Question about integrating differential forms

Question

Maybe it's stupid question, by why:

$$\int_S Fdx\wedge dy=\int_S Fdxdy$$

And is calculating a surface integral

$$\int_S Fdx\wedge dy+Gdy \wedge dz+H dz\wedge dx=\int_S Fdxdy+\int_SGdydz + \int_SHdzdx$$

equivalent to calculating it as

$$\int_S Fdx\wedge dy+Hdy \wedge dz+H dz\wedge dx=\int_A(G(T(s,t)),H(T(s,t)),F(T(s,t)))\cdot \vec{n} dA$$

Where $\vec{n}$ is a vector normal to surface $S$ and $T:A\rightarrow S$ is its parametrization.

(where the second integrals are common double integral).

I think it is equivalent knowing how components of normal vector looks like and that in the first integral we use change of variables theorem (also calculating jacobians).

Answer 1

1) It's mainly a matter of notation. When wedging the $1$-forms $dx$ and $dy$, one obtains the $2$-form $dx\wedge dy$. When integrating a real valued function on a domain in $\mathbb{R}^2$, it is customary to simply write $dxdy$.

2)Now we have a parametrization $T:A\to S$, where $A\subset \mathbb{R}^2,S\subset\mathbb{R}^3$, and consider for the moment the $2$-form $Gdy\wedge dz$. Pulling back one obtains $$T^*dy=\frac{\partial T_2}{\partial x}dx+\frac{\partial T_2}{\partial y}dy,\quad T^*dz=\frac{\partial T_3}{\partial x}dx+\frac{\partial T_3}{\partial y}dy,$$thus $$T^*dy\wedge dz=\left(\frac{\partial T_2}{\partial x}\cdot\frac{\partial T_3}{\partial y}-\frac{\partial T_2}{\partial y}\cdot\frac{\partial T_3}{\partial x}\right)dx\wedge dy,$$and $$T^*Gdy\wedge dz=G\left(\frac{\partial T_2}{\partial x}\cdot\frac{\partial T_3}{\partial y}-\frac{\partial T_2}{\partial y}\cdot\frac{\partial T_3}{\partial x}\right)dx\wedge dy.$$Note that the expression in brackets is no other than $N_1$, the first coordinate of the normal vector before normalizing. Performing a similar calculation for the two other $2$-forms yields $$T^*(Fdx\wedge dy+Gdy\wedge dz+Hdz\wedge dx)=(G,H,F)\cdot N,$$ and since $n$ is obtained by normalizing $N$,$$=(G,H,F)\cdot ndA.$$

Answer 2

It is not a matter of notation; it is a matter of convention.

The way a differential form integral is written can be understood as a shorthand. For instance, the integral

$$\int f \, \mathrm dx \wedge \mathrm dy$$

really means this:

$$\int f(x,y) \, (\mathrm dx \wedge \mathrm dy)(e_x \wedge e_y) \, dx \, dy$$

where $e_x$ and $e_y$ are the tangent vectors associated with $x$ and $y$. The choice to put $e_x \wedge e_y$ here is conventional: it is a conventional choice of orientation for the manifold we're integrating over. In problems in which the orientation of the manifold is specified regardless of the coordinate system used to parameterize it, this convention may give you the wrong answer, depending on the minus signs that came out.

Strictly speaking, no orientable manifold should be specified as the manifold of integration without an orientation specified. But, often we don't care what that orientation is, or that orientation is considered to be implied by the coordinate system, in which the convention is good.