Mixing
Redundancy in the definition of vector bundles?
$begingroup$
In John Lee's classic Introduction to Smooth Manifolds, the following definition of vector bundle is given.
Definition. Let $M$ be a topological space. A (real) vector bundle of rank $k$ over $M$ is a topological space $E$ together with a surjective continuous map $pi:Eto M$ satisfying the following conditions:
(i) For each $pin M$, the fiber $E_p=pi^{-1}(p)$ over $p$ is endowed with the structure of a $k$-dimensional real vector space.
(ii) For each $pin M$, there exist a neighbourhood $U$ of $p$ in $M$ and a homeomorphism $Phi:pi^{-1}(U)to UtimesBbb{R}^k$ (called a local trivialization of $E$ over $U$*), satisfying the following conditions:
$pi_UcircPhi=pi$ (where $pi_U:UtimesBbb{R}^kto U$ is the projection);
for each $qin U$, the restriction of $Phi$ to $E_q$ is a vector space isomorphism from $E_q$ to ${q}timesBbb{R}^kcongBbb{R}^k$.
But if we skip conditions (i) and 2, can't we just define the vector space structure on $E_p$ by using its set-theoric bijection with ${p}timesBbb{R}^k$?
In other words:
Question: Let $E$ and $M$ be topological spaces and $pi:Eto M$ a continuous map such that for each $pin M$ there exist a neighbourhood $U$ of $p$ in $M$ and a homeomorphism $Phi:pi^{-1}(U)to UtimesBbb{R}^k$ such that $pi_UcircPhi=pi$.
Is $E$ is vector bundle?
general-topology differential-geometry manifolds smooth-manifolds vector-bundles
asked Jan 3 '16 at 13:36
user302412user302412
211
$endgroup$
add a comment |
$begingroup$
In John Lee's classic Introduction to Smooth Manifolds, the following definition of vector bundle is given.
Definition. Let $M$ be a topological space. A (real) vector bundle of rank $k$ over $M$ is a topological space $E$ together with a surjective continuous map $pi:Eto M$ satisfying the following conditions:
(i) For each $pin M$, the fiber $E_p=pi^{-1}(p)$ over $p$ is endowed with the structure of a $k$-dimensional real vector space.
(ii) For each $pin M$, there exist a neighbourhood $U$ of $p$ in $M$ and a homeomorphism $Phi:pi^{-1}(U)to UtimesBbb{R}^k$ (called a local trivialization of $E$ over $U$*), satisfying the following conditions:
$pi_UcircPhi=pi$ (where $pi_U:UtimesBbb{R}^kto U$ is the projection);
for each $qin U$, the restriction of $Phi$ to $E_q$ is a vector space isomorphism from $E_q$ to ${q}timesBbb{R}^kcongBbb{R}^k$.
But if we skip conditions (i) and 2, can't we just define the vector space structure on $E_p$ by using its set-theoric bijection with ${p}timesBbb{R}^k$?
In other words:
Question: Let $E$ and $M$ be topological spaces and $pi:Eto M$ a continuous map such that for each $pin M$ there exist a neighbourhood $U$ of $p$ in $M$ and a homeomorphism $Phi:pi^{-1}(U)to UtimesBbb{R}^k$ such that $pi_UcircPhi=pi$.
Is $E$ is vector bundle?
general-topology differential-geometry manifolds smooth-manifolds vector-bundles
asked Jan 3 '16 at 13:36
user302412user302412
211
$endgroup$
3
$begingroup$
The problem is you don't know whether all trivializations will induce the same vector space structure on the fiber. For this to work, you'd have to use trivializations whose change of coordinates is linear on fibers.
$endgroup$
– Pedro
Jan 3 '16 at 13:59
add a comment |
$begingroup$
In John Lee's classic Introduction to Smooth Manifolds, the following definition of vector bundle is given.
Definition. Let $M$ be a topological space. A (real) vector bundle of rank $k$ over $M$ is a topological space $E$ together with a surjective continuous map $pi:Eto M$ satisfying the following conditions:
(i) For each $pin M$, the fiber $E_p=pi^{-1}(p)$ over $p$ is endowed with the structure of a $k$-dimensional real vector space.
(ii) For each $pin M$, there exist a neighbourhood $U$ of $p$ in $M$ and a homeomorphism $Phi:pi^{-1}(U)to UtimesBbb{R}^k$ (called a local trivialization of $E$ over $U$*), satisfying the following conditions:
$pi_UcircPhi=pi$ (where $pi_U:UtimesBbb{R}^kto U$ is the projection);
for each $qin U$, the restriction of $Phi$ to $E_q$ is a vector space isomorphism from $E_q$ to ${q}timesBbb{R}^kcongBbb{R}^k$.
