On a general quintic threefold $Y\subset \mathbb P^4$ there are $2875$ lines. The result is classical and one can obtain it via a Chern class computation. But $Y$ is a Calabi-Yau threefold, thus one can start the Donaldson-Thomas machinery, which is exactly a (virtual) count of curves on Calabi-Yau threefolds.

Question. Can we obtain the number $2875$ as the Donaldson-Thomas invariant of the moduli space of lines in $Y$?

I guess the answer has to be "yes", as the Donaldson-Thomas count refines the naive count of curves, giving results such as the quoted one.

All we have to do is to construct the correct moduli space. We want lines $C$ on $Y$, so we are looking at the moduli space $$M=M(1,0,-1,c_3)$$ of torsion-free sheaves with rank $1$, trivial determinant, $c_2=-1$, and $c_3=\chi(\mathscr O_C)$.

The Donaldson-Thomas invariant of $M$ is the degree of its virtual fundamental class $[M]^{\textrm{vir}}$. To rephrase the above question: how to see that $$\int_{[M]^{\textrm{vir}}}1=2875\,\,?$$

Thank you!


1 Answer 1


It is not true that $\int_{[M]^{vir}} 1 = 2875$ for all the moduli spaces you list. If we let $$N_n = \int_{[M(1,0,-1,n)]^{vir}} 1 $$ then $$Z_1(q) = \sum_{n=1}^\infty N_n q^n = 2875\cdot M(-q)^{e(Y)}\cdot \frac{q}{(1+q)^2}$$ where $$M(q) = \prod_{m=1}^\infty (1-q^m)^{-m}$$

What is going on here is that the moduli space $M(1,0,-1,n)$ is isomorphic to the Hilbert scheme of 1-dimensional subschemes $C$ in the class of a line with $\chi(\mathcal{O}_C)=n$. For $n=1$ this moduli space is 2875 points representing the 2875 lines, but for $n>1$, the subscheme will consist of a line and $n-1$ points (they could be far away from the lines or embedded into the lines). These moduli spaces are in general very complicated.

Note that by MNOP conjecture (recently proven for the quintic by Pandharipande-Pixton), the above formula for $Z_1(q)$ tells you what the degree 1 Gromov-Witten invariants are in all genus, not just genus 0.

So Donaldson-Thomas theory (the MNOP version using ideal sheaves of subschemes) is not a very efficient way of getting solely genus 0 information --- you have to compute for ideal sheaves of all euler characteristics, convert the corresponding (reduced) series in $q$ to a series in $\lambda$ via the substitution $q=-e^{i\lambda}$, pass from the disconnected to connected series, and then find the $\lambda^{-2}$ term in the expansion!

If you want purely genus 0 enumerative information from a Donaldson-Thomas like invariant, you could use Sheldon-Katz's proposed definition of genus 0 Gopakumar-Vafa invariants. He proposes $$n_{0,\beta}(X) = \int_{[M(0,0,\beta,1)]^{vir}} 1$$ where $X$ is a Calabi-Yau threefold, $\beta \in H_2(X)$ is a curve class, and $M(0,0,\beta,1)$ is the moduli space of stable pure sheaves with chern character $(0,0,\beta,1)$. In your case above, this space will be populated by only the structure sheaves of the 2875 lines and so indeed the genus zero Gopakumar-Vafa of the quintic in the class of the line is 2875.

Of course, none of this really addresses your real question which is how do we get the number 2875 from Donaldson-Thomas theory without a priori knowing the enumerative answer. The answer is perhaps disappointing in this case. There is no good computational way to see the 2875 lines in DT theory. In Gromov-Witten theory, one can relate genus 0 invariants of $Y$ to Gromov-Witten like invariants of the ambient space $\mathbb{CP}^4$ and in the case of the 2875, this can be reduced to classical intersection theory. DT theory is only defined for threefolds and so there is no known way to related the DT theory of $Y$ to something on the ambient space. The only available computational technique here for the quintic is degeneration (degenerating the quintic to two fano threefolds meeting along a K3 for example). This is what is employed by Pixton-Pandharipande to prove the MNOP conjecture, but it is rather complicated and probably impossible to excise the genus zero part of the computation (for the reasons explained above). Conceivably, one could try to apply degeneration techniques to the Katz definition of genus 0 GV invariants, but that hasn't been explored and also probably isn't easy (one would be forced into considering moduli spaces with strictly semi-stable objects and that causes difficulty in DT theory).

