Mixing
The fiber bundle given by a cocycle.
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I'm studing this notes, and I have a question about the contruction in the proposition 23.5, page 185. The statement is:
Let $g_αβ $ be a cocycle subordinates to an open cover $U_α $
of $M$ . There exists a vector bundle $ξ = (À, E, M )$, that admits a
trivialization $Æ_α $ for which the transition functions are the
$g_αβ $.
In the proof, he claims:
$E=cup_alpha in A(U_alphatimesmathbbR^r)Big/sim$ is a differentiable manifold with the relation gives by $(p,v)sim (q,w)$ iff $p=q$ and $exists$ $alpha, beta in A$ such that $g_αβw=v$.
My question is: Why $E$ is a manifold? Does Anyone have some hints about how can I find the charts?
differential-geometry manifolds vector-bundles
asked Aug 6 at 19:49
Leonardo Schultz
1085
add a comment |Â
up vote
2
down vote
favorite
I'm studing this notes, and I have a question about the contruction in the proposition 23.5, page 185. The statement is:
Let $g_αβ $ be a cocycle subordinates to an open cover $U_α $
of $M$ . There exists a vector bundle $ξ = (À, E, M )$, that admits a
trivialization $Æ_α $ for which the transition functions are the
$g_αβ $.
In the proof, he claims:
$E=cup_alpha in A(U_alphatimesmathbbR^r)Big/sim$ is a differentiable manifold with the relation gives by $(p,v)sim (q,w)$ iff $p=q$ and $exists$ $alpha, beta in A$ such that $g_αβw=v$.
My question is: Why $E$ is a manifold? Does Anyone have some hints about how can I find the charts?
differential-geometry manifolds vector-bundles
asked Aug 6 at 19:49
Leonardo Schultz
1085
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm studing this notes, and I have a question about the contruction in the proposition 23.5, page 185. The statement is:
Let $g_αβ $ be a cocycle subordinates to an open cover $U_α $
of $M$ . There exists a vector bundle $ξ = (À, E, M )$, that admits a
trivialization $Æ_α $ for which the transition functions are the
$g_αβ $.
In the proof, he claims:
$E=cup_alpha in A(U_alphatimesmathbbR^r)Big/sim$ is a differentiable manifold with the relation gives by $(p,v)sim (q,w)$ iff $p=q$ and $exists$ $alpha, beta in A$ such that $g_αβw=v$.
My question is: Why $E$ is a manifold? Does Anyone have some hints about how can I find the charts?
differential-geometry manifolds vector-bundles
asked Aug 6 at 19:49
Leonardo Schultz
1085
I'm studing this notes, and I have a question about the contruction in the proposition 23.5, page 185. The statement is:
Let $g_αβ $ be a cocycle subordinates to an open cover $U_α $
of $M$ . There exists a vector bundle $ξ = (À, E, M )$, that admits a
trivialization $Æ_α $ for which the transition functions are the
$g_αβ $.
In the proof, he claims:
$E=cup_alpha in A(U_alphatimesmathbbR^r)Big/sim$ is a differentiable manifold with the relation gives by $(p,v)sim (q,w)$ iff $p=q$ and $exists$ $alpha, beta in A$ such that $g_αβw=v$.
My question is: Why $E$ is a manifold? Does Anyone have some hints about how can I find the charts?
differential-geometry manifolds vector-bundles
asked Aug 6 at 19:49
Leonardo Schultz
1085
asked Aug 6 at 19:49
Leonardo Schultz
1085
asked Aug 6 at 19:49
Leonardo Schultz
1085
asked Aug 6 at 19:49
Leonardo Schultz
1085
1085
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
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accepted
Sure. Let $V^prime=coprod_alpha (alpha times U_alpha times mathbb R^k)/sim$ where $(beta,x, mathbb R^k) sim (alpha,x,g_alphabeta)$ and
$$q:coprod_alpha (alpha times U_alpha times mathbb R^k) to V^prime $$
be the quotient map.
Note that $q$ is a continuous open map and the restriction to each $(alpha times U_alpha times mathbb R^k)$ denoted $q_alpha$ is injective, we have that
$$left(q_alpha(alpha times U_alpha times mathbb R^k),q_alpha^-1right)$$
gives a chart.
details can actually be found in these set of notes actually (same notation is used.)
answered Aug 6 at 20:40
Andres Mejia
15.2k11444
Thanks! Is $q$ the quotient map instead of the projection?
– Leonardo Schultz
Aug 6 at 21:26
@LeonardoSchultz yes of course. Sorry about that!
