# Deciding whether a form in the exterior power $\bigwedge^k V$ is decomposable

Let $V$ be a vector space and $\bigwedge^kV$ be the $k$th exterior power. I'm trying to find a condition that characterizes when an element $\omega \in \bigwedge^kV$ is decomposable in the sense that $\omega = v_1 \wedge ... \wedge v_k$ for some $v_i \in V$.

Now if $\omega$ is decomposable, then $\omega^2 = 0$, and I wondered whether the converse holds in the general case? (Or perhaps for some restrictions on the dimension of $V$ or k?). This is trivially true for $k=1$ but I'm not sure about other cases.

• What textbook are you using to learn multilinear/exterior algebra by the way? I've tried to learn the subject several times, but it never seemed to click with me as much as did, say, functional analysis. Your help/recommendations would be greatly appreciated! May 13, 2016 at 23:01
– tom
May 14, 2016 at 1:48

For $$k < 2$$ and $$k > \dim V - 2$$ it's always true that a $$k$$-form $$\omega$$ both is decomposable and satisfies $$\omega \wedge \omega = 0$$.

For $$k = 2$$ (and provided the field underlying $$V$$ does not have characteristic $$2$$), the condition $$\omega \wedge \omega = 0$$ is both necessary and sufficient for decomposability. (Proving this is a nice exercise.)

The converse is not true in general, however. If $$k$$ is odd, then all $$k$$-forms $$\omega$$ satisfy $$\omega \wedge \omega = 0$$, but not all odd-degree multivectors (or, dually, forms) are decomposable:

Example If $$\dim V \geq 5$$, pick a basis $$(E_a)$$ and denote the dual basis by $$(e_a)$$. Then, the $$3$$-form $$\psi := (e^1 \wedge e^2 + e^3 \wedge e^4) \wedge e^5$$ satisfies $$\psi \wedge \psi = 0$$ but we can show that it is indecomposable: Contracting a vector into a decomposable form yields a decomposable form. On the other hand, $$\iota_{E^5} \psi = e^1 \wedge e^2 + e^3 \wedge e^4 ,$$ and computing gives $$(\iota_{E^5} \psi) \wedge (\iota_{E^5} \psi) \neq 0$$, so by the criterion for $$k = 2$$, $$\iota_{E^5} \psi$$ is indecomposable.

For an algorithm that checks decomposability of a general $$k$$-form, see this old question.

Remark For $$2 \leq k \leq \dim V - 2$$, most $$k$$-forms are not decomposable, and we can quantify this assertion: For $$0 \leq k \leq \dim V$$, any $$k$$-vector $$E_{a_1} \wedge \cdots \wedge E_{a_k}$$ determines a $$k$$-plane in $$V$$, namely, $$\langle E_{a_1}, \cdots, E_{a_k} \rangle$$, and any $$k$$-plane in $$V$$ determines an underlying form up to an overall nonzero multiplicative constant. So, we may regard the space $$D_k(V)$$ of (nonzero) decomposable $$k$$-forms as a (punctured) line bundle over the space of all $$k$$-planes in $$V$$; this latter space is called the Grassmannian (manifold), $$Gr(k, V)$$, and it has dimension $$k (\dim V - k)$$, so $$D_k(V)$$ is a smooth manifold of dimension $$k (\dim V - k) + 1$$. On the other hand, the space of all $$k$$-forms has dimension $$n \choose k$$, and for $$2 \leq k \leq \dim V - 2$$, $$\dim D_k(V) = k (\dim V - k) + 1 < {\dim V \choose k}$$ (but note that equality holds for $$k = 1, \dim V - 1$$).

Alternatively, the Plücker embedding realizes $$Gr(k, V)$$ as a projective variety in $$\Bbb P(\Lambda^k V)$$, and when $$2 \leq k \leq \dim V - 2$$ it is a proper subvariety, so its complement is nonempty and Zariski-open (and hence, when the underlying field is $$\Bbb R$$ or $$\Bbb C$$, dense with respect to the usual topology).

• That's great thanks for your answer and the counter example. I'm now trying to prove the first statement, namely the $k=2$ case, besides writing this out explicitly in a basis (which doesn't seem to help too much), I'm struggling where to start with that, do you have any hints? May 13, 2016 at 22:15
• You're welcome, I hope you found as useful. As for a hint: One can do this is an adapted basis, but it's faster to suppose that $\omega \wedge \omega = 0$, contract a vector $v$ into $\omega \wedge \omega$, and apply the Leibniz Rule for the wedge product. May 13, 2016 at 22:55
• @Wooster I've improved my answer, substantially, I think, to give some more intuition about how rare decomposable $k$-forms are. May 19, 2016 at 16:30
• @TravisWillse Can you please write an answer for this question of mine:math.stackexchange.com/questions/4443610/… I am a beginner in this field and i shall be really grateful if you can write an answer explaining the solution. I am badly struck on this.
– user775699
Jul 9, 2022 at 7:08
• @tparker That's a good question. Naively, if we fix $w \in \bigwedge^k V$, $x \mapsto w \wedge x$ is a linear map $T : V \mapsto \bigwedge^{k + 1} V$, so with the components of $w$ w.r.t. some basis you can use your favorite algorithm for computing the kernel of a matrix (specifically, the matrix representation $[T]$ of $T$). Jan 22, 2023 at 2:23