5
$\begingroup$

Let $G$ be a Grassmannian and $Q$ the tautological/universal quotient bundle of $G$. As far as I understand, the associated tautological quotient line bundle for the Plucker embedding of the Grassmannian $G$ is $\mathrm{det} Q$ and is isomorphic to the pullback of the very ample line bundle of the projective space (in which the Grassmannian is embedded). However, I could not find a reference where this is explained nicely. Could someone suggest a good reference for this.

$\endgroup$

2 Answers 2

6
$\begingroup$

Your claim descends directly from the definition of the Plucker embedding.

Let $G(n,h)$ be the Grassmannian parametrizing $h$-linear subspaces of $\mathbb{P}^n = \mathbb{P}(V)$. The fiber of the universal bundle $\mathcal{S}$ over $[L]\in G(n,h)$ is the vector space $W$, where $L = \mathbb{P}(W)$.

Now, the Plucker embedding $p:G(h,n)\rightarrow\mathbb{P}^N$ is intrinsically defined mapping $W\mapsto\bigwedge^{h+1}W$. Therefore, $p^{*}\mathcal{O}_{\mathbb{P}^N}(1)$ is a line bundle on $G(h,n)$ whose fiber over $[L]\in G(n,h)$ is the vector space $\bigwedge^{h+1}W$, where $L = \mathbb{P}(W)$, that is $$p^{*}\mathcal{O}_{\mathbb{P}^N}(1)= \bigwedge^{h+1}\mathcal{S}.$$

$\endgroup$
1
  • $\begingroup$ Don't you need to take the dual of $\mathcal{S}$? $\endgroup$
    – KS_
    Aug 7, 2020 at 0:33
2
$\begingroup$

Everything you've said sounds right, so I think you would be on track to work out the details on your own. One place I found with a quick Google is this book of Tevelev. There's also a reasonable discussion in these notes of Brion.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .