# Canonical spinors from gauge transformations

In this 2006 paper, http://arxiv.org/abs/hep-th/0610128, there is the concept of gauge transformation and how was it employed that I do not fully understand.

Note, what will be talked about below is related to the spinor's generic form:

$$\eta=\lambda 1 +\mu^ie^i+\sigma e^{12}\tag{2.4}$$ where $e^1$,$e^2$ are 1-forms and $e^{12}=e^1 \wedge e^2.$ Page 4, section 2.3 entitled "Gauge Transformations and Canonical Spinors. The authors say that

There are 2 types of gauge transformation which can be used to simplify the Killing spinors of this theory. First there are local $U(1)$ gauge transformations of the type $$\epsilon \rightarrow e^{i\theta}\epsilon\tag{2.13}$$ for real functions $\theta$, and there are also local Spin (3,1) gauge transformation of the form $$\epsilon \rightarrow e^{\frac{1}{2}f^{\mu\nu}\gamma_{\mu\nu}} \epsilon\tag{2.14}$$ for real functions $f^{\mu\nu}$.

They say:

Note in particular that $\gamma_{12}, \gamma_{13}$ and $\gamma_{23}$ generate $SU(2)$ transformations which act (simultaneously) on both $1, e^{12}$ and $e^1, e^2$.

In particular $\gamma_{13}$ acts via $$1 \rightarrow e^{i\theta}1, \hspace{1cm} e^1 \rightarrow e^{-i\theta}e^1, \hspace{1cm} e^2 \rightarrow e^{i\theta}e^2, \hspace{1cm} e^{12} \rightarrow e^{-i\theta}e^{12}\tag{2.15}$$ for $\theta$ belonging to $\mathbb{R}$.

My question is why was there an introduction of $\gamma_{13}$ and how did it act on any of $1, e^1, e^2, e^{12}$? In other words, how is this $\gamma_{13}$ related to the first quoted section above and how does it operate on the basis?

I edited your question (I moved one parenthesis so that your quote was not a misquote), and maybe that answered your question. In case not ...

If you had a Dirac spinor $\Psi$ then there is a transformation like $\Psi\mapsto e^{\theta\gamma_{13}/2}\Psi,$ where $\gamma_{13}=\gamma_1\gamma_3$ is a product of gamma matrices (or just the unit xz plane if you are familiar with geometric algebra or geometric calculus). And $\theta$ is a real scalar if this is at a point or a real function if this is for a field.

So we know exactly what the transformation does, it rotates by $\theta$ degrees in the $xz$ plane. But the paper you cite doesn't want to represent Dirac spinors as Dirac spinors it wants to represent them as complex elements of a 2d space of forms. Whatever. Its a four complex dimensional space, so it's not unimaginable to say that this four complex dimensional space is going to try to act like the other one (the space of Dirac spinors). But the Dirac spinors can be multiplied by gamma matrices (unit vectors) to get new spinors. If these complex forms are going to dress up and pretend to be Dirac spinors we have to say how they act when they meet a gamma matrix. This might be spelled out in equation 2.5 though I don't care much for the notation.

So they don't spell out (only cite) the correspondence. But they do try to say what multiplication by each of the $\gamma_\mu$ does. So in fact if you can read 2.5 then you know exactly what multiplying the Dirac spinor by $e^{\theta\gamma_{13}/2}$ should do.

Specifically $e^{\theta\gamma_{13}/2}=(\cos(\theta/2))+((\sin(\theta/2)(\gamma_1\gamma_3))$ so just take your spinor and make two copies multiple one copy by $\gamma_3$ on the left then by $\gamma_1$ on the left then by sine and multiply the other copy of the spinor by cosine then add both together. So you get a combination of the original and a rotated version, and you rotated by a half angle because that is how you rotate spinors (and in case you don't know why the geometric calculus view is that a spinor is basically an operator that does a two sided rotation on a fixed reference object so you multiply by a half angle so that the action of the spinor (being two sided) rotates the way it normally does but then also by that angle amount, and while I'm at it when I say dirac matrix or gamma matrix I just mean unit vector and I said left multiply because they is no reason to restrict yourself to just left operations just because people that like matrix representations like to restrict themselves that way).

OK so $e^{\theta\gamma_{13}/2}=(\cos(\theta/2))+((\sin(\theta/2)(\gamma_1\gamma_3))$ so all we have to do is take your complex element of the 2d space of forms and do the 2.5 action of $\gamma_3$ on it and then do the 2.5 action of $\gamma_1$ on it then multiply by sine then add that to the original element of the complex element of the 2d space of forms scaled by a cosine and we get the new complex element of the 2d space of forms.

So hopefully 2.5 is indeed consistent with what they got here. They were a bit vague when they said transformation generated by. So maybe they meant to act by $e^{\theta\gamma_{13}}$ not $e^{\theta\gamma_{13}/2}.$ You should look at the citation for the correspondence between the dirac spinors and the complex elements of the space of forms on 2d (I think I might have been saying that wrong before, but I'm at the stage where I delete tons every time I edit, so I should stop before the whole answer is deleted away into nothingness). After you verify the correspondence check 2.5 then you can check this part.

