- #1
Jim Kata
- 204
- 10
Hi I'm trying to understand very basic aspects of the WZW model, and would appreciate any help. The main thing i don't get is how you replace
[tex]
I = \frac{1}
{2}\int\limits_0^1 {\gamma _{\alpha \beta } } \frac{{dx^\alpha }}
{{d\tau }}\frac{{dx^\beta }}
{{d\tau }}d\tau
[/tex]
with
[tex]
I = \frac{{ - k}}
{4}\int\limits_0^1 {tr} (g^{ - 1} \frac{d}
{{d\tau }}g)^2 d\tau
[/tex]
for some group g
My thoughts are I don't know how the metric enters and the connection doesn't. For example consider
[tex]
{\mathbf{g}} = P\exp \int\limits_0^1 {\frac{{dx^\mu }}
{{d\tau }}} {\mathbf{A}}_\mu d\tau
[/tex]
then
[tex]
{\mathbf{g}}^{ - {\mathbf{1}}} \frac{d}
{{d\tau }}{\mathbf{g}} = - \frac{{dx^\mu }}
{{d\tau }}{\mathbf{A}}_\mu
[/tex]
so the connection enters but not sure how the metric does cos I'm not familiar with any identity that says
[tex]
tr\left( {{\mathbf{A}}_\mu {\mathbf{A}}_\tau } \right) = \gamma _{\mu \tau }
[/tex]
[tex]
I = \frac{1}
{2}\int\limits_0^1 {\gamma _{\alpha \beta } } \frac{{dx^\alpha }}
{{d\tau }}\frac{{dx^\beta }}
{{d\tau }}d\tau
[/tex]
with
[tex]
I = \frac{{ - k}}
{4}\int\limits_0^1 {tr} (g^{ - 1} \frac{d}
{{d\tau }}g)^2 d\tau
[/tex]
for some group g
My thoughts are I don't know how the metric enters and the connection doesn't. For example consider
[tex]
{\mathbf{g}} = P\exp \int\limits_0^1 {\frac{{dx^\mu }}
{{d\tau }}} {\mathbf{A}}_\mu d\tau
[/tex]
then
[tex]
{\mathbf{g}}^{ - {\mathbf{1}}} \frac{d}
{{d\tau }}{\mathbf{g}} = - \frac{{dx^\mu }}
{{d\tau }}{\mathbf{A}}_\mu
[/tex]
so the connection enters but not sure how the metric does cos I'm not familiar with any identity that says
[tex]
tr\left( {{\mathbf{A}}_\mu {\mathbf{A}}_\tau } \right) = \gamma _{\mu \tau }
[/tex]