# Extending the tautological bundle of $G(1,3)$?

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 $\mathbb P^5$?

## 2 Answers

I would say it doesn't. The tautological Chern classes generate $H^*$ of the Grassmanian, but the inclusion to $\mathbb{P}^4$ does not induce an epimorphism.

• (you mean P^5, right?) – Qfwfq Feb 21 '14 at 11:27
• Yes, sure. Actually, any $\mathbb{P}$ :) – Alex Degtyarev Feb 21 '14 at 11:29
• Alex, I think your reasoning is correct. Could you give me a reference to some nice book that explains that tautological Chern classes generate $H^*$ of the Grassmanian? – aglearner Feb 21 '14 at 11:31
• Over $\mathbb{Q}$ this is just comparing the dimensions (the keyword being "Schubert calculus"). A lot about Grassmanians (including everything about your favorite one) is found here: David Eisenbud and Joe Harris, 3264 & All That: Intersection Theory in Algebraic Geometry – Alex Degtyarev Feb 21 '14 at 11:43

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.