$omega$ 1-form, if $d(fomega)=0$ then $omegawedge domega=0$
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"Let $omega$ be a 1-form on a smooth manifold $M$ and let $f:Mrightarrow mathbb{R}$ be a function everywhere non-vanishing such that $d(fomega)=0$. Prove that $omegawedge domega=0$".
My idea is:
$d(fomega)=dfwedge omega+fdomega=0$ that means $fdomega=omegawedge df$. Now we consider $$fomegawedge domega=omegawedge fdomega=omegawedgeomegawedge df=0$$ since $omegawedgeomega=0$. Now since $f$ is a non-vanishing function, must be $omegawedge domega=0$. Does it work?
proof-verification differential-forms exterior-algebra
edited Nov 28 at 16:26
Mike Pierce
11.4k103583
asked Nov 30 '17 at 21:51
Andrew_Paste
1328
add a comment |
up vote
7
down vote
favorite
"Let $omega$ be a 1-form on a smooth manifold $M$ and let $f:Mrightarrow mathbb{R}$ be a function everywhere non-vanishing such that $d(fomega)=0$. Prove that $omegawedge domega=0$".
My idea is:
$d(fomega)=dfwedge omega+fdomega=0$ that means $fdomega=omegawedge df$. Now we consider $$fomegawedge domega=omegawedge fdomega=omegawedgeomegawedge df=0$$ since $omegawedgeomega=0$. Now since $f$ is a non-vanishing function, must be $omegawedge domega=0$. Does it work?
proof-verification differential-forms exterior-algebra
edited Nov 28 at 16:26
Mike Pierce
11.4k103583
asked Nov 30 '17 at 21:51
Andrew_Paste
1328
1
To be more precise, you take $omegawedge domega$, and multiply it with the smooth function $1=frac ff$ which doesn't change the form. Then your argument shows that the result is $0$. But yeah, it works.
– Arthur
Nov 30 '17 at 22:16
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
"Let $omega$ be a 1-form on a smooth manifold $M$ and let $f:Mrightarrow mathbb{R}$ be a function everywhere non-vanishing such that $d(fomega)=0$. Prove that $omegawedge domega=0$".
My idea is:
$d(fomega)=dfwedge omega+fdomega=0$ that means $fdomega=omegawedge df$. Now we consider $$fomegawedge domega=omegawedge fdomega=omegawedgeomegawedge df=0$$ since $omegawedgeomega=0$. Now since $f$ is a non-vanishing function, must be $omegawedge domega=0$. Does it work?
proof-verification differential-forms exterior-algebra
edited Nov 28 at 16:26
Mike Pierce
11.4k103583
asked Nov 30 '17 at 21:51
Andrew_Paste
1328
"Let $omega$ be a 1-form on a smooth manifold $M$ and let $f:Mrightarrow mathbb{R}$ be a function everywhere non-vanishing such that $d(fomega)=0$. Prove that $omegawedge domega=0$".
My idea is:
$d(fomega)=dfwedge omega+fdomega=0$ that means $fdomega=omegawedge df$. Now we consider $$fomegawedge domega=omegawedge fdomega=omegawedgeomegawedge df=0$$ since $omegawedgeomega=0$. Now since $f$ is a non-vanishing function, must be $omegawedge domega=0$. Does it work?
proof-verification differential-forms exterior-algebra
proof-verification differential-forms exterior-algebra
edited Nov 28 at 16:26
Mike Pierce
11.4k103583
asked Nov 30 '17 at 21:51
Andrew_Paste
1328
edited Nov 28 at 16:26
Mike Pierce
11.4k103583
asked Nov 30 '17 at 21:51
Andrew_Paste
1328
edited Nov 28 at 16:26
Mike Pierce
11.4k103583
edited Nov 28 at 16:26
Mike Pierce
11.4k103583
edited Nov 28 at 16:26
Mike Pierce
11.4k103583
11.4k103583
asked Nov 30 '17 at 21:51
Andrew_Paste
1328
asked Nov 30 '17 at 21:51
Andrew_Paste
1328
asked Nov 30 '17 at 21:51
Andrew_Paste
1328
1328
1
To be more precise, you take $omegawedge domega$, and multiply it with the smooth function $1=frac ff$ which doesn't change the form. Then your argument shows that the result is $0$. But yeah, it works.
– Arthur
Nov 30 '17 at 22:16
add a comment |
1
To be more precise, you take $omegawedge domega$, and multiply it with the smooth function $1=frac ff$ which doesn't change the form. Then your argument shows that the result is $0$. But yeah, it works.
– Arthur
Nov 30 '17 at 22:16
1
1
To be more precise, you take $omegawedge domega$, and multiply it with the smooth function $1=frac ff$ which doesn't change the form. Then your argument shows that the result is $0$. But yeah, it works.
– Arthur
Nov 30 '17 at 22:16
To be more precise, you take $omegawedge domega$, and multiply it with the smooth function $1=frac ff$ which doesn't change the form. Then your argument shows that the result is $0$. But yeah, it works.
– Arthur
Nov 30 '17 at 22:16
add a comment |
1 Answer
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From the comments above.
Yes, this works.
To be more precise, take the form $omega wedge domega$ and multiply it with the smooth function $1 = frac ff$. This can be done because $f$ is non-vanishing everywhere, and note that multiplying by $1$ does not change the form.
Now, your argument shows that $1 omega wedge domega = frac 1f (f omega wedge domega) = frac 1f 0 = 0.$ Hence, $omega wedge domega = 0$.
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From the comments above.
Yes, this works.
To be more precise, take the form $omega wedge domega$ and multiply it with the smooth function $1 = frac ff$. This can be done because $f$ is non-vanishing everywhere, and note that multiplying by $1$ does not change the form.
Now, your argument shows that $1 omega wedge domega = frac 1f (f omega wedge domega) = frac 1f 0 = 0.$ Hence, $omega wedge domega = 0$.
add a comment |
up vote
1
down vote
From the comments above.
Yes, this works.
To be more precise, take the form $omega wedge domega$ and multiply it with the smooth function $1 = frac ff$. This can be done because $f$ is non-vanishing everywhere, and note that multiplying by $1$ does not change the form.
Now, your argument shows that $1 omega wedge domega = frac 1f (f omega wedge domega) = frac 1f 0 = 0.$ Hence, $omega wedge domega = 0$.
add a comment |
up vote
1
down vote
up vote
1
down vote
From the comments above.
Yes, this works.
To be more precise, take the form $omega wedge domega$ and multiply it with the smooth function $1 = frac ff$. This can be done because $f$ is non-vanishing everywhere, and note that multiplying by $1$ does not change the form.
Now, your argument shows that $1 omega wedge domega = frac 1f (f omega wedge domega) = frac 1f 0 = 0.$ Hence, $omega wedge domega = 0$.
From the comments above.
Yes, this works.
To be more precise, take the form $omega wedge domega$ and multiply it with the smooth function $1 = frac ff$. This can be done because $f$ is non-vanishing everywhere, and note that multiplying by $1$ does not change the form.
Now, your argument shows that $1 omega wedge domega = frac 1f (f omega wedge domega) = frac 1f 0 = 0.$ Hence, $omega wedge domega = 0$.
answered Nov 5 at 9:11
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Brahadeesh
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To be more precise, you take $omegawedge domega$, and multiply it with the smooth function $1=frac ff$ which doesn't change the form. Then your argument shows that the result is $0$. But yeah, it works.
– Arthur
Nov 30 '17 at 22:16