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Non smooth topological vector bundle of rank 3 base 1?
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Does there exist a topological (not necessarily trivial) vector bundle $pi: Eto M$, with $dim(M)=1$ and $dimbig(pi^-1(x)big) =3$, that has no $C^1$ (or $C^2,C^infty$) structure?
I know that every topological manifold up to dimension 3 has a unique smooth structure and there are 4-manifolds with no smooth structure. So the question here is that can such a non-smooth 4-manifold be a rank 3 base 1 vector bundle? I am guessing that the local triviality should forbid the existence of such bundles but I do not know for sure.
$textbfAddition:$ It would also be sufficient to prove that a general topological vector bundle $pi:Eto M$ is automatically a $C^k$(or $C^infty$) bundle if the base manifold $M$ is $C^k$(or $C^infty$). In other words, to prove that it would be possible to construct an atlas of $E$ with bundle charts that are not only continuous but also has the same regularity of the charts of $M$.
Any references are warmly welcome. Thanks.
differential-geometry reference-request differential-topology smooth-manifolds vector-bundles
asked Jul 21 at 22:52
smiley06
2,5001323
 |Â
show 2 more comments
up vote
3
down vote
favorite
Does there exist a topological (not necessarily trivial) vector bundle $pi: Eto M$, with $dim(M)=1$ and $dimbig(pi^-1(x)big) =3$, that has no $C^1$ (or $C^2,C^infty$) structure?
I know that every topological manifold up to dimension 3 has a unique smooth structure and there are 4-manifolds with no smooth structure. So the question here is that can such a non-smooth 4-manifold be a rank 3 base 1 vector bundle? I am guessing that the local triviality should forbid the existence of such bundles but I do not know for sure.
$textbfAddition:$ It would also be sufficient to prove that a general topological vector bundle $pi:Eto M$ is automatically a $C^k$(or $C^infty$) bundle if the base manifold $M$ is $C^k$(or $C^infty$). In other words, to prove that it would be possible to construct an atlas of $E$ with bundle charts that are not only continuous but also has the same regularity of the charts of $M$.
Any references are warmly welcome. Thanks.
differential-geometry reference-request differential-topology smooth-manifolds vector-bundles
asked Jul 21 at 22:52
smiley06
2,5001323
2
The answer is no and it does not depend on dimension of either the base manifold or the vector bundle! (Well, you can achieve the same regularity as on the base manifold; more regularity doesn't make sense.) But every topological vector bundle over a smooth manifold has a unique structure as a smooth vector bundle up to isomorphism.
– Mike Miller
Jul 21 at 23:32
@MikeMiller So this smooth structure is induced from the local triviality, i.e. bundle charts. Is this right?
– smiley06
Jul 22 at 14:38
Yes, but you need to make sure the transition maps are smooth for the resulting vector bundle to naturally inherit the structure of a smooth manifold. This is the slightly delicate bit: you could try smoothing the maps $rho_alpha beta: U_alpha cap U_beta to GL(n)$, but you would probably lose the cocycle condition $rho_alpha beta rho_beta gamma = rho_alpha gamma$.
– Mike Miller
Jul 22 at 14:47
1
The easiest proof of the desired result is probably that on a compact manifold $M$, every rank $n$ vector bundle is given as the pullback of the tautological vector bundle over $textGr_n(Bbb R^n+k)$ for some sufficiently large $k$, which is a smooth manifold. The map $f: M to textGr$ is going to be continuous, but then you use the fact that any continuous map of smooth manifolds is homotopic to a smooth map (by an arbitrarily small homotopy).
– Mike Miller
Jul 22 at 14:47
@MikeMiller Can you give me a reference for the proof?
– smiley06
Jul 22 at 14:53
 |Â
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Does there exist a topological (not necessarily trivial) vector bundle $pi: Eto M$, with $dim(M)=1$ and $dimbig(pi^-1(x)big) =3$, that has no $C^1$ (or $C^2,C^infty$) structure?
I know that every topological manifold up to dimension 3 has a unique smooth structure and there are 4-manifolds with no smooth structure. So the question here is that can such a non-smooth 4-manifold be a rank 3 base 1 vector bundle? I am guessing that the local triviality should forbid the existence of such bundles but I do not know for sure.
$textbfAddition:$ It would also be sufficient to prove that a general topological vector bundle $pi:Eto M$ is automatically a $C^k$(or $C^infty$) bundle if the base manifold $M$ is $C^k$(or $C^infty$). In other words, to prove that it would be possible to construct an atlas of $E$ with bundle charts that are not only continuous but also has the same regularity of the charts of $M$.
