A coordinate-independent formula for a potential of an exact 2-formNatural and coordinate free definition for the Riemannian volume form?When can a functional be written as the integral of a 1-form?Change of Coordinate Formula for Differential FormsA necessary and sufficient condition for the admittance of integrating factorDoes every $C^1$ closed differential form differ from some $C^infty$ closed form by an exact form?Local condition on differential form satisfying Frobenius so that the quotient is Hausdorff.Existence of a potential for given exact form satisfying symmetry conditionsQuestion Regarding the Coordinate Independent Form of the Exterior DerivativeClosed and exact 1-form defined on $mathbbR^2$Finding an Antiderivative for an Exact Form
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A coordinate-independent formula for a potential of an exact 2-form
Natural and coordinate free definition for the Riemannian volume form?When can a functional be written as the integral of a 1-form?Change of Coordinate Formula for Differential FormsA necessary and sufficient condition for the admittance of integrating factorDoes every $C^1$ closed differential form differ from some $C^infty$ closed form by an exact form?Local condition on differential form satisfying Frobenius so that the quotient is Hausdorff.Existence of a potential for given exact form satisfying symmetry conditionsQuestion Regarding the Coordinate Independent Form of the Exterior DerivativeClosed and exact 1-form defined on $mathbbR^2$Finding an Antiderivative for an Exact Form
$begingroup$
If I somehow know that a given 1-form $omega$ on a contractible set $U$ on a manifold $M$ is exact (there exist a 0-form $f$ such that $omega=rmdf$), then I can find $f$ from the following coordinate-independent formula:
$$ f(x) = int_x_0^x omega$$
where point $x_0$ and integration path can be arbitrarily chosen, as long as they are contained in $U$.
Let as assume that $sigma$ is a exact 2-form on a contractible set $U$ on a manifold $M$. Is there an analogous coordinate-independent formula allowing to find its potential (the 1-form $omega$ such that $sigma=rmdomega$)? The only ones I've seen choose a specific coordinate system, or assume that $M=mathbbR^3$.
differential-geometry differential-forms
asked Mar 13 at 10:09
Adam LatosińskiAdam Latosiński
211
New contributor
Adam Latosiński is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If I somehow know that a given 1-form $omega$ on a contractible set $U$ on a manifold $M$ is exact (there exist a 0-form $f$ such that $omega=rmdf$), then I can find $f$ from the following coordinate-independent formula:
$$ f(x) = int_x_0^x omega$$
where point $x_0$ and integration path can be arbitrarily chosen, as long as they are contained in $U$.
Let as assume that $sigma$ is a exact 2-form on a contractible set $U$ on a manifold $M$. Is there an analogous coordinate-independent formula allowing to find its potential (the 1-form $omega$ such that $sigma=rmdomega$)? The only ones I've seen choose a specific coordinate system, or assume that $M=mathbbR^3$.
differential-geometry differential-forms
asked Mar 13 at 10:09
Adam LatosińskiAdam Latosiński
211
New contributor
Adam Latosiński is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Yes, $1$-forms are particularly easy. You can write down something coordinate-independent, but you do need to integrate out the $dt$ when the null-homotopy is given by varying $t$ from $0$ to $1$.
$endgroup$
– Ted Shifrin
Mar 14 at 17:39
$begingroup$
Can you show what you mean?
$endgroup$
– Adam Latosiński
20 hours ago
$begingroup$
Let $Fcolon Utimes [0,1]to U$ be the contraction, so $F(x,1)=x$ and $F(x,0)=x_0$ (fixed). Let $tin [0,1]$. The pullback $F^*sigma$ has some $dt$ terms; we can contract against $partial/partial t$ (this gives a $1$-form with coefficients that are functions of $x$ and $t$). It can be checked that if we set $$omega = int_0^1 (iota_partial/partial tsigma)dt,$$ then $domega = sigma$.
$endgroup$
– Ted Shifrin
13 hours ago
add a comment |
$begingroup$
If I somehow know that a given 1-form $omega$ on a contractible set $U$ on a manifold $M$ is exact (there exist a 0-form $f$ such that $omega=rmdf$), then I can find $f$ from the following coordinate-independent formula:
$$ f(x) = int_x_0^x omega$$
where point $x_0$ and integration path can be arbitrarily chosen, as long as they are contained in $U$.
Let as assume that $sigma$ is a exact 2-form on a contractible set $U$ on a manifold $M$. Is there an analogous coordinate-independent formula allowing to find its potential (the 1-form $omega$ such that $sigma=rmdomega$)? The only ones I've seen choose a specific coordinate system, or assume that $M=mathbbR^3$.
differential-geometry differential-forms
asked Mar 13 at 10:09
Adam LatosińskiAdam Latosiński
211
New contributor
Adam Latosiński is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If I somehow know that a given 1-form $omega$ on a contractible set $U$ on a manifold $M$ is exact (there exist a 0-form $f$ such that $omega=rmdf$), then I can find $f$ from the following coordinate-independent formula:
$$ f(x) = int_x_0^x omega$$
where point $x_0$ and integration path can be arbitrarily chosen, as long as they are contained in $U$.
