# Gamma matrices invariant under Lorentz transformation

I know this has been asked before but I just can't seem to get my head around it based on the answers I've read.

So the idea is that we have the gamma matrices $$\gamma^{\mu}$$. Now from my understanding, this $$\mu$$ corresponds to a space time index, whilst for each fixed value of $$\mu$$, $$\gamma^{\mu}$$ defines a $$4 \times 4$$ matrix which is defined in spinor space, thus more explicitly $$\gamma^{\mu}_{ab}$$ where $$a,b$$ denote the spinor indices.

Now usually when showing covariance of the Dirac equation, we take the gamma matrices to be scalars which thus don't transform. We then find that they must satisfy $$\gamma^{\mu}=S[\Lambda]\Lambda_{\nu}^{\mu} \gamma^{\nu} S[\Lambda]^{-1}$$ in order for covariance to be satisfied.

I just can't for the life of me understand why we would start by treating $$\gamma^{\mu}$$ as a scalar. To me it seems clear that it would transform as a field with two spinor indices and a vector index, that is $$\gamma^{\mu} \to \gamma^{\mu'}=S[\Lambda] \Lambda_{\nu}^{\mu} \gamma^{\nu} S[\Lambda]^{-1}$$

Why are we able to just ignore the spacetime and spinior indices associated to the gamma matrices whilst treating them like scalars?

Edit: I feel like this paper https://arxiv.org/abs/1309.7070 talks exactly about what I'm not really understanding on page 5, however it doesn't really give an answer to any of the problems it poses using that method, but instead switches to a different method/picture. "In addition, the mathematical origin of $$\gamma^{\mu}$$ matrices is explained, from which it becomes clear why they are fixed matrices which do not transform", something that explains this line would be super helpful!

• Actually the $\psi\dagger \gamma^0 \gamma^\mu \psi$ transform as a vector under Lorentz transformations. Commented Apr 25, 2018 at 22:10
• I deleted my answer because the paper you mentioned in the edit is saying the same thing that I have said in a more elegant and comprehensive way. I think the answer to your question lies on the expressed fact that gamma matrices can be treated separately as a vector and a spinor matrix. However these separated transformations are related by the equation (14). Commented Apr 25, 2018 at 22:36
• Actually, this issue become clearer when you work on the Dirac equation in curved spacetime (but to learn tetrad/vierbein formalism of gravity is the cost). Commented Apr 25, 2018 at 22:53
• In order to understand why this is the case you have to know the isomorphism theorem $\text{SO}^{\uparrow}(3,1)\cong \text{SL}(2,\mathbb{C})/\mathbb{Z}_2$. Then use the fact that the (reducible) dirac representation is the direct sum of two fundamental representations of $\text{SL}(2,\mathbb{C}$. The Pauli matrices (which have one vector, one dotted spinor and one undotted spinor indice) transform in a very similar way. They are the more fundamental object. Commented Apr 25, 2018 at 23:27
• It has the origin from general relativity. The invariance is resulted from the collusion of local Lorentz gauge symmetry and diffeomorphism symmetry. See here: physics.stackexchange.com/questions/532982/… Commented Jul 9, 2020 at 16:39

The gamma matrices are just a collection of numbers, they don't transform at all when you change the coordinate system. However they are a cleverly chosen collection of numbers, such that when you combine them with other quantities, the resulting quantity has a convenient behaviour when you transform the system.

This is exactly analogous to the $$\eta$$ matrix in special relativity. It is just a matrix with diagonals $$(-1,1,1,1)$$ and completely invariant under coordinate transformations. However when you consider the quantity $$\eta_{\mu \nu}V^\mu = V_\nu$$ for a contravariant 4-vector $$V^\mu$$, $$V_\nu$$ will transform like a covariant vector, because the $$\eta$$ has mixed its components in a particular way.

In the Dirac theory, the spinors $$\psi$$ transform via the spin-$$\frac{1}{2}$$ representation of the Lorentz group $$\psi \rightarrow \Lambda_\frac{1}{2}\psi.$$

If you "rearrange" the components of $$\psi$$ via the gamma matrices, the resulting quantity will transform under the "standard" representation like usual 4-vectors

$$\bar{\psi}\gamma^\mu\psi \rightarrow \Lambda^\mu{}_\nu\bar{\psi}\gamma^\nu\psi$$

The point you are missing is that the bi-spinor $$A_{\alpha\dot\alpha}$$ and a vector $$B_{m}$$ are equivalent representations of $$SO(1,3)$$, so it is possible to construct an object $$\sigma^{m}_{\alpha\dot\alpha}$$ that is invariant under $$SO(1,3)$$. You can use this object to translated one representation into another by:

$$A_{\alpha\dot\alpha} = A_{m}\sigma^{m}_{\alpha\dot\alpha}\qquad A_{m}=\frac{1}{2}\bar\sigma_{m}^{\dot\alpha\alpha}A_{\alpha\dot\alpha}$$

where $$\bar\sigma_{m}^{\dot\alpha\alpha}=\varepsilon^{\alpha\beta}\varepsilon^{\dot\alpha\dot\beta}\eta_{mn}\sigma^{n}_{\beta\dot\beta}$$. The same can be done for the $$B$$'s. You can check that this equations are right by transforming both sides in the equalities. Now, sometimes it is convenient to pretend that just one type of index in $$\sigma^{m}_{\alpha\dot\beta}$$ transform, such that $$\sigma^{m}_{\alpha\dot\beta}$$ will behave as a vector if we transform only $$m$$. If you turn transform just the spinorial indices than $$\sigma^{m}_{\alpha\dot\alpha}$$ will also transform as a vector but with the the inverse of the transformation since the combination of the transform in all indices should cancel.

Indeed an infinitesimal transformation on the spinorial indices is

$$\delta (\sigma^{m}_{\alpha\dot\alpha}) =\Theta_{np}(\sigma^{np})_{\alpha}\,^{\beta}\sigma^{m}_{\beta\dot\alpha}+\Theta_{np}(\sigma^{np})^{\dot\beta}\,_{\dot\alpha}\sigma^{m}_{\alpha\dot\beta} = -\Theta^{m}\,_{n}\sigma^{n}_{\alpha\dot\alpha}$$