Jtdylktuy

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.

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).