Is this section of the wikipedia page for gamma matrices wrong?

[samalkhaiat]

Would it be correct to say that this is the reason we must use tetrads to incorporate fermions in curved spacetime?

Correct. However, the tetrads (veirbein) are necessary objects for any differentiable manifold. They follow from the fact that the spacetime manifold, considered as a topological space, is locally (homeomorphic to $\mathbb{R}^{(1, n - 1)})$ flat Minkowskian.

Sam

 

[Demystifier]

Samalkhait, thank you for your answer which was quite clear to me. I have a few additional questions.

For this reason, we say that $GL(4)$ does not act in the index-space of spinors.

So does it mean that, under general coordinate transformations, spinors transform as scalars?

And if your answer is yes (which I think would be the correct answer), can we think of Lorentz transformations as nothing but a special case of general coordinate transformation?

And when we interpret Lorentz transformations in that way, then can we say that, in this interpretation at least, spinors transform as scalars under Lorentz transformations?

 

[samalkhaiat]

So does it mean that, under general coordinate transformations, spinors transform as scalars?

And if your answer is yes (which I think would be the correct answer),

In order to stress the fact that spinors do not belong to $GL(n)$, it is better to say that Lorentz spinor is treated as “scalar” in curved space.

can we think of Lorentz transformations as nothing but a special case of general coordinate transformation?

All real matrix groups are subgroups of $GL(n, \mathbb{R})$ including $\mbox{SL} ( n , \mathbb{R})$ and $\mbox{Spin} (n)$.

And when we interpret Lorentz transformations in that way, then can we say that, in this interpretation at least, spinors transform as scalars under Lorentz transformations?

You take spinor from flat space and treat it as “scalar” in curved space and then you conclude that spinor is Lorentz scalar?
This is false argument unless you tell people how you define “your” Lorentz scalar and spinor.

The usual definition of Lorentz scalar and Lorentz spionr are the following:

Lorentz scalar is an object with vanishing spin matrix, i.e. under Lorentz transformation, a scalar field transforms by the identity matrix:
$$\phi_{ i } ( x ) \leftarrow \bar{ \phi }_{ i } ( \bar{ x } ) = \delta^{ j }_{ i } \phi_{ j } ( x ) .$$
In the representation theory, we say that Lorentz scalar belongs to the representation space $V^{ ( 0 , 0 ) }$.

Lorentz (bi)spinor is an object with non-vanishing spin matrix $\Sigma^{ \mu \nu }$, i.e. under Lorentz transformation, Dirac spinor field transforms as
$$\psi ( x ) \rightarrow \psi^{ ' } ( \bar{ x } ) = \exp ( - \frac{ i }{ 2 } \omega_{ \mu \nu } \Sigma^{ \mu \nu } ) \psi ( x ) .$$
In the representation theory, we say that (bi)spinor belongs to the representation space $V^{ ( 0 , 1/2 ) } \oplus V^{ ( 1/2 , 0 ) } .$

Suppose (for the sake of argument) that $\psi$ is Lorentz scalar, then $\partial_{ a } \psi$ is Lorentz vector.

Now, Einstein’s EP tells you that the derivative of scalar is covariant vector in curved space. So, $\partial_{ a } \psi \rightarrow \partial_{ \mu } \psi$, i.e., there is no need for connection!

Suppose (as you say) that $\gamma^{ a }$ is Lorentz vector. Thus, in curved space, we will have the contra-variant vector $\gamma^{ \mu }$.

Thus, you would conclude (in 3 seconds) that Dirac equation $i \gamma^{ \mu } \partial_{ \mu } \psi = 0$, is generally covariant. Of course, we know this is not true.

We also know, It took 30 years to figure out the correct form of Dirac equation in GR.