Relation of complex line bundles and circle bundle
What is the relation of complex line bundles to circle bundles?
differential-geometry
asked Nov 20 at 16:56
Mustafa
12
migrated from mathoverflow.net Nov 20 at 17:45
This question came from our site for professional mathematicians.
add a comment |
What is the relation of complex line bundles to circle bundles?
differential-geometry
asked Nov 20 at 16:56
Mustafa
12
migrated from mathoverflow.net Nov 20 at 17:45
This question came from our site for professional mathematicians.
I am pretty sure that you should add more details. Anyway, the important relation is between complex line bundle and $S^2$-bundle. Indeed you get the second one via the one-point compactification of the first one.
– Vincenzo Zaccaro
Nov 20 at 17:12
1
Hi, welcome to MO. This is very broad, and presumably have many different answers. Try and add more details.
– Amir Sagiv
Nov 20 at 17:18
$S^1 = U(1)$ and Rank=n complex vector bundles with Hermitian metric <==> $U(n)$-principle bundles.
– Jake
Nov 21 at 0:45
add a comment |
What is the relation of complex line bundles to circle bundles?
differential-geometry
asked Nov 20 at 16:56
Mustafa
12
What is the relation of complex line bundles to circle bundles?
differential-geometry
differential-geometry
asked Nov 20 at 16:56
Mustafa
12
asked Nov 20 at 16:56
Mustafa
12
asked Nov 20 at 16:56
Mustafa
12
asked Nov 20 at 16:56
Mustafa
12
asked Nov 20 at 16:56
Mustafa
12
12
migrated from mathoverflow.net Nov 20 at 17:45
This question came from our site for professional mathematicians.
migrated from mathoverflow.net Nov 20 at 17:45
This question came from our site for professional mathematicians.
I am pretty sure that you should add more details. Anyway, the important relation is between complex line bundle and $S^2$-bundle. Indeed you get the second one via the one-point compactification of the first one.
– Vincenzo Zaccaro
Nov 20 at 17:12
1
Hi, welcome to MO. This is very broad, and presumably have many different answers. Try and add more details.
– Amir Sagiv
Nov 20 at 17:18
$S^1 = U(1)$ and Rank=n complex vector bundles with Hermitian metric <==> $U(n)$-principle bundles.
– Jake
Nov 21 at 0:45
add a comment |
I am pretty sure that you should add more details. Anyway, the important relation is between complex line bundle and $S^2$-bundle. Indeed you get the second one via the one-point compactification of the first one.
– Vincenzo Zaccaro
Nov 20 at 17:12
1
Hi, welcome to MO. This is very broad, and presumably have many different answers. Try and add more details.
– Amir Sagiv
Nov 20 at 17:18
$S^1 = U(1)$ and Rank=n complex vector bundles with Hermitian metric <==> $U(n)$-principle bundles.
– Jake
Nov 21 at 0:45
I am pretty sure that you should add more details. Anyway, the important relation is between complex line bundle and $S^2$-bundle. Indeed you get the second one via the one-point compactification of the first one.
– Vincenzo Zaccaro
Nov 20 at 17:12
I am pretty sure that you should add more details. Anyway, the important relation is between complex line bundle and $S^2$-bundle. Indeed you get the second one via the one-point compactification of the first one.
– Vincenzo Zaccaro
Nov 20 at 17:12
1
1
Hi, welcome to MO. This is very broad, and presumably have many different answers. Try and add more details.
– Amir Sagiv
Nov 20 at 17:18
Hi, welcome to MO. This is very broad, and presumably have many different answers. Try and add more details.
– Amir Sagiv
Nov 20 at 17:18
$S^1 = U(1)$ and Rank=n complex vector bundles with Hermitian metric <==> $U(n)$-principle bundles.
– Jake
Nov 21 at 0:45
$S^1 = U(1)$ and Rank=n complex vector bundles with Hermitian metric <==> $U(n)$-principle bundles.
– Jake
Nov 21 at 0:45
add a comment |
2 Answers
2
active
oldest
votes
A complex vector bundle of rank $n$ is equivalent to principal $GL_n(mathbb C)$ bundle. Since $S^1 hookrightarrow mathbb C^times$ is a homotopy equivalence, complex line bundles are equivalent to principal $S^1$-bundles. In this way you can build a circle bundle from a complex line bundle.
answered Nov 20 at 18:17
Ben
2,488616
add a comment |
Given a complex line bundle, picking any Hermitian metric on it, the set of unit length vectors, as measured in the Hermitian metric, is a circle bundle, principal as it is acted on by the unit length complex numbers.
