# Rigid curves, and the "richness" of their homology class

Let $X$ be a complex smooth projective variety, and $C\subset X$ a smooth curve. Then $C$ defines a cycle $$\beta=[C]\in H_2(X,\mathbb Z).$$ I have a very vague question about this situation:

Q. If $C$ is rigid in $X$, how far is this condition from $C$ being the unique curve on $X$ in class $\beta$?

I would not say that rigidity implies that $C$ is the only curve in class $[C]$ (what about the converse?), but somehow I cannot quite distinguish the two situations: I have a lack of examples in this sense, and this is what my question is really addressing. Also: does the answer to the above question depend on $X$? What if $X$ is a Calabi-Yau threefold?

Thanks!

PS. Feel free to improve the title...

Edit. By "rigid", I mean $H^0(C,N_{C/X})=0$, where $N_{C/X}$ is the normal bundle. I think this is equivalent to $C$ being an isolated point in the Chow variety of $1$-cycles in $X$ (but please correct me if I am wrong).

• What is your definition of rigid?
– abx
Jan 21 '15 at 19:11
• @abx: Thanks, I added the needed explanation. Jan 21 '15 at 19:15
• I'm not good enough to do the computations, but what about a ruled surface? Say, the projectivization of the sum of two distinct line bundles of degree $1$ over an elliptic curve? It seems to me that it has two sections, both rigid, in the same homology class. But this is to be confirmed by an expert :) Jan 21 '15 at 19:28

Let me spell out examples to keep in mind in both directions. I'll leave it to someone more clueful to give a better general answer.

1) Rigidity does not imply uniqueness in homology. Alex Degtyarev gave an example on ruled surfaces. Here's another. Let $C$ and $D$ be two curves in the same plane in $\mathbb P^3$, meeting transversally. Let $X$ be obtained by blowing up $C$, with exceptional divisor $E$, and let $Y$ be obtained by blowing up the strict transform of $D$ on $X$, with exceptional divisor $F$. Let $C_0$ be the any of the fibers of $E$ that lies above one of the points of intersection of $C$ and $D$, so that $C_0$ meets $F$. Then $C_0$ is equivalent in homology to all of the analogous curves in fibers over the other points of intersection, of which there are $(\deg C)(\deg D)$. This is basically Hironaka's construction of a smooth non-projective threefold (after we flop $C_0$); there's a nice picture in appendix B.3 of Hartshorne.

2) Non-rigidity does not imply non-uniqueness in homology. It's possible that $C$ is a rational curve in a threefold with $N_{C/X} = \mathcal O \oplus \mathcal O(-2)$. Then $H^0(C,N_{C/X}) = 1$, so it has a first-order infinitesimal deformation. However, this need not lift to an honest deformation, and so $C$ could be unique in its homology class. This situation is discussed in some detail in Reid's "Minimal models of canonical threefolds". (This is where the pictures of "Reid's pagoda" show up.)

• Whoops, I see you want a Calabi-Yau threefold. Problem (2) can occur verbatim in that case, but (1) can't since it involves blowing stuff up. However, I think you can get basically the same issue on a CY3 using other constructions: there can be a finite set of homologous rational curves, all with normal bundle $\mathcal O(-1) + \mathcal O(-1)$.
– user47305
Jan 21 '15 at 20:36

A Calabi-Yau example: A quintic threefold in $\mathbb P^4$ contains infinitely many rational curves $C$ with normal bundle $N_C=O(-1)+(-1)$ (thus they are rigid). However, the Picard number is 1, so all the curves are proportional in $H_2$.

• Ah, that'll do it! (though note that every other curve on $X$ is proportional in $H_2$ as well). Example 6.1 in Oguiso's arxiv.org/pdf/1206.1649v3.pdf is another: there you get a ray on $H_2$ that contains a finite set of rational curves, and nothing else. Actually, that example simultaneously shows that neither implication holds. Some of the curves have normal bundle $\mathcal O(-1) \oplus \mathcal O(-1)$, some have $\mathcal O \oplus \mathcal O(-2)$ (if memory serves), and they're all in a ray in $H_2$.
– user47305
Jan 22 '15 at 14:19
• Yes, that is a great example, thanks! As an aside: Is it true (it seems to be, according to your answer) that there is a rational curve $C\subset X$ of degree $d$ for every $d$? Jan 23 '15 at 20:05
• I think this is known, at least in many cases. You might want to google 'curve counting for quintic threefolds' and the paper and P. Candelas, X. C. de la Ossa, P. S. Green, and L. Parkes. A pair of Calabi-Yau manifolds as an exactly soluble superconformal theory. Jan 25 '15 at 15:12