Orientation for a vector bundle
$begingroup$
After the definition of orientation for a vector bundle$^{(1)}$ in "Characteristic Classes" by Milnor/Stasheff (page 96), the author make this comment:
The local compatibility condition implies that for every point in the base space there exists a neighborhood $N$ and a cohomology class $$u in H^n (pi^{-1}(N), pi^{-1}(N)_0;Z)$$ so that for every fiber $F$ over $N$ the restriction $$u | (F, F_0) in H^n(F, F_0; Z)$$ is equal to $u_F$.
Where $u_F$ is the generator associated to the orientation of the fiber F.
I understand the intuition behind this comment, but I don't see exactly where this $u$ comes from. My idea is that an orientation in $mathbb{R}^n$ gives a class in $H^n(mathbb{R}^n, mathbb{R}^n_0; Z)$ and this should give a class in $H^n(N times mathbb{R}^n, N times mathbb{R}^n_0; Z) cong H^n (pi^{-1}(N), pi^{-1}(N)_0;Z)$ somehow.
$^{(1)}$ The definition in the book is:
An orientation for $xi$ is a function which assigns an orientation to each fiber $F$ of $xi$, subjected to the following local compatibility condition. For every point $b_0$ in the base space there should exist a local coordinate system $(N, h)$, with $b_0 in N$ and $h: N times mathbb{R}^n to pi^{-1}(N)$, so that for each fiber $F = pi^{-1}(b)$ over N the homomorphism $x mapsto h(b, x)$ from $mathbb{R}^n$ to F is orientation preserving.
algebraic-topology vector-bundles characteristic-classes
asked Nov 29 '18 at 15:03
Karina LivramentoKarina Livramento
1136
$endgroup$
add a comment |
$begingroup$
After the definition of orientation for a vector bundle$^{(1)}$ in "Characteristic Classes" by Milnor/Stasheff (page 96), the author make this comment:
The local compatibility condition implies that for every point in the base space there exists a neighborhood $N$ and a cohomology class $$u in H^n (pi^{-1}(N), pi^{-1}(N)_0;Z)$$ so that for every fiber $F$ over $N$ the restriction $$u | (F, F_0) in H^n(F, F_0; Z)$$ is equal to $u_F$.
Where $u_F$ is the generator associated to the orientation of the fiber F.
I understand the intuition behind this comment, but I don't see exactly where this $u$ comes from. My idea is that an orientation in $mathbb{R}^n$ gives a class in $H^n(mathbb{R}^n, mathbb{R}^n_0; Z)$ and this should give a class in $H^n(N times mathbb{R}^n, N times mathbb{R}^n_0; Z) cong H^n (pi^{-1}(N), pi^{-1}(N)_0;Z)$ somehow.
$^{(1)}$ The definition in the book is:
An orientation for $xi$ is a function which assigns an orientation to each fiber $F$ of $xi$, subjected to the following local compatibility condition. For every point $b_0$ in the base space there should exist a local coordinate system $(N, h)$, with $b_0 in N$ and $h: N times mathbb{R}^n to pi^{-1}(N)$, so that for each fiber $F = pi^{-1}(b)$ over N the homomorphism $x mapsto h(b, x)$ from $mathbb{R}^n$ to F is orientation preserving.
algebraic-topology vector-bundles characteristic-classes
asked Nov 29 '18 at 15:03
Karina LivramentoKarina Livramento
1136
$endgroup$
add a comment |
$begingroup$
After the definition of orientation for a vector bundle$^{(1)}$ in "Characteristic Classes" by Milnor/Stasheff (page 96), the author make this comment:
The local compatibility condition implies that for every point in the base space there exists a neighborhood $N$ and a cohomology class $$u in H^n (pi^{-1}(N), pi^{-1}(N)_0;Z)$$ so that for every fiber $F$ over $N$ the restriction $$u | (F, F_0) in H^n(F, F_0; Z)$$ is equal to $u_F$.
Where $u_F$ is the generator associated to the orientation of the fiber F.
I understand the intuition behind this comment, but I don't see exactly where this $u$ comes from. My idea is that an orientation in $mathbb{R}^n$ gives a class in $H^n(mathbb{R}^n, mathbb{R}^n_0; Z)$ and this should give a class in $H^n(N times mathbb{R}^n, N times mathbb{R}^n_0; Z) cong H^n (pi^{-1}(N), pi^{-1}(N)_0;Z)$ somehow.
