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Preparation.

Let $R$ be a noetherian ring (perhaps we have to assume $2 \in R^*$), $1 \leq d \leq n$ and consider the Plücker embedding $\omega : G \hookrightarrow P$, where $G = \text{Grass}_d(R^n)$ and $P = \mathbb{P}(R^{\binom{n}{d}}) = \text{Proj}(R[\{x_{i_1,...,i_d}\}])$. On $X$-valued points points, this maps a quotient $s: \mathcal{O}_X^n \twoheadrightarrow \mathcal{F}$, where $\mathcal{F}$ is locally free of rank $d$, to the $d$th exterior power $x : \mathcal{O}_X^{\binom{n}{d}} \twoheadrightarrow \wedge^d \mathcal{F}$; see EGA I, § 9.

Let's make the convention that every expression of the form $p_{i_1,...,i_d}$ is meant to be alternating in the indices. Now $\omega$ is a closed immersion, which is cut out by the quasi-coherent ideal $I$ corresponding to the graded ideal of $R[\{x_{i_1,...,i_d}\}]$, generated by the Plücker relations

$\sum_{\lambda=1}^{d} (-1)^{\lambda} x_{i_1,...,i_{d-1},k_\lambda} x_{k_0,...,\hat{k_\lambda},...,k_d}$

for all $1 \leq i_1,...,i_{d-1},k_0,...,k_d \leq n$. Thus $\omega$ induces an isomorphism $G \cong V(I) \subseteq P$. This gives another interpretation (and also proof) of the Plücker embedding, see here for a related topic.

Now considering the identity of $G$ as a $G$-valued point of $G$, we get a (universal) locally free sheaf $\mathcal{F}$ of rank $d$ on $G$ (together with an epimorphism $s : \mathcal{O}^n \twoheadrightarrow \mathcal{F}$). Consider this on $V(I)$. Then $\omega_* \mathcal{F}$ is a coherent sheaf on $P$. If we apply Serre's Theorem twice, we see that $\omega_* \mathcal{F}$ fits into an exact sequence of the form

$\mathcal{O}_P(q_1)^{r_1} \to \mathcal{O}_P(q_2)^{r_2} \to \omega_\* \mathcal{F} \to 0.$

This may be viewn as a "twisted presentation". Here the map on the left is given by a $r_2 \times r_1$-matrix of global sections of $\mathcal{O}_P(q_2 - q_1)$, which are just homogenuous polynomials of degree $q_2 - q_1$ in the variables $x_{i_1,...,i_d}$.

Question. How does such a matrix look like explicitly?

Here is what I've done so far: Using the identifications, calculate explicitely $\mathcal{F}$ by means of cocycles on the covering of basic-opens $D_+(x_{i_1,...,i_d})$. This works. Then I wanted to calculate the graded module $\Gamma_*(\mathcal{F})$ and try to see twisted generators and relations. But it's all a big mess. By the way, I need this calculation for a sort of Plücker embedding in another context.

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Sasha's exact sequence below is correct, but there are also more elementary means to find it. –  Martin Brandenburg Jul 20 '11 at 12:38
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1 Answer

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To get a simpler answer it is better to fix a free $R$-module $V$ of rank $n$ so that $G = Gr(d,V)$ and $P = P(\Lambda^dV)$. First we have a tautological epimorphism $V\otimes O_G \to F$ on $G$ which by adjunction gives an epimorphism $V\otimes O_P \to i_*F$. Denote the kernel by $K$. Twisting the short exact sequence $$ 0 \to K \to V\otimes O_P \to i_*F \to 0 $$ by $O(1)$ and taking cohomology one gets $$ H^0(K) = Ker(V\otimes \Lambda^d V) \to \Sigma^{2,1^{d-1}}V), $$ where $\Sigma$ is the Schur functor. It follows that $H^0(K) = \Lambda^{d+1}V$, so we have $$ \Lambda^{d+1}V\otimes O_P(-1) \to V\otimes O_P \to i_*F \to 0. $$ I guess this is what you need.

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Thanks a lot! Twice you mean $K(1)$, right? You seem to use that $K(1)$ is generated by its global sections and that the first cohomology vanishes, why is that so? Also, I have to learn first what Schur functors are at all, could you please tell me what you have used here about their connection with cohomology? –  Martin Brandenburg Jul 6 '11 at 8:59
    
I've read some basics about Schur functors, but still don't understand your proof. Also, could you please indicate how the morphism $\wedge^{d+1} V \otimes O_P(-1) \to V \otimes O_P$ looks like? –  Martin Brandenburg Jul 7 '11 at 19:57
    
The map is the composition $\Lambda^{d+1}V\otimes O(-1) \subset V\otimes \Lambda^dV\otimes O(-1) \to V\otimes O$, the first map is the natural embedding and the second is the tautological map on $P$ tensored by $V$. –  Sasha Jul 7 '11 at 20:19
    
Thanks. Could you also explain the rest? We have an exact sequence $0 \to H^0(K(1)) \to H^0(V \otimes O_P(1)) \to H^0((i_* F)(1)) \to H^1(K(1)) \to H^1(V \otimes O_P(1))$. Here we have $H^1(V \otimes O_P(1))=0$ for $1<d<n$, and $H^0(V \otimes O_P(1)) = V \otimes \wedge^d(V)$. But what about $H^0((i_* F)(1)) \to H^1(K(1))$? –  Martin Brandenburg Jul 7 '11 at 20:57
    
$H^0(i_*F(1)) = H^0(F(1)) = \Sigma^{2,1^{d-1}}V$ by Theorem of Borel--Bott--Weil. The map $V\otimes \Lambda^dV \to \Sigma^{2,1^{d-1}}V$ is the canonical projection, so $H^1(K(1)) = 0$, $H^0(K(1)) = \Lambda^[d+1}V$. –  Sasha Jul 7 '11 at 21:33
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