This question was motivated by this related one: How "far" a differential form is from an exterior product .

Let $\mathbb{V}$ be a vector space of dimension $n$ with underlying field $\mathbb{F}$, and say (for lack of a better term) that the wedge rank of a $k$-form $$\phi \in \Lambda^k \mathbb{V}^*$$ is the minimum number $r$ for which there are exist wedge products $v_a^1 \wedge \cdots \wedge v_a^k$, $a = 1, \ldots, r$, of $1$-forms $v_a^b \in \mathbb{V}^*$, $b = 1, \ldots, k$, such that $$\phi = \sum_{a = 1}^r v_a^1 \wedge \cdots \wedge v_a^k.$$ (For convenience, we can declare the empty sum to have value the $0$ k-form, so that the wedge rank of $0$ is $0$.)

In general, given $\phi$, what is an effective way to determine its wedge rank $r$?

We can make a few obvious remarks: First, $r \leq \dim \Lambda^k \mathbb{V}^* = {{n}\choose{k}}$, but in general it is much smaller, and anyway for nonzero $0$-, $1$- and $n$-forms, $r = 1$, and exploiting the natural isomorphism $\Lambda^{n - 1} \mathbb{V}^* \cong \mathbb{V} \otimes \Lambda^n \mathbb{V}^*$ gives that the same applies to nonzero $(n - 1)$-forms.

For $k = 2$, a $2$-form $\phi$ has wedge rank $1$ (that is, it is decomposable) iff $\phi \wedge \phi = 0$, and we can exploit the isomorphism $\Lambda^{n - 2} \mathbb{V}^* \cong \Lambda^2 \mathbb{V} \otimes \Lambda^n \mathbb{V}^*$ to make an analogous statement about the $k = n - 2$ case. Furthermore, if $n$ is even, the wedge rank of $\phi$ is exactly $r$ iff $$\underbrace{\phi \wedge \cdots \wedge \phi}_r \neq 0 \qquad \text{but} \qquad \underbrace{\phi \wedge \cdots \wedge \phi}_{r + 1} = 0.$$ (Perhaps something similar holds for odd $n$?)

In higher tensor ranks, the story quickly becomes more complicated. For example, if $\dim \mathbb{V} = 7$ (and $\mathbb{F}$ perfect and $\text{char } \mathbb{F} \neq 2$), the tensor rank of a $3$-form $\phi$ is at most $5$ (this already seems nonobvious). It turns out (at least over $\mathbb{R}$ and $\mathbb{C}$) that $r = 5$ iff the $\Lambda^7 \mathbb{V}^*$-valued bilinear form $$(X, Y) \mapsto (i_X \phi) \wedge (i_Y \phi) \wedge \phi$$ is nondegenerate, but the rank of the bilinear form does not determine $r$ for all smaller values of $r$. Anyway, this particular property seems essentially unique to this $(n, k)$.

There's a further complication, namely that the wedge rank of a $k$-form need not remain the same under extension of the base field. This phenomenon already shows up in the smallest-dimensional case not covered by the above considerations: If $\mathbb{F} = \mathbb{R}$ and $\dim \Bbb V = 6$, there is a $3$-form whose stabilizer under the pullback action of $GL(\mathbb{V}) \cong GL(6, \mathbb{R})$ on $\Lambda^3 \mathbb{V}^*$ is exactly $SU(3)$, and any such $3$-form has wedge rank $4$ (in fact, there is a single $GL(\Bbb V)$-orbit of such $3$-forms, and it is open). When viewed as an element of the complex vector space $\mathbb{V} \otimes_{\mathbb{R}} \mathbb{C}$, however, any such $3$-form has wedge rank $2$. So, the structure of the underlying field $\mathbb{F}$ plays a (to me) subtle role, and quite possibly it turns out this question is easier to answer over algebraically closed fields.


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 $\min\{\text{rank}(A),\ A\in M_n(\mathbb{F}) \text{ and } A-A^t=PBP^t\}=s=\dfrac{rank(B)}{2}$ $\square$

  • $\begingroup$ Thanks, Daniel. Do you know whether this implies for odd-dimensional vector spaces that the wedge rank of $\omega$ is $p$ iff $\omega^p \neq 0$ but $\omega^{p + 1} = 0$? And do you have any insight into whether this approach can be used to give at least a partial answer for $k$-forms, $k \geq 3$? $\endgroup$ – Travis Sep 26 '14 at 16:04
  • 3
    $\begingroup$ @Travis, I don't know the answer for your first question. For the second I suspect the problem is really difficult. For the usual tensor products spaces with $k\geq 3$ the problem of computing tensor rank is NP-hard. See the answer provided by Nathaniel Johnston in this question. mathoverflow.net/questions/162116/… $\endgroup$ – Daniel Sep 26 '14 at 16:13

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