Are a sphere cross a circle ($S^2 times S$) and the 3-dimensional projective ($textbf{RP(3)}$) space...
$begingroup$
I have some questions about the space $textbf{SO(3)}$ (special orthogonal transformations in 3-dimensions). I understand that this transformations represent rotations in 3D space fixing the origin and also I understand that $textbf{SO(3)}$ is homeomorphic to $textbf{RP(3)}$ (real-projective 3D-space). A proof for this homeomorphism can be found in: prove $RP^3cong SO(3)$
Now imagine an orthogonal base at the origin ($e_1$, $e_2$, $e_3$). To specify a rotation you have to first choose where in the unit sphere you want to send $e_1$, once you choose that direction you have to choose where to send $e_2$ around a circle, and once you choose that you are done (because of the orthogonality). So with this analysis it seems that the group of rotations should be homeomorphic to $S^2 times S$ (sphere cross the cirle). But also $textbf{SO(3)}$ is homeomorphic to $textbf{RP(3)}$.
My question is: Are $S^2 times S$ and $RP^3$ really homeomorphic? and if not where is the mistake on my reasoning?, if they are, is there a more simple homeomorphism to show that (without the use of $textbf{SO(3)}$ and more in the spirit of cutting and gluing)?
Thanks in advance. :)
general-topology rotations spheres
asked Dec 1 '18 at 0:03
PonciopoPonciopo
624
$endgroup$
add a comment |
$begingroup$
I have some questions about the space $textbf{SO(3)}$ (special orthogonal transformations in 3-dimensions). I understand that this transformations represent rotations in 3D space fixing the origin and also I understand that $textbf{SO(3)}$ is homeomorphic to $textbf{RP(3)}$ (real-projective 3D-space). A proof for this homeomorphism can be found in: prove $RP^3cong SO(3)$
Now imagine an orthogonal base at the origin ($e_1$, $e_2$, $e_3$). To specify a rotation you have to first choose where in the unit sphere you want to send $e_1$, once you choose that direction you have to choose where to send $e_2$ around a circle, and once you choose that you are done (because of the orthogonality). So with this analysis it seems that the group of rotations should be homeomorphic to $S^2 times S$ (sphere cross the cirle). But also $textbf{SO(3)}$ is homeomorphic to $textbf{RP(3)}$.
My question is: Are $S^2 times S$ and $RP^3$ really homeomorphic? and if not where is the mistake on my reasoning?, if they are, is there a more simple homeomorphism to show that (without the use of $textbf{SO(3)}$ and more in the spirit of cutting and gluing)?
Thanks in advance. :)
general-topology rotations spheres
asked Dec 1 '18 at 0:03
PonciopoPonciopo
624
$endgroup$
3
$begingroup$
I would expect your argument shows that $RP^3$ is a fibered space over $S^2$ with fibers isomorphic to $S^1$; but I see no reason for this to be a globally trivial fibration.
$endgroup$
– Daniel Schepler
Dec 1 '18 at 0:11
add a comment |
$begingroup$
I have some questions about the space $textbf{SO(3)}$ (special orthogonal transformations in 3-dimensions). I understand that this transformations represent rotations in 3D space fixing the origin and also I understand that $textbf{SO(3)}$ is homeomorphic to $textbf{RP(3)}$ (real-projective 3D-space). A proof for this homeomorphism can be found in: prove $RP^3cong SO(3)$
Now imagine an orthogonal base at the origin ($e_1$, $e_2$, $e_3$). To specify a rotation you have to first choose where in the unit sphere you want to send $e_1$, once you choose that direction you have to choose where to send $e_2$ around a circle, and once you choose that you are done (because of the orthogonality). So with this analysis it seems that the group of rotations should be homeomorphic to $S^2 times S$ (sphere cross the cirle). But also $textbf{SO(3)}$ is homeomorphic to $textbf{RP(3)}$.
My question is: Are $S^2 times S$ and $RP^3$ really homeomorphic? and if not where is the mistake on my reasoning?, if they are, is there a more simple homeomorphism to show that (without the use of $textbf{SO(3)}$ and more in the spirit of cutting and gluing)?
Thanks in advance. :)
general-topology rotations spheres
asked Dec 1 '18 at 0:03
PonciopoPonciopo
624
$endgroup$
I have some questions about the space $textbf{SO(3)}$ (special orthogonal transformations in 3-dimensions). I understand that this transformations represent rotations in 3D space fixing the origin and also I understand that $textbf{SO(3)}$ is homeomorphic to $textbf{RP(3)}$ (real-projective 3D-space). A proof for this homeomorphism can be found in: prove $RP^3cong SO(3)$
Now imagine an orthogonal base at the origin ($e_1$, $e_2$, $e_3$). To specify a rotation you have to first choose where in the unit sphere you want to send $e_1$, once you choose that direction you have to choose where to send $e_2$ around a circle, and once you choose that you are done (because of the orthogonality). So with this analysis it seems that the group of rotations should be homeomorphic to $S^2 times S$ (sphere cross the cirle). But also $textbf{SO(3)}$ is homeomorphic to $textbf{RP(3)}$.
