Conformal invariance of gluon amplitudes

1. Nov 30, 2011

pkwei99

Hi,

I'm very ashamed to not understand how even the simplest gluon amplitudes are conformally invariant. See eg http://arxiv.org/abs/hep-th/0312171 pages 11-12.
$$M(1^-,2^-,3^+)=\delta(\sum_i \lambda_i\tilde{\lambda}_i)\frac{\langle12\rangle^4}{\langle12\rangle \langle 23\rangle\langle31\rangle}$$
the dilation operator is:
$$D\sim \lambda\frac{\partial}{\partial \lambda}+\tilde{\lambda}\frac{\partial}{\partial \tilde{\lambda}}+2$$
First, I assume the dilation operator contains a sum over all particles. Next, Witten says the delta function carries weight -4 under D. Ok. Then he says that $$\langle 12\rangle^4$$ has weight 4. This I don't get. Doesn't it have weight 4 under just eg $$\lambda_1\frac{\partial}{\partial \lambda_1}$$
So
$$D \langle12\rangle^4=\sum_i (\lambda\frac{\partial}{\partial \lambda}+\tilde{\lambda}\frac{\partial}{\partial \tilde{\lambda}}+2)\langle12\rangle^4=[(4+0+2)+(4+0+2)+(0+0+2) ] \langle12\rangle^4?=14\langle12\rangle^4$$

Thanks for any help:)

2. Nov 30, 2011

petergreat

Conformal invariance is obvious, because you're talking about the tree amplitude which simply reflects the classical behaviour of the theory. The Yang-Mills Lagrangian is classically conformal invariant. Only loop graphs break this symmetry. But if you want to explicitly verify conformal invariance from the MHV amplitude expression, I haven't studied enough to answer that.