# Conformal symmetry group for R(1,3)

1. Jun 18, 2009

### Jim Kata

I'm trying to derive the lie algebra for the conformal symmetry group and was curious if anyone could help?

I'll go through what I have and highlight the parts I"m unsure of

$$ds^2 = \bar g_{\alpha \beta } (\bar x)d\bar x^\alpha d\bar x^\beta = g_{\alpha \beta } (x)dx^\alpha dx^\beta$$

doing coordinate change $$\bar x^\alpha = x^\alpha + \varepsilon \xi ^\alpha (x)$$

where $$\varepsilon$$ is an infinitesimal and $$\xi$$ is a killing vector.

$$ds^2 = (g_{\alpha \beta } (x) + g_{\alpha \beta ,\gamma } \varepsilon \xi ^\gamma + \delta g_{\alpha \beta } )(\delta _\mu ^\alpha + \varepsilon \xi _\mu ^\alpha )(\delta _\tau ^\beta + \varepsilon \xi _\tau ^\beta )dx^\mu dx^\tau$$

if nothing is to change, you get the requirement

$$0 = \delta g_{\alpha \beta } dx^\alpha dx^\beta = - \varepsilon (\xi _{\alpha ;\beta } + \xi _{\beta ;\alpha } )$$

Assuming your on the light cone $$g_{\alpha \beta } dx^\alpha dx^\beta = 0$$, you can say $$\xi _{\alpha ;\beta } + \xi _{\beta ;\alpha } = \lambda g_{\alpha \beta }$$

( I'm not sure why this last equation is true. Wouldn't any symmetric rank two tensor on light cone work?)

This differential equation can be solved by looking at the power series and for dimensions greater or equal to 3 you get the answer:

$$\xi ^\alpha = x^\alpha + \omega ^{\alpha \beta } x_\beta + a^\alpha + \lambda x^\alpha + \rho ^\alpha x^\mu x_\mu - 2x^\alpha \rho ^\mu x_\mu$$

(I don't understand why the power series terminates after the second derivative term for dimensions 3 and higher?)

an infinitesimal group element is
$$U(1 + \omega ,a,\lambda ,\rho ) = 1 + (i/2)J^{\alpha \beta } \omega _{\alpha \beta } + iP^\alpha a_\alpha + i\lambda D + i\rho _\alpha K^\alpha$$

Now I should be able to use the composition of coordinates and infinitesimal group element to derive the lie algebra, but I can't seem to get it to work. I can do it for the Lorentz group and even the intrinsically projective Galilean group, but can't make this work. How do I go about it?

2. Jun 18, 2009

### samalkhaiat

3. Jul 14, 2009

### Jim Kata

great stuff sam,

I read your posts and they are a treasure trove, but I was still curious can you derive the lie algebra the unitary representation way without the postulate that the conformal generators transform like a lorentz vectors? I should be able to do it solely off of the composition law of the coordinate change right? But when I try to work it out, I run into problems with the conformal term. Any help?

4. Jul 18, 2009

### samalkhaiat

Lorentz covariace was an essential part of the algebriac method which I was doing, see EX(5.5). But yes, you should be able to do it, all you need is correct multiplication law. For the full conformal group, this law is a bit complicated because of the non-linear nature of the conformal transformation.This is why I was treating subgroups of of the conformal group. Any way, if you could show me your work, I might be able to tell you what to do about what.

regards

sam

5. Aug 23, 2009

### Jim Kata

Sam, let me ask a follow up question that I didn't see in your paper. Is there a central extension to the lie algebra for the conformal group? I was looking at Ferrara and Fayet and they mention this thing called the chiral charge, and it looks like a central extension. If there is one in which commutators does it appear?