Is it possible to define off-shell spinors? For the sake of concreteness, let us consider the Dirac spinor $u_s(\boldsymbol p)$. Is it possible to covariantly extend this to a function $u_s(p)$, such that it matches $u_s(\boldsymbol p)$ on the hyperboloid $p^2=m^2$?
I'm not sure what properties I would like $u_s(p)$ to have, so I leave it a bit arbitrary. For example, we may want to ask
$$
D(R)u(p)=u(Rp)
$$
for all $p\in\mathbb R^4$ and $R$ a rotation matrix (which may include space inversion?). We could even perhaps extend this to boosts as well as rotations, but the former mix the $s$ components and so I guess this gets trickier.
We want $u(p)$ to be a solution of $(\not p-m)u(p)=0$ whenever $p$ is on-shell (but not for general $p$, because $\text{det}(\not p-m)=(p^2-m^2)^2$, which means that for off-shell momenta, there are no non-trivial solutions).
Finally, I'm not sure if we could ask for $u_s(p)$ to be an helicity eigenstate $hu_\pm(p)=\pm \frac12 u_\pm(p)$, because for off-shell momenta, I don't think $h=\boldsymbol p\cdot\boldsymbol J/m$ is well-defined (and/or covariant).

Moreover, can we say something about higher spin polarisation vectors?
 A: The most natural extension of $u_s(p)$ off the mass shell is to define this object as the solution of
$$
\left(\not p+\sqrt{p\cdot p\vphantom{A}}\,1_{4\times4}\right)u_s(p)=0 \tag{1}
$$
which clearly has at least one solution, because
$$
\det\left(\not p+\sqrt{p\cdot p\vphantom{A}}\,1_{4\times4}\right)=0\tag{2}
$$
In fact, it is not difficult to show that
$$
\mathrm{rank} \left(\not p+\sqrt{p\cdot p\vphantom{A}}\,1_{4\times4}\right)=2\tag{3}
$$
and therefore $(1)$ has exactly two (orthogonal) solutions. The equation $(1)$ is covariant, and therefore its solutions depend covariantly on $p^\mu$, and agree with the standard definition of the Dirac spinors by letting $m^2\equiv p^2$ with $m>0$.
We note that this definition of $u_s(p)$ is no more nor less general than the standard definition: here $u$ depends on the four components $p^\mu$, while in the standard case it depends on $\boldsymbol p$, but also implicitly on $m$, and therefore both objects are actually the same thing (the former case emphasises the dependence on $p^0$, while the latter emphasises the dependence on $m$, but the relation $p^0=p^0(m)$ can always be inverted to yield $m=m(p^0)$ and viceversa).
One very important remark is that this "extension" of $u$ off the mass-shell has nothing to do with what Weinberg does to define the off-shell propagator. In particular, Weinberg's approach is based on the extension
$$
\sqrt{\boldsymbol p^2+m^2}\to p^0\tag{4}
$$
or, equivalently, at the level of the spin sums,
$$
\sum_s u(\boldsymbol p)\bar u_s(\boldsymbol p)\to \not p-m\tag{5}
$$
with $p^0$ an off-shell Fourier variable.
Weinberg's extension, and the one that we are considering here, are completely unrelated. My "extension" above is not to be used in any practical application. It is not the polarisation vector of an off-shell particle. It is not a part of an off-shell propagator. As I said in the OP, I just wanted a spinor (in the abstract, algebraic sense) that depends on the four variables $p^\mu$. I didn't have any particular motivation for this question; rather, I just wanted to have a simple spinor to play around with. In my opinion, this answer addresses precisely that, and nothing else. If someone thinks that this answer is wrong, then it is because they didn't understand what I was trying to ask in the OP (of course, I am to blame: I should have been more clear about what I had in mind). As far as I'm concerned, this post answers my original question.
Again, and to reiterate: in this answer I did not extend the propagator off the mass-shell (and in the OP I didn't ask that). My "extension" has no usefulness for actual computations in QFT.
