# I Spinorial identity

1. Oct 1, 2017

### Korybut

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. Oct 2, 2017

### Ben Niehoff

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.

3. Oct 2, 2017

### Korybut

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
4. Oct 14, 2017

### Korybut

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.