Projective bundle given by vanishing of a section

2012-11-16T06:43:48Z

This question might be tautological. It comes from a statement in the proof of the non-emptiness of the degeneracy loci of a vector bundle homomorphism that Prof. Lazarsfeld gives in his book "Positivity in AG II" (Theorem 7.2.1)

Take a homomorphism of vector bundles $v:E\rightarrow F$ with kernel $F=\ker v$ and image $K=Im v$ and consider the short exact sequence $$0\rightarrow N \rightarrow E \rightarrow K \rightarrow 0$$

The surjection $E^{\ast}\to N^{\ast}$ gives an embedding $\mathbb{P}(N^{\ast})\hookrightarrow \mathbb{P}(E^{\ast})$ and we seek to realize $\mathbb{P}(N^{\ast})$ as the zero locus of the section of some vector bundle.

The projectibve bundle $\pi:\mathbb{P}(E^{\ast})\rightarrow Y$ comes with a tautological surjection $$\pi^{\ast}E^{\ast}\rightarrow \mathcal{O}_{\mathbb{P}(E^{\ast})}(1) \rightarrow 0$$

(given by the identity $E^{\ast}\rightarrow E^{\ast}$) and the composition of its dual $\mathcal{O}_{\mathbb{P}(E^{\ast})}(-1) \rightarrow \pi^{\ast}E$ with the pullback homomorphism $\pi^{\ast}v:\pi^{\ast}E \rightarrow \pi^{\ast}K$ gives a section

$$s\in \Gamma(\mathbb{P}(E),\pi^{\ast}K \otimes \mathcal{O}_{\mathbb{P}(E^{\ast})}(1))$$

Then it is claimed that the zero-locus of this section gives precisely the subvariety $\mathbb{P}(N^{\ast})\hookrightarrow \mathbb{P}(E^{\ast})$.

I am wondering whether this is obvious or not, but this correspondence is not apparent to me.

Thanks in advance for any insight.

Answer by Will Sawin for Projective bundle given by vanishing of a section

2012-11-16T06:55:10Z

It's tautological. Remember that $\mathcal O_{\mathbb P(E^*)}(-1)$ , viewed as an actual bundle of lines, is the tautological bundle: It is $E$ with the origin blown up. The map to $\pi^* E$ is literally the map from $E$ with the origin blown up to $E$. So the map to $\pi^* K$ is that, composed with the projection down to $K$. The section is trivial at a point if the map is trivial in the corresponding line - that is, if the point of $\mathbb P(E^*)$ corresponds to a line lying in the part of $E$ that maps to $0$ in $K$ - that is, if the line lies in $N \subset E$.

But the points with lines lying in $N \subset E$ are of course just $\mathbb P(N^*) \subset \mathbb P(E^*)$ .

Projective bundle given by vanishing of a section - MathOverflow

Projective bundle given by vanishing of a section

Answer by Will Sawin for Projective bundle given by vanishing of a section