# Is the dual of the tautological subbundle on a Grassmannian ample?

Consider the Grassmannian $G(k,V_{n})$ of k-dimensional subspaces in an n-dimensional vector space $V_{n}$. We have the "restriction" map of vector bundles $V_n^* \rightarrow \mathcal{E}_k$, where $V_n^*$ is the trivial bundle with fiber $V_n^*$, and $\mathcal{E}_k$ is the dual of the tautological subbundle.

My question is: is $\mathcal{E}_k$ ($k\geq2$) ample (in the usual sense, of say Lazarsfeld Positivity II, Def. 6.1.1.)? What about its exterior powers, for example $\bigwedge^3(\mathcal{E}_6)$, where $(k,n) = (6,10)$? (Note: the top exterior power $\bigwedge^k(\mathcal{E}_k) = \mathcal{O}_{G(k,V_n)}(1)$, the pullback of $\mathcal{O}_{\mathbb{P}^N}(1), N = {n \choose k}$ under the Plucker embedding, and thus should be ample.)

Probably relevant: Lazarsfeld Examples 6.1.5 and 6.1.6, though he uses quotient bundles (?) and doesn't mention exterior powers. Thanks!

• Most books (for example Hartshorne's ample subvarieties) will prove these- exterior powers (upto the rank), symmetric powers $S^n, n>0$ of ample bundles are ample. Jun 5, 2016 at 17:21
• Yes, but I believe the bundle $\mathcal{E}_k$ is not ample to begin with. In fact, I think a negative answer for all exterior powers except the top can be given by Lazarsfeld's 6.1.5.(ii). Jun 5, 2016 at 17:49
• Of course, the tautological bundles of Grassmannians (except the projective space itself) are not ample. These contains lines in the Plucker embedding and the tautological bundle restricted to these lines splits as one copy of $\mathcal{O}(1)$ and the rest trivial bundles, since it is globally generated and determinant $\mathcal{O}(1)$. So, all the exterior powers except the top one are necessarily non-ample. Jun 5, 2016 at 19:16
• Could you explain that last comment? Maybe in an answer? Jun 5, 2016 at 22:27

Consider $G(k,n)$ and the usual surjection $\mathcal{O}_{G(k,n)}^n\to E$, where $E$ is the tautological bundle of rank $k$. If $k=1$, $G(k,n)$ is the projective space of dimension $n-1$, so assume that $k>1$. Then consider a general section of $E$ and let $X$ be the closed set where it vanishes. Over $X$, we have the natural restriction $\mathcal{O}_X^{n-1}\to E|_{X}$. Using this it is easy to check that $X=G(k,n-1)$ and by an easy induction, $X$ contains a line in the Plucker embedding (notice that $X$ is the intersection of $G(k,n)$ with a linear space in the Plucker embedding of $G(k,n)$), which of course is a line in $G(k,n)$. So, fix such a line $L\subset G(k,n)$. Then we have the surjection, by restricting, $\mathcal{O}_L^n\to E|_L$. Thus $E|_L=\oplus\mathcal{O}_L(a_i)$. Since $E$ is globally generated, we have $a_i\geq 0$ for all $i$. But $\det E=\mathcal{O}(1)$ implies, the only possibility is $a_i=0$ for all but one $i$ and the other is 1. I hope the rest is clear.