# Zero loci of sections of wedge product of bundles

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.

• A global section of $\wedge^2F$ is a bivector $\xi \in \wedge^2V$. The description of its zero locus is different in the case where the rank of $\xi$ is 4 or 2. Commented Nov 28, 2023 at 7:46
• @Sasha Thanks! So you mean the above description holds when $rk(\xi)\neq 2$ and $4$? How about $rk(\xi)= 2$ or $4$ cases? It seems like the zero locus is determined by $ker(\xi)$.
– Kim
Commented Nov 28, 2023 at 13:39
• No. I just want to say that in the case where $\dim V = 5$ there are two isomorphism classes of zero loci --- one (that corresponds to $\xi$ with $\mathrm{rank}(\xi) = 2$) can be indeed written as $G \cap \mathbb{P}(W_2 \wedge V)$ (where $W_2$ is the 2-dimensional subspace that orresponds to $\xi$) but the other (that corresponds to $\xi$ with $\mathrm{rank}(\xi) = 4$) is quite different. Commented Nov 28, 2023 at 17:49
• @Sasha Now I see what you mean. Could you briefly explain what happens for $\mathrm{rank}(\xi)=4$, or point out any reference?
– Kim
Commented Nov 29, 2023 at 1:20

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 $$\mathrm{LGr}(2,W_4) \cong Q^3.$$