But if we skip conditions (i) and 2, can't we just define the vector space structure on $E_p$ by using its set-theoric bijection with ${p}timesBbb{R}^k$?
In other words:
Question: Let $E$ and $M$ be topological spaces and $pi:Eto M$ a continuous map such that for each $pin M$ there exist a neighbourhood $U$ of $p$ in $M$ and a homeomorphism $Phi:pi^{-1}(U)to UtimesBbb{R}^k$ such that $pi_UcircPhi=pi$.
Is $E$ is vector bundle?
general-topology differential-geometry manifolds smooth-manifolds vector-bundles
asked Jan 3 '16 at 13:36
user302412user302412
211
$endgroup$
In John Lee's classic Introduction to Smooth Manifolds, the following definition of vector bundle is given.
Definition. Let $M$ be a topological space. A (real) vector bundle of rank $k$ over $M$ is a topological space $E$ together with a surjective continuous map $pi:Eto M$ satisfying the following conditions:
(i) For each $pin M$, the fiber $E_p=pi^{-1}(p)$ over $p$ is endowed with the structure of a $k$-dimensional real vector space.
(ii) For each $pin M$, there exist a neighbourhood $U$ of $p$ in $M$ and a homeomorphism $Phi:pi^{-1}(U)to UtimesBbb{R}^k$ (called a local trivialization of $E$ over $U$*), satisfying the following conditions:
$pi_UcircPhi=pi$ (where $pi_U:UtimesBbb{R}^kto U$ is the projection);
for each $qin U$, the restriction of $Phi$ to $E_q$ is a vector space isomorphism from $E_q$ to ${q}timesBbb{R}^kcongBbb{R}^k$.
But if we skip conditions (i) and 2, can't we just define the vector space structure on $E_p$ by using its set-theoric bijection with ${p}timesBbb{R}^k$?
In other words:
Question: Let $E$ and $M$ be topological spaces and $pi:Eto M$ a continuous map such that for each $pin M$ there exist a neighbourhood $U$ of $p$ in $M$ and a homeomorphism $Phi:pi^{-1}(U)to UtimesBbb{R}^k$ such that $pi_UcircPhi=pi$.
Is $E$ is vector bundle?
general-topology differential-geometry manifolds smooth-manifolds vector-bundles
general-topology differential-geometry manifolds smooth-manifolds vector-bundles
asked Jan 3 '16 at 13:36
user302412user302412
211
asked Jan 3 '16 at 13:36
user302412user302412
211
asked Jan 3 '16 at 13:36
user302412user302412
211
asked Jan 3 '16 at 13:36
user302412user302412
211
asked Jan 3 '16 at 13:36
user302412user302412
211
211
3
$begingroup$
The problem is you don't know whether all trivializations will induce the same vector space structure on the fiber. For this to work, you'd have to use trivializations whose change of coordinates is linear on fibers.
$endgroup$
– Pedro
Jan 3 '16 at 13:59
add a comment |
3
$begingroup$
The problem is you don't know whether all trivializations will induce the same vector space structure on the fiber. For this to work, you'd have to use trivializations whose change of coordinates is linear on fibers.
$endgroup$
– Pedro
Jan 3 '16 at 13:59
3
3
$begingroup$
The problem is you don't know whether all trivializations will induce the same vector space structure on the fiber. For this to work, you'd have to use trivializations whose change of coordinates is linear on fibers.
$endgroup$
– Pedro
Jan 3 '16 at 13:59
$begingroup$
The problem is you don't know whether all trivializations will induce the same vector space structure on the fiber. For this to work, you'd have to use trivializations whose change of coordinates is linear on fibers.
$endgroup$
– Pedro
Jan 3 '16 at 13:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No Look at $[0,1]times mathbb{R}$, now identify ${0} times mathbb{R}$ and ${1} times mathbb{R}$ via the map
$$f:{0} times mathbb{R}rightarrow {1} times mathbb{R}$$
$$f(0,x)=(1,x^3)$$
a non linear map. The base space is then $S^1$.
