Proving Shouten identity in QFT

[LAHLH]

Hi,

I'm trying to prove the Shouten identity for twistors:

$\langle pq\rangle\langle rs\rangle+\langle pr\rangle\langle sq\rangle+\langle ps\rangle\langle qr\rangle=0$

It's easy to show that the LHS here is cyclically symmetric under $q\to r\to s \to q$, and also completely antisymmetric in q,r,s (just use $\langle qr \rangle=-\langle rq \rangle$ etc)

But why does this imply the LHS must be zero?

 

[Bill_K]

Because q, r, s each take on only two values, so if you antisymmetrize on three of them you get zero.

 

[LAHLH]

Each twistor only has two spinorial components as far as I understand it. But then we are labeling each twistor by p,q,r,s and we have antisymmetry on q,r,s exchanges. I'm not sure I follow why this implies zero still, could you say some more please.

 

[Bill_K]

A twistor is a SU(2,2) spinor and is written as a pair of dotted and undotted Weyl spinors, Z^I = (λ^A, μ^A·). From the Weyl spinors we form Lorentz invariant products <λ_i, λ_j> = ε^AB λ_i^A λ_j^B and [μ_i, μ_j] = ε_A^·B^· μ_i^A· μ_j^B·. For short, <λ_i, λ_j> ≡ <i j> and [μ_i, μ_j] ≡ [i j].

There are two Schouten identities,

<i j><k l> + <i k><l j> + <i l> <j k> = 0
[i j][k l] + [i k][l j] + [i l][j k] = 0

They just express the fact that the λ's (and the μ's) are objects that live in a two-space, consequently any three of them must be linearly dependent, and if you try to antisymmetrize on all three of them you'll get zero.