13
$\begingroup$

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.

$\endgroup$
9
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.