Volume form on a compact manifold is not exact
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I am trying to show that a volume form $mu$ on a compact manifold $M$ is not exact, i.e. show there is no $alpha in Omega^{n-1}(M)$ such that $dalpha = mu$.
My attempt is the following: Suppose, as a contradiction, that $mu$ is exact. Then, there exist an $(n-1)$-form $alpha$ such that $dalpha = mu$. Then, since compact manifolds are manifolds without boundary, by stokes theorem we know that $int_M mu = int_M dalpha = int_{partial M} alpha = 0$ since $M$ has no boundary. From here I am not sure how to continue to end my contradiction. I would appreciate any hint or suggestions for the problem. Thanks!
differential-topology stokes-theorem
asked Dec 11 '18 at 15:48
BOlivianoperuano84BOlivianoperuano84
1778
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add a comment |
$begingroup$
I am trying to show that a volume form $mu$ on a compact manifold $M$ is not exact, i.e. show there is no $alpha in Omega^{n-1}(M)$ such that $dalpha = mu$.
My attempt is the following: Suppose, as a contradiction, that $mu$ is exact. Then, there exist an $(n-1)$-form $alpha$ such that $dalpha = mu$. Then, since compact manifolds are manifolds without boundary, by stokes theorem we know that $int_M mu = int_M dalpha = int_{partial M} alpha = 0$ since $M$ has no boundary. From here I am not sure how to continue to end my contradiction. I would appreciate any hint or suggestions for the problem. Thanks!
differential-topology stokes-theorem
asked Dec 11 '18 at 15:48
BOlivianoperuano84BOlivianoperuano84
1778
$endgroup$
$begingroup$
You argument is correct. Note that $int_M mu$ is positive from the definitions of volume form and integral.
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– Dante Grevino
Dec 11 '18 at 16:12
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Oh, ok so the contradiction comes from that fact of positivity. Thanks so much for the clarification!
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– BOlivianoperuano84
Dec 11 '18 at 17:09
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@BOlivianoperuano84 I recommend that you write an answer to your own question math.stackexchange.com/help/self-answer. You should also accept it math.stackexchange.com/help/accepted-answer. The benefit is that it becomes visible at first glance in the question list that is no longer open.
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– Paul Frost
Dec 11 '18 at 17:31
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Thanks Paul! I will write the answer!
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– BOlivianoperuano84
Dec 11 '18 at 18:53
add a comment |
$begingroup$
I am trying to show that a volume form $mu$ on a compact manifold $M$ is not exact, i.e. show there is no $alpha in Omega^{n-1}(M)$ such that $dalpha = mu$.
My attempt is the following: Suppose, as a contradiction, that $mu$ is exact. Then, there exist an $(n-1)$-form $alpha$ such that $dalpha = mu$. Then, since compact manifolds are manifolds without boundary, by stokes theorem we know that $int_M mu = int_M dalpha = int_{partial M} alpha = 0$ since $M$ has no boundary. From here I am not sure how to continue to end my contradiction. I would appreciate any hint or suggestions for the problem. Thanks!
differential-topology stokes-theorem
asked Dec 11 '18 at 15:48
BOlivianoperuano84BOlivianoperuano84
1778
$endgroup$
I am trying to show that a volume form $mu$ on a compact manifold $M$ is not exact, i.e. show there is no $alpha in Omega^{n-1}(M)$ such that $dalpha = mu$.
My attempt is the following: Suppose, as a contradiction, that $mu$ is exact. Then, there exist an $(n-1)$-form $alpha$ such that $dalpha = mu$. Then, since compact manifolds are manifolds without boundary, by stokes theorem we know that $int_M mu = int_M dalpha = int_{partial M} alpha = 0$ since $M$ has no boundary. From here I am not sure how to continue to end my contradiction. I would appreciate any hint or suggestions for the problem. Thanks!
differential-topology stokes-theorem
differential-topology stokes-theorem
asked Dec 11 '18 at 15:48
BOlivianoperuano84BOlivianoperuano84
1778
asked Dec 11 '18 at 15:48
BOlivianoperuano84BOlivianoperuano84
1778
asked Dec 11 '18 at 15:48
BOlivianoperuano84BOlivianoperuano84
1778
asked Dec 11 '18 at 15:48
BOlivianoperuano84BOlivianoperuano84
1778
asked Dec 11 '18 at 15:48
BOlivianoperuano84BOlivianoperuano84
1778
1778
$begingroup$
You argument is correct. Note that $int_M mu$ is positive from the definitions of volume form and integral.
$endgroup$
– Dante Grevino
Dec 11 '18 at 16:12
$begingroup$
Oh, ok so the contradiction comes from that fact of positivity. Thanks so much for the clarification!
$endgroup$
– BOlivianoperuano84
Dec 11 '18 at 17:09
$begingroup$
@BOlivianoperuano84 I recommend that you write an answer to your own question math.stackexchange.com/help/self-answer. You should also accept it math.stackexchange.com/help/accepted-answer. The benefit is that it becomes visible at first glance in the question list that is no longer open.
$endgroup$
– Paul Frost
Dec 11 '18 at 17:31
$begingroup$
Thanks Paul! I will write the answer!
$endgroup$
– BOlivianoperuano84
Dec 11 '18 at 18:53
add a comment |
$begingroup$
You argument is correct. Note that $int_M mu$ is positive from the definitions of volume form and integral.
