How do I verify the invariance on Yang-Mills' Lagrangian:

$$L = -\frac{1}{4} \sum_{a} \left(\partial_\mu A_\nu^a - \partial_\nu A_\mu^a + gf^{abc}A_\mu^bA_\nu^c \right)^2$$

under the transformation:

$$A_\mu^a (x) \rightarrow A_\mu^a (x) + \frac{1}{g} \partial_\mu \alpha^a(x) -f^{abc}\alpha^b(x)A_\mu^c(x)$$

The first term on the gauge transformation obviously recovers the old lagrangian, but the second and third terms are getting very problematic. Can someone help me with those? I'm note sure about my argument on terms $\partial_\nu \partial_\mu \alpha^a$ and $\partial_\mu \partial_\nu \alpha^a$, and the third one just get worse. I'm pretty sure that I must use the fact that the structure constant is totally antisymmetric, but it's not getting better.

  • $\begingroup$ Not totally sure what the issue is, but $\alpha$ is infinitesimal, so $\alpha^2$ terms should be dropped $\endgroup$ Commented May 21, 2019 at 18:55
  • 1
    $\begingroup$ Look here physics.stackexchange.com/q/456875 $\endgroup$
    – Vicky
    Commented May 21, 2019 at 19:00
  • $\begingroup$ The problem is entirely mathematical. The $\alpha^2$ terms will sure drop, but there are many other terms from the gauge transformation, mainly at $A_\mu^b A_\nu^c$, that i couldn't make it cancel with each other to show the lagrangian invariance. $\endgroup$
    – otto
    Commented May 21, 2019 at 19:26
  • $\begingroup$ @otto First, if you want to write an answer to a comment use '@nickname'. Otherwise the person you are answering will see nothing. Second, in the link I provided your question is solved in an 'exact' way, I mean without Taylor expansions. Make a full reading of it $\endgroup$
    – Vicky
    Commented May 24, 2019 at 11:17

