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Stokes theorem and Volume forms

2

$begingroup$

I have the following short argument which seems to say that there are no non-vanishing, (n-1)-forms on a closed manifold of dimension n but I am not very confident in my understanding of a "volume form" because I am used to only working with $textitthe$ volume form (induced by a metric) so I don't know if this applies:

Suppose that $M$ is an oriented, n-dimensional manifold which is also closed (i.e. $partial M$ is empty). Suppose also that $alpha$ is a non-vanishing (n-1)-form. Then $dalpha$ is an n-form.

$textbfQuestion 1:$ Does that mean that $dalpha$ is a "volume form"? Can I conclude that $int_Mdalpha$ is nonzero? Let's suppose the answer is yes. Let $dalpha = Vol_alpha$.

$textbfQuestion 2:$ In this case, it seems like Stokes theorem and the fact that $M$ is closed says:

$$0 neq int_MVol_alpha = int_Md alpha = int_partial M alpha = 0$$

Which is a contradiction.

If this fails, where does it fail? Are there any assumptions I can add to $alpha$ so that it would work?

linear-algebra differential-geometry riemannian-geometry stokes-theorem

share|cite|improve this question

edited Apr 3 at 14:20

pictorexcrucia

asked Apr 2 at 16:36

pictorexcruciapictorexcrucia

11310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You could have $dalpha=0$ even though $alpha$ is everywhere nonzero.
  $endgroup$
  – Lord Shark the Unknown
  Apr 2 at 16:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Even when $dalpha ne 0$, there is no need for it to be a constant multiple of the Volume form. What you have done is ruled out the possiblitity that $dalpha = k Vol$ for non-zero constant $k$.
  $endgroup$
  – achille hui
  Apr 2 at 16:46
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Here's a counterexample. Consider the torus $mathbb T^2=mathbb R^2 / mathbb Z^2$. This is a closed and compact $2$-manifold. You have the global coordinate system $(x_1, x_2)$, and $dx_1$ is a $1$-form that does not vanish at any point.
  $endgroup$
  – Giuseppe Negro
  Apr 2 at 16:48
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  (The name is Stokes, not Stoke.)
  $endgroup$
  – Hans Lundmark
  Apr 2 at 18:36
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The correct statement is that an exact $n$-form on a compact, oriented $n$-dimensional manifold with no boundary must vanish at some point.
  $endgroup$
  – Ted Shifrin
  Apr 2 at 22:52
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

I have the following short argument which seems to say that there are no non-vanishing, (n-1)-forms on a closed manifold of dimension n but I am not very confident in my understanding of a "volume form" because I am used to only working with $textitthe$ volume form (induced by a metric) so I don't know if this applies:

Suppose that $M$ is an oriented, n-dimensional manifold which is also closed (i.e. $partial M$ is empty). Suppose also that $alpha$ is a non-vanishing (n-1)-form. Then $dalpha$ is an n-form.

$textbfQuestion 1:$ Does that mean that $dalpha$ is a "volume form"? Can I conclude that $int_Mdalpha$ is nonzero? Let's suppose the answer is yes. Let $dalpha = Vol_alpha$.

$textbfQuestion 2:$ In this case, it seems like Stokes theorem and the fact that $M$ is closed says:

$$0 neq int_MVol_alpha = int_Md alpha = int_partial M alpha = 0$$

Which is a contradiction.

If this fails, where does it fail? Are there any assumptions I can add to $alpha$ so that it would work?

linear-algebra differential-geometry riemannian-geometry stokes-theorem

share|cite|improve this question

edited Apr 3 at 14:20

pictorexcrucia

asked Apr 2 at 16:36

pictorexcruciapictorexcrucia

11310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You could have $dalpha=0$ even though $alpha$ is everywhere nonzero.
  $endgroup$
  – Lord Shark the Unknown
  Apr 2 at 16:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Even when $dalpha ne 0$, there is no need for it to be a constant multiple of the Volume form. What you have done is ruled out the possiblitity that $dalpha = k Vol$ for non-zero constant $k$.
  $endgroup$
  – achille hui
  Apr 2 at 16:46
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Here's a counterexample. Consider the torus $mathbb T^2=mathbb R^2 / mathbb Z^2$. This is a closed and compact $2$-manifold. You have the global coordinate system $(x_1, x_2)$, and $dx_1$ is a $1$-form that does not vanish at any point.
  $endgroup$
  – Giuseppe Negro
  Apr 2 at 16:48
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  (The name is Stokes, not Stoke.)
  $endgroup$
  – Hans Lundmark
  Apr 2 at 18:36
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The correct statement is that an exact $n$-form on a compact, oriented $n$-dimensional manifold with no boundary must vanish at some point.
  $endgroup$
  – Ted Shifrin
  Apr 2 at 22:52
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I have the following short argument which seems to say that there are no non-vanishing, (n-1)-forms on a closed manifold of dimension n but I am not very confident in my understanding of a "volume form" because I am used to only working with $textitthe$ volume form (induced by a metric) so I don't know if this applies:

Suppose that $M$ is an oriented, n-dimensional manifold which is also closed (i.e. $partial M$ is empty). Suppose also that $alpha$ is a non-vanishing (n-1)-form. Then $dalpha$ is an n-form.

$textbfQuestion 1:$ Does that mean that $dalpha$ is a "volume form"? Can I conclude that $int_Mdalpha$ is nonzero? Let's suppose the answer is yes. Let $dalpha = Vol_alpha$.

$textbfQuestion 2:$ In this case, it seems like Stokes theorem and the fact that $M$ is closed says:

$$0 neq int_MVol_alpha = int_Md alpha = int_partial M alpha = 0$$

Which is a contradiction.

If this fails, where does it fail? Are there any assumptions I can add to $alpha$ so that it would work?

linear-algebra differential-geometry riemannian-geometry stokes-theorem

share|cite|improve this question

edited Apr 3 at 14:20

pictorexcrucia

asked Apr 2 at 16:36

pictorexcruciapictorexcrucia

11310

$endgroup$

share|cite|improve this question

edited Apr 3 at 14:20

pictorexcrucia

asked Apr 2 at 16:36

pictorexcruciapictorexcrucia

11310

share|cite|improve this question

edited Apr 3 at 14:20

pictorexcrucia

asked Apr 2 at 16:36

pictorexcruciapictorexcrucia

11310

asked Apr 2 at 16:36

pictorexcruciapictorexcrucia

11310

asked Apr 2 at 16:36

pictorexcruciapictorexcrucia

11310