Equivalence between Chern-Simons action and first order formalism

I can not derive second line from Chern-Simons action

\noindent $S_{cs}$=k$\int{Tr(A \wedge dA+\frac{2}{3}}A \wedge A\wedge A)$

\noindent =k$\int{\ e }^a\wedge$R[$\omega$]

we have to use gauge field below for ISO(2,1)

\noindent If :${\ A}_{\mu }$ =${\ e }^a_{\mu \ }P_{a\ }$+${\omega }^a_{\mu \ }J_{a\ }$

\noindent And for ISO(2,1) : [$\ P_{a\ ,}P_{b\ }]=0$ ; [$\ J_{a\ }{,P}_{b\ }]=\ {\varepsilon }_{abc}\ P^c\ \ \ \ \ \ \ \ \ ;$ [$\ J_{a,\ }{\ J}_{b\ }]=\ {\varepsilon }_{abc}\ J^c$

\noindent $\ Tr ( \ P_{a\ }, J_{a\ })$=$\ {\eta }_{ab}$

I can not solve it because Trace is for two generators not three.

• I think you could figure out the answer by looking at physics.stackexchange.com/q/119953 – ungerade Apr 22 '15 at 18:49
• i can not do it for forms. – Ali Apr 22 '15 at 21:17