Mixing
Given a differential form $omega$, is there a differential form $phi$ such that $omegawedgephi$ is closed?
$begingroup$
Let $M$ be a differential manifold and $Omega^p(M)$ the vector bundle
of $p$-forms. My question is:
Given a differential $p$-form $omega$, is there a differential $q$-form $phi$ such that $d(omegawedgephi)=0$?
I am excluding the trivial cases when $omega$ is already closed or when $(q+p)$ is larger or equal to the dimension of the cotangent space at a point.
My question is a generalization of the integrating factor problem, where $omega$ is a 1-form, $phi$ is a function and $d(fomega)$ should be exact and not only closed as I am requiring. In this question about the existence of integrating factor for 1-forms in two variables the answers say that the problem is difficult and still open, even in this simpler case.
I was unable to find any reference who could be of some help in answer my question, then I would really appreciate if someone can give me some directions in the literature and, if possible, discuss some special cases
where a solution is or is not possible.
differential-geometry smooth-manifolds closed-form vector-bundles exterior-derivative
asked Jan 24 at 15:58
jobejobe
1,101615
$endgroup$
add a comment |
$begingroup$
Let $M$ be a differential manifold and $Omega^p(M)$ the vector bundle
of $p$-forms. My question is:
Given a differential $p$-form $omega$, is there a differential $q$-form $phi$ such that $d(omegawedgephi)=0$?
I am excluding the trivial cases when $omega$ is already closed or when $(q+p)$ is larger or equal to the dimension of the cotangent space at a point.
My question is a generalization of the integrating factor problem, where $omega$ is a 1-form, $phi$ is a function and $d(fomega)$ should be exact and not only closed as I am requiring. In this question about the existence of integrating factor for 1-forms in two variables the answers say that the problem is difficult and still open, even in this simpler case.
I was unable to find any reference who could be of some help in answer my question, then I would really appreciate if someone can give me some directions in the literature and, if possible, discuss some special cases
where a solution is or is not possible.
differential-geometry smooth-manifolds closed-form vector-bundles exterior-derivative
asked Jan 24 at 15:58
jobejobe
1,101615
$endgroup$
add a comment |
$begingroup$
Let $M$ be a differential manifold and $Omega^p(M)$ the vector bundle
of $p$-forms. My question is:
Given a differential $p$-form $omega$, is there a differential $q$-form $phi$ such that $d(omegawedgephi)=0$?
I am excluding the trivial cases when $omega$ is already closed or when $(q+p)$ is larger or equal to the dimension of the cotangent space at a point.
My question is a generalization of the integrating factor problem, where $omega$ is a 1-form, $phi$ is a function and $d(fomega)$ should be exact and not only closed as I am requiring. In this question about the existence of integrating factor for 1-forms in two variables the answers say that the problem is difficult and still open, even in this simpler case.
I was unable to find any reference who could be of some help in answer my question, then I would really appreciate if someone can give me some directions in the literature and, if possible, discuss some special cases
where a solution is or is not possible.
differential-geometry smooth-manifolds closed-form vector-bundles exterior-derivative
asked Jan 24 at 15:58
jobejobe
1,101615
$endgroup$
Let $M$ be a differential manifold and $Omega^p(M)$ the vector bundle
of $p$-forms. My question is:
Given a differential $p$-form $omega$, is there a differential $q$-form $phi$ such that $d(omegawedgephi)=0$?
I am excluding the trivial cases when $omega$ is already closed or when $(q+p)$ is larger or equal to the dimension of the cotangent space at a point.
My question is a generalization of the integrating factor problem, where $omega$ is a 1-form, $phi$ is a function and $d(fomega)$ should be exact and not only closed as I am requiring. In this question about the existence of integrating factor for 1-forms in two variables the answers say that the problem is difficult and still open, even in this simpler case.
