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.

  • $\begingroup$ Is the notation $\mathbb{P}(V^*)$ a common notation in algebraic geometry for the space of $1$-dimensional subspaces of $V$ ? I've always seen it written as $\mathbb{P}(V)$. I checked the book and that is what he is doing although he calls $\mathbb{P}(V)$ the space of `$1$-dimensional quotients of $V$'. What is the definition of a $1$-dimensional quotient of $V$ ? $\endgroup$ – Michael Murray Nov 16 '12 at 12:59

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 $\mathbb P(N^*) \subset \mathbb P(E^*)$.

  • $\begingroup$ It should not be too hard to do this all with graded rings and graded modules instead. $\endgroup$ – Will Sawin Nov 16 '12 at 6:55

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