Let $\mathcal{F}$ be a locally free sheaf of rank $d$ on a scheme $X$ together with an epimorphism $\mathcal{O}_X^n \to \mathcal{F}$. Now due to abstract reasons (Plücker embedding, Serre's results on coherent sheaves etc.) there is a canonical exact sequence of the form $(\wedge^d \mathcal{F})^{\otimes k_2})^{r_2} \to (\wedge^d \mathcal{F})^{\otimes k_1})^{r_1} \to \mathcal{F}^* \to 0$. Here $\mathcal{F}^*$ is the dual of $\mathcal{F}$. Canonical means that the left morphism is a matrix which is built up out of exterior powers and tensor products of the given global sections of $\mathcal{F}$, in a way that does not depend on any other choices. But how can we get this sequence explicitly? I already know that there is a canonical exact sequence ${(\wedge^d \mathcal{F})^{\*}}^{\binom{n}{d+1}} \to \mathcal{O}_X^n \to \mathcal{F} \to 0$. The naive idea just to dualize it does not work since the arrows reverse their direction.

$\begingroup$ I don't understand what $F$ (not $\mathcal F$) is supposed to be in the exact sequence. $\endgroup$– Torsten EkedahlCommented Jul 24, 2011 at 6:16

$\begingroup$ It was a typo; now corrected. $\endgroup$– Martin BrandenburgCommented Jul 24, 2011 at 8:36
1 Answer
We have that $\mathcal F^\ast$ is, by the pairing induced by the exterior algebra, canonically isomorphic to $\Lambda^{d1}\mathcal F\bigotimes(\Lambda^d\mathcal F)^{1}$. Now, in general if $\mathcal H\to\mathcal G\to \mathcal F\to 0$ is exact then the kernel of the surjective map $\Lambda^\ast \mathcal G\to\Lambda^\ast\mathcal F$ is the ideal generated by the image of $\mathcal H\to\mathcal G$. Hence we get an exact sequence $\Lambda^{i1}\mathcal G\bigotimes \mathcal H \to \Lambda^i \mathcal H\to\Lambda^i\mathcal F\to0$. Applpy, this your second exact sequence and $i=d1$ gives a presentation of the desired type for $\Lambda^{d1}\mathcal F$ and then twist it by $(\Lambda^d\mathcal F)^{1}$.

$\begingroup$ I didn't know that $\mathcal{F}^* \cong \Lambda^{d1} \mathcal{F} \otimes (\Lambda^d \mathcal{F})^*$, this is very useful. Thanks! $\endgroup$ Commented Jul 24, 2011 at 10:15