# How to know which gauge transformation we should use?

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1. Jan 1, 2016

### samuelphysics

Spinors in $N=2, D=4$ supergravity can be simplified using gauge transformation and thus canonical spinors can be found. In the case of $N=2, D=4$ supergravity the gauge transformation Spin (3,1) is used. My question is how do we know which transformation can be used in a certain theory in order to simplify things?

Wish start of a good year!

2. Jan 1, 2016

### Lautaro

Samuel

A gauge transformation is given according to the group structure of the theory, in this case Spin(3,1). It must give to gauge independence.
Happy new year.

3. Jan 1, 2016

### haushofer

I'm not sure what you mean. Without cosmological constant spacetime locally looks like Minkowski, so one has SO(3,1) as Local Lorentz transformations (the homogeneous part of the isometry group of minkowski spacetime) and the gravitino sits in the spin 3/2 rep. of this group. But if you have a cosmological constant, your spacetime locally looks like AdS (dS is not compatible with pure SUGRA, as far as i know). This spacetime has SO(3,2) as isometry group. So spinors are defined as certain representations of this group, instead of SO(3,1).

In eqn.2.14 of http://arxiv.org/pdf/hep-th/0610128v3.pdf e.g. they use transformations on the Killing spinor epsilon to simplify stuff (I haven't really looked at the paper carefully). The 'gauge transformation' they talk about there is just the usual Lorentz transformation on the spinor, but this transformation depends on whether one has a cosmological constant or not (so(3,2) vs so(3,1) representations) .

i hope this helps.

4. Jan 1, 2016

### samuelphysics

@haushofer sure your answers always do help, I just want to make sure of some stuff going on in my head as I am trying to solve KSE's for the first time.
So why do people have the right to simplify a spinor that way?

Then in the paper you cited, you're saying that in the theory they're in (i.e. N=2, D=4) SUGRA, locally they can use the Spin (3,1) gauge transformation, my question was how as a student can I know which gauge transformation I can use? (I am having the thoughts that this gauge transformation might differ from one dimension to other and from certain supegravity theory to another, I might be hugely mistaken)

5. Jan 1, 2016

### haushofer

Well, that's the idea of (local) symmetries. E.g., if you may choose a vector in R3 for some calculation, a nice choice is often ~(0,0,1), so to make it point in the z-direction. The Killing-spinor equation is invariant under a local Lorentz transformation, so why not use this freedom to make your life a bit easier?

If you write down a theory like a sugra-theory, you know which symmetries you have. The specific form of a Local Lorentz transformation indeed changes per dimension, because the spinor representation differs per dimension (because the transformation is described with the Clifford algebra).

So my advice for students would be to find a textbook describing the sugra theory of your interest and check all the given symmetries and the closure of the algebra.

6. Jan 1, 2016

### samuelphysics

Oh ok, concerning what you said here
I was asking because if you go back to the paper you cited, specifically, you can see that the authors (upon the use of gauge transformation) turned the original very general $\epsilon =\lambda 1 +{\mu}^{i}e^i+\sigma e^{12}$ to 3 canonical forms which are $\epsilon=e^2$, $\epsilon=1+\alpha e^1$ and finally $\epsilon=1+\beta e^2$ (this last orbit represents the Killing spinor for the IWP metric which has a time-like Killing vector)
meanwhile the other two orbits correspond to plane-waves with null Killing vector. So, it looks like each canonical spinor that is resulting from Spin(3,1) will give us different solutions, isn't it?

7. Jan 2, 2016

### haushofer

Well, yes, but I also get different Killing vectors for my Schwarzschild background when I use different coordinates. A gct can then always make these components look more complex, but it's still the same vector.

I was thinking about your question about gauge transformations. Maybe it helps to know that GR can be obtained by gauging the Poincaré algebra. The gauge transformations one has are then local Lorentz transfo's, (LLT's), local translations and general coordinate transformations (because the gauge fields have a curved index). Imposing a curvature constraint effectively 'removes' this local translation, leaving you with just gct's and LLT's. For (A)dS algebra's I'm not sure if this works, because of the curvature constraint one imposes.

For N=1 Superpoincare one can do the same; now you obtain also local SUSY-transformations and an extra gauge field: the gravitino. All the SUSY transformations can be obtained by this gauging procedure; the gauge fields are as usual in the adjoint rep. of the algebra. Beyond N=1 this becomes impossible (afaik) because the multiplets become bigger and the extra gauge fields cannot be obtained by gauging the corresponding 'extra' spacetime transformations. One can gauge the R-symmetry group for SUSY algebra's, giving "gauged SUGRA's". But the SUSY-transformations for this field are not gauge transformations of the algebra.

8. Jan 2, 2016

### samuelphysics

Oh i see thanks for your efforts in explaining these, please if i could ask one more thing: when you replied by this

Could we know how this $\epsilon=1+\beta e^2$ represents the Killing spinor for the IWP metric which has a time-like Killing vector while the other two orbits correspond to plane-waves with null Killing vectors? What makes them time-like here but null there? The derivations are present in the paper you attached as I worked them out yesterday, but I didnt know what was the thing that let the others know that $\epsilon=1+\beta e^2$ corresponds to IWP metric which has a time-like Killing vector? I am thinking I have missed some hint or clue in the derivation.

thank you @haushofer

9. Jan 3, 2016

### haushofer

To answer that question I have to take a closer look at the paper; I encountered it by googling some terms from your OP ;) I'm not sure when I will be able to do so (I have to finish a PhD-thesis this month :P ) If I manage to take a closer look, I'll post an answer here.

10. Jan 3, 2016

### samuelphysics

good luck with your phd! dont worry about that last question, it was a result of curiosity when you shared that paper.. i have also to wory about m courses instead of drifting away. thanks for your discussion, @haushofer