# What's the relation between SUSY Generators and Covariant Derivatives?

1. Jan 4, 2009

### JosephButler

Hello once again. I'm trying to understand the relation between the superspace representation of the SUSY generators $Q_\alpha,\overline Q_{\dot\beta}$ and the covariant derivatives on superspaces $D_\alpha, \overline D_{\dot\beta}$:

$Q_\alpha = \frac{\partial}{\partial\theta^\alpha} - i\sigma^{\mu}_{\phantom\mu\alpha\dot\alpha}\overline\theta^{\dot\alpha}\partial_\mu$
$\overline Q_{\dot\alpha} = \frac{\partial}{\partial\overline\theta^{\dot\alpha}} - i\theta^\alpha\sigma^{\mu}_{\phantom\mu\alpha\dot\alpha}\partial_\mu$
$D_\alpha = \frac{\partial}{\partial\theta^\alpha} + i\sigma^{\mu}_{\phantom\mu\alpha\dot\alpha}\overline\theta^{\dot\alpha}\partial_\mu$
$\overline D_{\dot\alpha} = -\frac{\partial}{\partial\theta^\alpha} - i\sigma^{\mu}_{\phantom\mu\alpha\dot\alpha}\overline\theta^{\dot\alpha}\partial_\mu$

These look very similar. The difference between the $Q$s and the $D$ is the sign of their anticommutators:

$\{Q_\alpha,\overline Q_{\dot\alpha}\} = +2i\sigma^\mu_{\phantom\mu\alpha\dot\alpha}\partial_\mu = -2\sigma^{\mu}_{\phantom\mu\alpha\dot\alpha}P_\mu$
$\{D_\alpha,\overline D_{\dot\alpha}\} = -2i\sigma^\mu_{\phantom\mu\alpha\dot\alpha}\partial_\mu = +2\sigma^{\mu}_{\phantom\mu\alpha\dot\alpha}P_\mu$

My question is this: we have two pairs of differential operators which both induce a motion in superspace. Both pairs of operators transform appropriately under Lorentz transformations. Why is it, then, that one pair of operators get called the generators of the symmetry while the other set get called the covariant derivatives?

Could we have swapped them, so that we call the $Q$'s the covariant derivatives and the $D$'s the SUSY generators?

Wess & Bagger Chapter IV suggests that this has something to do with left- and right-multiplication... but it's not clear to me what the significance of this is.

I'm struggling to find an analogy for this. For gauge symmetries we never talk about the gauge covariant derivative and a differential representation of the gauge symmetry generators. For translations in flat space we never talk about the momentum generator versus the translation covariant derivative.

How are the $Q$s and $D$s related and why is one considered a generator while the other a covariant derivative? (And what does this all have to do with right- and left- multiplication?)

2. Jan 14, 2009

### samalkhaiat

1) Where do the operators ($D, \bar{D}$) come from?
2) Why are they called "covariant" derivatives?

I said SOON because the subject suffers from illness called "notational nightmare"
but I will try to be as careful as I possibly can.

regards

sam

3. Jan 18, 2009

### JosephButler

I understand now that the D's are 'covariant' in the sense of (anti)commuting with the Q's... but I don't understand what this means from a, say, Riemannian geometry perspective. (I.e. thinking of the covariant derivative in terms of parallel transport.)

-Joe

4. Feb 8, 2009

### JosephButler

Lykken's SUSY lectures (hep-th/9612114) say a few words about this, see equations (55) and (56) on page 17. It's still a little vague to me; the point seems that flat superspace may have no curvature, but it still has torsion. Thus the vielbeins are non-trivial even in the flat case. In this sense I suppose the `connection' terms in the SUSY covariant derivative are something like the spin connection?

(Though it's a bit confusing because these terms are still proportional to a spacetime derivative, so it doesn't appear to really be a connection term.)

Wess and Bagger also say a few words that seem related (page 102-103 and the following section), but I'm still having trouble applying this to the form of the covariant derivative.

In other words, why is it that in 8 dimensional superspace (4 spacetime dim, 4 grassmann dim) the covariant derivatives in the grassman directions contain spacetime derivatives as well. Are these terms connections? If so, why do they have spacetime derivatives? What does this have to do with the differential forms in Wess and Bagger?

Cheers,
Joe

5. Feb 9, 2009

### lbrits

When Q acts on a the most general function on superspace (superfield) $Y = \phi + \bar\theta \psi + \bar\theta\theta F$ then it interchanges $\phi$ and $\psi$ (I'm ignoring the auxiliary field now), which is a SUSY transformation if you build your action out of superfields. Since you want non-trivial dynamics, your action better have derivatives. And since you want to preserve the SUSY transformation, your derivatives better commute with SUSY transformations (I say commute, because a SUSY transformation takes a grassmann parameter multiplied by Q)

SUSY generators are a bit funny though, since they aren't really all that democratic.

6. Feb 9, 2009

### JosephButler

Thanks for the reply. There is further insight in on the nature of the SUSY covariant derivative in a few of the older books on SUSY/SUGRA, but it appears to be hidden in sections about differential forms. The point, I believe, that that the [differential] geometric interpretation of the SUSY covariant derivative is closely related to the vielbeins on flat superspace. This, in turn, is based on the formalism of differential forms which then translates to supergravity.

In an attempt to summarize, the point is that flat/rigid superspace has no curvature -- so there is no connection term in the covariant derivative. *However,* the space *does* have torsion, thus the vielbeins are nontrivial. (This is essentially what Lykken says in words.) Thus one has to consider the spin connection, where the vielbeins introduce the extra terms in the theta and theta-bar covariant derivatives.

References that I found useful are section 14.2 of West's SUSY & SUGRA book (that's relative to the 1st edition numbering), the analogous section for differential forms in Srivastava's book, and chapter 1 of Gieres' "Lecture Notes in Physics" book, "Geometry of Supersymmetric Gauge Theories."