Is there a group whose manifold is a fiber bundle with base is $S_1$ and fiber $mathbb{Z_2}?$
$begingroup$
Let's consider a fiber bundle with base $S_1$ and fiber $mathbb{Z}_2$. I want this manifold to be topologically non-trivial, the edge of the Möbius strip.
How do I know if is it possible to introduce a group structure on such a manifold? So that the manifold would turn into a principle bundle.
group-theory fiber-bundles principal-bundles
$endgroup$
add a comment |
$begingroup$
Let's consider a fiber bundle with base $S_1$ and fiber $mathbb{Z}_2$. I want this manifold to be topologically non-trivial, the edge of the Möbius strip.
How do I know if is it possible to introduce a group structure on such a manifold? So that the manifold would turn into a principle bundle.
group-theory fiber-bundles principal-bundles
$endgroup$
add a comment |
$begingroup$
Let's consider a fiber bundle with base $S_1$ and fiber $mathbb{Z}_2$. I want this manifold to be topologically non-trivial, the edge of the Möbius strip.
How do I know if is it possible to introduce a group structure on such a manifold? So that the manifold would turn into a principle bundle.
group-theory fiber-bundles principal-bundles
$endgroup$
Let's consider a fiber bundle with base $S_1$ and fiber $mathbb{Z}_2$. I want this manifold to be topologically non-trivial, the edge of the Möbius strip.
How do I know if is it possible to introduce a group structure on such a manifold? So that the manifold would turn into a principle bundle.
group-theory fiber-bundles principal-bundles
group-theory fiber-bundles principal-bundles
asked Dec 5 '18 at 21:23
mavzolejmavzolej
47028
47028
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$begingroup$
This manifold, considered independent of its fiber bundle structure, is just $S^1$ again, so it admits the same group structure as $S^1$. You can think of the resulting bundle as the short exact sequence
$$1 to mathbb{Z}_2 to S^1 xrightarrow{x mapsto x^2} S^1 to 1.$$
$endgroup$
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1 Answer
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1 Answer
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active
oldest
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active
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active
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$begingroup$
This manifold, considered independent of its fiber bundle structure, is just $S^1$ again, so it admits the same group structure as $S^1$. You can think of the resulting bundle as the short exact sequence
$$1 to mathbb{Z}_2 to S^1 xrightarrow{x mapsto x^2} S^1 to 1.$$
$endgroup$
add a comment |
$begingroup$
This manifold, considered independent of its fiber bundle structure, is just $S^1$ again, so it admits the same group structure as $S^1$. You can think of the resulting bundle as the short exact sequence
$$1 to mathbb{Z}_2 to S^1 xrightarrow{x mapsto x^2} S^1 to 1.$$
$endgroup$
add a comment |
$begingroup$
This manifold, considered independent of its fiber bundle structure, is just $S^1$ again, so it admits the same group structure as $S^1$. You can think of the resulting bundle as the short exact sequence
$$1 to mathbb{Z}_2 to S^1 xrightarrow{x mapsto x^2} S^1 to 1.$$
$endgroup$
This manifold, considered independent of its fiber bundle structure, is just $S^1$ again, so it admits the same group structure as $S^1$. You can think of the resulting bundle as the short exact sequence
$$1 to mathbb{Z}_2 to S^1 xrightarrow{x mapsto x^2} S^1 to 1.$$
answered Dec 5 '18 at 22:29
Qiaochu YuanQiaochu Yuan
278k32584919
278k32584919
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