Basically in a vector bundle the maps between the fibres must be linear.
answered Jan 3 '16 at 14:02
Rene SchipperusRene Schipperus
32.4k11960
$endgroup$
add a comment |
$begingroup$
As already pointed out, your definition is incomplete in that it lacks compatibility conditions among the different trivializations. The original definition guarantees this compatibility using a god-given vector space structure on the fibers, but it's true that instead you could take your new definition and add the following condition:
Given $U$ and $V$ admitting a trivialization, the map
$$Phi_U circ Phi_V^{-1}: (U cap V) times mathbb{R}^k to (U cap V) times mathbb{R}^k$$
is given by
$$(x,w) mapsto (x,g_{UV}(x)(w))$$
where $g_{UV}: U cap V to GL_n(mathbb{R})$ is smooth.
P.S.: I am not completely sure on how to prove that the original definition implies the smoothness of the $g_{UV}$... I hope that somebody will point it out. It should be the only point missing to show the equivalence of the two definitions.
Edit: As pointed out by Karl Kronenfeld in the comment, if we define smooth functions
$$ f_i: Ucap V to (U cap V) times mathbb{R}^k, x mapsto (x, e_i),$$
$$ g_j: (Ucap V)times mathbb{R}^k to mathbb{R}, (x,v) mapsto v_j,$$
then we obtain $g_{UV}^{ij}$ as the composition $g_j circ Phi_U circ Phi_V^{-1}circ f_i$.
answered Jan 28 at 15:05
57Jimmy57Jimmy
3,470422
$endgroup$
$begingroup$
As for smoothness of $g_{UV}$: the $i$th coordinate projection $(Ucap V)times Bbb R^ktoBbb R$ and the map $(Ucap V)to(Ucap V)timesBbb R^k$ sending $xmapsto (x,e_j)$ are smooth, so the $ij$th component of the matrix $g_{UV}(x)$ is smooth by composition.
$endgroup$
– Karl Kronenfeld
Jan 28 at 15:30
$begingroup$
Thanks, I've edited my answer to include this! :)
$endgroup$
– 57Jimmy
Jan 28 at 17:33
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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votes
$begingroup$
No Look at $[0,1]times mathbb{R}$, now identify ${0} times mathbb{R}$ and ${1} times mathbb{R}$ via the map
$$f:{0} times mathbb{R}rightarrow {1} times mathbb{R}$$
$$f(0,x)=(1,x^3)$$
a non linear map. The base space is then $S^1$.
Basically in a vector bundle the maps between the fibres must be linear.
answered Jan 3 '16 at 14:02
Rene SchipperusRene Schipperus
32.4k11960
$endgroup$
add a comment |
$begingroup$
No Look at $[0,1]times mathbb{R}$, now identify ${0} times mathbb{R}$ and ${1} times mathbb{R}$ via the map
$$f:{0} times mathbb{R}rightarrow {1} times mathbb{R}$$
$$f(0,x)=(1,x^3)$$
a non linear map. The base space is then $S^1$.
Basically in a vector bundle the maps between the fibres must be linear.
answered Jan 3 '16 at 14:02
Rene SchipperusRene Schipperus
32.4k11960
$endgroup$
add a comment |
$begingroup$
No Look at $[0,1]times mathbb{R}$, now identify ${0} times mathbb{R}$ and ${1} times mathbb{R}$ via the map
$$f:{0} times mathbb{R}rightarrow {1} times mathbb{R}$$
$$f(0,x)=(1,x^3)$$
a non linear map. The base space is then $S^1$.
Basically in a vector bundle the maps between the fibres must be linear.
answered Jan 3 '16 at 14:02
Rene SchipperusRene Schipperus
32.4k11960
$endgroup$
No Look at $[0,1]times mathbb{R}$, now identify ${0} times mathbb{R}$ and ${1} times mathbb{R}$ via the map
$$f:{0} times mathbb{R}rightarrow {1} times mathbb{R}$$
$$f(0,x)=(1,x^3)$$
a non linear map. The base space is then $S^1$.
Basically in a vector bundle the maps between the fibres must be linear.
answered Jan 3 '16 at 14:02
Rene SchipperusRene Schipperus
32.4k11960
answered Jan 3 '16 at 14:02
Rene SchipperusRene Schipperus
32.4k11960
answered Jan 3 '16 at 14:02
Rene SchipperusRene Schipperus
32.4k11960
answered Jan 3 '16 at 14:02
Rene SchipperusRene Schipperus
32.4k11960
32.4k11960
add a comment |
add a comment |
$begingroup$
As already pointed out, your definition is incomplete in that it lacks compatibility conditions among the different trivializations. The original definition guarantees this compatibility using a god-given vector space structure on the fibers, but it's true that instead you could take your new definition and add the following condition:
Given $U$ and $V$ admitting a trivialization, the map
$$Phi_U circ Phi_V^{-1}: (U cap V) times mathbb{R}^k to (U cap V) times mathbb{R}^k$$
is given by
$$(x,w) mapsto (x,g_{UV}(x)(w))$$
where $g_{UV}: U cap V to GL_n(mathbb{R})$ is smooth.
