It is well known that the Grassmanian of lines in $\mathbb P^3$ is isomorphic to a quadric in $\mathbb P^5$. I would like to ask if the tautological rank two bundle on the grassmanian extends to a rank two bundle on the whole of $\mathbb P^5$?
Here is an answer for this specific case. Let $Q$ be the tautological bundle on $X$, the Grassmannian. If it can be extended to say $E$ on 5-spcae, using the fact that $H^1_*(Q)=0$, one gets from the exact sequence, $0\to E(-2)\to E\to Q\to 0$, that $H^1_*(E)=0$ and thus $H^0(E)\to H^0(Q)$ is onto. In particular, you can lift the 4 sections of $Q$ to $E$ and since they generate $Q$, we see that they generate $E$ outside finite set of points. Restricting to a general hyperplane not passing through these points, we get a 4 generated rank 2 bundle on 4-space, which by a theorem of Faltings (see Inventiones, 1981), is trivial. Then the usual Horrocks theorem will say that $E$ itself is trivial and thus so is $Q$, which is absurd.