I was wondering if there is any general definition for twistors for spaces of any dimension with a definite (or indefinite) metric; for example, in $\mathbb{R}^3$.
Twistors are spinors of the compactificated minkowski space. Does this work in general? Are twistors the spinors of the compactification of a vectorial space? Thanks
"Twistors are spinors of the compactified Minkowski space" is not quite true. Twistors in Minkowski space are spinor fields which satisfy a particular PDE: the twistor spinor equation. Relative to flat coordinates and the corresponding global frame for the tangent bundle, a twistor spinor $\psi$ is given in terms of a pair of constant spinors $(\varphi,\eta)$ by $$ \psi(x) = \varphi + x \cdot \eta $$ where $\cdot$ is the Clifford action. Since the twistor spinor equation is conformally invariant, one can extend this to a spinor field on the conformal compactification of Minkowski space.
The situation in general is the following.
Let $(M,g)$ be a pseudoriemannian spin manifold. Let $\Sigma \to M$ be a bundle of modules of the Clifford bundle $Cl(TM)$. The spin connection defines a map on sections $$ \nabla : C^\infty(M;\Sigma) \to \Omega^1(M;\Sigma)~. $$ The Clifford action of 1-forms on spinor fields defines a bundle map $$ c: T^* M \otimes \Sigma \to \Sigma $$ which induces a map on sections $$ \Omega^1(M;\Sigma) \to C^\infty(M;\Sigma)~. $$ Composing this map with the covariant derivative above defines a differential operator $$ D : C^\infty(M;\Sigma) \to C^\infty(M;\Sigma) $$ This is the Dirac operator.
Now the kernel of $c : T^* M \otimes \Sigma \to \Sigma $ defines a sub-bundle $W \subset T^* M \otimes \Sigma$, so that $$ T^*M \otimes \Sigma \cong W \oplus \Sigma~.$$ Composing the above covariant derivative with the projection onto $W$ along $\Sigma$ defines a differential operator $$ P : C^\infty(M;\Sigma) \to C^\infty(M;W) $$ called the Penrose operator, whose kernel are called twistors.
In other words, a spinor field $\psi \in C^\infty(M;\Sigma)$ is a twistor (spinor) if $P\psi = 0$. We can write this equation more explicitly as follows: $$P_X \psi = \nabla_X \psi + \frac1n X \cdot D\psi~,$$ where $X \in C^\infty(M;TM)$, $\dim M = n$ and I have used the original Clifford conventions for the Clifford product, so that there is a minus sign in $$ X\cdot X \cdot \psi = - g(X,X) \psi~.$$
It is an instructive exercise to work out that when $(M,g)$ is $4$-dimensional Minkowski spacetime, the solutions to the twistor spinor equation are precisely given by the first formula in this answer.
