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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add a comment |
$begingroup$
Is $W$ a finite-dimensional vector space in Proposition $3.13$?
manifolds smooth-manifolds
$endgroup$
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@JackLee It seems there isn't enough information about $W$ in $textbf{ Proposition 3.13}$.
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– Born to be proud
Dec 13 '18 at 12:50
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The tangent space was only defined for finite-dimensional smooth manifolds, so there is no need to explicitely state this.
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– TheGeekGreek
Dec 13 '18 at 14:55
add a comment |
$begingroup$
Is $W$ a finite-dimensional vector space in Proposition $3.13$?
manifolds smooth-manifolds
$endgroup$
Is $W$ a finite-dimensional vector space in Proposition $3.13$?
manifolds smooth-manifolds
manifolds smooth-manifolds
edited Dec 13 '18 at 11:31
amWhy
1
1
asked Dec 13 '18 at 11:13
Born to be proudBorn to be proud
825510
825510
$begingroup$
@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
add a comment |
$begingroup$
@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
add a comment |
2 Answers
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The author seems to assume that $W$ is a vector space and also a smooth manifold. So yes, it is finite-dimensional.
$endgroup$
add a comment |
$begingroup$
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 $W.$
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add a comment |
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2 Answers
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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.
$endgroup$
add a comment |
$begingroup$
The author seems to assume that $W$ is a vector space and also a smooth manifold. So yes, it is finite-dimensional.
$endgroup$
add a comment |
$begingroup$
The author seems to assume that $W$ is a vector space and also a smooth manifold. So yes, it is finite-dimensional.
$endgroup$
The author seems to assume that $W$ is a vector space and also a smooth manifold. So yes, it is finite-dimensional.
answered Dec 13 '18 at 11:25
ChristophChristoph
12k1642
12k1642
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$begingroup$
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 $W.$
$endgroup$
add a comment |
$begingroup$
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 $W.$
$endgroup$
add a comment |
$begingroup$
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 $W.$
$endgroup$
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 $W.$
answered Dec 13 '18 at 14:27
positrón0802positrón0802
4,313520
4,313520
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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