Mixing
Metric on total space, connection and Hopf fibration
$begingroup$
As is well known that the three sphere $S^3$ may be viewed as a $S^1$ bundle over base space $S^2$. And it is more interesting, but likewise confusing to me that the metric on $S^3$ may be written as:
begin{equation}
ds^2 = (dpsi - cos theta dphi)^2 + dtheta^2 + sin^2 theta dphi^2
end{equation}
here $cos theta dphi$ is exactly the Ricci form connection on the base space $S^2$. It seems here that the non-triviality of the fibre bundle can be seen from the metric, constrasting to the idea that the bundle is always locally trivial.
Thus my question is (1) given a specific bundle how to construct the metric on the total space, and how the connection of the base might come into play? (2) given a specific metric, how could we tell if the underlying manifold exhibits a bundle structure? It would be very illustrative if examples may be given in the meantime (for example the Hopf fibration above).
It would also be great if reference on the related topic can be provided. I have been searching online, but when come to bundles most reference tends to fall into topological discussions.
metric-spaces fiber-bundles connections hopf-fibration
asked Sep 1 '14 at 17:41
Kevin YeKevin Ye
42829
$endgroup$
add a comment |
$begingroup$
As is well known that the three sphere $S^3$ may be viewed as a $S^1$ bundle over base space $S^2$. And it is more interesting, but likewise confusing to me that the metric on $S^3$ may be written as:
begin{equation}
ds^2 = (dpsi - cos theta dphi)^2 + dtheta^2 + sin^2 theta dphi^2
end{equation}
here $cos theta dphi$ is exactly the Ricci form connection on the base space $S^2$. It seems here that the non-triviality of the fibre bundle can be seen from the metric, constrasting to the idea that the bundle is always locally trivial.
Thus my question is (1) given a specific bundle how to construct the metric on the total space, and how the connection of the base might come into play? (2) given a specific metric, how could we tell if the underlying manifold exhibits a bundle structure? It would be very illustrative if examples may be given in the meantime (for example the Hopf fibration above).
It would also be great if reference on the related topic can be provided. I have been searching online, but when come to bundles most reference tends to fall into topological discussions.
metric-spaces fiber-bundles connections hopf-fibration
asked Sep 1 '14 at 17:41
Kevin YeKevin Ye
42829
$endgroup$
$begingroup$
I actually doubt this can be done. If there is any method of seeing the non-triviality from the metric, one has to also see the triviality of $ds^2=(dpsi -coshrho dphi)^2+drho^2+sinh^2rho dphi^2$, but this is far from being clear.
$endgroup$
– user110373
Aug 3 '17 at 19:43
add a comment |
$begingroup$
As is well known that the three sphere $S^3$ may be viewed as a $S^1$ bundle over base space $S^2$. And it is more interesting, but likewise confusing to me that the metric on $S^3$ may be written as:
begin{equation}
ds^2 = (dpsi - cos theta dphi)^2 + dtheta^2 + sin^2 theta dphi^2
end{equation}
here $cos theta dphi$ is exactly the Ricci form connection on the base space $S^2$. It seems here that the non-triviality of the fibre bundle can be seen from the metric, constrasting to the idea that the bundle is always locally trivial.
Thus my question is (1) given a specific bundle how to construct the metric on the total space, and how the connection of the base might come into play? (2) given a specific metric, how could we tell if the underlying manifold exhibits a bundle structure? It would be very illustrative if examples may be given in the meantime (for example the Hopf fibration above).
It would also be great if reference on the related topic can be provided. I have been searching online, but when come to bundles most reference tends to fall into topological discussions.
metric-spaces fiber-bundles connections hopf-fibration
asked Sep 1 '14 at 17:41
Kevin YeKevin Ye
42829
$endgroup$
As is well known that the three sphere $S^3$ may be viewed as a $S^1$ bundle over base space $S^2$. And it is more interesting, but likewise confusing to me that the metric on $S^3$ may be written as:
begin{equation}
ds^2 = (dpsi - cos theta dphi)^2 + dtheta^2 + sin^2 theta dphi^2
end{equation}
here $cos theta dphi$ is exactly the Ricci form connection on the base space $S^2$. It seems here that the non-triviality of the fibre bundle can be seen from the metric, constrasting to the idea that the bundle is always locally trivial.
Thus my question is (1) given a specific bundle how to construct the metric on the total space, and how the connection of the base might come into play? (2) given a specific metric, how could we tell if the underlying manifold exhibits a bundle structure? It would be very illustrative if examples may be given in the meantime (for example the Hopf fibration above).
It would also be great if reference on the related topic can be provided. I have been searching online, but when come to bundles most reference tends to fall into topological discussions.
metric-spaces fiber-bundles connections hopf-fibration
metric-spaces fiber-bundles connections hopf-fibration
asked Sep 1 '14 at 17:41
Kevin YeKevin Ye
42829
asked Sep 1 '14 at 17:41
Kevin YeKevin Ye
42829
asked Sep 1 '14 at 17:41
Kevin YeKevin Ye
42829
asked Sep 1 '14 at 17:41
Kevin YeKevin Ye
42829
asked Sep 1 '14 at 17:41
Kevin YeKevin Ye
42829
42829
$begingroup$
I actually doubt this can be done. If there is any method of seeing the non-triviality from the metric, one has to also see the triviality of $ds^2=(dpsi -coshrho dphi)^2+drho^2+sinh^2rho dphi^2$, but this is far from being clear.
