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Proving Shouten identity in QFT

  1. Nov 26, 2011 #1

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

    [itex] \langle pq\rangle\langle rs\rangle+\langle pr\rangle\langle sq\rangle+\langle ps\rangle\langle qr\rangle=0 [/itex]

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

    But why does this imply the LHS must be zero?
  2. jcsd
  3. Nov 27, 2011 #2


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    Because q, r, s each take on only two values, so if you antisymmetrize on three of them you get zero.
  4. Nov 29, 2011 #3
    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.
  5. Nov 29, 2011 #4


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    A twistor is a SU(2,2) spinor and is written as a pair of dotted and undotted Weyl spinors, ZI = (λA, μA·). From the Weyl spinors we form Lorentz invariant products <λi, λj> = εAB λiA λjB and [μi, μj] = εA·B· μiA· μjB·. For short, <λi, λj> ≡ <i j> and [μi, μj] ≡ [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.
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