– Andres Mejia
Aug 6 at 21:31
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Sure. Let $V^prime=coprod_alpha (alpha times U_alpha times mathbb R^k)/sim$ where $(beta,x, mathbb R^k) sim (alpha,x,g_alphabeta)$ and
$$q:coprod_alpha (alpha times U_alpha times mathbb R^k) to V^prime $$
be the quotient map.
Note that $q$ is a continuous open map and the restriction to each $(alpha times U_alpha times mathbb R^k)$ denoted $q_alpha$ is injective, we have that
$$left(q_alpha(alpha times U_alpha times mathbb R^k),q_alpha^-1right)$$
gives a chart.
details can actually be found in these set of notes actually (same notation is used.)
answered Aug 6 at 20:40
Andres Mejia
15.2k11444
Thanks! Is $q$ the quotient map instead of the projection?
– Leonardo Schultz
Aug 6 at 21:26
@LeonardoSchultz yes of course. Sorry about that!
– Andres Mejia
Aug 6 at 21:31
add a comment |Â
up vote
1
down vote
accepted
Sure. Let $V^prime=coprod_alpha (alpha times U_alpha times mathbb R^k)/sim$ where $(beta,x, mathbb R^k) sim (alpha,x,g_alphabeta)$ and
$$q:coprod_alpha (alpha times U_alpha times mathbb R^k) to V^prime $$
be the quotient map.
Note that $q$ is a continuous open map and the restriction to each $(alpha times U_alpha times mathbb R^k)$ denoted $q_alpha$ is injective, we have that
$$left(q_alpha(alpha times U_alpha times mathbb R^k),q_alpha^-1right)$$
gives a chart.
details can actually be found in these set of notes actually (same notation is used.)
answered Aug 6 at 20:40
Andres Mejia
15.2k11444
Thanks! Is $q$ the quotient map instead of the projection?
– Leonardo Schultz
Aug 6 at 21:26
@LeonardoSchultz yes of course. Sorry about that!
– Andres Mejia
Aug 6 at 21:31
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Sure. Let $V^prime=coprod_alpha (alpha times U_alpha times mathbb R^k)/sim$ where $(beta,x, mathbb R^k) sim (alpha,x,g_alphabeta)$ and
$$q:coprod_alpha (alpha times U_alpha times mathbb R^k) to V^prime $$
be the quotient map.
Note that $q$ is a continuous open map and the restriction to each $(alpha times U_alpha times mathbb R^k)$ denoted $q_alpha$ is injective, we have that
$$left(q_alpha(alpha times U_alpha times mathbb R^k),q_alpha^-1right)$$
gives a chart.
details can actually be found in these set of notes actually (same notation is used.)
answered Aug 6 at 20:40
Andres Mejia
15.2k11444
Sure. Let $V^prime=coprod_alpha (alpha times U_alpha times mathbb R^k)/sim$ where $(beta,x, mathbb R^k) sim (alpha,x,g_alphabeta)$ and
$$q:coprod_alpha (alpha times U_alpha times mathbb R^k) to V^prime $$
be the quotient map.
Note that $q$ is a continuous open map and the restriction to each $(alpha times U_alpha times mathbb R^k)$ denoted $q_alpha$ is injective, we have that
$$left(q_alpha(alpha times U_alpha times mathbb R^k),q_alpha^-1right)$$
gives a chart.
details can actually be found in these set of notes actually (same notation is used.)
answered Aug 6 at 20:40
Andres Mejia
15.2k11444
edited Aug 6 at 21:30
answered Aug 6 at 20:40
Andres Mejia
15.2k11444
answered Aug 6 at 20:40
Andres Mejia
15.2k11444
answered Aug 6 at 20:40
Andres Mejia
15.2k11444
15.2k11444
Thanks! Is $q$ the quotient map instead of the projection?
– Leonardo Schultz
Aug 6 at 21:26
@LeonardoSchultz yes of course. Sorry about that!
– Andres Mejia
Aug 6 at 21:31
add a comment |Â
Thanks! Is $q$ the quotient map instead of the projection?
– Leonardo Schultz
Aug 6 at 21:26
@LeonardoSchultz yes of course. Sorry about that!
– Andres Mejia
Aug 6 at 21:31
Thanks! Is $q$ the quotient map instead of the projection?
– Leonardo Schultz
Aug 6 at 21:26
Thanks! Is $q$ the quotient map instead of the projection?
– Leonardo Schultz
Aug 6 at 21:26
@LeonardoSchultz yes of course. Sorry about that!
– Andres Mejia
Aug 6 at 21:31
@LeonardoSchultz yes of course. Sorry about that!
– Andres Mejia
Aug 6 at 21:31
add a comment |Â
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