# Self-dual and Anti Self-dual antisymmetric tensor

1. Nov 24, 2012

### wphysics

I am working on Chapter 5.6 of Weinberg QFT book vol1.

In page 231 and 232, he said (1,0) or (0,1) field corresponds to an antisymmetric tensor
$F^{\mu\nu}$
that satisfies the further irreducibilitry 'duality' conditions
$F^{\mu\nu}= \pm {{i}\over{2}} \epsilon ^{\mu\nu\lambda\rho}F_{\lambda \rho}$
for (1,0) and (0,1) fields, respectively.

I check that Self-dual antisymmetrc tensor has three independent components and so does Anti Self-dual antisymmetric tensor. So, I can guess that there might be correspondence between Self-dual(or Anti Self-dual) antisymmetric tensor and (1,0) field (or (0,1)field ).

But, I could not show how they are related to each other mathematically.

I would like you guys to show me that how we correspond Self-dual antisymmetric tensor with (1,0) field and
Anti Self-dual antisymmetric tensor with (0,1) field.

Thank you so much.

2. Nov 24, 2012

### Bill_K

Given Fμν, we define its dual as Fμν* = ½εμνστFστ. The self-dual and anti-selfdual tensors are complex combinations,
SFμν = Fμν + iFμν* and AFμν = Fμν - iFμν*.

It's easier to represent the F's as a pair of 3-vectors, Fμν = (E, B). Then Fμν* = (B, -E), SFμν = (E + iB, B - iE), and AFμν = (E - iB, B + iE).

This shows that SFμν and AFμν each have only three complex components.

Under a space rotation, all of these 3-vectors behave the same way. Under an infinitesimal rotation about an axis with unit vector ω, a 3-vector V transforms as dV = ω x V, and this applies equally well to E, B and E ± iB. The generators for spatial rotations are denoted J.

Under a finite boost, E' = γ(E + β x B), B' = γ(B - β x E), and the infinitesimal form of this is dE = β x B, dB = - β x E. Then d(E + iB) = β x (B - iE) = -i β x (E + iB) and d(E - iB) = i β x (E - iB). The generators for boosts are denoted K.

The Lorentz group is a direct product of two groups generated by J + iK and J - iK. I'm getting tired of writing down all the details, but if you combine the above results and calculate how (E ± iB) transform under (J ± iK), you'll find the changes add in one case and cancel in the other, indicating that these quantities transform exactly as the (1,0) and (0,1) representations must.

Last edited: Nov 24, 2012
3. Nov 20, 2013

### m1rohit

Sir(Bill_K),
can you give me some reference on this topic.