An affine bundle has a global section?
$begingroup$
Let $X$ be a manifold. We say $pi: Y longrightarrow X$ is an rank $n$ affine bundle if there is an open cover ${ U_alpha }$ of $X$ such that
$ Y big|_{U_alpha} cong U_alpha times mathbb{R}^n $ and the transition function from $U_alpha$ to $U_beta$ is given by
$$ (x,v) mapsto (x, rho_{beta alpha }(x) v + u_{ beta alpha} (x)) $$ satisfying the cocycle condition
$ rho_{gamma alpha} (x) = rho_{gamma beta} (x) rho_{beta alpha } (x) $ and
$u_{gamma alpha}(x) = rho_{gamma beta} (x) u_{beta alpha} (x) + u_{gamma beta}(x)$.
Wikipedia claims that an affine bundle has a global section so it can be identified
with the vector bundle glued by the cocycles ${ rho_{gamma alpha} }$ in a non-canonical way.
How can we construct one exactly? Someone claimed that local sections exist so one can glue them to a global one by
standard partition of unity argument. Since multiplying by constant doesn't make sense for affine bundle, I cannot see why this is obvious.
geometry differential-geometry manifolds differential-topology vector-bundles
asked Apr 12 '18 at 17:31
Chris KuoChris Kuo
580210
$endgroup$
add a comment |
$begingroup$
Let $X$ be a manifold. We say $pi: Y longrightarrow X$ is an rank $n$ affine bundle if there is an open cover ${ U_alpha }$ of $X$ such that
$ Y big|_{U_alpha} cong U_alpha times mathbb{R}^n $ and the transition function from $U_alpha$ to $U_beta$ is given by
$$ (x,v) mapsto (x, rho_{beta alpha }(x) v + u_{ beta alpha} (x)) $$ satisfying the cocycle condition
$ rho_{gamma alpha} (x) = rho_{gamma beta} (x) rho_{beta alpha } (x) $ and
$u_{gamma alpha}(x) = rho_{gamma beta} (x) u_{beta alpha} (x) + u_{gamma beta}(x)$.
Wikipedia claims that an affine bundle has a global section so it can be identified
with the vector bundle glued by the cocycles ${ rho_{gamma alpha} }$ in a non-canonical way.
How can we construct one exactly? Someone claimed that local sections exist so one can glue them to a global one by
standard partition of unity argument. Since multiplying by constant doesn't make sense for affine bundle, I cannot see why this is obvious.
geometry differential-geometry manifolds differential-topology vector-bundles
asked Apr 12 '18 at 17:31
Chris KuoChris Kuo
580210
$endgroup$
add a comment |
$begingroup$
Let $X$ be a manifold. We say $pi: Y longrightarrow X$ is an rank $n$ affine bundle if there is an open cover ${ U_alpha }$ of $X$ such that
$ Y big|_{U_alpha} cong U_alpha times mathbb{R}^n $ and the transition function from $U_alpha$ to $U_beta$ is given by
$$ (x,v) mapsto (x, rho_{beta alpha }(x) v + u_{ beta alpha} (x)) $$ satisfying the cocycle condition
$ rho_{gamma alpha} (x) = rho_{gamma beta} (x) rho_{beta alpha } (x) $ and
$u_{gamma alpha}(x) = rho_{gamma beta} (x) u_{beta alpha} (x) + u_{gamma beta}(x)$.
Wikipedia claims that an affine bundle has a global section so it can be identified
with the vector bundle glued by the cocycles ${ rho_{gamma alpha} }$ in a non-canonical way.
How can we construct one exactly? Someone claimed that local sections exist so one can glue them to a global one by
standard partition of unity argument. Since multiplying by constant doesn't make sense for affine bundle, I cannot see why this is obvious.
geometry differential-geometry manifolds differential-topology vector-bundles
asked Apr 12 '18 at 17:31
Chris KuoChris Kuo
580210
$endgroup$
Let $X$ be a manifold. We say $pi: Y longrightarrow X$ is an rank $n$ affine bundle if there is an open cover ${ U_alpha }$ of $X$ such that
$ Y big|_{U_alpha} cong U_alpha times mathbb{R}^n $ and the transition function from $U_alpha$ to $U_beta$ is given by
$$ (x,v) mapsto (x, rho_{beta alpha }(x) v + u_{ beta alpha} (x)) $$ satisfying the cocycle condition
$ rho_{gamma alpha} (x) = rho_{gamma beta} (x) rho_{beta alpha } (x) $ and
$u_{gamma alpha}(x) = rho_{gamma beta} (x) u_{beta alpha} (x) + u_{gamma beta}(x)$.
Wikipedia claims that an affine bundle has a global section so it can be identified
with the vector bundle glued by the cocycles ${ rho_{gamma alpha} }$ in a non-canonical way.