Here they did not use U(1) gauge transformation but only Spin (3,1) gauge transformation. Do you have any clue on why might they have mentioned it in the first place?

They said that both types are useful for simplifying things later, maybe they use both types, but in different places later. For charged particles the local U(1) gauge be the same as picking a different electromagnetic gauge, in other words it is the phase and the potential together that is physical, neither by itself. So maybe they used it in a very subtle/implicit way if they just later picked a convenient electromagnetic gauge.

About the half when you said "they meant to act by $e^{\theta\gamma_{13}}$ instead of $e^{\theta\gamma_{13}/2}$". I have repeated the calculations and you are right, there is an extra 1/2 everywhere. Will this cause any problem?

No problems, they just said in words something like "when you generate with $\gamma_{13}$ you get X" and didn't say what they meant by generating with $\gamma_{13}$ if you figured out that the words meant $e^{\theta \gamma_{13}}$ then you know what they meant by those vague words. No problems.

You have explained in a way how $\gamma_{12}$, $\gamma_{13}$, and $\gamma_{23}$, generate SU(2) transformation.

Those are three generators for transformations on spinors. I'd use them to generate rotations in physical space, but you can do lots of things with them.

What is the necessity of mentioning $\gamma_{02}$ that appears too (eq 2.16)?

If $\gamma_{13}$ generates rotations in the $xz$ plane, then $\gamma_{02}$ generates velocity boosts in the $y$ direction. They are also good generators for transformations on spinors.

In the text they try to make a distinction between transformations generated by $\gamma_{12}$, $\gamma_{13}$, and $\gamma_{23}$ (the generators of spatial rotations) and say that they act on any vector in the $1,e^{12}$ plane (of forms) equally and act on any vector in the $e^1,e^2$ plane (of forms) equally (i.e. act on one forms equally).

They finally say that Spin(3,1) gauge transformation is generated by $\gamma_{01}-\gamma_{12}$ and $\gamma_{03}-\gamma_{23}$.

The authors are not saying that every gauge transformation is generated by $\gamma_{01}-\gamma_{12}$ and $\gamma_{03}-\gamma_{23}$ they are saying that first they want to make a gauge transformation to send a general spinor to be a complex element of the space of two forms that has no rank 2 element. Then, based on that and some adjustable parameters they consider the transformations generated by $\gamma_{01}-\gamma_{12}$ and $\gamma_{03}-\gamma_{23}.$ What do those look like? Each of those is also a plane, but they are not planes that you can rotate in or boost in, they are planes where the metric is degenerate when restricted to the planes, they are nilpotent planes. Specifically $e^{\alpha(\gamma_{01}-\gamma_{12})}=1+\alpha(\gamma_{01}-\gamma_{12})$ and $e^{\beta(\gamma_{03}-\gamma_{23})}=1+\beta(\gamma_{03}-\gamma_{23})$ for arbitrary real parameters $\alpha,$ and $\beta.$ They say you do that so that $1$ and $e^1$ are left unchanged, but someone how this affects the coefficient on $e^1$?? I'd check for a peer reviewed version. They said they were going to show how to use complex forms on a 2d space to represent Dirac spinors, but in my opinion they only cited it and didn't show it, and this looks like a typo and they didn't even say how they got rid of the $e^{12}$ part (I don't think they used the U(1) transformation because that scales the forms by a unit complex number). These are all things I would have complained about if I was the reviewer, maybe the published version addressed these.

The point is that they are using very specific transformations based on the given arbitrary form to try to reduce it to a specific form. And if these spinors/forms things are supposed to be varying from point to point, watch out that these transformations if they are using different transformations in different open sets you might need to remember that.

• thank you for your detailed answer as always, I appreciate it. The geometric perspective you gave is so useful. I have a few questions to get a better understanding, I list them: -1- Here they did not use U(1) gauge transformation but only Spin (3,1) gauge transformation. Do you have any clue on why might they have mentioned it in the first place? -2- About the half when you said "they meant to act by $e^{\theta\gamma_{13}}$ instead of $e^{\theta\gamma_{13}/2}$". I have repeated the calculations and you are right, there is an extra 1/2 everywhere. Will this cause any problem? Aug 3, 2015 at 14:36
• -3- You have explained in a way how $\gamma_{12}$, $\gamma_{13}$ and $\gamma_{23}$ generate SU(2) transformation. That is SU(2) transformation is generated by those. My question is what is the necessity of mentioning $\gamma_{02}$ that appears too (eq 2.16)? -4- They finally say that Spin(3,1) gauge transformation is generated by $\gamma_{01}-\gamma_{12}$ and $\gamma_{03}+\gamma_{23}$. How does one know that for sure? Is this something conventional? Aug 3, 2015 at 14:41
• pardon me for the too many questions. Thank you for your time. Aug 3, 2015 at 14:42
• @Beyond-formulas Edited Aug 4, 2015 at 4:01
• @Fluctuations As opposed to Spin(1,3)? There isn't much difference between Spin(1,3) and Spin(3,1) and usually you use the one the corresponds to the metric signature you selected. Dec 31, 2015 at 5:12