How to prove explicitly (i.e., to don't postulate it) that by including Lorentz indices $a$ the covariant derivative $D_{\mu}$ looks like $$ D_{\mu}A^{\nu a} = \partial_{\mu}A^{\nu a} + \Gamma^{\nu}_{\mu \alpha}A^{\alpha a} + (\omega_{\mu})^{a}_{\ b}A^{b \nu}? $$ Here $\omega_{\mu}^{ab}$ is spin connection, $\Gamma^{\alpha}_{\mu \nu}$ are Christoffel symbols.


closed as off-topic by Ryan Unger, ACuriousMind, Kyle Kanos, HDE 226868, Prahar Oct 28 '15 at 3:58

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    $\begingroup$ In general, the form for any covariant derivative is derived by postulating its transformation behaviour and making a more or less complete ansatz for the possible covariant derivatives. $\endgroup$ – ACuriousMind Feb 2 '15 at 18:32

Indeed as was commented before usually in physics these derivatives are 'derived' by postulating how the object transforms. Obviously there are more rigorous definitions, as is usually the case.

In this particular case you could try to find the structure which gives you these 'covariant derivatives' in the first place. In books they are often referred to as connections as well. To do this you will firstly want to find the structures that a) describe these 'vectors' and b) describe these 'spinors'. Vectors are quite easy, the most natural way to find these vectors is to construct the tangent bundle. Then you can find that there exists a unique 'compatible', 'torsionless' connection, of course this is what we call the Levi-Civita connection. What you probably already knew from your GR course. But you could also look at more general connections on more general bundles, then you go to vector bundles. If we take a vector bundle $E\rightarrow M$ and denote our smooth sections (functions from the manifold to our bundle, so for the tangent bundle these were just your vectorfields!) as $\Gamma(E)$ we define a connection as follows. In general a (linear) connection is a linear map \begin{align} \nabla :& \Gamma(E) \rightarrow \Omega^1(M) \times \Gamma(E) \end{align} such that \begin{equation} \nabla(f\eta) = df \otimes \eta + f \nabla \eta \end{equation} where f is a smooth function and $\eta \in \Gamma(E)$ If you want to look at spinors you will have to go to Clifford algebras, fortunately we can talk about Clifford algebras in a differential geometric context as well, these are called Clifford bundles. Which are quite easy to construct since on a Riemannian manifold we can construct a Clifford algebra in each point using the tangent space at that point, just as you construct a Clifford algebra starting from a vector space. When you talk about spinors in the context of Clifford algebras you know the spinor spaces are just the irreducible representations of these algebras (watch out since the word representation is used in several ways by different people). We want such a thing for our bundles as well. So we will say that a vector bundle 'S' is a spinor space if it is so that the elements in our Clifford bundle (just our Clifford algebra but now generalized to our manifold) are actually endomofisms of this spinor space. If such a spinor space exists we call our manifold $spin^c$. There is another requirement we can impose, which I won't tell you since I think there are enough technical things going on, but I'll say that when we can find the real parts out of our (in general complex) Clifford bundle we say that our manifold is 'spin'. Just as we defined a linear connection on a vector bundle before we can define a so called $spin^c$ connection on our $spin^c$ manifold by a compatible linear connection $\nabla^S$ which has the property that \begin{equation} \nabla^S(\hat{\alpha} \psi) = \hat{(\nabla \alpha)} \psi + \hat{\alpha} \nabla^S \psi \end{equation} where I use the hat to show I 'lift' the element to the Clifford bundle, and $\nabla$ is just the Levi-Civita connection. So if we have that $\alpha = dx^{\mu}$ we get after the lift that $\hat{\alpha} = \gamma^{\mu}$. Where these gamma matrices are as you know them, but now with the relation that $\{\gamma^{\mu},\gamma^{\nu}\} = 2g^{\mu\nu}$ ('position' dependent!). Hold on, we are almost there. So now we have a connection for our $spin^c$ structure. If our connection is also compatible with the spin structure, I mean that it commutes with the thing that allows us to pick out the real part (as I said before) then we call the connection a spin connection. Now comes the grand finale, On a manifold which I described now, and which is spin, there exists a unique spin connection which is just the one you know. And can actually be seen as the lift of the Levi-Civita connection on the tangent bundle to the Clifford bundle!

I was thinking that you actually have already seen your fair share of differential geometry, if not I'm sorry if I confused you! And if you have seen more then a share of differential geometry I'm guessing you already know all this and in that case I'm sorry for this useless answer... If you want to know more about these spin structures and want to look at it very rigorous I can suggest 'Spin Geometry' by Lawson and Michelsohn, which is a classic on the topic. If you want a lighter introduction into the subject I can suggest the master thesis of K. Dungen: The structure of gauge theories in almost commutative geometries, which follows the story I told you, also it gives you a nice introduction into non commutative geometry, if you're interested.

I guess you've noticed that I haven't really answered the your question (I think the answer is hidden in it already). I hoped what I tolled you gave an idea that these connections are actually very nice and not only some artificial thing to make the objects transform in a good way.

Concerning your question. You could actually say that this is just the spin connection, since this already follows this Leibniz rule which you need, as it preforms the Levi-Civita connection on the 1-forms! I must admit that I don't directly see the lift to the Clifford bundle in your question though... Sorry!