I have a problem understanding the general series representation of a p-form.
For 1-form things are pretty clear to me: For $h = (h_1, \dots, h_n)^T \in \mathbb{R}^n $ and $ h = \sum\limits_{i = 1}^{n} h_i e_i $ one has:
However one may generalizes this to p-forms and gets for: $h_j = (h_{1j}, \dots, h_{jn})$ and $ j = 1, \dots, p $ :
with the wegde product here $dx_{i_1} \wedge \dots \wedge dx_{i_p}$ usually referring to the determinants.
Choosing however $n = 3$ and $p = 2$ gives me the problem that $3 \times 2 $ matrix for which the determinant isn't defined. Thus either the series representation stated above is flawed (something I don't believe) or $n = 3$, $p =2 $ doesn't make any sense and/or my understanding is flawed.
Thus I would be very happy, if someone could point out my misconception. As always any constructive comments, answers are appreciated.
Even when $n=3$, you're still just dealing with a $2\times 2$ matrix, albeit now there are 3 of them!
Notice that the sum is over choices of numbers $1\le i_1 \le i_2 \le 3$ (in your case). Therefore, there's a term with $i_1=1$ and $i_2=2$, one with $i_1=2$ and $i_2=3$, and one with $i_1=1$ and $i_2=3$.
In each of the three terms, the determinant you take comes from the square matrix that you get by choosing just two of the three columns, and then idea is that any $p$-form can be represented as a combination of these forms. (These $p\times p$ "subdeterminants" of the $p\times n$ matrix are usually called its "maximal minors.")
When $p=1$, by the way, you can see that your matrix is just a long row, and you only get to "choose" one column. Therefore, you just choose a single one of your $x_i$'s. The maximal minors of a row vector are just each of its entries.