2 Answers 2


We know that the field strength $F^{a}{}_{\mu \nu }$ should transform like the matter field $\psi _{\ell }$ under its adjoint representation(remember $\delta \psi _{\ell }\rightarrow \mathrm{i} \alpha ^{a}( x) (T_{\alpha } ){_{\ell }}^{m} \psi _{m}( x)$), which means we should have \begin{equation} \delta F^{b}{}_{\nu \mu } =\mathrm{i} \alpha ^{a}( x) (T^{\mathrm{ad}}{}_{a} )^{b}{}_{c} F^{c}{}_{\nu \mu } =\alpha ^{a} f^{b}{}_{ca} F^{c}{}_{\nu \mu } , \end{equation} where $\alpha ^{a}( x)$ is the infinitesimal transformation parameter, $T^{\mathrm{ad}}{}_{a}$ is the adjoint representation of the generator $T_{a}$, which is defined by $(T^{\mathrm{ad}}{}_{a} )^{b}{}_{c} \equiv -\mathrm{i} f^{b}{}_{ca}$, and the definition of the Lie algebra is given by $[T_{a} ,T_{b} ]=\mathrm{i} f^{c}{}_{ab} T_{c}$. Now we try to prove the claim before. Define \begin{equation*} A^{a}{}_{\mu }\rightarrow A^{a}{}_{\mu } +\delta A^{a}{}_{\mu } , \end{equation*} and here we have \begin{equation*} \delta A^{b}{}_{\mu } =\partial _{\mu } \alpha ^{b} /g+\mathrm{i} \alpha ^{a} (T^{\mathrm{ad}}{}_{a} )^{b}{}_{c} A^{c}{}_{\mu } =\partial _{\mu } \alpha ^{b} /g+f^{b}{}_{ca} \alpha ^{a} A^{c}{}_{\mu } . \end{equation*} Then compute directly: enter image description here Here the red terms cancel out. Now according to the defintion of the structre constant $[T_{a} ,T_{b} ]=\mathrm{i} f^{c}{}_{ab} t_{c}$, it is obviously anti-symmetric in $a,b$: \begin{equation*} f^{c}{}_{ab} =-f^{c}{}_{ba} . \end{equation*} So renaming the green term gives: \begin{aligned} f^{b}{}_{cd} \partial _{\nu } \alpha ^{c} A_{\mu }^{d} +f^{b}{}_{cd} A_{\nu }^{c} \partial _{\mu } \alpha ^{d} & =f^{b}{}_{cd} \partial _{\mu } \alpha ^{d} A^{c}{}_{\nu } -f^{b}{}_{dc} \partial _{\nu } \alpha ^{c} A_{\mu }^{d}\\ & =f^{b}{}_{ca} (\partial _{\mu } \alpha ^{a} A^{c}{}_{\nu } -\partial _{\nu } \alpha ^{a} A^{c}{}_{\mu } ), \end{aligned} which cancels out the blue term. Finally we look at the orange term, renaming the dummy indices: \begin{equation*} gf^{b}{}_{cd} f^{c}{}_{ea} \alpha ^{a} A^{e}{}_{\nu } A^{d}{}_{\mu } +gf^{b}{}_{cd} f^{d}{}_{fe} \alpha ^{e} A^{c}{}_{\nu } A^{f}{}_{\mu } =(f^{c}{}_{ea} f^{b}{}_{cd} +f^{c}{}_{da} f^{b}{}_{ec} )g\alpha ^{a} A^{e}{}_{\nu } A^{d}{}_{\mu } . \end{equation*} Now the Jacobi identity is given by: \begin{equation*} f^{c}{}_{ea} f^{b}{}_{cd} +f^{c}{}_{de} f^{b}{}_{ca} +f^{c}{}_{ad} f^{b}{}_{ce} =0, \end{equation*} So this term becomes: \begin{equation*} gf^{b}{}_{cd} f^{c}{}_{ea} \alpha ^{a} A^{e}{}_{\nu } A^{d}{}_{\mu } +gf^{b}{}_{cd} f^{d}{}_{fe} \alpha ^{e} A^{c}{}_{\nu } A^{f}{}_{\mu } =gf^{b}{}_{ca} f^{c}{}_{ed} \alpha ^{a} A^{e}{}_{\nu } A^{d}{}_{\mu } . \end{equation*} Put the uncancelled term together, we have: \begin{equation*} \begin{aligned} \delta (F^{b}{}_{\nu \mu } ) & =f^{b}{}_{ca} \alpha ^{a} (\partial _{\nu } A^{c}{}_{\mu } -\partial _{\mu } A^{c}{}_{\nu } )+gf^{b}{}_{ca} f^{c}{}_{ed} \alpha ^{a} A^{e}{}_{\nu } A^{d}{}_{\mu }\\ & =f^{b}{}_{ca} \alpha ^{a} (\partial _{\nu } A^{c}{}_{\mu } -\partial _{\mu } A^{c}{}_{\nu } +gf^{c}{}_{ed} \alpha ^{a} A^{e}{}_{\nu } A^{d}{}_{\mu } )=\alpha ^{a} f^{b}{}_{ca} F^{c}{}_{\nu \mu } , \end{aligned} \end{equation*} which is exactly what we want. Now we look at the Larganian: \begin{equation*} \mathcal{L}_{\mathrm{YM}} =-\frac{1}{4}\sum _{a} (F^{a}{}_{\nu \mu } )^{2} , \end{equation*} Under the transformation law of $F$ we found above: \begin{equation*} \delta (\mathcal{L}_{\mathrm{YM}} )=-\frac{1}{2}\sum _{a} (F^{a}{}_{\nu \mu } )\delta F{_{a}}^{\nu \mu } =-\frac{1}{2}\sum _{a} F^{a}{}_{\nu \mu } \alpha ^{b} f_{abc} F^{c\nu \mu } \end{equation*} Now we should use the information from the Lagrangian. We assume that YM Lagrangian here have $SU( N)$ symmetry, then we can choose the "metric" as: \begin{equation*} \operatorname{tr} (T^{a} T^{b} )=\delta ^{ab} , \end{equation*} Then trace out $[T_{a} ,T_{b} ]T_{d} =\mathrm{i} f^{c}{}_{ab} T_{c} T_{d}$: \begin{equation*} \operatorname{tr} ([T_{a} ,T_{b} ]T_{d} )=\mathrm{i} f^{c}{}_{ab}\operatorname{tr}( T_{c} T_{d}) =\mathrm{i} f^{c}{}_{ab} \delta _{cd} =\mathrm{i} f_{dab} . \end{equation*} Thus \begin{equation*} \mathrm{i} f_{dab} =\operatorname{tr} ([T_{a} ,T_{b} ]T_{d} )=\operatorname{tr} (T_{a} T_{b} T_{d} -T_{b} T_{a} T_{d} )=\operatorname{tr} (T_{d} T_{a} T_{b} -T_{a} T_{d} T_{b} )=-\operatorname{tr}([ T_{a} ,T_{d}] T_{b}) =-\mathrm{i} f_{adb} , \end{equation*} This means $f_{dab} =f_{[ ad] b}$, together with $f_{dab} =f_{d[ ab]}$, we can see $f_{dab} =f_{[ dab]}$. Therefore \begin{equation*} \delta (\mathcal{L}_{\mathrm{YM}} )=-\frac{1}{2}\sum _{a} \alpha ^{b} f_{[ abc]} F^{( a}{}_{\nu \mu } F^{c) \nu \mu } =0. \end{equation*} Which means $\mathcal{L}_{\mathrm{YM}}$ is gauge invariant.


