Given two form $omega$, find $mu$ such that $dmu=omega$? [duplicate]
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This question already has an answer here:
Does there exist a $1$-form $alpha$ with $dalpha = omega$?
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I am reading a textbook on differential forms and there is a question of the form: given a $k$-form, $omega$, on say $mathbb{R^3}$, find $mu$ such that $dmu=omega$. For example, if $omega= (x_{1}^2 +x_{2}^2)dx_1 wedge dx_2$, what is $mu$?
I am looking for a systematic method to solve these types of questions.
Thanks
differential-forms
asked Jan 7 at 2:40
Ac711Ac711
345
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marked as duplicate by amd, Community♦ Jan 8 at 3:02
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
Does there exist a $1$-form $alpha$ with $dalpha = omega$?
2 answers
I am reading a textbook on differential forms and there is a question of the form: given a $k$-form, $omega$, on say $mathbb{R^3}$, find $mu$ such that $dmu=omega$. For example, if $omega= (x_{1}^2 +x_{2}^2)dx_1 wedge dx_2$, what is $mu$?
I am looking for a systematic method to solve these types of questions.
Thanks
differential-forms
asked Jan 7 at 2:40
Ac711Ac711
345
$endgroup$
marked as duplicate by amd, Community♦ Jan 8 at 3:02
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Does there exist a $1$-form $alpha$ with $dalpha = omega$?
2 answers
I am reading a textbook on differential forms and there is a question of the form: given a $k$-form, $omega$, on say $mathbb{R^3}$, find $mu$ such that $dmu=omega$. For example, if $omega= (x_{1}^2 +x_{2}^2)dx_1 wedge dx_2$, what is $mu$?
I am looking for a systematic method to solve these types of questions.
Thanks
differential-forms
asked Jan 7 at 2:40
Ac711Ac711
345
$endgroup$
This question already has an answer here:
Does there exist a $1$-form $alpha$ with $dalpha = omega$?
2 answers
I am reading a textbook on differential forms and there is a question of the form: given a $k$-form, $omega$, on say $mathbb{R^3}$, find $mu$ such that $dmu=omega$. For example, if $omega= (x_{1}^2 +x_{2}^2)dx_1 wedge dx_2$, what is $mu$?
I am looking for a systematic method to solve these types of questions.
Thanks
This question already has an answer here:
Does there exist a $1$-form $alpha$ with $dalpha = omega$?
2 answers
differential-forms
differential-forms
asked Jan 7 at 2:40
Ac711Ac711
345
asked Jan 7 at 2:40
Ac711Ac711
345
asked Jan 7 at 2:40
Ac711Ac711
345
asked Jan 7 at 2:40
Ac711Ac711
345
asked Jan 7 at 2:40
Ac711Ac711
345
345
marked as duplicate by amd, Community♦ Jan 8 at 3:02
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by amd, Community♦ Jan 8 at 3:02
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
1 Answer
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$dmu=omega$ implies that $d^2mu=domega=0$. The Poincare Lemma implies that $omega$ is exact, if you look at a standard proof of this lemma, it constructs $mu$ such that $dmu=omega$ given a closed form $omega$.
https://en.wikipedia.org/wiki/Closed_and_exact_differential_forms#Poincar%C3%A9_lemma
answered Jan 7 at 3:02
Tsemo AristideTsemo Aristide
59.9k11446
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$dmu=omega$ implies that $d^2mu=domega=0$. The Poincare Lemma implies that $omega$ is exact, if you look at a standard proof of this lemma, it constructs $mu$ such that $dmu=omega$ given a closed form $omega$.
https://en.wikipedia.org/wiki/Closed_and_exact_differential_forms#Poincar%C3%A9_lemma
answered Jan 7 at 3:02
Tsemo AristideTsemo Aristide
59.9k11446
$endgroup$
add a comment |
$begingroup$
$dmu=omega$ implies that $d^2mu=domega=0$. The Poincare Lemma implies that $omega$ is exact, if you look at a standard proof of this lemma, it constructs $mu$ such that $dmu=omega$ given a closed form $omega$.
https://en.wikipedia.org/wiki/Closed_and_exact_differential_forms#Poincar%C3%A9_lemma
answered Jan 7 at 3:02
Tsemo AristideTsemo Aristide
59.9k11446
$endgroup$
add a comment |
$begingroup$
$dmu=omega$ implies that $d^2mu=domega=0$. The Poincare Lemma implies that $omega$ is exact, if you look at a standard proof of this lemma, it constructs $mu$ such that $dmu=omega$ given a closed form $omega$.
https://en.wikipedia.org/wiki/Closed_and_exact_differential_forms#Poincar%C3%A9_lemma
answered Jan 7 at 3:02
Tsemo AristideTsemo Aristide
59.9k11446
$endgroup$
$dmu=omega$ implies that $d^2mu=domega=0$. The Poincare Lemma implies that $omega$ is exact, if you look at a standard proof of this lemma, it constructs $mu$ such that $dmu=omega$ given a closed form $omega$.
https://en.wikipedia.org/wiki/Closed_and_exact_differential_forms#Poincar%C3%A9_lemma
answered Jan 7 at 3:02
Tsemo AristideTsemo Aristide
59.9k11446
edited Jan 7 at 3:15
answered Jan 7 at 3:02
Tsemo AristideTsemo Aristide
59.9k11446
answered Jan 7 at 3:02
Tsemo AristideTsemo Aristide
59.9k11446
answered Jan 7 at 3:02
Tsemo AristideTsemo Aristide
59.9k11446
59.9k11446
add a comment |
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