Forms on vector bundles: vertically compactly supported
$begingroup$
Definition: Let $pi:V rightarrow M$ be a vector bundle.
$Omega^p_{cv}(V)$ is the sections of $p$ form on $V$, such that $pi^{-1}(K) cap supp , (w) $ for all $K subseteq M$ compact.
Definition: $Omega^p_c(V)$ is the set of sections $p$ forms on $V$ such that its support is compact.
It is claimed that
$Omega^p_c(V) = Omega^p_{cv}(V)$ when $M$ is compact.
How is this true? $subseteq $ is clear.
Source, Pg 84, Proof of Theorem 6.2.10, line +3
general-topology algebraic-topology differential-topology vector-bundles de-rham-cohomology
asked Dec 28 '18 at 2:08
CL.CL.
2,2402925
$endgroup$
add a comment |
$begingroup$
Definition: Let $pi:V rightarrow M$ be a vector bundle.
$Omega^p_{cv}(V)$ is the sections of $p$ form on $V$, such that $pi^{-1}(K) cap supp , (w) $ for all $K subseteq M$ compact.
Definition: $Omega^p_c(V)$ is the set of sections $p$ forms on $V$ such that its support is compact.
It is claimed that
$Omega^p_c(V) = Omega^p_{cv}(V)$ when $M$ is compact.
How is this true? $subseteq $ is clear.
Source, Pg 84, Proof of Theorem 6.2.10, line +3
general-topology algebraic-topology differential-topology vector-bundles de-rham-cohomology
asked Dec 28 '18 at 2:08
CL.CL.
2,2402925
$endgroup$
add a comment |
$begingroup$
Definition: Let $pi:V rightarrow M$ be a vector bundle.
$Omega^p_{cv}(V)$ is the sections of $p$ form on $V$, such that $pi^{-1}(K) cap supp , (w) $ for all $K subseteq M$ compact.
Definition: $Omega^p_c(V)$ is the set of sections $p$ forms on $V$ such that its support is compact.
It is claimed that
$Omega^p_c(V) = Omega^p_{cv}(V)$ when $M$ is compact.
How is this true? $subseteq $ is clear.
Source, Pg 84, Proof of Theorem 6.2.10, line +3
general-topology algebraic-topology differential-topology vector-bundles de-rham-cohomology
asked Dec 28 '18 at 2:08
CL.CL.
2,2402925
$endgroup$
Definition: Let $pi:V rightarrow M$ be a vector bundle.
$Omega^p_{cv}(V)$ is the sections of $p$ form on $V$, such that $pi^{-1}(K) cap supp , (w) $ for all $K subseteq M$ compact.
Definition: $Omega^p_c(V)$ is the set of sections $p$ forms on $V$ such that its support is compact.
It is claimed that
$Omega^p_c(V) = Omega^p_{cv}(V)$ when $M$ is compact.
How is this true? $subseteq $ is clear.
Source, Pg 84, Proof of Theorem 6.2.10, line +3
general-topology algebraic-topology differential-topology vector-bundles de-rham-cohomology
general-topology algebraic-topology differential-topology vector-bundles de-rham-cohomology
asked Dec 28 '18 at 2:08
CL.CL.
2,2402925
asked Dec 28 '18 at 2:08
CL.CL.
2,2402925
asked Dec 28 '18 at 2:08
CL.CL.
2,2402925
asked Dec 28 '18 at 2:08
CL.CL.
2,2402925
asked Dec 28 '18 at 2:08
CL.CL.
2,2402925
2,2402925
add a comment |
add a comment |
1 Answer
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Let $omega$ be a form on $V$ which is compactly supported on each fiber $V_x$. Consider $text{supp}(omega) subset V$; this is a closed subset of $V$ whose intersection with each fiber is compact. The goal is to show all such sets are compact themselves.
Put a Riemannian metric over V (just so that I may measure "size of vectors"). Then if $X subset V$ is closed, with the property that each $X cap V_x$ is compact, then we may define a function $sigma: M to Bbb R_{geq 0}$ given by sending $sigma(x)$ to the largest $|v|$ of any $v in X cap V_x$. This is well-defined because $X cap V_x$ is compact. The map $sigma$ may be seen to be continuous by brute-force; I do not want to do it here. (Intuitively, the nearby sets $X cap V_x$ are 'close', and hence should have 'nearby' size bounds.)
Thus $sigma$ is a continuous map from a compact space to $Bbb R$, and hence has an absolute maximum, say $L$. Thus the set $X$ is contained in the closed $L$-disc bundle of $V$, which is a compact space (given any sequence $(m_n, v_n)$ of norm bounded by $L$, choose a subsequence so that $m_n$ is convergent to $m$; working on a chart to identify all nearby fibers with $V_x$, use compactness of the $L$-disc ball to see that a further subsequence may be chosen so that $v_n$ converges). $X$, being a closed subspace of a compact Hausdorff space, is compact.
So any form with compact vertical support on a vector bundle $V$ over compact base $M$ has compact support, as desired.
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$begingroup$
Let $omega$ be a form on $V$ which is compactly supported on each fiber $V_x$. Consider $text{supp}(omega) subset V$; this is a closed subset of $V$ whose intersection with each fiber is compact. The goal is to show all such sets are compact themselves.
