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 $\mathbb{R}\mathbb{P}^3$?

  • $\begingroup$ Because $S^2$ is 2-dimensional so its unit tangent bundle should be dimension $2+(2-1) = 3$ :) $\endgroup$ – user98602 Dec 27 '18 at 22:58
  • $\begingroup$ 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! $\endgroup$ – Rubén Fernández Fuertes Dec 28 '18 at 10:31
  • $\begingroup$ @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. $\endgroup$ – mcwiggler Dec 28 '18 at 12:54
  • $\begingroup$ But if $B$ is a ball, $\partial B$ is a sphere, isn't it? And we are identifying antipodal points of $\partial B$... $\endgroup$ – Rubén Fernández Fuertes Dec 29 '18 at 16:23
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    $\begingroup$ 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. $\endgroup$ – mcwiggler Dec 30 '18 at 11:17

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