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

Question

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!

Answer 1

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.