$^{(1)}$ The definition in the book is:
An orientation for $xi$ is a function which assigns an orientation to each fiber $F$ of $xi$, subjected to the following local compatibility condition. For every point $b_0$ in the base space there should exist a local coordinate system $(N, h)$, with $b_0 in N$ and $h: N times mathbb{R}^n to pi^{-1}(N)$, so that for each fiber $F = pi^{-1}(b)$ over N the homomorphism $x mapsto h(b, x)$ from $mathbb{R}^n$ to F is orientation preserving.
algebraic-topology vector-bundles characteristic-classes
asked Nov 29 '18 at 15:03
Karina LivramentoKarina Livramento
1136
$endgroup$
After the definition of orientation for a vector bundle$^{(1)}$ in "Characteristic Classes" by Milnor/Stasheff (page 96), the author make this comment:
The local compatibility condition implies that for every point in the base space there exists a neighborhood $N$ and a cohomology class $$u in H^n (pi^{-1}(N), pi^{-1}(N)_0;Z)$$ so that for every fiber $F$ over $N$ the restriction $$u | (F, F_0) in H^n(F, F_0; Z)$$ is equal to $u_F$.
Where $u_F$ is the generator associated to the orientation of the fiber F.
I understand the intuition behind this comment, but I don't see exactly where this $u$ comes from. My idea is that an orientation in $mathbb{R}^n$ gives a class in $H^n(mathbb{R}^n, mathbb{R}^n_0; Z)$ and this should give a class in $H^n(N times mathbb{R}^n, N times mathbb{R}^n_0; Z) cong H^n (pi^{-1}(N), pi^{-1}(N)_0;Z)$ somehow.
$^{(1)}$ The definition in the book is:
An orientation for $xi$ is a function which assigns an orientation to each fiber $F$ of $xi$, subjected to the following local compatibility condition. For every point $b_0$ in the base space there should exist a local coordinate system $(N, h)$, with $b_0 in N$ and $h: N times mathbb{R}^n to pi^{-1}(N)$, so that for each fiber $F = pi^{-1}(b)$ over N the homomorphism $x mapsto h(b, x)$ from $mathbb{R}^n$ to F is orientation preserving.
algebraic-topology vector-bundles characteristic-classes
algebraic-topology vector-bundles characteristic-classes
asked Nov 29 '18 at 15:03
Karina LivramentoKarina Livramento
1136
asked Nov 29 '18 at 15:03
Karina LivramentoKarina Livramento
1136
asked Nov 29 '18 at 15:03
Karina LivramentoKarina Livramento
1136
asked Nov 29 '18 at 15:03
Karina LivramentoKarina Livramento
1136
asked Nov 29 '18 at 15:03
Karina LivramentoKarina Livramento
1136
1136
add a comment |
add a comment |
1 Answer
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As the cohomology of $(mathbb{R}^n, mathbb{R}^n_0)$ is torsion-free, and in particular,
$$ H^q( mathbb{R}^n, mathbb{R}^n_0) = begin{cases} mathbb{Z} & q = n \ 0 & q neq n end{cases},$$
by the Kunneth formula
$$ H^m(N times mathbb{R}^n, N times mathbb{R}^n_0) cong oplus_{p+q=m} H^p(N) otimes H^q(mathbb{R}^n,mathbb{R}^n_0) = H^{m-n}(N).$$
If $N$ is connected, the class you seek in $H^n(N times mathbb{R^n},N times mathbb{R}^n_0)$ is exactly the image of the generator in $H^0(N)$. Geometrically, the Kunneth isomorphism tells us this class in $H^n(N times mathbb{R}^n, N times mathbb{R}^n_0)$ is exactly a generator of $H^n(mathbb{R}^n,mathbb{R}^n_0)$ over every point in $N$.
