# non totally decomposability of vectors in Grassmannian

In "Algebraic geometry: a first course", by Harris, Grassmannian is described, under the Plucker embedding, as the locus of totally decomposable vectors in the projectivization of the exterior power $$\bigwedge^k V$$. Here

a characterization of $$m$$-decomposable vectors in $$\bigwedge^k V$$ is given; however, I am struggling to find any reference for the locus of $$m$$-decomposable vectors in $$\mathbb{P}(\bigwedge^k V)$$. I fell like it should be a variety, not necessarily smooth, that contains the Grassmannian $$G(k,V)$$ for every choice of $$m$$. I am particularly interested in the case of $$1$$-decomposable vectors in $$\mathbb{P}(\bigwedge^3 \mathbb{C}^5)$$, but I think there should exist a fancy treatment of the topic. Can you provide me some reference?

If a 2-vector is 1-decomposable, it is also 2-decomposable (more generally, if a $$k$$-vector is $$(k-1)$$-decomposable, it is $$k$$-decomposable). In particular, the locus of 1-decomposable vectors in $$\mathbb{P}(\wedge^2\mathbb{C}^5)$$ is $$G(2,5)$$.
EDIT. On the other hand, the action of the group $$\mathrm{PGL}(\mathbb{C}^5)$$ on $$\mathbb{P}(\wedge^2\mathbb{C}^5)$$ has only two orbits, the Grassmannian and its complement, and since the locus of 1-decomposable vectors is obviously $$\mathrm{PGL}(\mathbb{C}^5)$$-invariant, it follows that it is equal to the entire space $$\mathbb{P}(\wedge^2\mathbb{C}^5)$$.
• I am sorry, I wrote $\bigwedge^2 \mathbb{C}^5$ but I am really interested in $\bigwedge^3 \mathbb{C}^5$! Edited. Anyway, I am using the notation in the first answer of the question I linked, so that $G(3,\mathbb{C}^5)$ is the locus of $5$-decomposable vectors, and I am looking for the locus of vectors in the form $v\wedge \eta$ with $v\in \mathbb{C}^5$ and $\eta\in \bigwedge^2\mathbb{C}^5$. Is it clearer now? – Nutella Warrior Feb 16 at 20:57
• I see. So in general the locus of $m$-general vector in $\mathbb{P}(\bigwedge^k \mathbb{C}^n)$ correspond to the orbit under the action of PGL(\mathbb{C}^n) of a vector in the form $v_1\wedge \ldots\wedge v_m\wedge \eta$, for $v_i\in \mathbb{C}^n$ and $\eta\in \bigwedge^k \mathbb{C}^{n-m}$ – Nutella Warrior Feb 17 at 6:34
• @NutellaWarrior: A union of orbits, to be more precise. And $\eta \in \wedge^{k-m}\mathbb{C}^n$. – Sasha Feb 17 at 8:10