Forms on vector bundles: vertically compactly supported












0












$begingroup$


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







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






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      asked Dec 28 '18 at 2:08









      CL.CL.

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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.






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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.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 28 '18 at 7:27







                user98602





































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