Any references are warmly welcome. Thanks.
differential-geometry reference-request differential-topology smooth-manifolds vector-bundles
asked Jul 21 at 22:52
smiley06
2,5001323
Does there exist a topological (not necessarily trivial) vector bundle $pi: Eto M$, with $dim(M)=1$ and $dimbig(pi^-1(x)big) =3$, that has no $C^1$ (or $C^2,C^infty$) structure?
I know that every topological manifold up to dimension 3 has a unique smooth structure and there are 4-manifolds with no smooth structure. So the question here is that can such a non-smooth 4-manifold be a rank 3 base 1 vector bundle? I am guessing that the local triviality should forbid the existence of such bundles but I do not know for sure.
$textbfAddition:$ It would also be sufficient to prove that a general topological vector bundle $pi:Eto M$ is automatically a $C^k$(or $C^infty$) bundle if the base manifold $M$ is $C^k$(or $C^infty$). In other words, to prove that it would be possible to construct an atlas of $E$ with bundle charts that are not only continuous but also has the same regularity of the charts of $M$.
Any references are warmly welcome. Thanks.
differential-geometry reference-request differential-topology smooth-manifolds vector-bundles
asked Jul 21 at 22:52
smiley06
2,5001323
edited Jul 25 at 0:33
asked Jul 21 at 22:52
smiley06
2,5001323
asked Jul 21 at 22:52
smiley06
2,5001323
asked Jul 21 at 22:52
smiley06
2,5001323
2,5001323
2
The answer is no and it does not depend on dimension of either the base manifold or the vector bundle! (Well, you can achieve the same regularity as on the base manifold; more regularity doesn't make sense.) But every topological vector bundle over a smooth manifold has a unique structure as a smooth vector bundle up to isomorphism.
– Mike Miller
Jul 21 at 23:32
@MikeMiller So this smooth structure is induced from the local triviality, i.e. bundle charts. Is this right?
– smiley06
Jul 22 at 14:38
Yes, but you need to make sure the transition maps are smooth for the resulting vector bundle to naturally inherit the structure of a smooth manifold. This is the slightly delicate bit: you could try smoothing the maps $rho_alpha beta: U_alpha cap U_beta to GL(n)$, but you would probably lose the cocycle condition $rho_alpha beta rho_beta gamma = rho_alpha gamma$.
– Mike Miller
Jul 22 at 14:47
1
The easiest proof of the desired result is probably that on a compact manifold $M$, every rank $n$ vector bundle is given as the pullback of the tautological vector bundle over $textGr_n(Bbb R^n+k)$ for some sufficiently large $k$, which is a smooth manifold. The map $f: M to textGr$ is going to be continuous, but then you use the fact that any continuous map of smooth manifolds is homotopic to a smooth map (by an arbitrarily small homotopy).
– Mike Miller
Jul 22 at 14:47
@MikeMiller Can you give me a reference for the proof?
– smiley06
Jul 22 at 14:53
 |Â
show 2 more comments
2
The answer is no and it does not depend on dimension of either the base manifold or the vector bundle! (Well, you can achieve the same regularity as on the base manifold; more regularity doesn't make sense.) But every topological vector bundle over a smooth manifold has a unique structure as a smooth vector bundle up to isomorphism.
– Mike Miller
Jul 21 at 23:32
@MikeMiller So this smooth structure is induced from the local triviality, i.e. bundle charts. Is this right?
– smiley06
Jul 22 at 14:38
Yes, but you need to make sure the transition maps are smooth for the resulting vector bundle to naturally inherit the structure of a smooth manifold. This is the slightly delicate bit: you could try smoothing the maps $rho_alpha beta: U_alpha cap U_beta to GL(n)$, but you would probably lose the cocycle condition $rho_alpha beta rho_beta gamma = rho_alpha gamma$.
– Mike Miller
Jul 22 at 14:47
1
The easiest proof of the desired result is probably that on a compact manifold $M$, every rank $n$ vector bundle is given as the pullback of the tautological vector bundle over $textGr_n(Bbb R^n+k)$ for some sufficiently large $k$, which is a smooth manifold. The map $f: M to textGr$ is going to be continuous, but then you use the fact that any continuous map of smooth manifolds is homotopic to a smooth map (by an arbitrarily small homotopy).
– Mike Miller
Jul 22 at 14:47
@MikeMiller Can you give me a reference for the proof?
