Proyective space as the spherical tangent bundle of $S^2$.
$begingroup$
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 $B^2subset mathbb{R}^3$ 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 $mathbb{R}mathbb{P}^3$?
geometry differential-geometry manifolds
asked Dec 27 '18 at 22:48
Rubén Fernández FuertesRubén Fernández Fuertes
1107
$endgroup$
|
show 1 more comment
$begingroup$
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 $B^2subset mathbb{R}^3$ 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 $mathbb{R}mathbb{P}^3$?
geometry differential-geometry manifolds
asked Dec 27 '18 at 22:48
Rubén Fernández FuertesRubén Fernández Fuertes
1107
$endgroup$
$begingroup$
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
1
$begingroup$
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
|
show 1 more comment
$begingroup$
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 $B^2subset mathbb{R}^3$ 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 $mathbb{R}mathbb{P}^3$?
geometry differential-geometry manifolds
asked Dec 27 '18 at 22:48
Rubén Fernández FuertesRubén Fernández Fuertes
1107
$endgroup$
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 $B^2subset mathbb{R}^3$ 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 $mathbb{R}mathbb{P}^3$?
geometry differential-geometry manifolds
geometry differential-geometry manifolds
asked Dec 27 '18 at 22:48
Rubén Fernández FuertesRubén Fernández Fuertes
1107
asked Dec 27 '18 at 22:48
Rubén Fernández FuertesRubén Fernández Fuertes
1107
asked Dec 27 '18 at 22:48
Rubén Fernández FuertesRubén Fernández Fuertes
1107
asked Dec 27 '18 at 22:48
Rubén Fernández FuertesRubén Fernández Fuertes
1107
asked Dec 27 '18 at 22:48
Rubén Fernández FuertesRubén Fernández Fuertes
1107
1107
$begingroup$
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
1
$begingroup$
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
|
show 1 more comment
$begingroup$
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
1
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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
$begingroup$
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
1
1
$begingroup$
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
$begingroup$
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
|
show 1 more comment
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$begingroup$
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
1
$begingroup$
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