Grassmannian as a submanifold of $\Lambda^m(E)$.

2011-11-14T11:54:30Z

Let $E$ be a vector space of dimension $d\ge4$ over $K$, and $2\le m\le d$ be an integer. I am interested in the characterization of those elements $\omega$ of $\Lambda^m(E)$ that can be written in the simplest form $v_1\wedge\cdots\wedge v_m$. Equivalently, the line $K\omega$ is a point in the Grassmannian $G_{m,d}(K)$.

When $m=2$, the necessary and sufficient condition is that $\omega\wedge\omega=0$. When $m=d-2$, we may use a duality map $\Lambda^2(E)\leftrightarrow\Lambda^{d-2}(E)$ to derive a NSC in the form $\omega'\wedge\omega'=0$. In both cases, the characterization is given by polynomial equations in the coefficients of $\omega$ in some basis. I expect that such a characterization exists for every pair $(m,d)$, but I do not see what it must be. Clearly, a condition like $\omega\wedge\omega=0$ is not correct; there are several reasons for that. For instance, not all $\omega$ are elementary products when $d<2m$, yet we have $\omega\wedge\omega\equiv0$. Also, if $m=3$ and $d\ge5$, a non-zero element $\omega=\alpha\wedge v$ where $\alpha\in\Lambda^2(E)$ is not elementary, is not an elementary product, yet satisfies $\omega\wedge\omega=0$.

Answer by Ben McKay for Grassmannian as a submanifold of $\Lambda^m(E)$.

2011-11-14T12:01:54Z

Griffiths and Harris, p. 209: the conditions on a multivector that it be decomposable are precisely that its equals its annihilator under wedge product, and is equivalent to the vanishing of a collection of quadratic equations, given explicitly on p. 211, in terms of homogeneous coordinates.

Grassmannian as a submanifold of $\Lambda^m(E)$. - MathOverflow

Grassmannian as a submanifold of $\Lambda^m(E)$.

Answer by Ben McKay for Grassmannian as a submanifold of $\Lambda^m(E)$.