Zero loci of sections of wedge product of bundles

Question

Let $V$ be a $\mathbf{C}$-vector space of dimension $n$, and consider the Grassmannian $G:=Gr(2, V)$ of 2-dim subspaces of $V$. Then we have the tautological subbundle $E\subset V\otimes \mathcal{O}_G$ and the quotient bundle $V\otimes \mathcal{O}_G\twoheadrightarrow F$.

Classically, we know that the zero locus of a non-zero global section of $E^{\vee}$, if non-empty, is $G\cap \mathbf{P}(\wedge^2 W)$, where $W\subset V$ is a codimension one subspace and the intersection is taken under the Plucker embedding $G\hookrightarrow \mathbf{P}(\wedge^2 V)$. Similarly, if we use the natural identification between $Gr(2, V)$ and $Gr(3, V^{\vee})$, then one can see that the non-empty zero locus of a non-zero global section of $F$ is $G\cap \mathbf{P}(W_1\wedge W)$, where $W\subset V$ is a codimension one subspace and $W_1$ is a 1-dim subspace.

Question: Is there any similar description of the non-empty zero locus of a non-zero global section of $\wedge^2 F$?

When $\dim V=5$, I guess it is $G\cap \mathbf{P}(W_2\wedge V)$ (or a linear section of $G\cap \mathbf{P}(W_2\wedge V)$?) according to some lemmata in papers (without proofs), where $W_2$ is a 2-dim subspace of $V$, but I'm not sure how to get this.

Answer 1

A global section of $\wedge^2F$ is a bivector $\xi \in \wedge^2V$ and the zero locus of such a section is the scheme parameterizing all 2-dimensional subspaces $U \subset V$ such that the image of $\xi$ in $\wedge^2(V/U)$ is zero.

Assume $\dim(V) = 5$. Then the action of $\mathrm{GL}(V)$ on $\wedge^2V$ has two orbits:

• bivectors of rank 2, i.e., $\xi = v_1 \wedge v_2$,

• bivectors of rank 4, i.e., $\xi = v_1 \wedge v_2 + v_3 \wedge v_4$,

where $v_1,v_2,v_3,v_4$ are linearly independent vectors of $V$.

If the rank of $\xi$ is 2, the condition that the image of $\xi$ in $\wedge^2(V/U)$ is zero is equivalent to the condition that a linear combination of $v_1$ and $v_2$ is contained in $U$, and then $$ \wedge^2U \subset v_1 \wedge V + v_2 \wedge V = W_2 \wedge V, $$ where $W_2$ is the linear span of $v_1$ and $v_2$. It is easy to see that this is the cone over $$ \mathbb{P}(W_2) \times \mathbb{P}(V/W_2) \cong \mathbb{P}^1 \times \mathbb{P}^2 $$ with vertex at the point $\mathbb{P}(\wedge^2W_2)$.

If the rank of $\xi$ is 4, the condition that the image of $\xi$ in $\wedge^2(V/U)$ is zero implies that $$ U \subset W_4 := \langle v_1, v_2, v_3, v_4 \rangle $$ and when this holds, it is equivalent to the condition that $U$ is isotropic with respect to the symplectic form of $W_4$ associated with $\xi$. Thus, in this case the zero locus is $$ \mathrm{LGr}(2,W_4) \cong Q^3. $$