Nontriviality of the Hopf Fibration
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A simple question how to understand why even though locally $S^3$ is homeomorphic to $S^2times S^1$, how do you see that globally this is not true?
differential-geometry fiber-bundles fibration hopf-fibration
asked Dec 8 '18 at 13:42
BunnehBunneh
557
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add a comment |
$begingroup$
A simple question how to understand why even though locally $S^3$ is homeomorphic to $S^2times S^1$, how do you see that globally this is not true?
differential-geometry fiber-bundles fibration hopf-fibration
asked Dec 8 '18 at 13:42
BunnehBunneh
557
$endgroup$
$begingroup$
Do you mean $S^2 times S^1$?
$endgroup$
– Rolf Hoyer
Dec 8 '18 at 13:48
$begingroup$
Yes sorry! In my head I was thinking about the wrong theorem showing that the quotient of two lie groups forming a bundle with the quotient being the fibres! I'll amend the question!
$endgroup$
– Bunneh
Dec 9 '18 at 12:31
add a comment |
$begingroup$
A simple question how to understand why even though locally $S^3$ is homeomorphic to $S^2times S^1$, how do you see that globally this is not true?
differential-geometry fiber-bundles fibration hopf-fibration
asked Dec 8 '18 at 13:42
BunnehBunneh
557
$endgroup$
A simple question how to understand why even though locally $S^3$ is homeomorphic to $S^2times S^1$, how do you see that globally this is not true?
differential-geometry fiber-bundles fibration hopf-fibration
differential-geometry fiber-bundles fibration hopf-fibration
asked Dec 8 '18 at 13:42
BunnehBunneh
557
asked Dec 8 '18 at 13:42
BunnehBunneh
557
edited Dec 9 '18 at 12:32
Bunneh
asked Dec 8 '18 at 13:42
BunnehBunneh
557
asked Dec 8 '18 at 13:42
BunnehBunneh
557
asked Dec 8 '18 at 13:42
BunnehBunneh
557
557
$begingroup$
Do you mean $S^2 times S^1$?
$endgroup$
– Rolf Hoyer
Dec 8 '18 at 13:48
$begingroup$
Yes sorry! In my head I was thinking about the wrong theorem showing that the quotient of two lie groups forming a bundle with the quotient being the fibres! I'll amend the question!
$endgroup$
– Bunneh
Dec 9 '18 at 12:31
add a comment |
$begingroup$
Do you mean $S^2 times S^1$?
$endgroup$
– Rolf Hoyer
Dec 8 '18 at 13:48
$begingroup$
Yes sorry! In my head I was thinking about the wrong theorem showing that the quotient of two lie groups forming a bundle with the quotient being the fibres! I'll amend the question!
$endgroup$
– Bunneh
Dec 9 '18 at 12:31
$begingroup$
Do you mean $S^2 times S^1$?
$endgroup$
– Rolf Hoyer
Dec 8 '18 at 13:48
$begingroup$
Do you mean $S^2 times S^1$?
$endgroup$
– Rolf Hoyer
Dec 8 '18 at 13:48
$begingroup$
Yes sorry! In my head I was thinking about the wrong theorem showing that the quotient of two lie groups forming a bundle with the quotient being the fibres! I'll amend the question!
$endgroup$
– Bunneh
Dec 9 '18 at 12:31
$begingroup$
Yes sorry! In my head I was thinking about the wrong theorem showing that the quotient of two lie groups forming a bundle with the quotient being the fibres! I'll amend the question!
$endgroup$
– Bunneh
Dec 9 '18 at 12:31
add a comment |
1 Answer
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oldest
votes
$begingroup$
A standard way to show that $S^3$ is not homeomorphic to $S^2 times S_1$ is to look at its fundamental group: $S^3$ is simply connected (that is, $pi_1$ is trivial), but $pi_1(S^2 times S^1) simeq pi_1(S^1) = Bbb{Z}$.
More intuitively, try this: in $S^3$, any imbedded 2-sphere will separate the space into two connected components. But in $S^2 times S^1$, a 2-sphere $S^2 times {x}$ does not separate the space.
answered Dec 8 '18 at 14:16
Hew WolffHew Wolff
2,245716
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1 Answer
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$begingroup$
A standard way to show that $S^3$ is not homeomorphic to $S^2 times S_1$ is to look at its fundamental group: $S^3$ is simply connected (that is, $pi_1$ is trivial), but $pi_1(S^2 times S^1) simeq pi_1(S^1) = Bbb{Z}$.
More intuitively, try this: in $S^3$, any imbedded 2-sphere will separate the space into two connected components. But in $S^2 times S^1$, a 2-sphere $S^2 times {x}$ does not separate the space.
answered Dec 8 '18 at 14:16
Hew WolffHew Wolff
2,245716
$endgroup$
add a comment |
$begingroup$
A standard way to show that $S^3$ is not homeomorphic to $S^2 times S_1$ is to look at its fundamental group: $S^3$ is simply connected (that is, $pi_1$ is trivial), but $pi_1(S^2 times S^1) simeq pi_1(S^1) = Bbb{Z}$.
More intuitively, try this: in $S^3$, any imbedded 2-sphere will separate the space into two connected components. But in $S^2 times S^1$, a 2-sphere $S^2 times {x}$ does not separate the space.
answered Dec 8 '18 at 14:16
Hew WolffHew Wolff
2,245716
$endgroup$
add a comment |
$begingroup$
A standard way to show that $S^3$ is not homeomorphic to $S^2 times S_1$ is to look at its fundamental group: $S^3$ is simply connected (that is, $pi_1$ is trivial), but $pi_1(S^2 times S^1) simeq pi_1(S^1) = Bbb{Z}$.
More intuitively, try this: in $S^3$, any imbedded 2-sphere will separate the space into two connected components. But in $S^2 times S^1$, a 2-sphere $S^2 times {x}$ does not separate the space.
answered Dec 8 '18 at 14:16
Hew WolffHew Wolff
2,245716
$endgroup$
A standard way to show that $S^3$ is not homeomorphic to $S^2 times S_1$ is to look at its fundamental group: $S^3$ is simply connected (that is, $pi_1$ is trivial), but $pi_1(S^2 times S^1) simeq pi_1(S^1) = Bbb{Z}$.
More intuitively, try this: in $S^3$, any imbedded 2-sphere will separate the space into two connected components. But in $S^2 times S^1$, a 2-sphere $S^2 times {x}$ does not separate the space.
answered Dec 8 '18 at 14:16
Hew WolffHew Wolff
2,245716
answered Dec 8 '18 at 14:16
Hew WolffHew Wolff
2,245716
answered Dec 8 '18 at 14:16
Hew WolffHew Wolff
2,245716
answered Dec 8 '18 at 14:16
Hew WolffHew Wolff
2,245716
2,245716
add a comment |
add a comment |
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$begingroup$
Do you mean $S^2 times S^1$?
$endgroup$
– Rolf Hoyer
Dec 8 '18 at 13:48
$begingroup$
Yes sorry! In my head I was thinking about the wrong theorem showing that the quotient of two lie groups forming a bundle with the quotient being the fibres! I'll amend the question!
$endgroup$
– Bunneh
Dec 9 '18 at 12:31