P.S.: I am not completely sure on how to prove that the original definition implies the smoothness of the $g_{UV}$... I hope that somebody will point it out. It should be the only point missing to show the equivalence of the two definitions.
Edit: As pointed out by Karl Kronenfeld in the comment, if we define smooth functions
$$ f_i: Ucap V to (U cap V) times mathbb{R}^k, x mapsto (x, e_i),$$
$$ g_j: (Ucap V)times mathbb{R}^k to mathbb{R}, (x,v) mapsto v_j,$$
then we obtain $g_{UV}^{ij}$ as the composition $g_j circ Phi_U circ Phi_V^{-1}circ f_i$.
answered Jan 28 at 15:05
57Jimmy57Jimmy
3,470422
$endgroup$
$begingroup$
As for smoothness of $g_{UV}$: the $i$th coordinate projection $(Ucap V)times Bbb R^ktoBbb R$ and the map $(Ucap V)to(Ucap V)timesBbb R^k$ sending $xmapsto (x,e_j)$ are smooth, so the $ij$th component of the matrix $g_{UV}(x)$ is smooth by composition.
$endgroup$
– Karl Kronenfeld
Jan 28 at 15:30
$begingroup$
Thanks, I've edited my answer to include this! :)
$endgroup$
– 57Jimmy
Jan 28 at 17:33
add a comment |
$begingroup$
As already pointed out, your definition is incomplete in that it lacks compatibility conditions among the different trivializations. The original definition guarantees this compatibility using a god-given vector space structure on the fibers, but it's true that instead you could take your new definition and add the following condition:
Given $U$ and $V$ admitting a trivialization, the map
$$Phi_U circ Phi_V^{-1}: (U cap V) times mathbb{R}^k to (U cap V) times mathbb{R}^k$$
is given by
$$(x,w) mapsto (x,g_{UV}(x)(w))$$
where $g_{UV}: U cap V to GL_n(mathbb{R})$ is smooth.
P.S.: I am not completely sure on how to prove that the original definition implies the smoothness of the $g_{UV}$... I hope that somebody will point it out. It should be the only point missing to show the equivalence of the two definitions.
Edit: As pointed out by Karl Kronenfeld in the comment, if we define smooth functions
$$ f_i: Ucap V to (U cap V) times mathbb{R}^k, x mapsto (x, e_i),$$
$$ g_j: (Ucap V)times mathbb{R}^k to mathbb{R}, (x,v) mapsto v_j,$$
then we obtain $g_{UV}^{ij}$ as the composition $g_j circ Phi_U circ Phi_V^{-1}circ f_i$.
answered Jan 28 at 15:05
57Jimmy57Jimmy
3,470422
$endgroup$
$begingroup$
As for smoothness of $g_{UV}$: the $i$th coordinate projection $(Ucap V)times Bbb R^ktoBbb R$ and the map $(Ucap V)to(Ucap V)timesBbb R^k$ sending $xmapsto (x,e_j)$ are smooth, so the $ij$th component of the matrix $g_{UV}(x)$ is smooth by composition.
$endgroup$
– Karl Kronenfeld
Jan 28 at 15:30
$begingroup$
Thanks, I've edited my answer to include this! :)
$endgroup$
– 57Jimmy
Jan 28 at 17:33
add a comment |
$begingroup$
As already pointed out, your definition is incomplete in that it lacks compatibility conditions among the different trivializations. The original definition guarantees this compatibility using a god-given vector space structure on the fibers, but it's true that instead you could take your new definition and add the following condition:
Given $U$ and $V$ admitting a trivialization, the map
$$Phi_U circ Phi_V^{-1}: (U cap V) times mathbb{R}^k to (U cap V) times mathbb{R}^k$$
is given by
$$(x,w) mapsto (x,g_{UV}(x)(w))$$
where $g_{UV}: U cap V to GL_n(mathbb{R})$ is smooth.
P.S.: I am not completely sure on how to prove that the original definition implies the smoothness of the $g_{UV}$... I hope that somebody will point it out. It should be the only point missing to show the equivalence of the two definitions.
