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Oct 18, 2023 at 18:21 vote accept user267839
Oct 18, 2023 at 15:16 comment added Will Sawin @DamianRössler Ha, I didn't think to check the publication date. In any case part of the point of that book is to unify and make rigorous some of the things people were doing before in an ad hoc or even non-rigorous manner, so I think it still makes sense to read these kinds of things via the modern definition.
Oct 18, 2023 at 12:51 history edited user267839 CC BY-SA 4.0
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Oct 18, 2023 at 9:50 comment added user267839 I see that in projective setting ( eg Hartshorne, the chapter on ruled surfaces, say if we embed $L$ in $\mathbb{P}(\mathcal{L} \oplus \mathcal{O})$ as explained in Edit# above) that is what the calculation gives at the end of the day. But I'm wondering if if there is a concrete geometric picture keeping this identification of normal bundle with $\mathcal{L}$ to be plausible on intuitive level
Oct 18, 2023 at 9:39 comment added user267839 @CraniumClamp: could you elaborate a bit more concretely why it is "intuitively" obvious that the normal bundle of the zero section $C_0$ is $\mathcal{L}$? Just heuristically, how to see it, ie which "picture" one should have in mind in order to regard it as "geometrically plausible"?
Oct 18, 2023 at 9:33 vote accept user267839
Oct 18, 2023 at 9:40
Oct 18, 2023 at 6:16 comment added Damian Rössler @Will Savin. I see what you mean (at the level of coherent sheaves, this amounts to saying that for all $k\geq 0$, $Tor^k_{\cal L}(O_{C_0},O_{C_0})$ has a support, which is complete). I am not sure that in the reference, the author would have been aware of Fulton’s theory though.
Oct 18, 2023 at 4:51 comment added Cranium Clamp (1) The total space needs to have a dual on the line bundle $ \mathcal{L} $. (2) The normal bundle of the zero section of C in L is $ \mathcal{L} $ itself - this is intuitively obvious from a mental picture.
Oct 17, 2023 at 19:11 answer added Will Sawin timeline score: 4
Oct 17, 2023 at 18:59 comment added Will Sawin @DamianRössler No, I am referring to the intersection theory as in Fulton's book, where one does not define the intersection product of two rational equivalence classes of cycles but instead really of two cycles. In this theory the intersection is an equivalence class on the intersection of the supports of the original cycles, which is well-defined since we are intersecting honest cycles. This construction will agree with the intersection number in a compactified setting.
Oct 17, 2023 at 17:52 comment added Damian Rössler @Will Savin. The intersection between two cycles is only a rational equivalence class of cycles, it does not have a support on the nose. Perhaps you mean that in this case, since it is a self-intersection, one may simply define it as the degree of the normal bundle of $C_0$ in $L$, which does give the required formula.
Oct 17, 2023 at 16:37 history edited user267839 CC BY-SA 4.0
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Oct 17, 2023 at 16:31 history edited user267839 CC BY-SA 4.0
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Oct 17, 2023 at 16:21 comment added user267839 of this line bundle... otherwise without compactification as you said one might run into troubles with the well definedness of intersection numbers...
Oct 17, 2023 at 16:19 comment added user267839 @DamianRössler: Thanks, a good catch, I see to problem, because otherwise the intersection might be not well defined, since some points could "vanish" if we pass the an equivalent divisor. Hmm, the source where I found this example is this paper example 2.5 at page 425. I'm not sure precisely what Wagreich meant there by a line bundle $L \to X$ where he defined this intersection number. Maybe - that's just a guess of mine - he wanted to consider tacitly the "compactified" version $L=\mathbb{P}(\mathcal{L} \oplus \mathcal{O}_X)$
Oct 17, 2023 at 16:17 comment added Will Sawin @DamianRössler It seems pretty clear to me: The intersection is well-defined as a cycle class on the set-theoretic intersection. The set-theoretic intersection is complete so you can take the degree.
Oct 17, 2023 at 15:53 comment added Damian Rössler The self-intersection is only defined in general if the ambient variety is complete. However $L$ is not complete so you should explain what you want to compute.
Oct 17, 2023 at 15:25 history edited user267839 CC BY-SA 4.0
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Oct 17, 2023 at 14:53 history asked user267839 CC BY-SA 4.0