# Proving Shouten identity in QFT

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.

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