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Question on the proof of Stokes' Theorem in Spivak
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The following is a quick outline of the proof of Stokes' Theorem as found in a Comprehensive Introduction to Differential Geometry Vol. 1 by Spivak.
Theorem (Local Stokes' Theorem). Let $M$ be a smooth manifold, $c$ a singular $k$-chain and $omega$ a $k - 1$-form on $M$. Then $$int_c domega = int_{partial c} omega.$$
Theorem (Stokes' Theorem). Let $M^n$ be an oriented smooth manifold with boundary and $omega in Omega^{n - 1}_c(M)$. Then $$int_M domega = int_{partial M}omega$$ where $partial M$ is given the induced orientation.
Proof. Suppose that the support of $omega$ is contained in the interior of some positively oriented singular cube $c$ with $operatorname{im} c cap partial M = varnothing$. Then we can apply the local Stokes' theorem to conclude. Indeed, we have that $$int_M domega = int_c domega = int_{partial c}omega = 0.$$
Shouldn't it really be $operatorname{int} M$ in the first integral instead of just $M$? Because the local Stokes' theorem only applies for manifolds with boundaries. However, the next step is to consider a singular cube $c$ such that $partial M cap operatorname{im}c = operatorname{im}F_1c$, where $F_1c$ is the first front face. Spivak then proceeds again by using the local version of Stokes' theorem: $$int_M domega = int_c domega = int_{partial c}omega = ...$$ Why can we use the local version here? Again, I mean, the local version applies for manifolds without boundary.
proof-verification differential-geometry stokes-theorem
asked Jan 30 at 9:27
TheGeekGreekTheGeekGreek
5,11131036
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add a comment |
$begingroup$
The following is a quick outline of the proof of Stokes' Theorem as found in a Comprehensive Introduction to Differential Geometry Vol. 1 by Spivak.
Theorem (Local Stokes' Theorem). Let $M$ be a smooth manifold, $c$ a singular $k$-chain and $omega$ a $k - 1$-form on $M$. Then $$int_c domega = int_{partial c} omega.$$
Theorem (Stokes' Theorem). Let $M^n$ be an oriented smooth manifold with boundary and $omega in Omega^{n - 1}_c(M)$. Then $$int_M domega = int_{partial M}omega$$ where $partial M$ is given the induced orientation.
Proof. Suppose that the support of $omega$ is contained in the interior of some positively oriented singular cube $c$ with $operatorname{im} c cap partial M = varnothing$. Then we can apply the local Stokes' theorem to conclude. Indeed, we have that $$int_M domega = int_c domega = int_{partial c}omega = 0.$$
Shouldn't it really be $operatorname{int} M$ in the first integral instead of just $M$? Because the local Stokes' theorem only applies for manifolds with boundaries. However, the next step is to consider a singular cube $c$ such that $partial M cap operatorname{im}c = operatorname{im}F_1c$, where $F_1c$ is the first front face. Spivak then proceeds again by using the local version of Stokes' theorem: $$int_M domega = int_c domega = int_{partial c}omega = ...$$ Why can we use the local version here? Again, I mean, the local version applies for manifolds without boundary.
proof-verification differential-geometry stokes-theorem
asked Jan 30 at 9:27
TheGeekGreekTheGeekGreek
5,11131036
$endgroup$
2
$begingroup$
A side remark: The name of the guy was Stokes, not Stoke.
$endgroup$
– Hans Lundmark
Jan 30 at 10:14
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@HansLundmark Oh dear...that was a terrible mistake, sorry!
$endgroup$
– TheGeekGreek
Jan 30 at 10:18
$begingroup$
sites.math.washington.edu/~morrow/335_17/…
$endgroup$
– Charlie Frohman
Feb 3 at 19:02
add a comment |
$begingroup$
The following is a quick outline of the proof of Stokes' Theorem as found in a Comprehensive Introduction to Differential Geometry Vol. 1 by Spivak.
Theorem (Local Stokes' Theorem). Let $M$ be a smooth manifold, $c$ a singular $k$-chain and $omega$ a $k - 1$-form on $M$. Then $$int_c domega = int_{partial c} omega.$$
Theorem (Stokes' Theorem). Let $M^n$ be an oriented smooth manifold with boundary and $omega in Omega^{n - 1}_c(M)$. Then $$int_M domega = int_{partial M}omega$$ where $partial M$ is given the induced orientation.
Proof. Suppose that the support of $omega$ is contained in the interior of some positively oriented singular cube $c$ with $operatorname{im} c cap partial M = varnothing$. Then we can apply the local Stokes' theorem to conclude. Indeed, we have that $$int_M domega = int_c domega = int_{partial c}omega = 0.$$
Shouldn't it really be $operatorname{int} M$ in the first integral instead of just $M$? Because the local Stokes' theorem only applies for manifolds with boundaries. However, the next step is to consider a singular cube $c$ such that $partial M cap operatorname{im}c = operatorname{im}F_1c$, where $F_1c$ is the first front face. Spivak then proceeds again by using the local version of Stokes' theorem: $$int_M domega = int_c domega = int_{partial c}omega = ...$$ Why can we use the local version here? Again, I mean, the local version applies for manifolds without boundary.
proof-verification differential-geometry stokes-theorem
asked Jan 30 at 9:27
TheGeekGreekTheGeekGreek
5,11131036
$endgroup$
The following is a quick outline of the proof of Stokes' Theorem as found in a Comprehensive Introduction to Differential Geometry Vol. 1 by Spivak.
