How can I determine the number of wedge products of $1$-forms needed to express a $k$-form as a sum of such?

If $char(\mathbb{F})\neq 2$ and $V=\mathbb{F}^n$ then it is simple to compute the wedge rank in $\mathbb{F}^n\wedge\mathbb{F}^n$. It does not depend on the field.

• Definition: If $w= r\wedge s\neq 0$ and $r,s\in \mathbb{F}^n$ then we say that $w$ has wedge rank $1$. The wedge rank of an arbitrary $0\neq w\in \mathbb{F}^n\wedge\mathbb{F}^n$ is the minimal number of elements of $\mathbb{F}^n\wedge\mathbb{F}^n$ with wedge rank 1 that must be added to obtain $w$.

Let $char(\mathbb{F})\neq 2$ and consider $A_n(\mathbb{F})=\{A\in M_n(\mathbb{F}), -A=A^t\}$ and the following isomorphism $$G:\mathbb{F}^n\wedge\mathbb{F}^n\rightarrow A_n(\mathbb{F}), \ \ \ G(\sum_{i=1}^mv_i\wedge w_i)=\sum_{i=1}^m v_iw_i^t-\sum_{i=1}^m w_iv_i^t.$$

• Thm 1: Suppose $char(\mathbb{F})\neq 2$. The wedge rank of $w\in \mathbb{F}^n\wedge\mathbb{F}^n $ is $$\min\{rank(A),\ A\in M_n(\mathbb{F}) \text{ and } A-A^t=G(w)\}.$$

• Thm 2: Suppose $char(\mathbb{F})\neq 2$. Let $B\in A_n(F)$. Then $$\min\{\text{rank}(A),\ A\in M_n(\mathbb{F}) \text{ and } A-A^t=B\}=\frac{rank(B)}{2}$$

• Corollary: If $char(\mathbb{F})\neq 2$ then the wedge rank of $w\in \mathbb{F}^n\wedge\mathbb{F}^n $ is $\dfrac{rank(G(w))}{2}$.

  Thus, the wedge rank depends only on the rank of $G(w)$, which does not depend on the field.

Proof of Thm 1: For every decomposition of $w$ as $\sum_{i=1}^mv_i\wedge w_i$ we obtain a matrix $A=\sum_{i=1}^mv_iw_i^t$ such that $G(w)=A-A^t$ and $rank(A)\leq m$.

Thus, $\min\{rank(A),\ A\in M_n(\mathbb{F}) \text{ and } A-A^t=G(w)\}\leq wedge\ rank(w).$

Next, for every matrix $A$ such that $A-A^t=G(w)$, we can write $A=\sum_{i=1}^mv_iw_i^t$, where m is the rank of $A$. Thus, $w=G^{-1}(\sum_{i=1}^mv_iw_i^t-\sum_{i=1}^mw_iv_i^t)=\sum_{i=1}^mv_i\wedge w_i$.

Thus, $wedge\ rank(w)\leq \min\{rank(A),\ A\in M_n(\mathbb{F}) \text{ and } A-A^t=G(w)\}$.

Finally, $wedge\ rank(w)= \min\{rank(A),\ A\in M_n(\mathbb{F}) \text{ and } A-A^t=G(w)\}$. $\square$

Proof of Thm 2: If $char(\mathbb{F})\neq 2$, we know that exists a invertible matrix $P\in M_n(\mathbb{F})$ such that $PBP^t=\left( \begin{array}{ccc} 0_{s\times s} & I_{s\times s} & 0_{s\times n-2s} \\ -I_{s\times s} & 0_{s\times s} & 0_{s\times n-2s} \\ 0_{n-2s\times s} & 0_{n-2s\times s} & 0_{n-2s \times n-2s} \end{array}\right)$, where 2s is the rank of $B$.

Now, notice that $\min\{\text{rank}(A),\ A\in M_n(\mathbb{F}) \text{ and } A-A^t=B\}=$ $$=\min\{\text{rank}(A),\ A\in M_n(\mathbb{F}) \text{ and } A-A^t=PBP^t\}.$$

Now, for every $A\in M_n(\mathbb{F})$ such that $A-A^t=PBP^t$, we must have $rank(A)\geq s$, because $2s=rank(PBP^t)\leq 2rank(A)$. Thus, $\min\{\text{rank}(A),\ A\in M_n(\mathbb{F}) \text{ and } A-A^t=PBP^t\}\geq s$.

Finally, $A=\left( \begin{array}{ccc} 0_{s\times s} & I_{s\times s} & 0_{s\times n-2s} \\ 0_{s\times s} & 0_{s\times s} & 0_{s\times n-2s} \\ 0_{n-2s\times s} & 0_{n-2s\times s} & 0_{n-2s \times n-2s} \end{array}\right)$ satisfy $A-A^t=PBP^t$ and $A$ has rank $s$. Therefore, $\min\{\text{rank}(A),\ A\in M_n(\mathbb{F}) \text{ and } A-A^t=PBP^t\}\leq s$, which implies $\min\{\text{rank}(A),\ A\in M_n(\mathbb{F}) \text{ and } A-A^t=PBP^t\}=s=\dfrac{rank(B)}{2}$ $\square$