I was unable to find any reference who could be of some help in answer my question, then I would really appreciate if someone can give me some directions in the literature and, if possible, discuss some special cases
where a solution is or is not possible.
differential-geometry smooth-manifolds closed-form vector-bundles exterior-derivative
differential-geometry smooth-manifolds closed-form vector-bundles exterior-derivative
asked Jan 24 at 15:58
jobejobe
1,101615
asked Jan 24 at 15:58
jobejobe
1,101615
edited Jan 24 at 16:06
jobe
asked Jan 24 at 15:58
jobejobe
1,101615
asked Jan 24 at 15:58
jobejobe
1,101615
asked Jan 24 at 15:58
jobejobe
1,101615
1,101615
add a comment |
add a comment |
1 Answer
1
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$begingroup$
a) If $omegainOmega^{p}(M)$ for $p$ odd, then $omegawedgeomega=0$, so certainly $d(omegawedgeomega)=0$.
b) If $omegainOmega^{p}(M)$ for $p$ even, then $d(omegawedge domega)=domegawedge domega=0$.
answered Jan 24 at 16:21
studiosusstudiosus
2,174715
$endgroup$
$begingroup$
Great! But I am feeling dumb... It seems that my "generalization" is trivial and not harder than the original problem for the integrating factor. Thank you!
$endgroup$
– jobe
Jan 24 at 16:27
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
a) If $omegainOmega^{p}(M)$ for $p$ odd, then $omegawedgeomega=0$, so certainly $d(omegawedgeomega)=0$.
b) If $omegainOmega^{p}(M)$ for $p$ even, then $d(omegawedge domega)=domegawedge domega=0$.
answered Jan 24 at 16:21
studiosusstudiosus
2,174715
$endgroup$
$begingroup$
Great! But I am feeling dumb... It seems that my "generalization" is trivial and not harder than the original problem for the integrating factor. Thank you!
$endgroup$
– jobe
Jan 24 at 16:27
add a comment |
$begingroup$
a) If $omegainOmega^{p}(M)$ for $p$ odd, then $omegawedgeomega=0$, so certainly $d(omegawedgeomega)=0$.
b) If $omegainOmega^{p}(M)$ for $p$ even, then $d(omegawedge domega)=domegawedge domega=0$.
answered Jan 24 at 16:21
studiosusstudiosus
2,174715
$endgroup$
$begingroup$
Great! But I am feeling dumb... It seems that my "generalization" is trivial and not harder than the original problem for the integrating factor. Thank you!
$endgroup$
– jobe
Jan 24 at 16:27
add a comment |
$begingroup$
a) If $omegainOmega^{p}(M)$ for $p$ odd, then $omegawedgeomega=0$, so certainly $d(omegawedgeomega)=0$.
b) If $omegainOmega^{p}(M)$ for $p$ even, then $d(omegawedge domega)=domegawedge domega=0$.
answered Jan 24 at 16:21
studiosusstudiosus
2,174715
$endgroup$
a) If $omegainOmega^{p}(M)$ for $p$ odd, then $omegawedgeomega=0$, so certainly $d(omegawedgeomega)=0$.
b) If $omegainOmega^{p}(M)$ for $p$ even, then $d(omegawedge domega)=domegawedge domega=0$.
answered Jan 24 at 16:21
studiosusstudiosus
2,174715
answered Jan 24 at 16:21
studiosusstudiosus
2,174715
answered Jan 24 at 16:21
studiosusstudiosus
2,174715
answered Jan 24 at 16:21
studiosusstudiosus
2,174715
2,174715
$begingroup$
Great! But I am feeling dumb... It seems that my "generalization" is trivial and not harder than the original problem for the integrating factor. Thank you!
$endgroup$
– jobe
Jan 24 at 16:27
add a comment |
$begingroup$
Great! But I am feeling dumb... It seems that my "generalization" is trivial and not harder than the original problem for the integrating factor. Thank you!
$endgroup$
– jobe
Jan 24 at 16:27
$begingroup$
Great! But I am feeling dumb... It seems that my "generalization" is trivial and not harder than the original problem for the integrating factor. Thank you!
$endgroup$
– jobe
Jan 24 at 16:27
$begingroup$
Great! But I am feeling dumb... It seems that my "generalization" is trivial and not harder than the original problem for the integrating factor. Thank you!
$endgroup$
– jobe
Jan 24 at 16:27
add a comment |
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