SO(3) diffeomorphic to RP^3

SO(3)\cong\mathbb{R}P^3}

Proof:

We consider SO(3) as the group of all rotations about the origin of \mathbb{R}^3 under the operation of composition. Every non-trivial rotation is determined by its axis of rotation (a line through the origin) and its angle of rotation.

We consider \mathbb{R}P^3 as the unit 3-sphere S^3 with antipodal points identified.

Consider the map from the unit ball D in \mathbb{R}^3 to SO(3), which maps (x,y,z) to the rotation about (x,y,z) through the angle \pi\sqrt{x^2+y^2+z^2} (and maps (0,0,0) to the identity rotation). This mapping is clearly smooth and surjective. Its restriction to the interior of D is injective since on the interior \pi\sqrt{x^2+y^2+z^2}<\pi. On the boundary of D, two rotations through \pi and through -\pi are the same. Hence the mapping induces a smooth bijective map from D/\sim, with antipodal points on the boundary identified, to SO(3). The inverse of this map, \displaystyle ((x,y,z),\pi\sqrt{x^2+y^2+z^2})\mapsto(x,y,z) is also smooth. (To see that the inverse is smooth, write \theta=\pi\sqrt{x^2+y^2+z^2}. Then x=\sqrt{\frac{\theta^2}{\pi^2}-y^2-z^2}, and so \frac{\partial^k x}{\partial\theta^k} exists and is continuous for all orders k. Similar results hold for the variables y and z, and also mixed partials. By multivariable chain rule, one can see that all component functions are indeed smooth, so the inverse is smooth as claimed.)

Hence SO(3)\cong D/\sim, the unit ball D in \mathbb{R}^3 with antipodal points on the boundary identified.

Next, the mapping \displaystyle (x,y,z)\mapsto (x,y,z,\sqrt{1-x^2-y^2-z^2}) is a diffeomorphism between D/\sim and the upper unit hemisphere of S^3 with antipodal points on the equator identified. The latter space is clearly diffeomorphic to \mathbb{R}P^3. Hence, we have shown \displaystyle SO(3)\cong D/\sim\cong\mathbb{R}P^3.

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