My question is: Are $S^2 times S$ and $RP^3$ really homeomorphic? and if not where is the mistake on my reasoning?, if they are, is there a more simple homeomorphism to show that (without the use of $textbf{SO(3)}$ and more in the spirit of cutting and gluing)?
Thanks in advance. :)
general-topology rotations spheres
general-topology rotations spheres
asked Dec 1 '18 at 0:03
PonciopoPonciopo
624
asked Dec 1 '18 at 0:03
PonciopoPonciopo
624
asked Dec 1 '18 at 0:03
PonciopoPonciopo
624
asked Dec 1 '18 at 0:03
PonciopoPonciopo
624
asked Dec 1 '18 at 0:03
PonciopoPonciopo
624
624
3
$begingroup$
I would expect your argument shows that $RP^3$ is a fibered space over $S^2$ with fibers isomorphic to $S^1$; but I see no reason for this to be a globally trivial fibration.
$endgroup$
– Daniel Schepler
Dec 1 '18 at 0:11
add a comment |
3
$begingroup$
I would expect your argument shows that $RP^3$ is a fibered space over $S^2$ with fibers isomorphic to $S^1$; but I see no reason for this to be a globally trivial fibration.
$endgroup$
– Daniel Schepler
Dec 1 '18 at 0:11
3
3
$begingroup$
I would expect your argument shows that $RP^3$ is a fibered space over $S^2$ with fibers isomorphic to $S^1$; but I see no reason for this to be a globally trivial fibration.
$endgroup$
– Daniel Schepler
Dec 1 '18 at 0:11
$begingroup$
I would expect your argument shows that $RP^3$ is a fibered space over $S^2$ with fibers isomorphic to $S^1$; but I see no reason for this to be a globally trivial fibration.
$endgroup$
– Daniel Schepler
Dec 1 '18 at 0:11
add a comment |
1 Answer
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They are not even homotopy equivalent. We have $pi_1(textbf{RP(3)})= mathbb{Z}_2$ (see Sammy Black's comment to An intuitive idea about fundamental group of $mathbb{RP}^2$), but $pi_1(S^2 times S^1 )= mathbb{Z}$.
answered Dec 1 '18 at 0:20
Paul FrostPaul Frost
9,8753732
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$begingroup$
They are not even homotopy equivalent. We have $pi_1(textbf{RP(3)})= mathbb{Z}_2$ (see Sammy Black's comment to An intuitive idea about fundamental group of $mathbb{RP}^2$), but $pi_1(S^2 times S^1 )= mathbb{Z}$.
answered Dec 1 '18 at 0:20
Paul FrostPaul Frost
9,8753732
$endgroup$
add a comment |
$begingroup$
They are not even homotopy equivalent. We have $pi_1(textbf{RP(3)})= mathbb{Z}_2$ (see Sammy Black's comment to An intuitive idea about fundamental group of $mathbb{RP}^2$), but $pi_1(S^2 times S^1 )= mathbb{Z}$.
answered Dec 1 '18 at 0:20
Paul FrostPaul Frost
9,8753732
$endgroup$
add a comment |
$begingroup$
They are not even homotopy equivalent. We have $pi_1(textbf{RP(3)})= mathbb{Z}_2$ (see Sammy Black's comment to An intuitive idea about fundamental group of $mathbb{RP}^2$), but $pi_1(S^2 times S^1 )= mathbb{Z}$.
answered Dec 1 '18 at 0:20
Paul FrostPaul Frost
9,8753732
$endgroup$
They are not even homotopy equivalent. We have $pi_1(textbf{RP(3)})= mathbb{Z}_2$ (see Sammy Black's comment to An intuitive idea about fundamental group of $mathbb{RP}^2$), but $pi_1(S^2 times S^1 )= mathbb{Z}$.
answered Dec 1 '18 at 0:20
Paul FrostPaul Frost
9,8753732
answered Dec 1 '18 at 0:20
Paul FrostPaul Frost
9,8753732
answered Dec 1 '18 at 0:20
Paul FrostPaul Frost
9,8753732
answered Dec 1 '18 at 0:20
Paul FrostPaul Frost
9,8753732
9,8753732
add a comment |
add a comment |
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I would expect your argument shows that $RP^3$ is a fibered space over $S^2$ with fibers isomorphic to $S^1$; but I see no reason for this to be a globally trivial fibration.
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– Daniel Schepler
Dec 1 '18 at 0:11