I have a question regarding differential 2-forms in $\mathbb R^3$.
Consider the $\Phi: \mathbb R^2 \to \mathbb R^3$ given by $\Phi(x,y):= (x,y,0)$
EDIT: The full example is here.
Is it essential that set $A:=[-1,1]\times[-1,1]$, would $[0,1]\times[0,1]$ not work?
Also why is the 2-form $a^2(dx\wedge dy)+b^2(dx\wedge dz)$? I don't understand why $a$ and $b$ have been squared in this formula?
Apologies if this question is completely trivial. Any help would be great!
I'm guessing (lacking other information) that the author chose $a^2$ and $b^2$ so that they were both nonnegative. It's just a shorthand way of saying it. It may also be that the author wants things whose derivatives are basically $a$ and $b$ (up to a constant). Hard to know without further context.
As for choosing the set $A$, you could choose any subset of the plane and define the 2-form, but presumably in the lines that follow your excerpt, the author will be doing something with that subset. For instance, it may matter that the subset includes the origin in its interior, and $[0, 1] \times [0, 1]$ wouldn't have that property. Without further information, I can't help you.
But as long as $A$ is a subset of the plane, the function $\omega$ can be defined on $A$.