Infinitesimal SUSY transformation of SYM lagrangian

[karlzr]

I tried to verify that the SYM lagrangian is invariant under SUSY transformation, but it turned out there is a term that doesn't vanish.
The SYM lagrangian is:
$$\mathscr{L}_{SYM}=-\frac{1}{4}F^{a\mu\nu}F^a_{\mu\nu}+i\lambda^{\dagger a}\bar{\sigma}^\mu D_\mu \lambda^a+\frac{1}{2}D^a D^a$$
the infinitesimal SUSY transformations of the gauge multiplet are:
$$\delta A^a_\mu =-\frac{1}{\sqrt{2}}(\epsilon^\dagger \bar{\sigma}_\mu \lambda^a+\lambda^{\dagger a} \bar{\sigma}_\mu \epsilon)=-\frac{1}{\sqrt{2}}\epsilon^\dagger \bar{\sigma}_\mu \lambda^a+h.c.$$
$$\delta\lambda^a_\alpha =-\frac{i}{2\sqrt{2}}(\sigma^\mu \bar{\sigma}^\nu\epsilon)_\alpha F^a_{\mu\nu}+\frac{1}{\sqrt{2}}\epsilon_\alpha D^a$$
$$\delta \lambda^{\dagger a}_{\dot{\alpha}} =\frac{i}{2\sqrt{2}}(\epsilon^\dagger \bar{\sigma}^\nu\sigma^\mu)_{\dot{\alpha}} F^a_{\mu\nu}+\frac{1}{\sqrt{2}}\epsilon^\dagger_{\dot{\alpha}} D^a$$
$$\delta D^a =\frac{-i}{\sqrt{2}}(\epsilon^\dagger \bar{\sigma}^\mu D_\mu \lambda^a-D_\mu \lambda^{\dagger a}\bar{\sigma}^\mu \epsilon)=\frac{-i}{\sqrt{2}}\epsilon^\dagger \bar{\sigma}^\mu D_\mu \lambda^a+h.c.$$
I derived that
$$\delta F^a_{\mu\nu}= -\frac{1}{\sqrt{2}}\epsilon^\dagger(\bar{\sigma}_\nu D_\mu-\bar{\sigma}_\mu D_\nu)\lambda^a+h.c.$$
Then when I calculated the variation of the total lagrangian, I obtained that the $D^a$ and $F^a_{\mu\nu}$ terms cancel out separately. But there is a term from the variation of $A^b_\mu$ in the definition of covariant derivative in $D_\mu \lambda^a$ that remains after all the other cancellation. It looks like
$$i g f^{abc}\lambda^{\dagger a}\bar{\sigma}^\nu \lambda^c (\epsilon^\dagger \bar{\sigma}_\nu \lambda^b+h.c.)$$
This is the only term with three $\lambda$s. So it seems there is no term to cancel this one. I have checked my calculation over and over again but couldn't find the mistake. So any idea what 's wrong?
Thanks a lot,

 

[Korybut]

Does the extra term vanish on the equations of motion?

 

[karlzr]

Does the extra term vanish on the equations of motion?

That term is proportional to the current of the gaugino. But since all $D_\mu F^{a\mu\nu}$ terms have already been canceled, this current term doesn't vanish by itself.

 

[karlzr]

Does the extra term vanish on the equations of motion?

Actually I do have this question. Is it ok if the infinitesimal transformation of lagrangian vanishes only after we apply the equations of motion to it? since I don't think so. I expect it vanishes off-shell.

 

[Korybut]

Lagrangian should be invariant off-shell, you are right.

I am not an expert in this. But probably you need to introduce new field with purely algebraic equations of motion, like F in WZ model, just to cancel this term.

 

[karlzr]

Lagrangian should be invariant off-shell, you are right.

I am not an expert in this. But probably you need to introduce new field with purely algebraic equations of motion, like F in WZ model, just to cancel this term.

I think the $D^a$ is introduced to do the same thing as $F$ in WZ model. But this $D^a$ doesn't solve this problem. Many textbooks I know discussed the infinitesimal SUSY transformation of WZ lagrangian but not this SYM lagrangian.
Thanks though.
I guess I will check my calculation again. I couldn't think of a way to cancel the term containing three $\lambda$s.

 

[karlzr]

I think I know what's going on there. The remaining term vanishes by itself by using Fierz identity.

 

[Jim Kata]

I think the answer to your question is related to the normed division algebras $$tri(\lambda)$$ law. It is also the reason SYM only takes place in dimensions 3,4,6,10 same with super strings and why there only superbranes in 4 different dimensions due to the fact there are only 4 normed division algebras. Check out this paper.
http://math.ucr.edu/home/baez/sym.pdf