# Proyective space as the spherical tangent bundle of $S^2$.

I have a problem to identify $$\mathbb{R}\mathbb{P}^3$$ with the spherical tangent bundle of $$S^2$$, $$ST(S^2)\cong SO(3)$$. The spherical tangent bundle is just a sub bundle of the fiber bundle consisting of vectors of norm 1.

The book I am reading takes a ball $$B^2\subset \mathbb{R}^3$$ of radius $$\pi$$. For each point $$x$$ of $$B^3$$ we can associate the rotation of axis $$ox$$ and angle $$|x|$$, and since antipodal points of $$\partial B^3$$ rotate $$\mathbb{R}^3$$ around the same axis and with angle $$\pi$$, they define the same rotation of $$\mathbb{R}^3$$, so we can identify them.

Finally, the author said that this is a $$3$$-manifold obtained from this ball by indentifying antipodal points of its boundary.

It is probably a stupid question, and sorry for that, but, why isn't it $$\mathbb{R}\mathbb{P}^2$$ instead of $$\mathbb{R}\mathbb{P}^3$$?

• Because $S^2$ is 2-dimensional so its unit tangent bundle should be dimension $2+(2-1) = 3$ :) – user98602 Dec 27 '18 at 22:58
• Yeah, that is what I thought, but the construction of identifying antipodal points of a ball is the same of the projective plane, isn't it? That is why I don't understand why is $\mathbb{R}\mathbb{P}^3$... Thanks! – Rubén Fernández Fuertes Dec 28 '18 at 10:31
• @RubénFernándezFuertes Not quite, the projective space is constructed by identifying antipodal points on a sphere, not a ball. I suppose you can think about the construction of projective 3-space as gluing an open 2-ball onto the projective plane. – mcwiggler Dec 28 '18 at 12:54
• But if $B$ is a ball, $\partial B$ is a sphere, isn't it? And we are identifying antipodal points of $\partial B$... – Rubén Fernández Fuertes Dec 29 '18 at 16:23
• Yes, but the points in the interior of the ball are still left. Projective space consists of only the equivalence classes of the points in the boundary sphere. – mcwiggler Dec 30 '18 at 11:17