# Koszul complex on Grassmannian

We are on a Grassmannian $G = G(r,n)$ with tautological subbundle $\mathscr{F}$ of rank $n-r$ and quotient bundle $\mathscr{Q}$ of rank $r$. These vector bundles occur by the following exact sequence $$0 \to \mathscr{F}^* \to V \to \mathscr{Q} \to 0,$$ where $V$ is the trivial bundle on $G$. How do we obtain the following Koszul complex? $S^{a}(\mathscr{F})$ denotes the a-th symmetric power and $\mathcal{O}_G^{\oplus m}$ denotes some copies of the trivial line bundle. Fulton: Intersection Theory, P431, B.3.4 doesn't help either.

$$0 \to \wedge^a\mathscr{Q}^* \to \mathcal{O}^{\oplus \binom{n}{a}}_G \to \mathcal{O}^{\oplus \binom{n}{a-1}}_G \otimes \mathscr{F} \to \dots \to \mathcal{O}^{\oplus \binom{n}{1}}_G \otimes S^{a-1}(\mathscr{F}) \to S^a(\mathscr{F}) \to 0.$$

• Are you sure this is correct? For instance, with $r=n-1$, so $\mathscr{F} = \mathcal{O}(-1)$, the last map is (a sum of maps) $\mathcal{O}(-a+1) \to \mathcal{O}(-a)$. But there are no such maps. Perhaps you mean to use $\mathscr{F}^*$? Sep 20, 2016 at 14:10
• Yes, I edited it. Sep 21, 2016 at 18:44

You should think of the trivial bundles along the sequence as $\bigwedge^i V^*$ and also the correct sequence should use $\mathrm{Sym}^j \mathscr{F}^*$. So, the maps are now

$$\bigwedge^i V^* \otimes \mathrm{Sym}^j \mathscr{F}^* \to \bigwedge^{i-1} V^* \otimes \mathrm{Sym}^{j+1} \mathscr{F}^*.$$

This is given by factoring out a vector from the $\bigwedge V^*$ term and multiplying it onto the $\mathrm{Sym}\mathscr{F}^*$ term (using the surjection $V^* \twoheadrightarrow \mathscr{F}^*$). In coordinates:

$$v_1^* \wedge \cdots \wedge v_i^* \otimes \phi^* \mapsto \sum_\ell (-1)^\ell v_1^* \wedge \cdots \wedge \widehat{v_\ell^*} \wedge \cdots \wedge v_i^* \otimes v_\ell^* \cdot \phi^*.$$

Here I am using $\phi^*$ to denote the monomial, the section of $\mathscr{F}^*$, and the notation $\widehat{x}$ means "omit the factor $x$".

In other words, this is the canonical inclusion $\bigwedge^i V^* \hookrightarrow \bigwedge^{i-1} V^* \otimes V^*$, followed by applying the extracted tensor factor $V^*$ in the composition $V^* \otimes \mathrm{Sym}^j \mathscr{F}^* \twoheadrightarrow \mathscr{F}^* \otimes \mathrm{Sym}^j \mathscr{F}^* \twoheadrightarrow \mathrm{Sym}^{j+1} \mathscr{F}^*.$

Incidentally the dual map is cleaner to write down (no signs needed): it is

$$\bigwedge^{i-1} V \otimes \mathrm{Sym}^{j+1} \mathscr{F} \to \bigwedge^i V \otimes \mathrm{Sym}^j \mathscr{F},$$ $$\alpha \otimes f_1 \cdots f_{j+1} \mapsto \sum_\ell \alpha \wedge f_\ell \otimes \frac{f_1 \cdots f_{j+1}}{f_\ell}.$$

I'm not going to check the rest of the claims carefully, though it is easy to check that this gives a complex of vector bundles. Exactness should follow from the same reason ordinary Koszul complexes are exact: $\bigwedge^a \mathscr{Q}$ is a quotient of $\bigwedge^a V$ by a regular sequence (locally), namely all those wedges that contain vectors from $\mathscr{F}$.

• Thanks for your comprehensive answer! The first part might be clear to me, only the last paragraph seems quite unclear. What do you mean by the last sentence? A filtration sequence considering $\wedge^aV$? Sep 22, 2016 at 7:43
• Maybe? Like I said, I haven't really thought through this part, it's just how I would expect it to go. All I meant is that, locally, we can choose bases for $\mathscr{F}$ and $\mathscr{Q}$, thereby getting a basis for $V$. Then $\bigwedge^aV \twoheadrightarrow \bigwedge^a \mathscr{Q}$ is obtained by modding out by all the wedges $v_1 \wedge \cdots \wedge v_a$ where some $v_i$ is a basis element from $\mathscr{F}$. In other words, the sequence $\mathscr{F} \otimes \bigwedge^{a-1}V \to \bigwedge^a V \to \bigwedge^a \mathscr{Q} \to 0$ is right exact. Sep 22, 2016 at 15:17
• (And it looks like the beginning of a Koszul complex resulting from modding out by a regular sequence.) Sep 22, 2016 at 15:19
• Yep, I think that was what I needed to know. Thank you! Sep 23, 2016 at 7:05