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Let $Gr(k,n)$ be the Grassmannian of $k$-dimensional vector subspaces $H^k$ of an $n$-dimensional vector space $V$.

Let us fix an $h$-dimensional vector subspace $\Gamma\subset V$ with $h\leq k$, and let $X\subseteq Gr(k,n)$ be the variety parametrizing the $H^k$'s containing $\Gamma$. Then $X\cong Gr(n-k,n-h)$.

Let $E$ the the rank $k-h$ vector bundle on $X$ whose fiber over a point $H^k\in X$ is $H^k/\Gamma$. Does there exists a rank $k-h$ vector bundle $F$ on $Gr(k,n)$ whose restriction to $X$ is $E$ ?

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take the flag variety F(h,k,n) whose elements are of the form $H\subset K\subset V$. We have maps $p_1:F(h,k,n)\rightarrow Gr(k,n)$ and $p_2:F(h,k,n)\rightarrow Gr(h,n)$. Take ${p_1}_*{p_2}^* \mathcal{O}(1)$. This is the desired vector bundle.

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  • $\begingroup$ I am confused. Take $h=0$, then $p_1:F(0,k,n)\rightarrow Gr(k,n)$ is an isomorphism, and $p_{1*}\mathcal{O}(1)\cong\mathcal{O}(1)$ is a line bundle. However in this case the desired vector bundle should be the universal bundle on $Gr(k,n)$. $\endgroup$
    – Sabina
    Jun 26 '17 at 21:57
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This is an interesting question. Suppose that $ h = k-1$. Then the answer appears to be yes. Consider the determinant line bundle $L $ on $Gr(k,n) $; its fibre at $ H $ is $ \det H $.

Now let us pick some a point $ H \in X$. We have the short exact sequence $$ 0 \rightarrow \Gamma \rightarrow H \rightarrow H / \Gamma \rightarrow 0 $$ and thus $ \det \Gamma \otimes \det H/\Gamma \cong \det H $. As $ \Gamma $ is a fixed vector space, $ \det \Gamma $ is a trivial line bundle and thus $ H/\Gamma \cong \det H $.

Hence $ L $ restricts to $ E $ as desired.

However, I don't see how to generalize this idea.

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