Twistors for spaces of $n-$dimensions 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
 A: "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.
