# $\mathbb{RP}^3 \rightarrow \mathbb{CP}^1$ defines a principal U(1)-bundle

I have to show that the map $\pi: (x_o : x_1 : x_2 : x_3) \in \mathbb{RP}^3 \rightarrow (x_0 + i x_1) : (x_2+ ix_3) \in \mathbb{CP}^1$ defines a principal U(1)-bundle. The two standard coordinate patches (call these $U_1,U_2$) on $\mathbb{CP}^1$ may be taken as trivializing neighbourhoods, and the transition functions are then given by $\phi(z:1) = (z/|z|)^2$.

I started by noting that $\pi^{-1}((x_0 + i x_1) : (x_2+ ix_3))$ is the set $\{(x_0 \cos{\theta} - x_1 \sin{\theta}:x_0 + x_1 \cos{\theta}: x_2 \cos{\theta}-x_3 \sin{\theta}: x_2 \sin{\theta} + x_3 \cos{\theta} \,|\, \theta \in S^1 \}$, since $(x_0 + i x_1) : (x_2+ ix_3)$ is really the equivalence class of points in $\mathbb{C}^2$ of the form $\lambda(x_0 + i x_1, x_2+ ix_3)$ for $\lambda \in \mathbb{C}$, so simultaneous rotations of the first two coordinates and last two coordinates are in the same equivalence class. This set seems the same as $S^1 = U(1)$, which is the typical fibre. However, to show that the bundle trivializes locally I need to find a diffeomorphism $\pi^{-1}(U_i) \rightarrow U_i \times U(1)$. But given a point in $\pi^{-1}(U_i) \in \mathbb{RP}^3$, there is a set which looks like $S^1$ that gets mapped to the same point in $U_i$, and it is not clear how to choose the representative of this set (which gets mapped to the corresponding element in $\mathbb{CP}^1$ by $\pi$) so that the required diffeomorphism becomes a smooth map. Also, without explicit map for the trivializing neighbourhoods I cannot find the transition functions.

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