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Twistor space, as complex projective space $\mathbb{CP}^3$, is related to Minkowski 4-D space time (metric $1, -1,-1,-1$), by the incidence relation.

Let $Z = (v_a, u^{\dot{a}})$ a point in $\mathbb{CP}^3$, where $v_a$ and $u^{\dot{a}}$ are 2-complex components spinors.

Let $x_{a\dot{a}} = \sum x^\mu (\sigma_\mu)_{a\dot{a}}$, where $\sigma_\mu$ are the Pauli matrices, and $x^\mu$ a point in Minkowski 4-D space-time.

The incidence relation is then $v_a = x_{a\dot{a}} u^{\dot{a}}$

The representation of a space-time point, in the twistor space $\mathbb{CP}^3$, is a complex line $\mathbb{CP}^1$.

So, $\mathbb{CP}^3$ may be seen as a bundle space, with base Minkowski space-time, and fiber $\mathbb{CP}^1$.

But, is the inverse true, that is: could we see $\mathbb{CP}^3$ as a bundle space, with base $\mathbb{CP}^1$, and as fiber the 4-D Minkowski space-time?

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The Penrose twistor fibration is $\mathbb{CP}^1 \to \mathbb{CP}^3 \to S^4$. You seem to be asking the following question: is there a fibration $S^4 \to \mathbb{CP}^3 \to \mathbb{CP}^1$?

The long exact sequence in homotopy applied to this fibration would give

$\dots \to \pi_4(\mathbb{CP}^3) \to \pi_4(\mathbb{CP}^1) \to \pi_3(S^4) \to \dots$

From the long exact sequence in homotopy applied to the fibration $S^1 \to S^{2n+1} \to \mathbb{CP}^n$, we see that $\pi_i(\mathbb{CP}^n) = \pi_i(S^{2n+1})$ for all $i > 2$, in particular $\pi_4(\mathbb{CP}^3) = \pi_4(S^7) = 0$. As $\pi_3(S^4) = 0$, it follows that $\pi_4(\mathbb{CP}^1)$ must be zero, but this is not true: $\pi_4(\mathbb{CP}^1) = \pi_4(S^2) = \mathbb{Z}_2$.

Therefore, there is no fibration $S^4 \to \mathbb{CP}^3 \to \mathbb{CP}^1$.