In twistor theory, given the complexified Minkowski space $CM$ and the projective twistor space $PT$, an $\alpha-$plane is defined as the correspondence in $CM$ wit a point $Z \in PT$.

But I found two versions of it.

(a) One in "Twistor geometry and field theory" by Ward and Wells, where $\alpha-$plane is a projective complex plane in $CM$ corresponding to a point $Z \in PT$ without mentioning if it's self-dual or not.

(b) Another is by Penrose that $\alpha-$plane is a two-complex-dimensional locus of points in $CM$ which is incident to $Z$ and also it's self-dual.

I am confused if the two versions are different.

(1) Are the two definitions mean the same thing? i.e., every point of the projective complex plane in (a) is incident to $Z$ as defined in (b)?

(2) Is the definition by Penrose in (b) a subset of the former in (a), so that only a subset of the points of the $\alpha-$plane in (a) is self-dual? Since according to (a), the $\alpha-$plane is a PROJECTIVE complex plane for any point $Z \in PT$. It seems that any point in $CM$ must fall on at least one $\alpha-$plane corresponding to some point $Z' \in PT$. Then the self-duality of Penrose should not hold for the definition (a) since not every point in $CM$ leads to self-duality property.

(3) If (b) is really a subset of (a), then not every point in $CM$ falls on an $\alpha-$plane and the union of all $\alpha-$planes is just a subset of $CM$, this sounds reasonable .

To explain my question, I put the definitions (a)(b) in figures. For definition (a), $P$ is in fact just $PT$ and the $\alpha-$plane is given by $\tilde{p}$ as given in Figure 1. Figure 2 and 3 show the schemes of (a) and (b).

I am new on this topic. Can anybody help to clarify the concept?

Definition (a) Figure 1. Definition (a) of Ward & Wells [Twistor geometry and field theory]

Figure of definition (a) Figure 2. Definition (a) of Ward & Wells, figure from Dorje C. Brody and Lane P. Hughston [2005].

Figure of definition (b) Figure 3. Definition (b) of Penrose [The road to reality]



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.