Wedge products of lenght $r$ generate the $r$-forms

Question

I've read some book that use the wedge products of $dx_i$ to define $r$-forms in $\mathbb{R}^n$. However, I have a professor that defines the $r$-forms just using that are $r$-linear and antisymmetric.

How can I get from here that the wedge products of lenght $r$ generate the differential $r$-forms?

Many thanks in advance.

Answer 1

This is a fundamental fact from multilinear algebra. I'll follow Tu's "Introduction to Manifolds," and present it from a more general perspective.

First, say we have a basis $e_1,\cdots e_n$ for a real vector space $V$, and let $\alpha^1,\cdots,\alpha^n$ be the dual basis for $V'.$ First, one proves a lemma

Lemma. If $I=(i_1,i_2,\cdots,i_k)$ and $J=(j_1,j_2,\cdots,j_k)$ are multi-indices with $1\leq i_1<\cdots<i_k\leq n$ and $1\leq j_1<\cdots<j_k\leq n$, then $$\alpha^I(e_J)=\delta_J^I,$$ where $e_J=(e_{i_1},\cdots, e_{i_k})$ and $\alpha^I=\alpha^{i_1}\wedge\cdots\wedge \alpha^{i_k}.$

The proof is not hard, and it uses the relationship between evaluating a wedge of $k$ covectors and the determinant. Using such a lemma, one can proceed to show that

Proposition. The set $\{\alpha^{i_1}\wedge\cdots\wedge \alpha^{i_k}: 1\leq i_1<\cdots <i_k\leq n\}$ forms a basis for $A_k(V),$ the space of alternating $k$-linear functions on $V$.

The lemma is used to prove linear independence. Showing that they span is straightforward. For details, I recommend reading Tu (pages $31$-$32$).

In your situation, $V=T_p(\mathbb{R}^n),$ and $e_j=\partial/\partial_{x_j}|_p,$ $\alpha^j=dx^j|_p.$