Showing closure of the SUSY algebra of a free abelian gauge multiplet

Question

Given the complete supersymmetric lagrangian of a free abelian gauge multiplet $$ \mathcal{L} = -\frac{1}{4} F_{\mu\nu} F^{\mu\nu} + i \bar{\lambda} \bar{\sigma}^\mu \partial_\mu \lambda + \frac{1}{2} D^2 + \xi D $$ With the following variations $$ \delta_\zeta A_\mu = \bar{\zeta} \bar{\sigma}^\mu \lambda + \bar{\lambda} \bar{\sigma}^\mu \zeta $$ $$ \delta_\zeta \lambda_\alpha = \frac{i}{2} F_{\mu\nu} (\sigma^\mu \bar{\sigma}^\nu \zeta)_\alpha + D \zeta_\alpha $$ $$ \delta_\zeta D = -i \bar{\zeta} \bar{\sigma}^\mu \partial_\mu \lambda + i \partial_\mu \bar{\lambda} \bar{\sigma}^\mu \zeta $$

Where $\zeta_\alpha$ is an infinitesimal Grassmann number.

Now to show that the algebra closes I took the commutator of the variations, acting on each of the fields.

Showing that the algebra closes for $\lambda$ and $D$ is relatively start forward algebra, resulting in $$ [\delta_\psi, \delta_\zeta] \lambda_\alpha = -2i(\bar{\zeta} \bar{\sigma}^\mu \psi - \bar{\psi} \bar{\sigma}^\mu \zeta)\partial_\mu \lambda_\alpha $$

$$ [\delta_\psi, \delta_\zeta] D = -2i(\bar{\zeta} \bar{\sigma}^\mu \psi - \bar{\psi} \bar{\sigma}^\mu \zeta)\partial_\mu D $$

However when I do it for the gauge fields I find, $$ [\delta_\psi, \delta_\zeta] A_\mu = -2i(\bar{\zeta} \bar{\sigma}^\alpha \psi - \bar{\psi} \bar{\sigma}^\alpha \zeta)(\partial_\alpha A_\mu - \partial_\mu A_\alpha) $$

Which almosts looks like the translation that one would expect. However it has that extra piece there. Is there something that makes that $\partial_\mu A_\alpha$ term zero that I am missing?

In http://arxiv.org/abs/hep-th/0109172 Jose suggests (pg 12) that the auxiliary fields are added to close the algebra off-shell, which in the case of the gauginos it did. He also mentions on (pg 21) that in the case of having gauge fields the algebra will only close up to a gauge transformation. Now he did that before adding auxiliary fields, so maybe it isn't true...

However if this is true, it seems odd to me that the algebra would only close up to a gauge transformation. To me that would imply that gauge fixing is required and that doesn't really seem kosher. For we would need to choose a gauge for the theory to be valid, otherwise we don't have a true gauge symmetry. How do we get around that problem?

EDIT:

@Qmechanic I believe a good reference is http://arxiv.org/abs/hep-ph/9709356, just a note in advanced he uses the mostly plus convention where I use the mostly minus, essentially the translations and sigma matrices just differ by an overall minus sign, but the spirit is there. I would look at sections 2 and 3 in particular section 3.3 (for the abelian case set a=1 and the covariant derivatives just become standard derivatives). Also note that I absorbed the factor $\frac{1}{2}$ into the infinitesimal parameter and he's choice of sign is different on the variation of the gauge field.

Notes of Convention: My conventions are nearly identical to the SUSY Primer reference except for the metric. $$ \eta^{\mu\nu} = diag(+ - - -) $$

Clarification of the Question Provided that my arithmetic is correct, how can one see that the $\partial_\mu A_\alpha$ term dies?

After that, even without checking my work (thank you if you do!), why is it for the non-abelian case equation (3.3.9), do we see a covariant derivative? If the algebra were to truly be closed I would expect that to be a partial derivative which are the spacetime translations that we expect to result (see eq 3.1.29). If we should see a covariant derivative, then why?

Additional Comment

It turns out that the turn doesn't die and is indeed correct. It gives the form $$ [\delta_\psi, \delta_\zeta] A_\mu^a = -2i(\bar{\zeta} \bar{\sigma}^\alpha \psi - \bar{\psi} \bar{\sigma}^\alpha \zeta)F_{\alpha\mu}^a $$

Which holds in the case of the Non-Abelian case (the other terms go from being of the form in (3.3.9) of SUSY Primer)

In Freedman's and De Wit's paper journals.aps.org/prd/abstract/10.1103/PhysRevD.12.2286 they discuss the implications of what having a gauge symmetry does to the commutator of two covariant SUSY transformations. It now generates a new transformation which can be described as a translation plus a gauge transformation, or what they call a gauge-covariant translation.

How is this considered closing the graded lie algebra, which should consist of Translations, Boosts/Rotations, and a SUSY transformation?

Answer 1

The bracket you have written is of the form $$[\delta,\delta] A_\mu = v^\nu \partial_\nu A_\mu + \partial_\mu (v^\nu A_\nu).$$ As you pointed out, the first term corresponds to a translation by $v^\mu$. The second term corresponds to a gauge transformation $\delta A_\mu = \partial_\mu \lambda$ with $\lambda = v^\nu A_\nu$. So the algebra closes up to a translation and a gauge transformation, taking the form $[\delta,\delta]A_\mu = v^\mu F_{\mu\nu}$, which is typical of these kinds of calculations.

The non-abelian case is similar. The commutator of two supersymmetry transformations on, say, a scalar $\phi$ in the vector multiplet gives a translation $v^\nu\partial_\nu \phi$ plus a gauge transformation $[v^\nu A_\nu,\phi]$ (since fields in the vector multiplet transform in the adjoint of the gauge group). Thus, $$[\delta,\delta]\phi = v^\nu \partial_\nu \phi + [v^\nu A_\nu, \phi] = v^\nu D_\nu \phi,$$ where $D_\nu = \partial_\nu + [A,\cdot]$.

To answer your follow-up question: Consider the vector field $A_\mu$. It belongs to the space $\mathcal{A}$ of gauge fields on $\mathbb{R}^{1,3}$. One would like to write down a representation of the supersymmetry algebra on the space of functionals of the fields $\{A_\mu,\ldots\}$, where the $\ldots$ denote other fields in the multiplet like scalars and fermions. But this space does not give a representation of the algebra. That's not surprising, since field configurations which differ by a (small) gauge transformation are physically equivalent. Thus, to be more precise, we should say that the physical gauge field corresponds to an equivalence class $[A_\mu]$ in $\mathcal{A}/\mathcal{G}$, where $\mathcal{G}$ denotes the space of small gauge transformations. It is the functionals of $\{[A_\mu],\ldots\}$ which gives a representation of the algebra, which requires that $$\big[[\delta,\delta]A_\mu\big] = \big[v^\sigma \partial_\sigma A_\mu\big].$$ In terms of fields in $\mathcal{A}$, this equality of classes in $\mathcal{A}/\mathcal{G}$ means that there exists some $\lambda$ such that $$[\delta,\delta]A_\mu - v^\sigma \partial_\sigma A_\mu = \partial_\mu \lambda,$$ which is precisely what we showed earlier with $\lambda = v^\mu A_\mu$.