answered Nov 29 '18 at 19:01
Joshua MundingerJoshua Mundinger
2,5141026
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$begingroup$
As the cohomology of $(mathbb{R}^n, mathbb{R}^n_0)$ is torsion-free, and in particular,
$$ H^q( mathbb{R}^n, mathbb{R}^n_0) = begin{cases} mathbb{Z} & q = n \ 0 & q neq n end{cases},$$
by the Kunneth formula
$$ H^m(N times mathbb{R}^n, N times mathbb{R}^n_0) cong oplus_{p+q=m} H^p(N) otimes H^q(mathbb{R}^n,mathbb{R}^n_0) = H^{m-n}(N).$$
If $N$ is connected, the class you seek in $H^n(N times mathbb{R^n},N times mathbb{R}^n_0)$ is exactly the image of the generator in $H^0(N)$. Geometrically, the Kunneth isomorphism tells us this class in $H^n(N times mathbb{R}^n, N times mathbb{R}^n_0)$ is exactly a generator of $H^n(mathbb{R}^n,mathbb{R}^n_0)$ over every point in $N$.
answered Nov 29 '18 at 19:01
Joshua MundingerJoshua Mundinger
2,5141026
$endgroup$
add a comment |
$begingroup$
As the cohomology of $(mathbb{R}^n, mathbb{R}^n_0)$ is torsion-free, and in particular,
$$ H^q( mathbb{R}^n, mathbb{R}^n_0) = begin{cases} mathbb{Z} & q = n \ 0 & q neq n end{cases},$$
by the Kunneth formula
$$ H^m(N times mathbb{R}^n, N times mathbb{R}^n_0) cong oplus_{p+q=m} H^p(N) otimes H^q(mathbb{R}^n,mathbb{R}^n_0) = H^{m-n}(N).$$
If $N$ is connected, the class you seek in $H^n(N times mathbb{R^n},N times mathbb{R}^n_0)$ is exactly the image of the generator in $H^0(N)$. Geometrically, the Kunneth isomorphism tells us this class in $H^n(N times mathbb{R}^n, N times mathbb{R}^n_0)$ is exactly a generator of $H^n(mathbb{R}^n,mathbb{R}^n_0)$ over every point in $N$.
answered Nov 29 '18 at 19:01
Joshua MundingerJoshua Mundinger
2,5141026
$endgroup$
add a comment |
$begingroup$
As the cohomology of $(mathbb{R}^n, mathbb{R}^n_0)$ is torsion-free, and in particular,
$$ H^q( mathbb{R}^n, mathbb{R}^n_0) = begin{cases} mathbb{Z} & q = n \ 0 & q neq n end{cases},$$
by the Kunneth formula
$$ H^m(N times mathbb{R}^n, N times mathbb{R}^n_0) cong oplus_{p+q=m} H^p(N) otimes H^q(mathbb{R}^n,mathbb{R}^n_0) = H^{m-n}(N).$$
If $N$ is connected, the class you seek in $H^n(N times mathbb{R^n},N times mathbb{R}^n_0)$ is exactly the image of the generator in $H^0(N)$. Geometrically, the Kunneth isomorphism tells us this class in $H^n(N times mathbb{R}^n, N times mathbb{R}^n_0)$ is exactly a generator of $H^n(mathbb{R}^n,mathbb{R}^n_0)$ over every point in $N$.
answered Nov 29 '18 at 19:01
Joshua MundingerJoshua Mundinger
2,5141026
$endgroup$
As the cohomology of $(mathbb{R}^n, mathbb{R}^n_0)$ is torsion-free, and in particular,
$$ H^q( mathbb{R}^n, mathbb{R}^n_0) = begin{cases} mathbb{Z} & q = n \ 0 & q neq n end{cases},$$
by the Kunneth formula
$$ H^m(N times mathbb{R}^n, N times mathbb{R}^n_0) cong oplus_{p+q=m} H^p(N) otimes H^q(mathbb{R}^n,mathbb{R}^n_0) = H^{m-n}(N).$$
If $N$ is connected, the class you seek in $H^n(N times mathbb{R^n},N times mathbb{R}^n_0)$ is exactly the image of the generator in $H^0(N)$. Geometrically, the Kunneth isomorphism tells us this class in $H^n(N times mathbb{R}^n, N times mathbb{R}^n_0)$ is exactly a generator of $H^n(mathbb{R}^n,mathbb{R}^n_0)$ over every point in $N$.
answered Nov 29 '18 at 19:01
Joshua MundingerJoshua Mundinger
2,5141026
answered Nov 29 '18 at 19:01
Joshua MundingerJoshua Mundinger
2,5141026
answered Nov 29 '18 at 19:01
Joshua MundingerJoshua Mundinger
2,5141026
answered Nov 29 '18 at 19:01
Joshua MundingerJoshua Mundinger
2,5141026
2,5141026
add a comment |
add a comment |
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