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Why is not every vectorbundle trivial and we only need local trivialization?
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Why is should the trivialization only need to be local for vectorbundles? For example the tangent bundle. If the tangent space at a point p of a manifold M is identified with let's say $mathbbR^n$ , why can't we just identify the total space of the vectorbundle with M x $mathbbR^n$ ? My questions could be formulated as, why is not every vectorbundle a trivial one? Since I have the feeling that all the fibers can be identified with $mathbbR^n$ and therefore you don't have only local trivialization, but everywhere.
differential-geometry vector-bundles
asked 3 hours ago
user3397129
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up vote
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Why is should the trivialization only need to be local for vectorbundles? For example the tangent bundle. If the tangent space at a point p of a manifold M is identified with let's say $mathbbR^n$ , why can't we just identify the total space of the vectorbundle with M x $mathbbR^n$ ? My questions could be formulated as, why is not every vectorbundle a trivial one? Since I have the feeling that all the fibers can be identified with $mathbbR^n$ and therefore you don't have only local trivialization, but everywhere.
differential-geometry vector-bundles
asked 3 hours ago
user3397129
1517
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Why is should the trivialization only need to be local for vectorbundles? For example the tangent bundle. If the tangent space at a point p of a manifold M is identified with let's say $mathbbR^n$ , why can't we just identify the total space of the vectorbundle with M x $mathbbR^n$ ? My questions could be formulated as, why is not every vectorbundle a trivial one? Since I have the feeling that all the fibers can be identified with $mathbbR^n$ and therefore you don't have only local trivialization, but everywhere.
differential-geometry vector-bundles
asked 3 hours ago
user3397129
1517
Why is should the trivialization only need to be local for vectorbundles? For example the tangent bundle. If the tangent space at a point p of a manifold M is identified with let's say $mathbbR^n$ , why can't we just identify the total space of the vectorbundle with M x $mathbbR^n$ ? My questions could be formulated as, why is not every vectorbundle a trivial one? Since I have the feeling that all the fibers can be identified with $mathbbR^n$ and therefore you don't have only local trivialization, but everywhere.
differential-geometry vector-bundles
differential-geometry vector-bundles
asked 3 hours ago
user3397129
1517
asked 3 hours ago
user3397129
1517
asked 3 hours ago
user3397129
1517
asked 3 hours ago
user3397129
1517
asked 3 hours ago
user3397129
1517
1517
add a comment |Â
add a comment |Â
1 Answer
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The same reason not all manifolds are copies of $mathbbR^n$ --- just because we're gluing things together to locally preserve structure, that doesn't mean the global result has that same structure. (If it did, then differential topology wouldn't be such an interesting area of study!)
Here are two examples of nontrivializable vector bundles.
Mobius band. This is a 1-dimensional real vector bundle over $mathbbS^1$ and it is not orientable, hence not diffeomorphic to a cylinder (and therefore not trivializable).
Tangent bundle to $mathbbS^2$. By the hairy ball theorem, this has no globally nonzero sections, hence it cannot be trivializable.
answered 3 hours ago
Neal
23.1k23582
When showing locally trivialization formally, should one make use of charts in general?
– user3397129
3 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
The same reason not all manifolds are copies of $mathbbR^n$ --- just because we're gluing things together to locally preserve structure, that doesn't mean the global result has that same structure. (If it did, then differential topology wouldn't be such an interesting area of study!)
Here are two examples of nontrivializable vector bundles.
Mobius band. This is a 1-dimensional real vector bundle over $mathbbS^1$ and it is not orientable, hence not diffeomorphic to a cylinder (and therefore not trivializable).
Tangent bundle to $mathbbS^2$. By the hairy ball theorem, this has no globally nonzero sections, hence it cannot be trivializable.
answered 3 hours ago
Neal
23.1k23582
When showing locally trivialization formally, should one make use of charts in general?
– user3397129
3 hours ago
add a comment |Â
up vote
6
down vote
The same reason not all manifolds are copies of $mathbbR^n$ --- just because we're gluing things together to locally preserve structure, that doesn't mean the global result has that same structure. (If it did, then differential topology wouldn't be such an interesting area of study!)
Here are two examples of nontrivializable vector bundles.
Mobius band. This is a 1-dimensional real vector bundle over $mathbbS^1$ and it is not orientable, hence not diffeomorphic to a cylinder (and therefore not trivializable).
Tangent bundle to $mathbbS^2$. By the hairy ball theorem, this has no globally nonzero sections, hence it cannot be trivializable.
answered 3 hours ago
Neal
23.1k23582
When showing locally trivialization formally, should one make use of charts in general?
– user3397129
3 hours ago
add a comment |Â
up vote
6
down vote
up vote
6
down vote
The same reason not all manifolds are copies of $mathbbR^n$ --- just because we're gluing things together to locally preserve structure, that doesn't mean the global result has that same structure. (If it did, then differential topology wouldn't be such an interesting area of study!)
Here are two examples of nontrivializable vector bundles.
Mobius band. This is a 1-dimensional real vector bundle over $mathbbS^1$ and it is not orientable, hence not diffeomorphic to a cylinder (and therefore not trivializable).
Tangent bundle to $mathbbS^2$. By the hairy ball theorem, this has no globally nonzero sections, hence it cannot be trivializable.
answered 3 hours ago
Neal
23.1k23582
The same reason not all manifolds are copies of $mathbbR^n$ --- just because we're gluing things together to locally preserve structure, that doesn't mean the global result has that same structure. (If it did, then differential topology wouldn't be such an interesting area of study!)
Here are two examples of nontrivializable vector bundles.
Mobius band. This is a 1-dimensional real vector bundle over $mathbbS^1$ and it is not orientable, hence not diffeomorphic to a cylinder (and therefore not trivializable).
Tangent bundle to $mathbbS^2$. By the hairy ball theorem, this has no globally nonzero sections, hence it cannot be trivializable.
answered 3 hours ago
Neal
23.1k23582
answered 3 hours ago
Neal
23.1k23582
answered 3 hours ago
Neal
23.1k23582
answered 3 hours ago
Neal
23.1k23582
23.1k23582
When showing locally trivialization formally, should one make use of charts in general?
– user3397129
3 hours ago
add a comment |Â
When showing locally trivialization formally, should one make use of charts in general?
– user3397129
3 hours ago
When showing locally trivialization formally, should one make use of charts in general?
– user3397129
3 hours ago
When showing locally trivialization formally, should one make use of charts in general?
– user3397129
3 hours ago
add a comment |Â
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