Let $X$ be a projective variety and let $\mathcal{E}$ be a vector bundle of rank $r$ on $X$ which is generated by its global sections $V=\Gamma(X,\mathcal{E})$. Recall that this gives us a map $f:X\to\textrm{Gr}(r,V^*)$ to the Grassmannian as follows: To every point $x\in X$ we associate the orthogonal complement of the space of all sections $s\in V$ that vanish at $x$. On the Grassmannian $\textrm{Gr}(r,V^*)$ we have the universial sub and quotient bundles $\mathcal{S}$ and $\mathcal{Q}$. We have $\mathcal{E}=f^*(\mathcal{S}^*)$. On the other hand we can define $\mathcal{F}=f^*\mathcal{Q}$. The bundle $\mathcal{F}$ has been defined in a canonical way from the bundle $\mathcal{E}$. This leads to my question:

Can $\mathcal{F}$ can be constructed from $\mathcal{E}$ "directly on $X$" without taking this detour along the Grassmannian?


1 Answer 1


Of course, $\mathcal{F}$ is the dual of the kernel of the evaluation map $$ V \otimes \mathcal{O}_X \to \mathcal{E}. $$

  • $\begingroup$ Thanks, but can you maybe say a few words on why this is the case? $\endgroup$
    – Hans
    Oct 22, 2018 at 16:58
  • $\begingroup$ Just pullback the tautological exact sequence from the Grassmannian. $\endgroup$
    – Sasha
    Oct 23, 2018 at 17:06

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