## 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 $\displaystyle SO(3)\cong D/\sim\cong\mathbb{R}P^3.$