How "far" a differential form is from an exterior product

Question

Consider two differential manifolds $X$ and $Y$. Consider now a differential form (of any order) $\omega$ on $X\times Y$. The easiest example is taking $\omega=\xi\wedge\eta$, where $\xi$ is a differential form on $X$ and $\eta$ is a differential form on $Y$.

In general, any form $\omega$ is not such an exterior product. Is there a way to "measure" how much $\omega$ differs from being such a decomposition?

I'm looking for something in the "spirit" of:

The rank of a matrix measures how much the matrix differs from being a tensor product of two vectors.

Is there an analogous quantity for differential forms?

Thanks.

Answer 1

Here's a partial answer, which gives a more direct analogue of the matrix rank example but (like my earlier comment) ignores the product structure.

It follows partially from the subadditivity property $$\text{rank}(A + B) \leq \text{rank} A + \text{rank} B$$ for $m \times n$ matrices $A, B$ that the rank of a matrix $A$ is the minimum number $r$ for which there exist rank-$1$ matrices $M_1, \ldots M_r$ such that $$A = \sum_{a = 1}^r M_a;$$ in the language of covariant tensors, we can say that for a tensor $A \in \mathbb{V}^* \otimes \mathbb{W}^*$ this quantity is the smallest number of simple elements $v_a \otimes w_a \in \mathbb{V}^* \otimes \mathbb{W}^*$ such that $$A = \sum_{i = 1}^r v_a \otimes w_a.$$ (Unfortunately, the term tensor rank is already used for the number of arguments of a tensor when viewed as an $\mathbb{F}$-valued multilinear map.)

Similarly, for a vector space $\mathbb{V}$ over a field $\mathbb{F}$, we could define the "wedge rank" of a $k$-form $$\phi \in \Lambda^k \mathbb{V}^*$$ (for lack of a more inspired name) to be the minimum number $r$ for which there are simple wedge products $v_a^1 \wedge \cdots \wedge v_a^k$ for which $$\phi = \sum_{i = 1}^r v_a^1 \wedge \cdots \wedge v_a^k.$$

For a particular vector space dimension $n$ and rank $k$, obviously the wedge rank of any element is no more than $\dim \Lambda^k \mathbb{V} = {{n}\choose{k}}$, but in general it will be much smaller than this. Also, note that wedge rank is invariant under the natural action of $GL(\mathbb{V})$ on $\Lambda^k \mathbb{V}^*$ by pullback.

Note too the nonobvious (to me) fact wedge rank of a form can change when one extends the base field $\mathbb{F}$. For example, if $\dim \mathbb{V} = 6$ and $\mathbb{F} = \mathbb{R}$, there is an open set of $3$-forms $\phi$ whose stabilizers under the $GL(\mathbb{V})$-action are isomorphic to $SU(3)$, and the wedge rank for each of these is (I believe) $4$. However, if we view such any $\phi$ as an element of $\mathbb{V} \otimes \mathbb{C}$, its wedge rank is $2$.

The concept of wedge rank turns out to occur in at least few interesting places:

• For $n$ even and $\mathbb{F} = \mathbb{R}$ or $\mathbb{F} = \mathbb{C}$, the wedge rank of any $2$-form $\phi$ is $\leq \frac{n}{2}$, and equality holds precisely when $\phi$ is a symplectic form on $\mathbb{V}$, that is, when $$\underbrace{\phi \wedge \cdots \wedge \phi}_{n/2} \neq 0.$$ As mentioned in my earlier comment, a $2$-form $\phi$ has wedge rank $1$ iff $\phi \wedge \phi = 0$. In particular, if $\dim \mathbb{V} = 3$, all nonzero $2$-forms have wedge rank $1$, and when $\dim \mathbb{V} = 4$, these results together give an immediate test for determining the "wedge rank" of a $2$-form.

• For $n = 7$ and $\mathbb{F}$ perfect (including $\mathbb{R}$ and $\mathbb{C}$) and $\text{char } \mathbb{F} \neq 2$, the wedge rank of any $3$-form $\phi$ is $\leq 5$. If $\mathbb{F} = \mathbb{C}$, then equality holds iff the stabilizer of $\phi$ under the natural action of $GL(\mathbb{V}) \cong GL(7, \mathbb{C})$ is isomorphic to the exceptional simple complex Lie group $G_2$ (!), and such a $3$-form is equivalent to a choice of $7$-dimensional cross product structure on $\mathbb{V}$. If $\mathbb{F} = \mathbb{R}$, the stabilizer in $GL(\mathbb{V}) \cong GL(7, \mathbb{R})$ of a $3$-form of wedge rank $5$ is either of two the real forms of $G_2$. Relatedly, such a $3$-form can be used to construct a $7$-dimensional cross product on $\mathbb{V}$, which can in turn be used to construct the algebra of the octonions (or its somehow weirder cousin, the split-octonions).

Edit I've asked a new followup question about effectively determining wedge rank for a general $k$-form: How can I determine the number of wedge products of $1$-forms needed to express a $k$-form as a sum of such?

Answer 2

Let $V$ denote an $n$-dim vector space (over $\mathbb{R}$, say). Recall that a $p$-form $\alpha \in \Lambda^p(V^*)$ is decomposable (or simple) iff it can be written as a wedge product of $1$-forms -- i.e., $\alpha = \omega_1 \wedge \cdots \wedge \omega_p$, with each $\omega_i \in \Lambda^1(V^*) = V^*$.

Definition: Let $\alpha \in \Lambda^p(V^*)$ be a $p$-form. The space of linear divisors of $\alpha$ is $$L_\alpha := \{\omega \in V^* \colon \omega \wedge \alpha = 0\} \subset V^*.$$

Exercise: Let $\alpha \in \Lambda^p(V^*)$ be a $p$-form. Then:

(a) $\dim(L_\alpha) \leq p$.

(b) $\dim(L_\alpha) = p$ $\iff$ $\alpha$ is decomposable.

The reason for the name "linear divisors" has to do with the following proposition and corollary:

Prop: Let $\alpha \in \Lambda^p(V^*)$. Let $\{\omega^1, \ldots, \omega^q\}$ be a basis of $L_\alpha$. Then we can write $$\alpha = \omega^1 \wedge \cdots \wedge \omega^q \wedge \pi$$ for some $\pi \in \Lambda^{p-q}(V^*)$.

That is, we can "divide" $\alpha$ by one-forms $q$ times. Said another way, $q = \dim(L_\alpha)$ tells us how many times we can divide $\alpha$ by one-forms.

Cor: Let $\alpha \in \Lambda^p(V^*)$. Let $\omega \in V^*$, $\omega \neq 0$. Then $\alpha = \omega \wedge \psi$ for some $\psi \in \Lambda^{p-1}(V^*)$ if and only if $\alpha \wedge \omega = 0$.

That is, $\omega$ "divides" $\alpha$ if and only if $\omega \in L_\alpha$. This explains the name.

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References

Agricola, Friedrich, "Global Analysis: Differential Forms in Analysis, Geometry, and Physics."

Bryant, Chern, Gardner, Goldschmidt, Griffiths, "Exterior Differential Systems."