TransWikia.com

Kinematics of Scattering Amplitudes in $left(2, 2right)$ Signature within the Amplituhedron

Physics Asked by petermailpan on August 10, 2021

I am just working my way through the concepts of Amplituhedron and often stumble across the phrase

[…] in $left(2,2right)$ signature $lambda$, $tilde{lambda}$ are real and independent […]

in various references (Jaroslav Trnka, 2014; page 7, Nima Arkani-Hamed, 2014; page 24 or Livia Ferro, 2020; page 3). What I don’t quite understand is the following: If we take the following momentum four-vector:
begin{equation}
p_mu =
begin{pmatrix}
-1, 0, 1, 0
end{pmatrix}
,
end{equation}

and contract this with the Pauli Matrices in $left(2, 2right)$ signature (I am very sorry that I may not use the right terminology here), we obtain the following matrix:
begin{equation}
P_{alphadot{alpha}} =frac{1}{2}
begin{pmatrix}
p_0 + p_3, p_1 – p_2
p_1 + p_2, p_0 – p_3
end{pmatrix}
=
begin{pmatrix}
-frac{1}{2}, -frac{1}{2}
frac{1}{2}, -frac{1}{2}
end{pmatrix}
.
end{equation}

However, unlike in $left(3, 1right)$ signature, the determinant of the matrix is not $0$, but $frac{1}{2}$.

In all the YouTube videos and references I found, this $left(2, 2right)$ signature is never properly explained or referenced. Do you have any sources about the signature that can help me identify my mistakes or that can help me understand this concept better?

One Answer

The main points are:

  1. There is bijective isometry from the split-signature space $(mathbb{R}^{2,2},||⋅||^2)$ to the space of $2times 2$ real matrices $({rm Mat}_{2times 2}(mathbb{R}),det(⋅))$, where $$ ||p||^2~=~(p^0)^2-(p^1)^2+(p^2)^2-(p^3)^2~=~det(p), $$ $$ {rm Mat}_{2times 2}(mathbb{R})~=~{rm span}_{mathbb{R}}{ sigma_0,sigma_1,isigma_2,sigma_3} .$$

  2. There is a bilinear map $$ mathbb{R}^2times mathbb{R}^2~ni~(lambda,tilde{lambda})quadmapstoquad p~:=~lambdatilde{lambda}^T~in~ {rm Mat}_{2times 2} (mathbb{R}),$$ from 2 Weyl spinors $lambda,tilde{lambda}$ to a rank-1 operator $p$, which necessarily has vanishing determinant, i.e. the corresponding momentum is light-like/null.

Correct answer by Qmechanic on August 10, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP