# Infinitesimal SUSY transformation of SYM lagrangian

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,

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Does the extra term vanish on the equations of motion?

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.

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.

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.

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.

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

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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