Example for a Vector Bundle That is Not of Bounded Geometry
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Here is the definition of vector bundle of bounded geometry from Shubin: Spectral theory of elliptic operators on non-compact manifolds, page 29:
Let $E$ be a complex vector bundle on a manifold $X$.
we shall say that $E$ is a bundle of bounded geometry if it is supplied by an additional structure:
trivializations of $E$ on every canonical coordinate neighbourhood $E$ such that the corresponding matrix transition functions $g_UU'$ on all intersections $E cap U'$ of such neighbourhoods are $C^infty$-bounded, i.e. all their derivatives $partial ^alpha_y g_UU'(y)$ with respect to the canonical coordinates are bounded with bounds $C_alpha$ which do not depend on the chosen pair $U$, $U'$.
The same definition with different wording can be found in Eldering: Normally Hyperbolic Invariant Manifolds: The Noncompact Case.
Can anyone give an example of a vector bundle that is not of bounded geometry?
I have trouble thinking of one, as rescaling a trivialization of my bundle gives another trivialization smaller norm transition functions.
differential-geometry vector-bundles
asked Aug 21 at 12:59
user538044
1758
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Here is the definition of vector bundle of bounded geometry from Shubin: Spectral theory of elliptic operators on non-compact manifolds, page 29:
Let $E$ be a complex vector bundle on a manifold $X$.
we shall say that $E$ is a bundle of bounded geometry if it is supplied by an additional structure:
trivializations of $E$ on every canonical coordinate neighbourhood $E$ such that the corresponding matrix transition functions $g_UU'$ on all intersections $E cap U'$ of such neighbourhoods are $C^infty$-bounded, i.e. all their derivatives $partial ^alpha_y g_UU'(y)$ with respect to the canonical coordinates are bounded with bounds $C_alpha$ which do not depend on the chosen pair $U$, $U'$.
The same definition with different wording can be found in Eldering: Normally Hyperbolic Invariant Manifolds: The Noncompact Case.
Can anyone give an example of a vector bundle that is not of bounded geometry?
I have trouble thinking of one, as rescaling a trivialization of my bundle gives another trivialization smaller norm transition functions.
differential-geometry vector-bundles
asked Aug 21 at 12:59
user538044
1758
1
Interesting question. Consider the trivial complex line bundle on $Bbb R$ given by transition functions $g_n,n+1(x) = e^x$ with respect to the open covering $U_n: ninBbb Z$ of $Bbb R$, with $U_n = (n,n+2)$. Is there a way to give a bounded geometry presentation of this?
– Ted Shifrin
Aug 21 at 17:16
@TedShifrin It is true that the trivialization functions that you write are not bounded. However, of bounded geometry should be a property of the bundle itself, and not of a bundle with given trivializations. (That this is the correct interpretation is clear from other places where this definition has been reproduced) The trivial bundle from your example has other trivializations with bounded transition functions, like all trivial bundles. For any trivial bundle $M times E$ the "obvious" trivializations have the identity as the transition maps, which is bounded.
– user538044
Aug 22 at 8:57
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here is the definition of vector bundle of bounded geometry from Shubin: Spectral theory of elliptic operators on non-compact manifolds, page 29:
Let $E$ be a complex vector bundle on a manifold $X$.
we shall say that $E$ is a bundle of bounded geometry if it is supplied by an additional structure:
trivializations of $E$ on every canonical coordinate neighbourhood $E$ such that the corresponding matrix transition functions $g_UU'$ on all intersections $E cap U'$ of such neighbourhoods are $C^infty$-bounded, i.e. all their derivatives $partial ^alpha_y g_UU'(y)$ with respect to the canonical coordinates are bounded with bounds $C_alpha$ which do not depend on the chosen pair $U$, $U'$.
The same definition with different wording can be found in Eldering: Normally Hyperbolic Invariant Manifolds: The Noncompact Case.
Can anyone give an example of a vector bundle that is not of bounded geometry?
I have trouble thinking of one, as rescaling a trivialization of my bundle gives another trivialization smaller norm transition functions.
differential-geometry vector-bundles
asked Aug 21 at 12:59
user538044
1758
Here is the definition of vector bundle of bounded geometry from Shubin: Spectral theory of elliptic operators on non-compact manifolds, page 29:
Let $E$ be a complex vector bundle on a manifold $X$.
we shall say that $E$ is a bundle of bounded geometry if it is supplied by an additional structure:
trivializations of $E$ on every canonical coordinate neighbourhood $E$ such that the corresponding matrix transition functions $g_UU'$ on all intersections $E cap U'$ of such neighbourhoods are $C^infty$-bounded, i.e. all their derivatives $partial ^alpha_y g_UU'(y)$ with respect to the canonical coordinates are bounded with bounds $C_alpha$ which do not depend on the chosen pair $U$, $U'$.