Let as assume that $sigma$ is a exact 2-form on a contractible set $U$ on a manifold $M$. Is there an analogous coordinate-independent formula allowing to find its potential (the 1-form $omega$ such that $sigma=rmdomega$)? The only ones I've seen choose a specific coordinate system, or assume that $M=mathbbR^3$.
differential-geometry differential-forms
differential-geometry differential-forms
asked Mar 13 at 10:09
Adam LatosińskiAdam Latosiński
211
New contributor
Adam Latosiński is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 13 at 10:09
Adam LatosińskiAdam Latosiński
211
New contributor
Adam Latosiński is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 13 at 10:09
Adam LatosińskiAdam Latosiński
211
New contributor
Adam Latosiński is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 13 at 10:09
Adam LatosińskiAdam Latosiński
211
asked Mar 13 at 10:09
Adam LatosińskiAdam Latosiński
211
211
New contributor
Adam Latosiński is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Adam Latosiński is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Adam Latosiński is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Yes, $1$-forms are particularly easy. You can write down something coordinate-independent, but you do need to integrate out the $dt$ when the null-homotopy is given by varying $t$ from $0$ to $1$.
$endgroup$
– Ted Shifrin
Mar 14 at 17:39
$begingroup$
Can you show what you mean?
$endgroup$
– Adam Latosiński
20 hours ago
$begingroup$
Let $Fcolon Utimes [0,1]to U$ be the contraction, so $F(x,1)=x$ and $F(x,0)=x_0$ (fixed). Let $tin [0,1]$. The pullback $F^*sigma$ has some $dt$ terms; we can contract against $partial/partial t$ (this gives a $1$-form with coefficients that are functions of $x$ and $t$). It can be checked that if we set $$omega = int_0^1 (iota_partial/partial tsigma)dt,$$ then $domega = sigma$.
$endgroup$
– Ted Shifrin
13 hours ago
add a comment |
$begingroup$
Yes, $1$-forms are particularly easy. You can write down something coordinate-independent, but you do need to integrate out the $dt$ when the null-homotopy is given by varying $t$ from $0$ to $1$.
$endgroup$
– Ted Shifrin
Mar 14 at 17:39
$begingroup$
Can you show what you mean?
$endgroup$
– Adam Latosiński
20 hours ago
$begingroup$
Let $Fcolon Utimes [0,1]to U$ be the contraction, so $F(x,1)=x$ and $F(x,0)=x_0$ (fixed). Let $tin [0,1]$. The pullback $F^*sigma$ has some $dt$ terms; we can contract against $partial/partial t$ (this gives a $1$-form with coefficients that are functions of $x$ and $t$). It can be checked that if we set $$omega = int_0^1 (iota_partial/partial tsigma)dt,$$ then $domega = sigma$.
$endgroup$
– Ted Shifrin
13 hours ago
$begingroup$
Yes, $1$-forms are particularly easy. You can write down something coordinate-independent, but you do need to integrate out the $dt$ when the null-homotopy is given by varying $t$ from $0$ to $1$.
$endgroup$
– Ted Shifrin
Mar 14 at 17:39
$begingroup$
Yes, $1$-forms are particularly easy. You can write down something coordinate-independent, but you do need to integrate out the $dt$ when the null-homotopy is given by varying $t$ from $0$ to $1$.
$endgroup$
– Ted Shifrin
Mar 14 at 17:39
$begingroup$
Can you show what you mean?
$endgroup$
– Adam Latosiński
20 hours ago
$begingroup$
Can you show what you mean?
$endgroup$
– Adam Latosiński
20 hours ago
$begingroup$
Let $Fcolon Utimes [0,1]to U$ be the contraction, so $F(x,1)=x$ and $F(x,0)=x_0$ (fixed). Let $tin [0,1]$. The pullback $F^*sigma$ has some $dt$ terms; we can contract against $partial/partial t$ (this gives a $1$-form with coefficients that are functions of $x$ and $t$). It can be checked that if we set $$omega = int_0^1 (iota_partial/partial tsigma)dt,$$ then $domega = sigma$.
$endgroup$
– Ted Shifrin
13 hours ago
$begingroup$
Let $Fcolon Utimes [0,1]to U$ be the contraction, so $F(x,1)=x$ and $F(x,0)=x_0$ (fixed). Let $tin [0,1]$. The pullback $F^*sigma$ has some $dt$ terms; we can contract against $partial/partial t$ (this gives a $1$-form with coefficients that are functions of $x$ and $t$). It can be checked that if we set $$omega = int_0^1 (iota_partial/partial tsigma)dt,$$ then $domega = sigma$.
