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Premise: I have asked the same question on math.stackexchange about a week ago, without receiving answer. Therefore I've decided to ask it also here. If this violates the rules, I apologize and I'll delete this question.

We work over ${\mathbb C}$. Let $Y$ be a smooth projective variety, and let $\mathcal{E}$ be a locally free sheaf of rank $r$ over $Y$. Define $X:= \mathbb{P}_Y(\mathcal{E})$ to be the projective bundle over $Y$ (following Hartshorne's book notation), with projection map $\pi:X\to Y$.

I know that sections $s:Y\to X$ of $\pi$ are in correspondence with quotient invertible sheaves $\mathcal{E}\to \mathcal{L}\to 0$. This suggests me that the quotient invertible sheaf $\mathcal{L}$ tells something about the image of section $s$, and I would like to understand it with explicit examples.

For instance, suppose I have that I have the following quotients:

  1. $\mathcal{E}\to \mathcal{O}_Y\to 0$
  2. $\mathcal{E}\to \mathcal{O}_Y(1)\to 0$
  3. $\mathcal{E}\to \mathcal{O}_Y(-1)\to 0$

Thanks to Harthsorne, chap. II, exercise 7.8, I know that I obtain sections $s: Y\to X$ such that, in each of the above quotients, we get that:

  1. $s^*\mathcal{O}_{X}(1)=\mathcal{O}_Y$
  2. $s^*\mathcal{O}_X(1)=\mathcal{O}_Y(1)$
  3. $s^*\mathcal{O}_X(1)=\mathcal{O}_Y(-1)$

I would expect that these quotients give rise to different sections. Is there a way to understand the geometry of the associated sections? For instance, am I contracting the image of the section? Is it isomorphic to $Y$?

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    $\begingroup$ Of course, $\pi $ induces an isomorphism of $s(Y)$ onto $X$. And I don't understand what you mean by "am I contracting the image of the section?". $\endgroup$
    – abx
    Commented Dec 6 at 6:31
  • $\begingroup$ Dear @abx, thank you very much for your answer. I guess I am quite confused about all of this, I thought the sections do not need to be isomorphic to $Y$, and my guess was that we constructed the section as the projective spectra of the section rings $\bigoplus_{m\geq 0} H^0(Y,ms^*\mathcal{O}_Y(1))$ (and in case 1., this was the meaning of "we contract the section") $\endgroup$
    – YetAnotherMathStudent
    Commented Dec 9 at 6:59
  • $\begingroup$ But for instance, consider a cone $X=\mathbb{P}(\mathcal{E})$ over a smooth projective variety $Y$, which has a structure of $\mathbb{P}^1$-bundle over $Y$. I thought that I could construct a section $s: Y\to X$ such that $s(Y)$ was the vertex of the cone, and such section was determined by a quotient of the vector bundle $\mathcal{E}\to L\to 0$. $\endgroup$
    – YetAnotherMathStudent
    Commented Dec 9 at 7:04
  • $\begingroup$ I don't understand what you say. $X$ is not a cone. $\endgroup$
    – abx
    Commented Dec 9 at 9:54

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