The same definition with different wording can be found in Eldering: Normally Hyperbolic Invariant Manifolds: The Noncompact Case.
Can anyone give an example of a vector bundle that is not of bounded geometry?
I have trouble thinking of one, as rescaling a trivialization of my bundle gives another trivialization smaller norm transition functions.
differential-geometry vector-bundles
asked Aug 21 at 12:59
user538044
1758
asked Aug 21 at 12:59
user538044
1758
asked Aug 21 at 12:59
user538044
1758
asked Aug 21 at 12:59
user538044
1758
1758
1
Interesting question. Consider the trivial complex line bundle on $Bbb R$ given by transition functions $g_n,n+1(x) = e^x$ with respect to the open covering $U_n: ninBbb Z$ of $Bbb R$, with $U_n = (n,n+2)$. Is there a way to give a bounded geometry presentation of this?
– Ted Shifrin
Aug 21 at 17:16
@TedShifrin It is true that the trivialization functions that you write are not bounded. However, of bounded geometry should be a property of the bundle itself, and not of a bundle with given trivializations. (That this is the correct interpretation is clear from other places where this definition has been reproduced) The trivial bundle from your example has other trivializations with bounded transition functions, like all trivial bundles. For any trivial bundle $M times E$ the "obvious" trivializations have the identity as the transition maps, which is bounded.
– user538044
Aug 22 at 8:57
add a comment |Â
1
Interesting question. Consider the trivial complex line bundle on $Bbb R$ given by transition functions $g_n,n+1(x) = e^x$ with respect to the open covering $U_n: ninBbb Z$ of $Bbb R$, with $U_n = (n,n+2)$. Is there a way to give a bounded geometry presentation of this?
– Ted Shifrin
Aug 21 at 17:16
@TedShifrin It is true that the trivialization functions that you write are not bounded. However, of bounded geometry should be a property of the bundle itself, and not of a bundle with given trivializations. (That this is the correct interpretation is clear from other places where this definition has been reproduced) The trivial bundle from your example has other trivializations with bounded transition functions, like all trivial bundles. For any trivial bundle $M times E$ the "obvious" trivializations have the identity as the transition maps, which is bounded.
– user538044
Aug 22 at 8:57
1
1
Interesting question. Consider the trivial complex line bundle on $Bbb R$ given by transition functions $g_n,n+1(x) = e^x$ with respect to the open covering $U_n: ninBbb Z$ of $Bbb R$, with $U_n = (n,n+2)$. Is there a way to give a bounded geometry presentation of this?
– Ted Shifrin
Aug 21 at 17:16
Interesting question. Consider the trivial complex line bundle on $Bbb R$ given by transition functions $g_n,n+1(x) = e^x$ with respect to the open covering $U_n: ninBbb Z$ of $Bbb R$, with $U_n = (n,n+2)$. Is there a way to give a bounded geometry presentation of this?
– Ted Shifrin
Aug 21 at 17:16
@TedShifrin It is true that the trivialization functions that you write are not bounded. However, of bounded geometry should be a property of the bundle itself, and not of a bundle with given trivializations. (That this is the correct interpretation is clear from other places where this definition has been reproduced) The trivial bundle from your example has other trivializations with bounded transition functions, like all trivial bundles. For any trivial bundle $M times E$ the "obvious" trivializations have the identity as the transition maps, which is bounded.
– user538044
Aug 22 at 8:57
@TedShifrin It is true that the trivialization functions that you write are not bounded. However, of bounded geometry should be a property of the bundle itself, and not of a bundle with given trivializations. (That this is the correct interpretation is clear from other places where this definition has been reproduced) The trivial bundle from your example has other trivializations with bounded transition functions, like all trivial bundles. For any trivial bundle $M times E$ the "obvious" trivializations have the identity as the transition maps, which is bounded.
– user538044
Aug 22 at 8:57
add a comment |Â
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Interesting question. Consider the trivial complex line bundle on $Bbb R$ given by transition functions $g_n,n+1(x) = e^x$ with respect to the open covering $U_n: ninBbb Z$ of $Bbb R$, with $U_n = (n,n+2)$. Is there a way to give a bounded geometry presentation of this?
– Ted Shifrin
Aug 21 at 17:16
@TedShifrin It is true that the trivialization functions that you write are not bounded. However, of bounded geometry should be a property of the bundle itself, and not of a bundle with given trivializations. (That this is the correct interpretation is clear from other places where this definition has been reproduced) The trivial bundle from your example has other trivializations with bounded transition functions, like all trivial bundles. For any trivial bundle $M times E$ the "obvious" trivializations have the identity as the transition maps, which is bounded.
– user538044
Aug 22 at 8:57