# Equality between k and p forms

1. Nov 3, 2014

### davi2686

if i have $\int_{\partial S} \omega=0$ by stokes theorem $\int_{S} d \omega=0$, can i say $d \omega=0$? even 0 as a scalar is a 0-form?

2. Nov 3, 2014

### ShayanJ

Consider $d\omega=x^3 dx$ integrated over $S=(-a,a)$. The integral gives zero but the integrand is zero in only one point of the region of integration. So this is a counterexample to $\int_S d\omega=0 \Rightarrow d\omega=0$.

3. Nov 3, 2014

### davi2686

thanks, but have no problem with 0 is a 0-form and $d\omega$ a k-form? so can i work with something like $d\omega=4$?

4. Nov 3, 2014

### HallsofIvy

Staff Emeritus
I have no idea what "$d\omega= 4$", a differential form equal to a number, could even mean. Could you please explain that?

5. Nov 3, 2014

### davi2686

my initial motivation is in Gauss's Law, $\int_{\partial V} \vec{E}\cdot d\vec{S}$=$\int_V \frac{\rho}{\epsilon_0}dV$, i rewrite the left side with differential forms, $\int_{\partial V} \star\vec{E}^{\flat}=\int_V \frac{\rho}{\epsilon_0}dV$ which by the Stokes Theorem $\int_{V} d(\star\vec{E}^{\flat})=\int_V \frac{\rho}{\epsilon_0}dV\Rightarrow d(\star\vec{E}^{\flat})=\frac{\rho}{\epsilon_0}$, if i dont make something wrong in these steps, in left side we get a n-form and right side a 0-form, and that i dont know if i can do.

6. Nov 3, 2014

### ShayanJ

Its correct that 0 is a 0-form but by a zero 1-form we actually mean $\omega= 0 dx$ and write it as $\omega= 0$ when there is no chance of confusion.

You missed something. You should have written $d(\star \vec E ^\flat)=\frac{\rho}{\epsilon_0} dV$.(What's $\flat$ anyway?)

7. Nov 5, 2014

### davi2686

Thanks i did not know that.

That is musical isomorphism $\flat:M \mapsto M^*$, in fact i understand it works like a lower indice, $\vec{B}^{\flat}$ give me a co-variant B or it related 1-form.