Theorem (Local Stokes' Theorem). Let $M$ be a smooth manifold, $c$ a singular $k$-chain and $omega$ a $k - 1$-form on $M$. Then $$int_c domega = int_{partial c} omega.$$
Theorem (Stokes' Theorem). Let $M^n$ be an oriented smooth manifold with boundary and $omega in Omega^{n - 1}_c(M)$. Then $$int_M domega = int_{partial M}omega$$ where $partial M$ is given the induced orientation.
Proof. Suppose that the support of $omega$ is contained in the interior of some positively oriented singular cube $c$ with $operatorname{im} c cap partial M = varnothing$. Then we can apply the local Stokes' theorem to conclude. Indeed, we have that $$int_M domega = int_c domega = int_{partial c}omega = 0.$$
Shouldn't it really be $operatorname{int} M$ in the first integral instead of just $M$? Because the local Stokes' theorem only applies for manifolds with boundaries. However, the next step is to consider a singular cube $c$ such that $partial M cap operatorname{im}c = operatorname{im}F_1c$, where $F_1c$ is the first front face. Spivak then proceeds again by using the local version of Stokes' theorem: $$int_M domega = int_c domega = int_{partial c}omega = ...$$ Why can we use the local version here? Again, I mean, the local version applies for manifolds without boundary.
proof-verification differential-geometry stokes-theorem
proof-verification differential-geometry stokes-theorem
asked Jan 30 at 9:27
TheGeekGreekTheGeekGreek
5,11131036
asked Jan 30 at 9:27
TheGeekGreekTheGeekGreek
5,11131036
edited Jan 30 at 10:18
TheGeekGreek
asked Jan 30 at 9:27
TheGeekGreekTheGeekGreek
5,11131036
asked Jan 30 at 9:27
TheGeekGreekTheGeekGreek
5,11131036
asked Jan 30 at 9:27
TheGeekGreekTheGeekGreek
5,11131036
5,11131036
2
$begingroup$
A side remark: The name of the guy was Stokes, not Stoke.
$endgroup$
– Hans Lundmark
Jan 30 at 10:14
$begingroup$
@HansLundmark Oh dear...that was a terrible mistake, sorry!
$endgroup$
– TheGeekGreek
Jan 30 at 10:18
$begingroup$
sites.math.washington.edu/~morrow/335_17/…
$endgroup$
– Charlie Frohman
Feb 3 at 19:02
add a comment |
2
$begingroup$
A side remark: The name of the guy was Stokes, not Stoke.
$endgroup$
– Hans Lundmark
Jan 30 at 10:14
$begingroup$
@HansLundmark Oh dear...that was a terrible mistake, sorry!
$endgroup$
– TheGeekGreek
Jan 30 at 10:18
$begingroup$
sites.math.washington.edu/~morrow/335_17/…
$endgroup$
– Charlie Frohman
Feb 3 at 19:02
2
2
$begingroup$
A side remark: The name of the guy was Stokes, not Stoke.
$endgroup$
– Hans Lundmark
Jan 30 at 10:14
$begingroup$
A side remark: The name of the guy was Stokes, not Stoke.
$endgroup$
– Hans Lundmark
Jan 30 at 10:14
$begingroup$
@HansLundmark Oh dear...that was a terrible mistake, sorry!
$endgroup$
– TheGeekGreek
Jan 30 at 10:18
$begingroup$
@HansLundmark Oh dear...that was a terrible mistake, sorry!
$endgroup$
– TheGeekGreek
Jan 30 at 10:18
$begingroup$
sites.math.washington.edu/~morrow/335_17/…
$endgroup$
– Charlie Frohman
Feb 3 at 19:02
$begingroup$
sites.math.washington.edu/~morrow/335_17/…
$endgroup$
– Charlie Frohman
Feb 3 at 19:02
add a comment |
1 Answer
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The local version of Stokes Theorem is one of those results which can be generalised to hold for the case of a manifold with boundary with pretty low effort.
answered Feb 9 at 21:11
TheGeekGreekTheGeekGreek
5,11131036
$endgroup$
add a comment |
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The local version of Stokes Theorem is one of those results which can be generalised to hold for the case of a manifold with boundary with pretty low effort.
answered Feb 9 at 21:11
TheGeekGreekTheGeekGreek
5,11131036
$endgroup$
add a comment |
$begingroup$
The local version of Stokes Theorem is one of those results which can be generalised to hold for the case of a manifold with boundary with pretty low effort.
answered Feb 9 at 21:11
TheGeekGreekTheGeekGreek
5,11131036
$endgroup$
add a comment |
$begingroup$
The local version of Stokes Theorem is one of those results which can be generalised to hold for the case of a manifold with boundary with pretty low effort.
answered Feb 9 at 21:11
TheGeekGreekTheGeekGreek
5,11131036
$endgroup$
The local version of Stokes Theorem is one of those results which can be generalised to hold for the case of a manifold with boundary with pretty low effort.
answered Feb 9 at 21:11
TheGeekGreekTheGeekGreek
5,11131036
answered Feb 9 at 21:11
TheGeekGreekTheGeekGreek
5,11131036
answered Feb 9 at 21:11
TheGeekGreekTheGeekGreek
5,11131036
answered Feb 9 at 21:11
TheGeekGreekTheGeekGreek
5,11131036
5,11131036
add a comment |
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2
$begingroup$
A side remark: The name of the guy was Stokes, not Stoke.
$endgroup$
– Hans Lundmark
Jan 30 at 10:14
$begingroup$
@HansLundmark Oh dear...that was a terrible mistake, sorry!
$endgroup$
– TheGeekGreek
Jan 30 at 10:18
$begingroup$
sites.math.washington.edu/~morrow/335_17/…
$endgroup$
– Charlie Frohman
Feb 3 at 19:02