Is $W$ a finite-dimensional vector space in Proposition 3.13?












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Is $W$ a finite-dimensional vector space in Proposition $3.13$?







manifolds smooth-manifolds







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  • $begingroup$
    @JackLee It seems there isn't enough information about $W$ in $textbf{ Proposition 3.13}$.
    $endgroup$
    – Born to be proud
    Dec 13 '18 at 12:50










  • $begingroup$
    The tangent space was only defined for finite-dimensional smooth manifolds, so there is no need to explicitely state this.
    $endgroup$
    – TheGeekGreek
    Dec 13 '18 at 14:55


















0












$begingroup$


enter image description here




Is $W$ a finite-dimensional vector space in Proposition $3.13$?







manifolds smooth-manifolds







share|cite|improve this question











$endgroup$












  • $begingroup$
    @JackLee It seems there isn't enough information about $W$ in $textbf{ Proposition 3.13}$.
    $endgroup$
    – Born to be proud
    Dec 13 '18 at 12:50










  • $begingroup$
    The tangent space was only defined for finite-dimensional smooth manifolds, so there is no need to explicitely state this.
    $endgroup$
    – TheGeekGreek
    Dec 13 '18 at 14:55
















0












0








0





$begingroup$


enter image description here




Is $W$ a finite-dimensional vector space in Proposition $3.13$?







manifolds smooth-manifolds







share|cite|improve this question











$endgroup$




enter image description here




Is $W$ a finite-dimensional vector space in Proposition $3.13$?







manifolds smooth-manifolds




manifolds smooth-manifolds






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 11:31









amWhy

1




1










asked Dec 13 '18 at 11:13









Born to be proudBorn to be proud

854510




854510












  • $begingroup$
    @JackLee It seems there isn't enough information about $W$ in $textbf{ Proposition 3.13}$.
    $endgroup$
    – Born to be proud
    Dec 13 '18 at 12:50










  • $begingroup$
    The tangent space was only defined for finite-dimensional smooth manifolds, so there is no need to explicitely state this.
    $endgroup$
    – TheGeekGreek
    Dec 13 '18 at 14:55




















  • $begingroup$
    @JackLee It seems there isn't enough information about $W$ in $textbf{ Proposition 3.13}$.
    $endgroup$
    – Born to be proud
    Dec 13 '18 at 12:50










  • $begingroup$
    The tangent space was only defined for finite-dimensional smooth manifolds, so there is no need to explicitely state this.
    $endgroup$
    – TheGeekGreek
    Dec 13 '18 at 14:55


















$begingroup$
@JackLee It seems there isn't enough information about $W$ in $textbf{ Proposition 3.13}$.
$endgroup$
– Born to be proud
Dec 13 '18 at 12:50




$begingroup$
@JackLee It seems there isn't enough information about $W$ in $textbf{ Proposition 3.13}$.
$endgroup$
– Born to be proud
Dec 13 '18 at 12:50












$begingroup$
The tangent space was only defined for finite-dimensional smooth manifolds, so there is no need to explicitely state this.
$endgroup$
– TheGeekGreek
Dec 13 '18 at 14:55






$begingroup$
The tangent space was only defined for finite-dimensional smooth manifolds, so there is no need to explicitely state this.
$endgroup$
– TheGeekGreek
Dec 13 '18 at 14:55












2 Answers
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$begingroup$

The author seems to assume that $W$ is a vector space and also a smooth manifold. So yes, it is finite-dimensional.






share|cite|improve this answer









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    1












    $begingroup$

    Yes it is, and there is really no need to write down that explicitly.
    The theorem states that there is a canonical isomorphism $Vcong T_aV$ when $V$ is finite-dimensional, so in writing $Wcong T_{La}W$ it is implicitly assumed that $W$ is finite-dimensional as well as it is using this canonical isomorphism for $W.$






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






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      1












      $begingroup$

      The author seems to assume that $W$ is a vector space and also a smooth manifold. So yes, it is finite-dimensional.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The author seems to assume that $W$ is a vector space and also a smooth manifold. So yes, it is finite-dimensional.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The author seems to assume that $W$ is a vector space and also a smooth manifold. So yes, it is finite-dimensional.






          share|cite|improve this answer









          $endgroup$



          The author seems to assume that $W$ is a vector space and also a smooth manifold. So yes, it is finite-dimensional.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 11:25









          ChristophChristoph

          12.4k1642




          12.4k1642























              1












              $begingroup$

              Yes it is, and there is really no need to write down that explicitly.
              The theorem states that there is a canonical isomorphism $Vcong T_aV$ when $V$ is finite-dimensional, so in writing $Wcong T_{La}W$ it is implicitly assumed that $W$ is finite-dimensional as well as it is using this canonical isomorphism for $W.$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Yes it is, and there is really no need to write down that explicitly.
                The theorem states that there is a canonical isomorphism $Vcong T_aV$ when $V$ is finite-dimensional, so in writing $Wcong T_{La}W$ it is implicitly assumed that $W$ is finite-dimensional as well as it is using this canonical isomorphism for $W.$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Yes it is, and there is really no need to write down that explicitly.
                  The theorem states that there is a canonical isomorphism $Vcong T_aV$ when $V$ is finite-dimensional, so in writing $Wcong T_{La}W$ it is implicitly assumed that $W$ is finite-dimensional as well as it is using this canonical isomorphism for $W.$






                  share|cite|improve this answer









                  $endgroup$



                  Yes it is, and there is really no need to write down that explicitly.
                  The theorem states that there is a canonical isomorphism $Vcong T_aV$ when $V$ is finite-dimensional, so in writing $Wcong T_{La}W$ it is implicitly assumed that $W$ is finite-dimensional as well as it is using this canonical isomorphism for $W.$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 14:27









                  positrón0802positrón0802

                  4,468520




                  4,468520






























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