mjhjmtu

Is spinor $psi$ actually the sum of scalar, vector, bi-vector, ..., pseudo-scalar?

Before talking about spinors, we have to differentiate two kinds of spacetime, demonstrated with the example of spin connection 1-form $omega$ in Einstein-Cartan gravity
$$
omega = frac14omega^ab_mugamma_agamma_bdx^mu.
$$
The spin connection 1-form $omega$ is a

1. Vector in spacetime manifold characterized by differential vector base $dx^mu$.


2. Bi-vector in internal (local, tangential, or spinor bundle) spacetime characterized by anti-symmetric $gamma_agamma_b$, where $gamma_a$ are Dirac vector basis.

Now back to our initial assertion: a spinor is a

1. 
   Scalar in spacetime manifold characterized by differential vector base $dx^mu$. The spinor $psi$ is a $0$-form.


2. 
   Sum of scalar, vectors, bi-vectors, ..., pseudo-scalar, in internal (spinor bundle) spacetime characterized by Dirac vector base $gamma_a$: $psi = xi_s + xi_v^agamma_a + xi_bv^abgamma_agamma_b + ...$. The individual coefficients like $xi_s$ and $xi_v^a$ are just numbers (Grassmann odd though), not columns. In other words, a Dirac spinor (e.g., a neutrino and an electron putting together as a whole) spans the whole space of the 16 elements of the "Dirac algebra"! Not just some sub-space of it. It's partially evidenced via indirectly projecting a spinor to the measureable components such as scalar bi-linear $tr(barpsipsi)$, vector bi-linear (current) $tr(barpsigamma_apsi)$, bi-vector bi-linear $tr(barpsigamma_agamma_bpsi)$, etc. Note that, here a Dirac spinor is not a 4*1 column anymore, rather, it lives in the same operator space (4*4 matrices, if you will, that is why we have to take traces tr(...) for Spinor bilinears above) spanned by the Dirac vector base $gamma_a$. The conventional column spinor is just an idempotent projection (an individual left ideal like a neutrino or an electron) of the matrix spinor.

Accordingly, there are two kinds (external and internal) of Lorentz transformations:

1. External Lorentz transformation (global rotation portion of local diffeomorphism) of electromagnetic gauge field 1-form $A = A_mudx^mu$ is performed in the spacetime manifold characterized by differential vector base $dx^mu$.


2. Internal Lorentz transformation of a spinor field $psi$ is performed in the internal spacetime characterized by Dirac vector base $gamma_a$.

Most of times, we are sloppy (and get away with it) with going back/forth between these two (external and internal) spaces without explicitly mentioning it, thanks to the delta function-like soldering 1-form $e$ (vielbein/frame/tetrad, a vector in spacetime manifold characterized by differential vector base $dx^mu$, as well as a vector in the internal/spinor bundle spanned by Dirac vector base $gamma_a$) in flat spacetime:
$$
e = e^a_mugamma_adx^mu = delta^a_mugamma_adx^mu,
$$
which nicely bridges the two spaces, "soldering" $gamma_a$ and $dx^mu$ via delta function $delta^a_mu$ and producing Minkowskian metric
$$eta_munu sim tr(e^a_mugamma_ae^b_nugamma_b) = tr(gamma_mu gamma_nu).
$$

1. There are no "two spacetimes". There is a single spacetime $M$, which is a four-dimensional Lorentzian manifold, and there's a bunch of bundles over it.




2. The spin connection is formally a connection form on a $mathrmSO(1,3)$-principal bundle over $M$ that can act on anything that transforms in a representation of the Lorentz algebra. It is not a bivector, it is a local 1-form that is $mathfrakso(1,3)$-valued. So as a 1-form it has components $omega_mu$ and as a $mathfrakso(1,3)$-valued object we may expand it in a basis of the Lorentz algebra. The commutators of the $gamma$-matrices form a basis of $mathfrakso(1,3)$, so we get your expansion $omega = omega_mumathrmdx^mu = omega_mu^ab[gamma_a, gamma_b] mathrmdx^mu$. For more on the spin connection, see also this answer of mine.




3. A spinor field is now a function that takes values in some spinorial representation of $mathfrakso(1,3)$, e.g. that labeled by $(1/2,1/2)$, the Dirac representation. Therefore, the spin connection can act on it. Put another way, a spinor field is a section of an associated bundle (the spinor bundle) to the bundle the spin connection lives on, other equivalently, the spin connection lives in the spinor bundle's frame bundle. Note that this is a spinor field. A "spinor" would more commonly be understood to not live on a manifold at all and just be a single vectors in the $(1/2,1/2)$-representation space.

However, all this talk of bundles is overkill for many physical applications, though it is good to be aware of its existence. Most of the time, it suffices to take the local viewpoint (in which the bundles are products), so that the spin connection is simply a $mathfrakso(1,3)$-valued field that acts on other fields that take values in some representation of $mathfrakso(1,3)$.

The expression
$$ psi = psi_s + psi_v^a gamma_a +dots$$
you write down makes no sense in any of these contexts. You can pick a basis of the representation vector space the spinor takes values in and expand it in terms of it, but the $gamma$-matrices act on the spinor, they are not a basis for its space.

I think that you actually mean "element of the Dirac algebra" where you say "Dirac spinor".

The Dirac algebra is the Clifford algebra of Minkowski space. This is the smallest algebra that contains Minkowski space $mathbb R^1,3$ itself, and so that for elements $v,win mathbb R^1,3$ we have $vw = langle v,wrangle$, where $vw$ is the product in the Dirac algebra, and $langle v,wrangle$ is the inner product of the two, which is a scalar, and as such an element of the Dirac algebra as well. This has a representation as an algebra of $4times 4$ matrices, as you know well.

If we write $gamma^0,gamma^1,gamma^2,gamma^3$ for the image of an orthonormal basis of $mathbb R^1,3$ in the Dirac algebra, then it is not hard to see that elements of the form $gamma_i_1cdotsgamma_i_n$ indeed generate the Dirac algebra over $mathbb R$ (or $mathbb C$ if we look at the complex Dirac algebra, as we do in many situations), in other words, every element of the Dirac algebra can be written in the form that you wrote down.

The invertible elements of the Dirac algebra contain an important subgroup, namely $textSpin(mathbb R^1,3)$ of elements of norm 1. Elements of the spin group are not called spinors either. This group has a unique complex irreducible representation (up to complex conjugation). It is elements of this representation that are almost spinors: namely, if we lift the metric structure of the tangent bundle of Minkowski space to a Spin structure, the sections are what are called Dirac spinors. This is what ACuriousMind was talking about.