Why are the 4-vector and bispinor representation of the Lorentz algebra in particular so related?

Question

When learning about the Dirac equation, there are several indications that the fundamental (4-vector) representation and the bispinor representations are connected in some way. To give an example, the Dirac equation itself contains both 4-vector indices and spinor indices:

$$(i\gamma^\mu_{ab} \partial_\mu-m)\psi^b(x)=0 \tag{1}.$$

We also have the following identity which is often cited as allowing us to treat the "4-vector indices" on the $\gamma$ matrices as proper 4-vector indices:

$$S^{-1}[\Lambda]\gamma^\mu S[\Lambda]={\Lambda^\mu}_\nu\gamma^\nu \tag{2},$$

which seems to suggest some kind of non-trivial relationship between these two representations.

My question is, is there actually anything to this? If I had to hazard a guess at an answer I'd say there might be something about the Clifford algebra (of which $\gamma^\mu$ are the generators) which interlinks these two representations? But I'm not really sure, am I getting at anything significant?

There is some useful discussion here that I have read and sort of understood, this seems to be a representation theoretic relationship. I hope this is not considered simply a duplicate.

Answer 1

$\gamma_a$ lives in the spin/Clifford bundle denoted by Roman indices $a$, while spacetime $dx^\mu$ is characterized by Greek indices $\mu$.

Romans and Greeks shake hands with the help of tetrad (or vierbein) $$ e^a_\mu \gamma_a dx^\mu $$ The flat Minkowsky space (with metric $g_{\mu\nu} = \eta_{ab}e^a_\mu e^a_\nu = \eta_{\mu\nu}$) is characterized by $$ e^a_\mu = \delta^a_\mu $$ Therefore $$ e^a_\mu \gamma_a dx^\mu = \delta^a_\mu\gamma_a dx^\mu = \gamma_\mu dx^\mu $$ The delta function $\delta^a_\mu$ is kinda of "soldering" the Roman and Greek indices together, hence the perfect linkage between the fundamental (4-vector) representation and the bispinor (sandwiching $\gamma_\mu$ between two spinors $\bar{\psi}\gamma_\mu\psi$) representations.

In curved spacetime, the simple correspondence $e^a_\mu = \delta^a_\mu$ is not true anymore. Therefore you have to be fluent in both Roman and Greek languages and use $e^a_\mu$ as a dictionary.