Understanding De Rham's Period and Stokes Theorem

[Silviu]

Hello! I am reading this paper and on page 9 it defines the De Rham's period as $\int_C \omega = <C,\omega>$, where C is a cycle and $\omega$ is a closed one form i.e. $d\omega = 0$. The author says that $<C,\omega>:\Omega^p(M) \times C_p(M) \to R$. I am a bit confused by this, as $\omega$ is an one-form so in order to give a real number it needs a vector, while here it receives a cycle, which I am not sure it is a vector. Does the author mean by this that you apply $\omega$ to the vector tangent at the cycle C on the manifold at each point and add up the values? Also he then uses this in association with Stokes theorem. However the $\omega$ appearing in the Stokes theorem is not necessary closed ($d\omega$ is), so why can he still use the De Rham period there? Thank you!

 

[lavinia]

Hello! I am reading this paper and on page 9 it defines the De Rham's period as $\int_C \omega = <C,\omega>$, where C is a cycle and $\omega$ is a closed one form i.e. $d\omega = 0$. The author says that $<C,\omega>:\Omega^p(M) \times C_p(M) \to R$. I am a bit confused by this, as $\omega$ is an one-form so in order to give a real number it needs a vector, while here it receives a cycle, which I am not sure it is a vector. Does the author mean by this that you apply $\omega$ to the vector tangent at the cycle C on the manifold at each point and add up the values? Also he then uses this in association with Stokes theorem. However the $\omega$ appearing in the Stokes theorem is not necessary closed ($d\omega$ is), so why can he still use the De Rham period there? Thank you!

He means to integrate the 1 form over the cycle.

 

[Silviu]

He means to integrate the 1 form over the cycle.

But how can you integrate a one form, without a vector? A one-form is a function, so in order to integrate it, you need to give it some values. Where does it takes the vectors from? Are the tangent vectors to the cycle?

 

[lavinia]

But how can you integrate a one form, without a vector? A one-form is a function, so in order to integrate it, you need to give it some values. Where does it takes the vectors from? Are the tangent vectors to the cycle?

Evaluating the 1 form one the tangent vectors to the cycle. This is just a line integral.

 

[Orodruin]

But how can you integrate a one form, without a vector? A one-form is a function, so in order to integrate it, you need to give it some values. Where does it takes the vectors from? Are the tangent vectors to the cycle?

In general, if you can parametrise a $p$-dimensional sub-manifold $M$ with $p$ parameters $t_1$ to $t_p$, the integral of the $p$-form $\omega$ over that sub-manifold is given by
$$
\int_M \omega = \int_{M^*} \omega(\dot\gamma_1, \dot \gamma_2, \ldots, \dot\gamma_p) dt_1 \ldots dt_p,
$$
where $\dot\gamma_i$ is the tangent vector to the coordinate line of $t_i$ and $M^*$ is the region in the parameter space that maps to $M$.

 

[mathwonk]

I personally recommend you make a more thorough study of calculus of diffferential forms than the very brief sketch in this paper. you won't regret it. here is a free course on it from a professor at cornell. at least read the section on integration of one forms.

http://www.math.cornell.edu/~sjamaar/manifolds/manifold.pdf