It is important to differentiate between helicity and chirality. Helicity is the spin angular momentum of a particle projected onto its direction of motion. For a massive particle this quantity is frame dependent. Furthermore, since angular momentum is conserved, as a particle propagates helicity is conserved.
On the other hand, chirality is an innate property of a particle and doesn't change with frame. However, the mass term for a Dirac particle is,
\begin{equation}
-m(\psi_L^\dagger \psi_R + \psi_R^\dagger\psi_L)
\end{equation}
(in this notation the Dirac spinor is $\Psi= (\psi_L , \psi_R) ^T $). This term can be thought of as an interaction term in the Lagrangian which switches the chirality of a particle (e.g. a left chiral particle can spontaneously turn into a right chiral particle)
For a massless particle, chirality is equal to helicity.
With that background we can finally address your questions.
- Both helicity and chirality definitely make sense for a massive
Dirac spinor. However, that doesn't mean that a Dirac spinor is a
helicity and chirality eigenstate. In the same sense that energy
makes sense for a particle, but it may not be an energy eigenstate.
- As you mention the left chiral and right chiral fields can't be
decoupled from each other due to the mass term. The mass term can
always switch a right handed field to a left handed field and vice
versa.
- As I said above, the helicity of an electron is indeed frame
dependent. So it may look like a left or right helicity electron
depending on the frame, however its chirality is not frame
dependent.
- If we write the Dirac Lagrangian in terms of chirality eignestates
then we have, \begin{equation} {\cal L} _D = i \psi _L ^\dagger
\sigma ^\mu \partial _\mu \psi _L + i \psi _R ^\dagger \bar{\sigma}
^\mu \partial _\mu \psi _R - m \psi _L ^\dagger \psi _R - m \psi _R
^\dagger \psi _L \end{equation} Then we can think of $\psi_L$ (left
chiral particle) and $\psi_R$ (right chiral particle) as two
different particles that can turn into each other spontaneously
through a mass term. Putting them together, into a Dirac spinor
masks this property. However, they are still well defined
separately.
You are looking for a unitary representation of parity on spinors. That it should be unitary can be seen from the fact, that parity commutes with the Hamiltonian. Compare this to time-reversal and charge conjugation, which anticommute with $P^0$ and hence need be antiunitary and antilinear. They involve complex conjugation.
As demonstrated parity transforms a $(\frac{1}{2},0)$ into a $(0,\frac{1}{2})$ representation. Hence it cannot act on any such representation alone in a meaningful way. The Dirac-spinors in the Weyl-basis on the other hand contain a left- and right-handed component
$$ \Psi = \begin{pmatrix}
\chi_L \\ \xi_R
\end{pmatrix} $$
As a linear operator on those spinors - a matrix in a chosen basis - it mixes up the spinor components. After what has been said before, left- and right-handed components should transform into each other. The only matrix one can write down that does this is $\gamma^0$. There could in principle be a phase factor. In a theory with global $U(1)$-symmetry this may be set to one however.
Edit:
Statements like $\chi_L \rightarrow P\chi_L=\chi_R $ for a Weyl-Spinor $\chi_L$ are not sensible. The Weyl-spinors are reps. of $\mathrm{Spin(1,3)}$, whereas $P\in \mathrm{Pin(1,3)}$. One cannot expect that some representation is also a representation of a larger group. Dirac-spinors on the other hand are precisely irreps. of $\mathrm{Spin(1,3)}$ including parity, which cannot act in any other sensible way than by exchanging the chiral components.
Think of what representation means. It's a homomorphism from a group to the invertible linear maps on a vector space. $$ \rho: G \rightarrow GL(V)$$
Particularly, for any $g\in G$ and $v\in V$, $\rho(g)v\in V$. Now set $V$ to be the space of say left-handed Weyl-spinors and $g=P\in\mathrm{Pin(1,3)}$ the parity operation. As you have shown above, the image of a potential $\rho(P)$ is not a left-handed Weyl-spinor, hence is not represented.
Best Answer
From the relativistic covariance of the Dirac equation (see Section 2.1.3 in the QFT book of Itzykson and Zuber for a derivation. I also more or less follow their notation.), you know how a Dirac spinor transforms. One has $$\psi'(x')=S(\Lambda)\ \psi(x)$$ under the Lorentz transformation $$x'^\mu= {\Lambda^\mu}_\nu\ x^\nu= {\exp(\lambda)^\mu}_\nu\ x^\nu=(I + {\lambda^\mu}_\nu+\cdots)\ x^\nu\ .$$ Explicitly, one has $S(\Lambda)=\exp\left(\tfrac{1}8[\gamma_\mu,\gamma_\nu]\ \lambda^{\mu\nu}\right)$.
To show reducibility, all you need is to find a basis for the gamma matrices (as well as Dirac spinors) such that $[\gamma_\mu,\gamma_\nu]$ is block diagonal with two $2\times 2$ blocks. Once this is shown, it proves the reducibility of Dirac spinors under Lorentz transformations since $S(\Lambda)$ is also block-diagonal. Such a basis is called the chiral basis. It is also important to note that a mass term in the Dirac term mixes the Weyl spinors in the Dirac equation but that is not an issue for reducibility.
While this derivation does not directly use representation theory of the Lorentz group, it does use the Lorentz covariance of the Dirac equation. I don't know if this is what you wanted.
(I am not interested in your bounty -- please don't award me anything.)