Given any principal circle bundle $B$, the associated bundle $B times^{rho} mathbb{C}$ (defined as the quotient of $B times mathbb{C}$ by the action $s(b,c)=(sb,sc)$ for $s in S^1$, $b in B$ and $c in mathbb{C}$) associated to any representation $rho colon S^1 to mathbb{C}^{times}=GL(1,mathbb{C})$ is a complex line bundle.
answered Nov 20 at 18:50
Ben McKay
1,009512
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
A complex vector bundle of rank $n$ is equivalent to principal $GL_n(mathbb C)$ bundle. Since $S^1 hookrightarrow mathbb C^times$ is a homotopy equivalence, complex line bundles are equivalent to principal $S^1$-bundles. In this way you can build a circle bundle from a complex line bundle.
answered Nov 20 at 18:17
Ben
2,488616
add a comment |
A complex vector bundle of rank $n$ is equivalent to principal $GL_n(mathbb C)$ bundle. Since $S^1 hookrightarrow mathbb C^times$ is a homotopy equivalence, complex line bundles are equivalent to principal $S^1$-bundles. In this way you can build a circle bundle from a complex line bundle.
answered Nov 20 at 18:17
Ben
2,488616
add a comment |
A complex vector bundle of rank $n$ is equivalent to principal $GL_n(mathbb C)$ bundle. Since $S^1 hookrightarrow mathbb C^times$ is a homotopy equivalence, complex line bundles are equivalent to principal $S^1$-bundles. In this way you can build a circle bundle from a complex line bundle.
answered Nov 20 at 18:17
Ben
2,488616
A complex vector bundle of rank $n$ is equivalent to principal $GL_n(mathbb C)$ bundle. Since $S^1 hookrightarrow mathbb C^times$ is a homotopy equivalence, complex line bundles are equivalent to principal $S^1$-bundles. In this way you can build a circle bundle from a complex line bundle.
answered Nov 20 at 18:17
Ben
2,488616
answered Nov 20 at 18:17
Ben
2,488616
answered Nov 20 at 18:17
Ben
2,488616
answered Nov 20 at 18:17
Ben
2,488616
2,488616
add a comment |
add a comment |
Given a complex line bundle, picking any Hermitian metric on it, the set of unit length vectors, as measured in the Hermitian metric, is a circle bundle, principal as it is acted on by the unit length complex numbers.
Given any principal circle bundle $B$, the associated bundle $B times^{rho} mathbb{C}$ (defined as the quotient of $B times mathbb{C}$ by the action $s(b,c)=(sb,sc)$ for $s in S^1$, $b in B$ and $c in mathbb{C}$) associated to any representation $rho colon S^1 to mathbb{C}^{times}=GL(1,mathbb{C})$ is a complex line bundle.
answered Nov 20 at 18:50
Ben McKay
1,009512
add a comment |
Given a complex line bundle, picking any Hermitian metric on it, the set of unit length vectors, as measured in the Hermitian metric, is a circle bundle, principal as it is acted on by the unit length complex numbers.
Given any principal circle bundle $B$, the associated bundle $B times^{rho} mathbb{C}$ (defined as the quotient of $B times mathbb{C}$ by the action $s(b,c)=(sb,sc)$ for $s in S^1$, $b in B$ and $c in mathbb{C}$) associated to any representation $rho colon S^1 to mathbb{C}^{times}=GL(1,mathbb{C})$ is a complex line bundle.
answered Nov 20 at 18:50
Ben McKay
1,009512
add a comment |
Given a complex line bundle, picking any Hermitian metric on it, the set of unit length vectors, as measured in the Hermitian metric, is a circle bundle, principal as it is acted on by the unit length complex numbers.
Given any principal circle bundle $B$, the associated bundle $B times^{rho} mathbb{C}$ (defined as the quotient of $B times mathbb{C}$ by the action $s(b,c)=(sb,sc)$ for $s in S^1$, $b in B$ and $c in mathbb{C}$) associated to any representation $rho colon S^1 to mathbb{C}^{times}=GL(1,mathbb{C})$ is a complex line bundle.
answered Nov 20 at 18:50
Ben McKay
1,009512
Given a complex line bundle, picking any Hermitian metric on it, the set of unit length vectors, as measured in the Hermitian metric, is a circle bundle, principal as it is acted on by the unit length complex numbers.
Given any principal circle bundle $B$, the associated bundle $B times^{rho} mathbb{C}$ (defined as the quotient of $B times mathbb{C}$ by the action $s(b,c)=(sb,sc)$ for $s in S^1$, $b in B$ and $c in mathbb{C}$) associated to any representation $rho colon S^1 to mathbb{C}^{times}=GL(1,mathbb{C})$ is a complex line bundle.
answered Nov 20 at 18:50
Ben McKay
1,009512
answered Nov 20 at 18:50
Ben McKay
1,009512
answered Nov 20 at 18:50
Ben McKay
1,009512
answered Nov 20 at 18:50
Ben McKay
1,009512
1,009512
add a comment |
add a comment |
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I am pretty sure that you should add more details. Anyway, the important relation is between complex line bundle and $S^2$-bundle. Indeed you get the second one via the one-point compactification of the first one.
– Vincenzo Zaccaro
Nov 20 at 17:12
1
Hi, welcome to MO. This is very broad, and presumably have many different answers. Try and add more details.
– Amir Sagiv
Nov 20 at 17:18
$S^1 = U(1)$ and Rank=n complex vector bundles with Hermitian metric <==> $U(n)$-principle bundles.
– Jake
Nov 21 at 0:45