How can we construct one exactly? Someone claimed that local sections exist so one can glue them to a global one by
standard partition of unity argument. Since multiplying by constant doesn't make sense for affine bundle, I cannot see why this is obvious.
geometry differential-geometry manifolds differential-topology vector-bundles
geometry differential-geometry manifolds differential-topology vector-bundles
asked Apr 12 '18 at 17:31
Chris KuoChris Kuo
580210
asked Apr 12 '18 at 17:31
Chris KuoChris Kuo
580210
asked Apr 12 '18 at 17:31
Chris KuoChris Kuo
580210
asked Apr 12 '18 at 17:31
Chris KuoChris Kuo
580210
asked Apr 12 '18 at 17:31
Chris KuoChris Kuo
580210
580210
add a comment |
add a comment |
1 Answer
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Multiplication by functions doesn’t in general make sense, but affine combinations of sections do.
Specifically, if $sigma,tauinGamma(pi)$ are sections and $f, ginmathrm C^infty(X)$ are smooth functions on the base summing to unity, then the obvious definition of $fsigma + gtau$ is independent of trivialization if (!) $f(x)+g(x) = 1$. This defines affine combinations only for a finite number of terms, but you can of course also do that for an arbitrary one, as long as only a finite number of coefficients are non-zero at any given point.
Now choose a (locally finite) trivializing open cover of the bundle, a partition of unity corresponding to that cover and an arbitrary local section over every neighbourhood. Then the affine combination of these sections with coefficients given by this partition is a well-defined global section.
answered Jan 8 at 11:56
Alex ShpilkinAlex Shpilkin
349315
$endgroup$
add a comment |
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$begingroup$
Multiplication by functions doesn’t in general make sense, but affine combinations of sections do.
Specifically, if $sigma,tauinGamma(pi)$ are sections and $f, ginmathrm C^infty(X)$ are smooth functions on the base summing to unity, then the obvious definition of $fsigma + gtau$ is independent of trivialization if (!) $f(x)+g(x) = 1$. This defines affine combinations only for a finite number of terms, but you can of course also do that for an arbitrary one, as long as only a finite number of coefficients are non-zero at any given point.
Now choose a (locally finite) trivializing open cover of the bundle, a partition of unity corresponding to that cover and an arbitrary local section over every neighbourhood. Then the affine combination of these sections with coefficients given by this partition is a well-defined global section.
answered Jan 8 at 11:56
Alex ShpilkinAlex Shpilkin
349315
$endgroup$
add a comment |
$begingroup$
Multiplication by functions doesn’t in general make sense, but affine combinations of sections do.
Specifically, if $sigma,tauinGamma(pi)$ are sections and $f, ginmathrm C^infty(X)$ are smooth functions on the base summing to unity, then the obvious definition of $fsigma + gtau$ is independent of trivialization if (!) $f(x)+g(x) = 1$. This defines affine combinations only for a finite number of terms, but you can of course also do that for an arbitrary one, as long as only a finite number of coefficients are non-zero at any given point.
Now choose a (locally finite) trivializing open cover of the bundle, a partition of unity corresponding to that cover and an arbitrary local section over every neighbourhood. Then the affine combination of these sections with coefficients given by this partition is a well-defined global section.
answered Jan 8 at 11:56
Alex ShpilkinAlex Shpilkin
349315
$endgroup$
add a comment |
$begingroup$
Multiplication by functions doesn’t in general make sense, but affine combinations of sections do.
Specifically, if $sigma,tauinGamma(pi)$ are sections and $f, ginmathrm C^infty(X)$ are smooth functions on the base summing to unity, then the obvious definition of $fsigma + gtau$ is independent of trivialization if (!) $f(x)+g(x) = 1$. This defines affine combinations only for a finite number of terms, but you can of course also do that for an arbitrary one, as long as only a finite number of coefficients are non-zero at any given point.
Now choose a (locally finite) trivializing open cover of the bundle, a partition of unity corresponding to that cover and an arbitrary local section over every neighbourhood. Then the affine combination of these sections with coefficients given by this partition is a well-defined global section.
answered Jan 8 at 11:56
Alex ShpilkinAlex Shpilkin
349315
$endgroup$
Multiplication by functions doesn’t in general make sense, but affine combinations of sections do.
Specifically, if $sigma,tauinGamma(pi)$ are sections and $f, ginmathrm C^infty(X)$ are smooth functions on the base summing to unity, then the obvious definition of $fsigma + gtau$ is independent of trivialization if (!) $f(x)+g(x) = 1$. This defines affine combinations only for a finite number of terms, but you can of course also do that for an arbitrary one, as long as only a finite number of coefficients are non-zero at any given point.
Now choose a (locally finite) trivializing open cover of the bundle, a partition of unity corresponding to that cover and an arbitrary local section over every neighbourhood. Then the affine combination of these sections with coefficients given by this partition is a well-defined global section.
answered Jan 8 at 11:56
Alex ShpilkinAlex Shpilkin
349315
edited Jan 8 at 12:02
answered Jan 8 at 11:56
Alex ShpilkinAlex Shpilkin
349315
answered Jan 8 at 11:56
Alex ShpilkinAlex Shpilkin
349315
answered Jan 8 at 11:56
Alex ShpilkinAlex Shpilkin
349315
349315
add a comment |
add a comment |
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