kinematic configurations imply that $p_1 p_2 = p_2 p_3 = p_3 p_1 = 0$ and require complex momenta so that $\lambda_i$ and $\tilde{\lambda}_i$ are independent variables.

I understand how (3.2) and (3.3) are obtained, and I also understand the impication $p_1 p_2 = p_2 p_3 = p_3 p_1 = 0$ in the claim, the latter following straightforwardly from the massless momentum conservation. I do not see why complex momenta are required, nor is it immediate to me why this would further imply that $\lambda_i$ and $\lambda_i$ must be independent variables.
Separately, the immediate next claim is,

Without loss of generality, the three particle amplitude in the holomorphic kinematic configuration takes the form
\begin{align}
A(1^{h_1}2^{h_2}3^{h3}) = \langle 12 \rangle^{n_3} \langle 23\rangle^{n_1}\langle 31 \rangle^{n_2}. \tag{3.4}
\end{align}

These $n_i$ exponents can be written in terms of the helicities, $h_i$, which is done in the ensuing text and which I understand. I don't understand why this above claim opens with "Without loss of generality", as it seems we're restricting to one of the kinematic configurations here and specifically one that involves the "angle" bracket quantities rather than the "square" bracket ones. Is the statement saying just that the analogous expression is valid, with "square" brackets appropriately substituted, for the other kinematic configuration?
Furthermore, the simplicity of this expression is very much non-obvious to me. Is the author making a leap here to be later explained? Or is the simplicity of this tree-level amplitude something that should be more apparent to me if I better recalled the basics of scattering amplitude calculation?
 A: Let's look at real massless 4-momenta. Let us parameterize this as
$$
p_i^\mu = \omega_i ( 1 + z_i {\bar z}_i , z_i + {\bar z}_i , - i ( z_i - {\bar z}_i ) , 1 - z_i {\bar z}_i ), \qquad z_i \in {\mathbb C}
$$
Even though $z_i$ is a complex variable, note that $p_i^\mu$ is completely real.  From this, we find that
$$
p_i \cdot p_j = - \omega_i \omega_j | z_{ij}|^2 .
$$
We then see that if $p_i \cdot p_j = 0$, then we must have either $z_i=z_j$ (which automatically implies ${\bar z}_i = {\bar z}_j$) or that $\omega_i = 0$ or $\omega_j = 0$. The latter two cases implies that $p^\mu_i = 0$ (or $p^\mu_j = 0$) which is not appropriate for a scattering amplitude. Consequently, we must conclude that $z_i = z_j$ AND ${\bar z}_i = {\bar z}_j$.
Now, consider the square and angle brackets. An appropriate choice for this is
$$
| i ] = \sqrt{\omega_i} \begin{pmatrix} 1 \\ z_i \end{pmatrix} , \qquad | i \rangle = \sqrt{\omega_i} \begin{pmatrix} 1 \\ {\bar z}_i \end{pmatrix} , \qquad | i ]^* = | i \rangle . 
$$
Consequently, we have
$$
[ij] = \sqrt{\omega_i\omega_j} z_{ij} , \qquad \langle ij \rangle = \sqrt{\omega_i \omega_j } {\bar z}_{ij} . 
$$
Now, for a 3-point amplitude, basic kinematics implies that $p_i\cdot p_j = 0$ for all $i,j=1,2,3$. Consequently, we must have that $z_1=z_2=z_3$ AND ${\bar z}_1 = {\bar z}_2 = {\bar z}_3$ which then implies that $\langle ij \rangle = [ij] = 0$ for all $i,j=1,2,3$. Putting all of this together, we find that that a 3-point amplitude MUST vanish (since all the things it could possibly depend on are zero) if the momenta are all real.
One way around this issue is to take $z_i$ and ${\bar z}_i$ to be independent variables that are NOT complex conjugates of each other. Of course, such a choice immediately makes the 4-momenta $p^\mu_i$ complex. This also implies that $|i]$ and $|i\rangle$ are now independent spinors.
If $z_i$ and ${\bar z}_i$ are independent variables, then $p_i \cdot p_j = 0$ implies that EITHER $z_i = z_j$ OR ${\bar z}_i = {\bar z}_j$ (we no longer need to assume that both are true simultaneously). For instance, in the 3-pt amplitude case, the kinematics are satisfied if we assume that $z_1=z_2=z_3$. In this case, all the square brackets vanish identically $[ij]=0$, but $\langle ij \rangle \neq 0$. Consequently, a 3-pt amplitude has the general form
$$
A_3(1^{h_1}2^{h_2}3^{h_3} ) = f ( \langle 12 \rangle  , \langle 23 \rangle  , \langle 31 \rangle  ) 
$$
The precise form of the function $f$ can then be fixed by dimensional analysis and the little group transformation properties of the scattering amplitude.
A: *

*Ref. 1 is considering 3+1D massless kinematics. By complex momenta is meant non-real momenta.


*If the momenta are real, then $$\tilde{\lambda}~=~\pm(\lambda)^{\ast}.\tag{2.5}$$
Hence since at least one of the conditions (3.2) and (3.3) are satisfied, both are. This in turn means that the momenta $$p_1~\propto~p_2~\propto~p_3$$
are collinear, corresponding to special kinematics. And it means that all angle and square brackets (and hence all Mandelstam variables, and ultimately the 3-amplitudes) vanish.


*So we will treat the null-4-momentum
$$p~=~\lambda\tilde{\lambda}^t\tag{2.4}$$
as complex, i.e. as having 3 complex DOF. The LHS of eq. (2.4) is matched by treating $\lambda$ and $\tilde{\lambda}$ as having 2+2=4 independent complex DOF, keeping in mind a complexified little group symmetry that removes 1 complex DOF on the RHS of eq. (2.4).


*Little group scaling arguments fix the form of eq. (3.4).
References:

*

*C. Cheung, TASI Lectures on Scattering Amplitudes, arXiv:1708.03872.