Edit: As pointed out by Karl Kronenfeld in the comment, if we define smooth functions
$$ f_i: Ucap V to (U cap V) times mathbb{R}^k, x mapsto (x, e_i),$$
$$ g_j: (Ucap V)times mathbb{R}^k to mathbb{R}, (x,v) mapsto v_j,$$
then we obtain $g_{UV}^{ij}$ as the composition $g_j circ Phi_U circ Phi_V^{-1}circ f_i$.
answered Jan 28 at 15:05
57Jimmy57Jimmy
3,470422
$endgroup$
As already pointed out, your definition is incomplete in that it lacks compatibility conditions among the different trivializations. The original definition guarantees this compatibility using a god-given vector space structure on the fibers, but it's true that instead you could take your new definition and add the following condition:
Given $U$ and $V$ admitting a trivialization, the map
$$Phi_U circ Phi_V^{-1}: (U cap V) times mathbb{R}^k to (U cap V) times mathbb{R}^k$$
is given by
$$(x,w) mapsto (x,g_{UV}(x)(w))$$
where $g_{UV}: U cap V to GL_n(mathbb{R})$ is smooth.
P.S.: I am not completely sure on how to prove that the original definition implies the smoothness of the $g_{UV}$... I hope that somebody will point it out. It should be the only point missing to show the equivalence of the two definitions.
Edit: As pointed out by Karl Kronenfeld in the comment, if we define smooth functions
$$ f_i: Ucap V to (U cap V) times mathbb{R}^k, x mapsto (x, e_i),$$
$$ g_j: (Ucap V)times mathbb{R}^k to mathbb{R}, (x,v) mapsto v_j,$$
then we obtain $g_{UV}^{ij}$ as the composition $g_j circ Phi_U circ Phi_V^{-1}circ f_i$.
answered Jan 28 at 15:05
57Jimmy57Jimmy
3,470422
edited Jan 28 at 17:32
answered Jan 28 at 15:05
57Jimmy57Jimmy
3,470422
answered Jan 28 at 15:05
57Jimmy57Jimmy
3,470422
answered Jan 28 at 15:05
57Jimmy57Jimmy
3,470422
3,470422
$begingroup$
As for smoothness of $g_{UV}$: the $i$th coordinate projection $(Ucap V)times Bbb R^ktoBbb R$ and the map $(Ucap V)to(Ucap V)timesBbb R^k$ sending $xmapsto (x,e_j)$ are smooth, so the $ij$th component of the matrix $g_{UV}(x)$ is smooth by composition.
$endgroup$
– Karl Kronenfeld
Jan 28 at 15:30
$begingroup$
Thanks, I've edited my answer to include this! :)
$endgroup$
– 57Jimmy
Jan 28 at 17:33
add a comment |
$begingroup$
As for smoothness of $g_{UV}$: the $i$th coordinate projection $(Ucap V)times Bbb R^ktoBbb R$ and the map $(Ucap V)to(Ucap V)timesBbb R^k$ sending $xmapsto (x,e_j)$ are smooth, so the $ij$th component of the matrix $g_{UV}(x)$ is smooth by composition.
$endgroup$
– Karl Kronenfeld
Jan 28 at 15:30
$begingroup$
Thanks, I've edited my answer to include this! :)
$endgroup$
– 57Jimmy
Jan 28 at 17:33
$begingroup$
As for smoothness of $g_{UV}$: the $i$th coordinate projection $(Ucap V)times Bbb R^ktoBbb R$ and the map $(Ucap V)to(Ucap V)timesBbb R^k$ sending $xmapsto (x,e_j)$ are smooth, so the $ij$th component of the matrix $g_{UV}(x)$ is smooth by composition.
$endgroup$
– Karl Kronenfeld
Jan 28 at 15:30
$begingroup$
As for smoothness of $g_{UV}$: the $i$th coordinate projection $(Ucap V)times Bbb R^ktoBbb R$ and the map $(Ucap V)to(Ucap V)timesBbb R^k$ sending $xmapsto (x,e_j)$ are smooth, so the $ij$th component of the matrix $g_{UV}(x)$ is smooth by composition.
$endgroup$
– Karl Kronenfeld
Jan 28 at 15:30
$begingroup$
Thanks, I've edited my answer to include this! :)
$endgroup$
– 57Jimmy
Jan 28 at 17:33
$begingroup$
Thanks, I've edited my answer to include this! :)
$endgroup$
– 57Jimmy
Jan 28 at 17:33
add a comment |
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The problem is you don't know whether all trivializations will induce the same vector space structure on the fiber. For this to work, you'd have to use trivializations whose change of coordinates is linear on fibers.
$endgroup$
– Pedro
Jan 3 '16 at 13:59