  1. Here we use the "operator" $\delta $ to simplify the calculation, but you can also pulg in the tranformed $A_{\mu }$ directly, and get the same result. However, the only thing different is, in this way you may encounter the term of $\alpha ( x)$ in second order. Remember it is an infinitesimal transformation and just drop it.
  2. Till we derive the transformation rule of $F^{a}{}_{\nu \mu }$, we didn't use the condition that $f_{abc}$ is totally anti-symmetric. Note that not all structure constants of Lie algebra are totally anti-symmetric. However, not only for YM theory,as long as we want to choose our Lagrangian to be real, gauge invariant and be the positive-definte quadric form of the field $F^{a}{}_{\mu \nu }$, we can prove our structre constants are totally anti-symmetric(See Weinberg, The quantum theory of fields, Vol II, Chap 15.2.). Of course, the $SU( N)$ symmetry here is due to the matter field $\psi _{\ell }$, which doesn't occur in the Lagrangian here. So, you cannot solve this problem through pure math without any physical assumption, i.e. without knowing any information about the Lie algebra. We construct such a Lagrangian because the gauge invariance should be satisfied.

As already mentioned in the comment, the exact version was proven in the link. However, if you insisted to show it infinitesimally in your notation, here is how.

First, consider the variation (warning: sloppy of notation use in the following but it is clear in the context I guess)

\begin{equation} \label{delL} \delta L = -\frac{1}{2} \sum_a \left( \partial_\mu A_\nu^a - \partial_\nu A^a_\mu + gf^{abc}A^b_\mu A^c_\nu \right) \left( \partial_\mu \delta A^a_\nu + gf^{abc}A^b_\mu \delta A^c_\nu - (\mu \leftrightarrow \nu) \right), \end{equation} where \begin{equation} \label{delA} \delta A^a_\mu = \frac{1}{g} \partial_\mu A^a - f^{abc} \alpha^b A^c_\mu. \end{equation} Now, if you substitute \tag{delA} into \eqref{delL} and patiently listed every terms, you will encounter the terms you noticed already: \begin{equation} \delta L = -\frac{1}{2}\sum_a F_{\mu\nu}^a \left[ \partial_\mu \left( \frac{1}{g}\partial_\nu \alpha^a - f^{abc}\alpha^b A^c_\nu \right) + f^{abc}A^b_\mu \left( \frac{1}{g}\partial_\nu\alpha^c - f^{cde}\alpha^d A^e_\nu \right) - (\mu\leftrightarrow\nu) \right] \end{equation} I will not show in detail every step, but describe the computation. Expanding the first big round bracket inside the big square bracket, you will have the derivative hitting on the (infinitesimal) gauge parameter, $ \frac{1}{g}\partial_\mu \partial_\nu \alpha^a - (\mu\leftrightarrow\nu). $ The above two terms (notice the term with $\mu,\nu$ exchanged actually cancel each other, because \begin{equation} \partial_\mu \partial_\nu \alpha^a = \partial_\nu \partial_\mu \alpha^a, \end{equation} i.e. partial derivatives commute. There are also terms, $ -f^{abc} (\partial_\mu \alpha^b) A^c_\nu + f^{abc}A^b_\mu \partial_\nu \alpha^c - (\mu\leftrightarrow\nu), $ where the above first term comes from the first round bracket and the above second term from the second round bracket inside the big square bracket. One can easily show that they cancel each other after renaming the indices and considering the anti-symmetric properties of the spacetime indices $\mu,\nu$ and the internal indices $a,b,c$.

Finally, the most tricky part is the remaining terms, \begin{equation} -f^{abc} \alpha^b \partial_\mu A^c_\nu - f^{abc}f^{cde} \alpha^d A^b_\mu A^e_\nu - (\mu \leftrightarrow \nu). \end{equation} The second term above together with their $- (\mu \leftrightarrow \nu)$ counter part can be rewritten using Jacobi identity and then combine with the rest term to get \begin{equation} f^{abc}\alpha^b F_{\mu\nu}^c \end{equation} Substitute this back to the original variation equation \eqref{delL}, you will get \begin{equation} f^{abc} \alpha^b F_{\mu\nu}^c F^{\mu\nu\, a}. \end{equation} You may want to verify that the last pieces in this form vanish (relatively) manifest.

  • $\begingroup$ Anyone has a clue how Latex equation reference is done here? I tried \eqref{} and \label{} but it does not work. $\endgroup$
    – chichi
    Commented Aug 13, 2019 at 0:50
  • $\begingroup$ Hmm...both \ eqref and \ tag do not work, I gave up. $\endgroup$
    – chichi
    Commented Aug 13, 2019 at 3:08

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