This is what I understand in the case of finite dimensions:
Consider $1-$vectors in $R^2$. Given two $1-$vectors, $v_1$ and $v_2$, we can form a wedge product $v_1 \wedge v_2$, and if we are given $w_1$ and $w_2$ in the dual space of $R^2$, we can project $v_1$ and $v_2$ on $w_1$ and $w_2$ and take the wedge product (i.e., determinant) for the resulting two vectors, thus making $w_1$ and $w_2$ as a $1-$form. And a collection of all such $1-$forms at every point of $R^2$, will constitute a differential $1-$form in $R^2$.
Now moving on to infinite dimensions:
Consider all functions $f$ in $C[0,1]$. The dual space will consist of different sets of linear functionals. By which, I mean $\int_0^1 fdx$ could be one set of linear functionals, Dirac function $\delta_0(f) = f(0)$, could be another set, etc. I get that $dx$ is a $1-$form since $fdx$ can be imagined as a projection of $f$ on $dx$ and taking the wedge product would correspond to integrating this projection since we are computing the area. And if we do this for every function we get the differential $1-$form in $C[0,1]$.
First, I would like to know if I am right?!
If so, what would be the infinite dimensional analogue for the $2-$form in $R^3$? In other words, what are $2-$vectors in the case infinite dimensional vector space? If its the wegde product of two functions, how it is defined? Thank you very much.