$endgroup$
– Dante Grevino
Dec 11 '18 at 16:12
$begingroup$
Oh, ok so the contradiction comes from that fact of positivity. Thanks so much for the clarification!
$endgroup$
– BOlivianoperuano84
Dec 11 '18 at 17:09
$begingroup$
@BOlivianoperuano84 I recommend that you write an answer to your own question math.stackexchange.com/help/self-answer. You should also accept it math.stackexchange.com/help/accepted-answer. The benefit is that it becomes visible at first glance in the question list that is no longer open.
$endgroup$
– Paul Frost
Dec 11 '18 at 17:31
$begingroup$
Thanks Paul! I will write the answer!
$endgroup$
– BOlivianoperuano84
Dec 11 '18 at 18:53
$begingroup$
You argument is correct. Note that $int_M mu$ is positive from the definitions of volume form and integral.
$endgroup$
– Dante Grevino
Dec 11 '18 at 16:12
$begingroup$
You argument is correct. Note that $int_M mu$ is positive from the definitions of volume form and integral.
$endgroup$
– Dante Grevino
Dec 11 '18 at 16:12
$begingroup$
Oh, ok so the contradiction comes from that fact of positivity. Thanks so much for the clarification!
$endgroup$
– BOlivianoperuano84
Dec 11 '18 at 17:09
$begingroup$
Oh, ok so the contradiction comes from that fact of positivity. Thanks so much for the clarification!
$endgroup$
– BOlivianoperuano84
Dec 11 '18 at 17:09
$begingroup$
@BOlivianoperuano84 I recommend that you write an answer to your own question math.stackexchange.com/help/self-answer. You should also accept it math.stackexchange.com/help/accepted-answer. The benefit is that it becomes visible at first glance in the question list that is no longer open.
$endgroup$
– Paul Frost
Dec 11 '18 at 17:31
$begingroup$
@BOlivianoperuano84 I recommend that you write an answer to your own question math.stackexchange.com/help/self-answer. You should also accept it math.stackexchange.com/help/accepted-answer. The benefit is that it becomes visible at first glance in the question list that is no longer open.
$endgroup$
– Paul Frost
Dec 11 '18 at 17:31
$begingroup$
Thanks Paul! I will write the answer!
$endgroup$
– BOlivianoperuano84
Dec 11 '18 at 18:53
$begingroup$
Thanks Paul! I will write the answer!
$endgroup$
– BOlivianoperuano84
Dec 11 '18 at 18:53
add a comment |
1 Answer
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By definition, the integral over a manifold of a top (volume) form must be positive. Thus, since we have that $int_M mu = 0$, this is the contradiction we are looking for and so we deduce that a volume form on a compact manifold is not exact.
answered Dec 12 '18 at 18:39
BOlivianoperuano84BOlivianoperuano84
1778
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add a comment |
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$begingroup$
By definition, the integral over a manifold of a top (volume) form must be positive. Thus, since we have that $int_M mu = 0$, this is the contradiction we are looking for and so we deduce that a volume form on a compact manifold is not exact.
answered Dec 12 '18 at 18:39
BOlivianoperuano84BOlivianoperuano84
1778
$endgroup$
add a comment |
$begingroup$
By definition, the integral over a manifold of a top (volume) form must be positive. Thus, since we have that $int_M mu = 0$, this is the contradiction we are looking for and so we deduce that a volume form on a compact manifold is not exact.
answered Dec 12 '18 at 18:39
BOlivianoperuano84BOlivianoperuano84
1778
$endgroup$
add a comment |
$begingroup$
By definition, the integral over a manifold of a top (volume) form must be positive. Thus, since we have that $int_M mu = 0$, this is the contradiction we are looking for and so we deduce that a volume form on a compact manifold is not exact.
answered Dec 12 '18 at 18:39
BOlivianoperuano84BOlivianoperuano84
1778
$endgroup$
By definition, the integral over a manifold of a top (volume) form must be positive. Thus, since we have that $int_M mu = 0$, this is the contradiction we are looking for and so we deduce that a volume form on a compact manifold is not exact.
answered Dec 12 '18 at 18:39
BOlivianoperuano84BOlivianoperuano84
1778
answered Dec 12 '18 at 18:39
BOlivianoperuano84BOlivianoperuano84
1778
answered Dec 12 '18 at 18:39
BOlivianoperuano84BOlivianoperuano84
1778
answered Dec 12 '18 at 18:39
BOlivianoperuano84BOlivianoperuano84
1778
1778
add a comment |
add a comment |
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$begingroup$
You argument is correct. Note that $int_M mu$ is positive from the definitions of volume form and integral.
$endgroup$
– Dante Grevino
Dec 11 '18 at 16:12
$begingroup$
Oh, ok so the contradiction comes from that fact of positivity. Thanks so much for the clarification!
$endgroup$
– BOlivianoperuano84
Dec 11 '18 at 17:09
$begingroup$
@BOlivianoperuano84 I recommend that you write an answer to your own question math.stackexchange.com/help/self-answer. You should also accept it math.stackexchange.com/help/accepted-answer. The benefit is that it becomes visible at first glance in the question list that is no longer open.
$endgroup$
– Paul Frost
Dec 11 '18 at 17:31
$begingroup$
Thanks Paul! I will write the answer!
$endgroup$
– BOlivianoperuano84
Dec 11 '18 at 18:53