$endgroup$
– Ted Shifrin
13 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can use the generalized Stoke's theorem:
$$int_partial M omega = int_M sigma.$$
Where $partial M$ is the boundary of the manifold $M$.
answered Mar 13 at 14:35
setnosetsetnoset
36519
$endgroup$
1
$begingroup$
And how does it help me to define $omega$ in any specific point? Or, considering that its value in one point can be arbitrary, how does it help me to define $omega$ in the vicinity of any specific point?
$endgroup$
– Adam Latosiński
Mar 13 at 15:22
add a comment |
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1 Answer
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$begingroup$
You can use the generalized Stoke's theorem:
$$int_partial M omega = int_M sigma.$$
Where $partial M$ is the boundary of the manifold $M$.
answered Mar 13 at 14:35
setnosetsetnoset
36519
$endgroup$
1
$begingroup$
And how does it help me to define $omega$ in any specific point? Or, considering that its value in one point can be arbitrary, how does it help me to define $omega$ in the vicinity of any specific point?
$endgroup$
– Adam Latosiński
Mar 13 at 15:22
add a comment |
$begingroup$
You can use the generalized Stoke's theorem:
$$int_partial M omega = int_M sigma.$$
Where $partial M$ is the boundary of the manifold $M$.
answered Mar 13 at 14:35
setnosetsetnoset
36519
$endgroup$
1
$begingroup$
And how does it help me to define $omega$ in any specific point? Or, considering that its value in one point can be arbitrary, how does it help me to define $omega$ in the vicinity of any specific point?
$endgroup$
– Adam Latosiński
Mar 13 at 15:22
add a comment |
$begingroup$
You can use the generalized Stoke's theorem:
$$int_partial M omega = int_M sigma.$$
Where $partial M$ is the boundary of the manifold $M$.
answered Mar 13 at 14:35
setnosetsetnoset
36519
$endgroup$
You can use the generalized Stoke's theorem:
$$int_partial M omega = int_M sigma.$$
Where $partial M$ is the boundary of the manifold $M$.
answered Mar 13 at 14:35
setnosetsetnoset
36519
answered Mar 13 at 14:35
setnosetsetnoset
36519
answered Mar 13 at 14:35
setnosetsetnoset
36519
answered Mar 13 at 14:35
setnosetsetnoset
36519
36519
1
$begingroup$
And how does it help me to define $omega$ in any specific point? Or, considering that its value in one point can be arbitrary, how does it help me to define $omega$ in the vicinity of any specific point?
$endgroup$
– Adam Latosiński
Mar 13 at 15:22
add a comment |
1
$begingroup$
And how does it help me to define $omega$ in any specific point? Or, considering that its value in one point can be arbitrary, how does it help me to define $omega$ in the vicinity of any specific point?
$endgroup$
– Adam Latosiński
Mar 13 at 15:22
1
1
$begingroup$
And how does it help me to define $omega$ in any specific point? Or, considering that its value in one point can be arbitrary, how does it help me to define $omega$ in the vicinity of any specific point?
$endgroup$
– Adam Latosiński
Mar 13 at 15:22
$begingroup$
And how does it help me to define $omega$ in any specific point? Or, considering that its value in one point can be arbitrary, how does it help me to define $omega$ in the vicinity of any specific point?
$endgroup$
– Adam Latosiński
Mar 13 at 15:22
add a comment |
Adam Latosiński is a new contributor. Be nice, and check out our Code of Conduct.
Adam Latosiński is a new contributor. Be nice, and check out our Code of Conduct.
Adam Latosiński is a new contributor. Be nice, and check out our Code of Conduct.
Adam Latosiński is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Yes, $1$-forms are particularly easy. You can write down something coordinate-independent, but you do need to integrate out the $dt$ when the null-homotopy is given by varying $t$ from $0$ to $1$.
$endgroup$
– Ted Shifrin
Mar 14 at 17:39
$begingroup$
Can you show what you mean?
$endgroup$
– Adam Latosiński
20 hours ago
$begingroup$
Let $Fcolon Utimes [0,1]to U$ be the contraction, so $F(x,1)=x$ and $F(x,0)=x_0$ (fixed). Let $tin [0,1]$. The pullback $F^*sigma$ has some $dt$ terms; we can contract against $partial/partial t$ (this gives a $1$-form with coefficients that are functions of $x$ and $t$). It can be checked that if we set $$omega = int_0^1 (iota_partial/partial tsigma)dt,$$ then $domega = sigma$.
$endgroup$
– Ted Shifrin
13 hours ago