- #1
Jim Kata
- 197
- 10
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
<br /> ds^2 = \bar g_{\alpha \beta } (\bar x)d\bar x^\alpha d\bar x^\beta = g_{\alpha \beta } (x)dx^\alpha dx^\beta <br />
doing coordinate change <br /> \bar x^\alpha = x^\alpha + \varepsilon \xi ^\alpha (x)<br />
where \varepsilon is an infinitesimal and \xi is a killing vector.
<br /> 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 <br />
if nothing is to change, you get the requirement
<br /> 0 = \delta g_{\alpha \beta } dx^\alpha dx^\beta = - \varepsilon (\xi _{\alpha ;\beta } + \xi _{\beta ;\alpha } )<br />
Assuming your on the light cone <br /> g_{\alpha \beta } dx^\alpha dx^\beta = 0<br />, you can say <br /> \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:
<br /> \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 <br />
(I don't understand why the power series terminates after the second derivative term for dimensions 3 and higher?)
an infinitesimal group element is
<br /> 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 <br />
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?
I'll go through what I have and highlight the parts I"m unsure of
<br /> ds^2 = \bar g_{\alpha \beta } (\bar x)d\bar x^\alpha d\bar x^\beta = g_{\alpha \beta } (x)dx^\alpha dx^\beta <br />
doing coordinate change <br /> \bar x^\alpha = x^\alpha + \varepsilon \xi ^\alpha (x)<br />
where \varepsilon is an infinitesimal and \xi is a killing vector.
<br /> 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 <br />
if nothing is to change, you get the requirement
<br /> 0 = \delta g_{\alpha \beta } dx^\alpha dx^\beta = - \varepsilon (\xi _{\alpha ;\beta } + \xi _{\beta ;\alpha } )<br />
Assuming your on the light cone <br /> g_{\alpha \beta } dx^\alpha dx^\beta = 0<br />, you can say <br /> \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:
<br /> \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 <br />
(I don't understand why the power series terminates after the second derivative term for dimensions 3 and higher?)
an infinitesimal group element is
<br /> 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 <br />
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?
