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In Penrose's construction of the twistor space of Minkowski spacetime $\mathbb R^{1,3}$, we first complexify $\mathbb R^{1,3}$ to $\mathbb C^4$ and then think of points in it as matrices acting on $\mathbb C^2$ via a tensor decomposition $\mathbb C^4 = \mathbb C^2\otimes \mathbb C^2=\mathrm{End}(\mathbb C^2)$. Such a matrix can be depicted as a graph in $\mathbb C^2\oplus \mathbb C^2$, which Penrose calls the twistor space. It makes sense to quotient out by a $\mathbb C^\times$ action, following which we get the projective twistor space. In particular, points in the complexified Minkowski space correspond to embeddings of $\mathbb{CP}^1$ into the projective twistor space.

Salamon, on the other hand, defines the twistor space of a general (possibly Ricci flat) quaternionic (pseudo-)Kähler manifold $M$ as the bundle of almost complex structures in the quaternionic subbundle of $\mathrm{End}(TM)$. The fibre over a point is a sphere which may be identified with $\mathbb {CP}^1$. So there is a $\mathbb{CP}^1$ for every point in $M$.

Particularly, in Penrose's construction, these spheres are supposed to intersect whenever two points in Minkowski spacetime are null-separated, whereas in Salamon's consruction, the spheres are all separate fibres over the points of $M$.

[EDIT: My question was originally about what happens when we take $M$ to be the Minkowski spacetime, but as Deane Yang pointed out in the comments, this is not possible.]

Question. Is there a general definition of a twistor space on a manifold $M$ which specialises to the Penrose (resp. Salamon) twistor space in the case that $M$ is the Minkowski spacetime (resp. quaternionic (pseudo-)Kähler)? In particular, why are almost Hermitian structures the natural analogues of null rays?

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  • $\begingroup$ crossposted: math.stackexchange.com/questions/2575968/… $\endgroup$ Commented Dec 24, 2017 at 16:10
  • $\begingroup$ Your question seems to assume that Minkowski spacetime is a quaternionic (pseudo-)Kähler manifold. Is that the case? $\endgroup$
    – Deane Yang
    Commented Dec 24, 2017 at 16:37
  • $\begingroup$ Ah, sorry, you're right. In order to be quaternionic (pseudo-)Kähler, the signature of a metric has to be of the form $(4p,4q)$. I will edit the question accordingly. $\endgroup$
    – Arpan Saha
    Commented Dec 25, 2017 at 0:25

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