One computational advantage that DT theory does have over GW theory is that it can be computed motivically: the integral over the virtual class is given by a weighted Euler characteristic and the euler characteristic of a space can be computed by summing the Euler characteristic of the strata in a stratification. This actually does provide a close connection between the DT theory (really PT theory --- DT theory's little brother) and the enumerative geometry of surfaces. The paper of Kool-Shende-Thomas use this idea to prove Gottsche's conjecture which gives a universal formula for the enumerative geometry of surfaces. The subsequent papers of Kool-Thomas make this connection between enumerative geometry of a surface $S$ and the PT theory of the threefold $K_S$ more explicit. Gottsche and Shende go further and use refined PT theory to obtain a refined version of Gottsche's conjecture which contains even subtler enumerative informations (like the number of real curves on a surface).

To summarize this long winded answer: DT theory is not very good at directly seeing the enumerative geometry of threefolds like the quintic (in particular the 2875 lines), but it is very good for understanding the enumerative geometry of surfaces.

Edit: here are some links to papers discussed above:

Pandharipande-Pixton prove the MNOP conjecture in a series of papers. This is the final one (which references the earlier ones)): http://arxiv.org/abs/1206.5490

Kool-Shende-Thomas proof of the Gottsche conjecture and subsequent papers by Kool-Thomas: http://arxiv.org/abs/1010.3211 http://arxiv.org/abs/1112.3069 http://arxiv.org/abs/1112.3070

Gottsche-Shende's papers on refined Gottsche conjecture: http://arxiv.org/abs/1208.1973 http://arxiv.org/abs/1307.4316

Brenin asked for a reference for the computation of $Z_1(q)$. Here is a direct computation in DT theory: http://arxiv.org/abs/math/0601203 , but it is easier to understand this computation in PT theory: see the example on page 28 in this beautiful expository paper: http://arxiv.org/pdf/1111.1552.pdf

  • 1
    $\begingroup$ This answer is amazing! Actually I thought that virtual counts somehow disregarded the possibly nasty scheme structure of the moduli spaces, and would give the "correct" answer even when these are bad. Now I learnt that I was wrong! Could you please give me a reference for the formula of $Z_1(q)$? I am also quite curious of seeing how Sheldon-Katz came up with considering sheaves of rank $0$ instead of rank $1$. In any case, thank you for taking the time to write up this beautiful answer. $\endgroup$
    – Brenin
    Mar 20, 2014 at 19:41
  • $\begingroup$ You're welcome! The virtual machinery does correctly take into account the singularities of the moduli space, but for CY3 geometry, you are forced (in some sense) to consider moduli spaces which include points as well as curves. Here is a really good expository paper which discusses this and much more: arxiv.org/abs/1111.1552. I'll add some references to the above answer. $\endgroup$
    – Jim Bryan
    Mar 20, 2014 at 19:52
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    $\begingroup$ Sheldon Katz's idea for genus zero GV invariants comes from the geometric description of GV invariants in physics. In physics, these numbers come from the cohomology of D2-branes. The higher genus numbers come from an $sl_2$ action on the cohomology, but the genus 0 numbers are an euler characteristic. The moduli space of D2-branes is mathematically given by sheaves of dimension 1. Since the associated DT invariants are given by a weighted euler characteristic, this becomes a reasonable mathematical interpretation of the physics definition. $\endgroup$
    – Jim Bryan
    Mar 21, 2014 at 17:02
  • $\begingroup$ I now have many things to learn from your answer (this was the goal). Thank you very much! also for the references. $\endgroup$
    – Brenin
    Mar 23, 2014 at 10:20
  • $\begingroup$ Dear @JimBryan, sorry to bother you again. I am reading the paper you quoted on super-rigid curves, and I was wondering: the very first sentence in the proof of Lemma 2.9 says: "The generating function for the number of 3d partitions with one infinite leg of shape $\lambda$...". Is this infinite leg arbitrary (one of the $3$ axes, no matter which one), or is it a specified one? At first sight I thought it was an arbitrary one, but now I start thinking it may be a particular one. Looking at the reference [ORV] did not really help, as that paper is very difficult for me at the moment. Regards, $\endgroup$
    – Brenin
    Apr 28, 2014 at 17:04

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