I am reading the chapter of non-abelian gauge invariance from Peskin and Schroeder. Why is the term $-\frac{1}{4}(L_{\mu\nu}^i)^{2} $ gauge invariant?
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what's $L^i_{\mu\nu}$, field strength? If so, it's "square" -what you wrote- is proportional to Killing form (which is essentially a trace); this is gauge invariant since $L$ transforms like $L\mapsto gLg^{-1} $ under gauge transformations.
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$\begingroup$ Could I ask: what is the link between "proportional to Killing form" and "L transforms like L $\rightarrow g^{-1}Lg$"? Thanks. $\endgroup$ Commented Jun 5, 2012 at 23:50
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1$\begingroup$ The gauge transformation on the field strength looks rather like $L\mapsto gLg^{-1}=:Ad_g(L)=L'$. The Killing or Killing-Cartan is a nondegenerate form is defined as $K(X,Y)=Trace(\mathrm{ad}_X\circ\mathrm{ad}_Y)=X^aY^bk_{ab}$, where $k_{ab}$ are determined by the structure constants of the algebra and the generators are usually chosen sothat $k_{ab}=\delta_{ab}$. $K$ is Ad invariant, by cyclicity of the trace. Therefore, if $X=Y=L^a t_a$ (being $t_a$ the aforementioned normalized generators), one gets the gauge invariance for the term $L_aL^a=K(L,L)=K(Ad_gL,Ad_gL)=K(L',L')=L'_a(L')^a.$ $\endgroup$– c.p.Commented Jun 6, 2012 at 4:00
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$\begingroup$ Hey Thanks a lot for the answer. But, please could you help me prove that $K(X,Y)=K(Ad_gX,Ad_gY)$. I have never seen this property before. $\endgroup$– user7757Commented Jun 6, 2012 at 4:44
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$\begingroup$ $G$, the Lie group, acts on itself by Ad$_g$, and this action is well defined on the Lie algebra, $Ad_g X=gXg^{-1} (X\in\mathfrak{g})$ One has the action of the Lie algebra $\mathfrak{g}=$Lie($G$) on itself given by $ad_X(Y)=[X,Y]$. Now you can easily prove that $ad_{Ad_gX} Y=Ad_g ad_X Ad_{g^{-1}} Y$ for all $X,Y\in\mathfrak{g}$, whence, by cyclicity of Tr $K(Ad_gX, Ad_gY)=Trace(Ad_gad_XAd_{g^{-1}}Ad_gad_YAd_{g^{-1}})=Trace(ad_Xad_Y)=K(X,Y)$. $\endgroup$– c.p.Commented Jun 6, 2012 at 5:31
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$\begingroup$ Actually, Gregory Naber refers directly to the Killing-Cartan form as $K(A,B)=-2\mathrm{Tr}(AB)$. This is evidently invariant under the Adjoint action $K(Ad_gA,Ad_gB)=-2\mathrm{Tr}(Ad_gA \,Ad_gB)=-2\mathrm{Tr}(gAg^{-1} gBg^{-1})=-2\mathrm{Tr}(gABg^{-1})=K(A,B)$, by cyclicity of the trace. Perhaps that was the easy answer. (the factor "-2" is a matter of [perhaps physical] convention) $\endgroup$– c.p.Commented Jun 6, 2012 at 5:40