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A smooth right group action is a smooth map $(p,g)\mapsto p\cdot g$ from $P\times G$ to $P$, with $p\cdot e=p$ and $(p\cdot g)\cdot h=p\cdot (gh)$. It is a free action of a group if $p\cdot g=p$ implies $g=e$.

A smooth principal bundle consists of such a free action and a surjective map $\pi :P\to B$ whose fibres are the orbits, together with equivariant local trivializationswhere $(x,g)\cdot h=(x,gh)$. The trivializations are diffeomorphisms; in particular the bundle projection is a submersion.

For two such trivializations, $H={\Phi }_2{\Phi }_1^{-1}$ fixes the base coordinate and is equivariant. Write $H(x,e)=(x,\psi (x))$. Equivariance immediately givesThe function $\psi$ is smooth because $x\mapsto (x,e)$ and $H$ are smooth. This proves the general transition function of a principal bundle formula; the multiplier is on the left, even though the original action is on the right.

For the explicit bundle, represent a point of ${\mathbb{RP}}^3$ by a nonzero vector $w=(w_1,w_2)\in {\mathbb{C}}^2$ modulo nonzero real scaling. The map sends $[w]$ to the complex line $[w_1:w_2]$, so it is well-defined and surjective. Its smoothness follows from the local ratio charts below. Ordinary scalar multiplication by $u\in U(1)$ would not be free, since $u=-1$ acts trivially on real projective classes. Instead defineThe two possible roots differ by a real sign, so give the same class. This is independent of the real representative of $[w]$. Products of chosen roots show the right action law, and the identity acts trivially. Locally on $U(1)$ a smooth square-root branch exists; these local definitions agree projectively and prove global smoothness of the action.

If $[\lambda w]=[w]$, then $\lambda w=rw$ for some nonzero real $r$. Since $w\ne 0$ and $|\lambda |=1$, this forces $\lambda =\pm 1$ and hence $u=1$. The action is free. Two points in the same fibre have representatives $w^{\prime }=cw$ for some $c\in {\mathbb{C}}^{\ast }$; absorb $|c|$ into the real scaling and use $u=(c/|c|{)}^2$. Thus the orbits are exactly the fibres.

Let $N_1=\{[z:1]\}$ and $N_2=\{[1:\zeta ]\}$. Using these base coordinates, defineBoth formulas are invariant under real nonzero scaling, including negative scaling. Their inverses areAgain the sign of the root does not matter. Direct substitution shows they are two-sided inverses. The forward maps are smooth, and local square-root branches prove that the inverses are smooth. Under the right action, each squared phase is multiplied by $u$, so the maps are equivariant. They are therefore genuine principal trivializations of the principal circle bundle on real projective three-space.

On the overlap $z\ne 0$, we have $\zeta =1/z$ andConsequentlyWhen the base point itself, rather than its chart coordinate, is written as $x$, its coordinate remains $x$ in the general transition formula. The opposite transition has reciprocal multiplier. The doubled phase is essential to freeness and is not the ordinary Hopf-bundle phase.

Write a point of Real projective space as a nonzero vector $(w_1,w_2)\in {\mathbb{C}}^2$ modulo real nonzero scaling. The map to Complex projective space sends its class to the complex line through that vector. A free right action of the circle group is $[w]\cdot u=[\lambda w]$ where ${\lambda }^2=u$; the two possible roots differ by a real sign and therefore give the same projective point. Local smooth square-root branches prove smoothness. Over $[z:1]$, the fibre coordinate is $(w_2/|w_2|{)}^2$; over $[1:\zeta ]$, it is $(w_1/|w_1|{)}^2$. On the overlap $\zeta =1/z$, the transition function of a principal bundle is $u\mapsto (z/\overline{z})u$. Its phase winds twice. Ordinary scalar multiplication by $u$ instead has a kernel of order two and is not a free action of a group.