$endgroup$
– user110373
Aug 3 '17 at 19:43
add a comment |
$begingroup$
I actually doubt this can be done. If there is any method of seeing the non-triviality from the metric, one has to also see the triviality of $ds^2=(dpsi -coshrho dphi)^2+drho^2+sinh^2rho dphi^2$, but this is far from being clear.
$endgroup$
– user110373
Aug 3 '17 at 19:43
$begingroup$
I actually doubt this can be done. If there is any method of seeing the non-triviality from the metric, one has to also see the triviality of $ds^2=(dpsi -coshrho dphi)^2+drho^2+sinh^2rho dphi^2$, but this is far from being clear.
$endgroup$
– user110373
Aug 3 '17 at 19:43
$begingroup$
I actually doubt this can be done. If there is any method of seeing the non-triviality from the metric, one has to also see the triviality of $ds^2=(dpsi -coshrho dphi)^2+drho^2+sinh^2rho dphi^2$, but this is far from being clear.
$endgroup$
– user110373
Aug 3 '17 at 19:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No metric related the total space since we consider only configuration of points and line as property of projective geometry.
edited Aug 23 '17 at 1:22
Parcly Taxel
41.8k1372101
answered Aug 22 '17 at 21:07
Mohamed Farah idrisMohamed Farah idris
1
$endgroup$
$begingroup$
You've chosen to respond to an older Question (nearly three years old), so brevity and haste are of no merit. At best you've claimed that neither of the problems posed in the Question can be answered, but it is not a convincing demonstration for either part.
$endgroup$
– hardmath
Aug 22 '17 at 21:51
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
No metric related the total space since we consider only configuration of points and line as property of projective geometry.
edited Aug 23 '17 at 1:22
Parcly Taxel
41.8k1372101
answered Aug 22 '17 at 21:07
Mohamed Farah idrisMohamed Farah idris
1
$endgroup$
$begingroup$
You've chosen to respond to an older Question (nearly three years old), so brevity and haste are of no merit. At best you've claimed that neither of the problems posed in the Question can be answered, but it is not a convincing demonstration for either part.
$endgroup$
– hardmath
Aug 22 '17 at 21:51
add a comment |
$begingroup$
No metric related the total space since we consider only configuration of points and line as property of projective geometry.
edited Aug 23 '17 at 1:22
Parcly Taxel
41.8k1372101
answered Aug 22 '17 at 21:07
Mohamed Farah idrisMohamed Farah idris
1
$endgroup$
$begingroup$
You've chosen to respond to an older Question (nearly three years old), so brevity and haste are of no merit. At best you've claimed that neither of the problems posed in the Question can be answered, but it is not a convincing demonstration for either part.
$endgroup$
– hardmath
Aug 22 '17 at 21:51
add a comment |
$begingroup$
No metric related the total space since we consider only configuration of points and line as property of projective geometry.
edited Aug 23 '17 at 1:22
Parcly Taxel
41.8k1372101
answered Aug 22 '17 at 21:07
Mohamed Farah idrisMohamed Farah idris
1
$endgroup$
No metric related the total space since we consider only configuration of points and line as property of projective geometry.
edited Aug 23 '17 at 1:22
Parcly Taxel
41.8k1372101
answered Aug 22 '17 at 21:07
Mohamed Farah idrisMohamed Farah idris
1
edited Aug 23 '17 at 1:22
Parcly Taxel
41.8k1372101
edited Aug 23 '17 at 1:22
Parcly Taxel
41.8k1372101
edited Aug 23 '17 at 1:22
Parcly Taxel
41.8k1372101
41.8k1372101
answered Aug 22 '17 at 21:07
Mohamed Farah idrisMohamed Farah idris
1
answered Aug 22 '17 at 21:07
Mohamed Farah idrisMohamed Farah idris
1
answered Aug 22 '17 at 21:07
Mohamed Farah idrisMohamed Farah idris
1
1
$begingroup$
You've chosen to respond to an older Question (nearly three years old), so brevity and haste are of no merit. At best you've claimed that neither of the problems posed in the Question can be answered, but it is not a convincing demonstration for either part.
$endgroup$
– hardmath
Aug 22 '17 at 21:51
add a comment |
$begingroup$
You've chosen to respond to an older Question (nearly three years old), so brevity and haste are of no merit. At best you've claimed that neither of the problems posed in the Question can be answered, but it is not a convincing demonstration for either part.
$endgroup$
– hardmath
Aug 22 '17 at 21:51
$begingroup$
You've chosen to respond to an older Question (nearly three years old), so brevity and haste are of no merit. At best you've claimed that neither of the problems posed in the Question can be answered, but it is not a convincing demonstration for either part.
$endgroup$
– hardmath
Aug 22 '17 at 21:51
$begingroup$
You've chosen to respond to an older Question (nearly three years old), so brevity and haste are of no merit. At best you've claimed that neither of the problems posed in the Question can be answered, but it is not a convincing demonstration for either part.
$endgroup$
– hardmath
Aug 22 '17 at 21:51
add a comment |
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$begingroup$
I actually doubt this can be done. If there is any method of seeing the non-triviality from the metric, one has to also see the triviality of $ds^2=(dpsi -coshrho dphi)^2+drho^2+sinh^2rho dphi^2$, but this is far from being clear.
$endgroup$
– user110373
Aug 3 '17 at 19:43