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.

  • $\begingroup$ 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! $\endgroup$ May 13, 2016 at 23:01
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    $\begingroup$ math.stackexchange.com/questions/341540/… might be helpful. $\endgroup$
    – tom
    May 14, 2016 at 1:48

1 Answer 1


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).

  • $\begingroup$ 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? $\endgroup$
    – Wooster
    May 13, 2016 at 22:15
  • $\begingroup$ 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. $\endgroup$ May 13, 2016 at 22:55
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    $\begingroup$ @Wooster I've improved my answer, substantially, I think, to give some more intuition about how rare decomposable $k$-forms are. $\endgroup$ May 19, 2016 at 16:30
  • $\begingroup$ @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. $\endgroup$
    – user775699
    Jul 9, 2022 at 7:08
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    $\begingroup$ @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 $T$). $\endgroup$ Jan 22, 2023 at 2:23

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