I Solve Exercise 2.4 in Supergravity by Freedman & Van Proeyen

  • I
  • Thread starter Thread starter Korybut
  • Start date Start date
  • Tags Tags
    Identity
Korybut
Messages
71
Reaction score
3
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?
 
Physics news on Phys.org
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.
 
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:
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 really drive someone insane when one sees the expression like ## (M^T)_\alpha {}^\beta## without any explanation on the meaning of this notation.
 
https://arxiv.org/pdf/2503.09804 From the abstract: ... Our derivation uses both EE and the Newtonian approximation of EE in Part I, to describe semi-classically in Part II the advection of DM, created at the level of the universe, into galaxies and clusters thereof. This advection happens proportional with their own classically generated gravitational field g, due to self-interaction of the gravitational field. It is based on the universal formula ρD =λgg′2 for the densityρ D of DM...
Back
Top