Put a Riemannian metric over V (just so that I may measure "size of vectors"). Then if $X subset V$ is closed, with the property that each $X cap V_x$ is compact, then we may define a function $sigma: M to Bbb R_{geq 0}$ given by sending $sigma(x)$ to the largest $|v|$ of any $v in X cap V_x$. This is well-defined because $X cap V_x$ is compact. The map $sigma$ may be seen to be continuous by brute-force; I do not want to do it here. (Intuitively, the nearby sets $X cap V_x$ are 'close', and hence should have 'nearby' size bounds.)
Thus $sigma$ is a continuous map from a compact space to $Bbb R$, and hence has an absolute maximum, say $L$. Thus the set $X$ is contained in the closed $L$-disc bundle of $V$, which is a compact space (given any sequence $(m_n, v_n)$ of norm bounded by $L$, choose a subsequence so that $m_n$ is convergent to $m$; working on a chart to identify all nearby fibers with $V_x$, use compactness of the $L$-disc ball to see that a further subsequence may be chosen so that $v_n$ converges). $X$, being a closed subspace of a compact Hausdorff space, is compact.
So any form with compact vertical support on a vector bundle $V$ over compact base $M$ has compact support, as desired.
$endgroup$
add a comment |
$begingroup$
Let $omega$ be a form on $V$ which is compactly supported on each fiber $V_x$. Consider $text{supp}(omega) subset V$; this is a closed subset of $V$ whose intersection with each fiber is compact. The goal is to show all such sets are compact themselves.
Put a Riemannian metric over V (just so that I may measure "size of vectors"). Then if $X subset V$ is closed, with the property that each $X cap V_x$ is compact, then we may define a function $sigma: M to Bbb R_{geq 0}$ given by sending $sigma(x)$ to the largest $|v|$ of any $v in X cap V_x$. This is well-defined because $X cap V_x$ is compact. The map $sigma$ may be seen to be continuous by brute-force; I do not want to do it here. (Intuitively, the nearby sets $X cap V_x$ are 'close', and hence should have 'nearby' size bounds.)
Thus $sigma$ is a continuous map from a compact space to $Bbb R$, and hence has an absolute maximum, say $L$. Thus the set $X$ is contained in the closed $L$-disc bundle of $V$, which is a compact space (given any sequence $(m_n, v_n)$ of norm bounded by $L$, choose a subsequence so that $m_n$ is convergent to $m$; working on a chart to identify all nearby fibers with $V_x$, use compactness of the $L$-disc ball to see that a further subsequence may be chosen so that $v_n$ converges). $X$, being a closed subspace of a compact Hausdorff space, is compact.
So any form with compact vertical support on a vector bundle $V$ over compact base $M$ has compact support, as desired.
$endgroup$
add a comment |
$begingroup$
Let $omega$ be a form on $V$ which is compactly supported on each fiber $V_x$. Consider $text{supp}(omega) subset V$; this is a closed subset of $V$ whose intersection with each fiber is compact. The goal is to show all such sets are compact themselves.
Put a Riemannian metric over V (just so that I may measure "size of vectors"). Then if $X subset V$ is closed, with the property that each $X cap V_x$ is compact, then we may define a function $sigma: M to Bbb R_{geq 0}$ given by sending $sigma(x)$ to the largest $|v|$ of any $v in X cap V_x$. This is well-defined because $X cap V_x$ is compact. The map $sigma$ may be seen to be continuous by brute-force; I do not want to do it here. (Intuitively, the nearby sets $X cap V_x$ are 'close', and hence should have 'nearby' size bounds.)
Thus $sigma$ is a continuous map from a compact space to $Bbb R$, and hence has an absolute maximum, say $L$. Thus the set $X$ is contained in the closed $L$-disc bundle of $V$, which is a compact space (given any sequence $(m_n, v_n)$ of norm bounded by $L$, choose a subsequence so that $m_n$ is convergent to $m$; working on a chart to identify all nearby fibers with $V_x$, use compactness of the $L$-disc ball to see that a further subsequence may be chosen so that $v_n$ converges). $X$, being a closed subspace of a compact Hausdorff space, is compact.
So any form with compact vertical support on a vector bundle $V$ over compact base $M$ has compact support, as desired.
$endgroup$
Let $omega$ be a form on $V$ which is compactly supported on each fiber $V_x$. Consider $text{supp}(omega) subset V$; this is a closed subset of $V$ whose intersection with each fiber is compact. The goal is to show all such sets are compact themselves.
Put a Riemannian metric over V (just so that I may measure "size of vectors"). Then if $X subset V$ is closed, with the property that each $X cap V_x$ is compact, then we may define a function $sigma: M to Bbb R_{geq 0}$ given by sending $sigma(x)$ to the largest $|v|$ of any $v in X cap V_x$. This is well-defined because $X cap V_x$ is compact. The map $sigma$ may be seen to be continuous by brute-force; I do not want to do it here. (Intuitively, the nearby sets $X cap V_x$ are 'close', and hence should have 'nearby' size bounds.)
Thus $sigma$ is a continuous map from a compact space to $Bbb R$, and hence has an absolute maximum, say $L$. Thus the set $X$ is contained in the closed $L$-disc bundle of $V$, which is a compact space (given any sequence $(m_n, v_n)$ of norm bounded by $L$, choose a subsequence so that $m_n$ is convergent to $m$; working on a chart to identify all nearby fibers with $V_x$, use compactness of the $L$-disc ball to see that a further subsequence may be chosen so that $v_n$ converges). $X$, being a closed subspace of a compact Hausdorff space, is compact.
So any form with compact vertical support on a vector bundle $V$ over compact base $M$ has compact support, as desired.
answered Dec 28 '18 at 7:27
user98602
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