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I Spinorial identity

  1. Oct 1, 2017 #1
    Hi there!

    I am reading textbook "Supergravity" by Freedman and Van Proeyen and got stuck on a simple exercise (Ex 2.4). Usually I would proceed further marking it as a typo but I've checked the errata list on the website and didn't find this exercise there

    Exercise 2.4 Show that ## A\bar{\sigma}_\mu A^\dagger=\bar{\sigma}_\nu \Lambda^{-1}{}^\nu{}_\mu## and ##A^\dagger \sigma A=\sigma_\nu \Lambda^\nu{}_\mu## . This gives precise meaning to the statement that the matrices ##\bar{\sigma}_\mu## and ##\sigma_\nu## are 4-vectors

    Sigma matrices in this book are defined as

    $$\sigma_\mu=\left(-\mathbb{1},\sigma_i\right),\;\;\; \bar{\sigma}_\mu=\sigma^\mu=\left(\mathbb{1},\sigma_i\right).$$

    And SL(2,C) transformations is defined as
    $$ \mathbf{x}^\prime \equiv A \mathbf{x} A^\dagger$$

    The first identity in the exercise is kinda straightforward and it is also easy to see that the second one holds for barred sigma-matrices. But I didn't managed to work out the identity in the form written in the exercise

    Is it a typo or I missing something very deep?
     
  2. jcsd
  3. Oct 2, 2017 #2

    Ben Niehoff

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    It looks like the only difference between the barred and unbarred sigma matrices is that ##\sigma_0 = -1## while ##\bar \sigma_0 = 1##. Since 1 is just the identity matrix, I don't see how you could have gotten different results for the barred vs. unbarred case.
     
  4. Oct 2, 2017 #3
    Since it is involved in the contraction on r.h.s. of the identity it makes difference.

    This identity is true
    $$ A^\dagger \bar{\sigma}_\mu A=\bar{\sigma}_\nu\Lambda^\nu{}_\mu.$$
    And it is kinda obvious. When I apply inverse SL(2,C) transformation it should result in inverse Lorentz transformation.

    Also one may show the following

    $$A^\dagger \sigma_\nu A=\sum_\mu \sigma_\mu \Lambda_\mu{}^\nu$$
    (I know that it looks weird from tensor calculus perspective. By writing this I mean one should extract exact values (for particular ##\mu## and ##\nu##)in some referrence frame)

    The initial identity is proven if ## \Lambda_\mu{}^\nu=\Lambda^\mu{}_\nu##, but it is not true
     
    Last edited: Oct 2, 2017
  5. Oct 14, 2017 #4
    After some thinking and asking I believe that this identity may be true due different index structure of sigma matrices

    $$ \sigma_{\mu \alpha \dot{\alpha}}, \bar{\sigma}_\mu {}^{\dot{\alpha} \alpha}$$

    If someone has a nice comprehensive refference on this spinor algebra issues I would be very thankful.

    P.S. It can realy drive someone insane when one sees the expression like ## (M^T)_\alpha {}^\beta## without any explanation on the meaning of this notation.
     
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