– smiley06
Jul 22 at 14:53
2
2
The answer is no and it does not depend on dimension of either the base manifold or the vector bundle! (Well, you can achieve the same regularity as on the base manifold; more regularity doesn't make sense.) But every topological vector bundle over a smooth manifold has a unique structure as a smooth vector bundle up to isomorphism.
– Mike Miller
Jul 21 at 23:32
The answer is no and it does not depend on dimension of either the base manifold or the vector bundle! (Well, you can achieve the same regularity as on the base manifold; more regularity doesn't make sense.) But every topological vector bundle over a smooth manifold has a unique structure as a smooth vector bundle up to isomorphism.
– Mike Miller
Jul 21 at 23:32
@MikeMiller So this smooth structure is induced from the local triviality, i.e. bundle charts. Is this right?
– smiley06
Jul 22 at 14:38
@MikeMiller So this smooth structure is induced from the local triviality, i.e. bundle charts. Is this right?
– smiley06
Jul 22 at 14:38
Yes, but you need to make sure the transition maps are smooth for the resulting vector bundle to naturally inherit the structure of a smooth manifold. This is the slightly delicate bit: you could try smoothing the maps $rho_alpha beta: U_alpha cap U_beta to GL(n)$, but you would probably lose the cocycle condition $rho_alpha beta rho_beta gamma = rho_alpha gamma$.
– Mike Miller
Jul 22 at 14:47
Yes, but you need to make sure the transition maps are smooth for the resulting vector bundle to naturally inherit the structure of a smooth manifold. This is the slightly delicate bit: you could try smoothing the maps $rho_alpha beta: U_alpha cap U_beta to GL(n)$, but you would probably lose the cocycle condition $rho_alpha beta rho_beta gamma = rho_alpha gamma$.
– Mike Miller
Jul 22 at 14:47
1
1
The easiest proof of the desired result is probably that on a compact manifold $M$, every rank $n$ vector bundle is given as the pullback of the tautological vector bundle over $textGr_n(Bbb R^n+k)$ for some sufficiently large $k$, which is a smooth manifold. The map $f: M to textGr$ is going to be continuous, but then you use the fact that any continuous map of smooth manifolds is homotopic to a smooth map (by an arbitrarily small homotopy).
– Mike Miller
Jul 22 at 14:47
The easiest proof of the desired result is probably that on a compact manifold $M$, every rank $n$ vector bundle is given as the pullback of the tautological vector bundle over $textGr_n(Bbb R^n+k)$ for some sufficiently large $k$, which is a smooth manifold. The map $f: M to textGr$ is going to be continuous, but then you use the fact that any continuous map of smooth manifolds is homotopic to a smooth map (by an arbitrarily small homotopy).
– Mike Miller
Jul 22 at 14:47
@MikeMiller Can you give me a reference for the proof?
– smiley06
Jul 22 at 14:53
@MikeMiller Can you give me a reference for the proof?
– smiley06
Jul 22 at 14:53
 |Â
show 2 more comments
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2
The answer is no and it does not depend on dimension of either the base manifold or the vector bundle! (Well, you can achieve the same regularity as on the base manifold; more regularity doesn't make sense.) But every topological vector bundle over a smooth manifold has a unique structure as a smooth vector bundle up to isomorphism.
– Mike Miller
Jul 21 at 23:32
@MikeMiller So this smooth structure is induced from the local triviality, i.e. bundle charts. Is this right?
– smiley06
Jul 22 at 14:38
Yes, but you need to make sure the transition maps are smooth for the resulting vector bundle to naturally inherit the structure of a smooth manifold. This is the slightly delicate bit: you could try smoothing the maps $rho_alpha beta: U_alpha cap U_beta to GL(n)$, but you would probably lose the cocycle condition $rho_alpha beta rho_beta gamma = rho_alpha gamma$.
– Mike Miller
Jul 22 at 14:47
1
The easiest proof of the desired result is probably that on a compact manifold $M$, every rank $n$ vector bundle is given as the pullback of the tautological vector bundle over $textGr_n(Bbb R^n+k)$ for some sufficiently large $k$, which is a smooth manifold. The map $f: M to textGr$ is going to be continuous, but then you use the fact that any continuous map of smooth manifolds is homotopic to a smooth map (by an arbitrarily small homotopy).
– Mike Miller
Jul 22 at 14:47
@MikeMiller Can you give me a reference for the proof?
– smiley06
Jul 22 at 14:53