3
$\begingroup$

One can consider the Calabi-Yau threefold K3$\times E$ where the Donaldson-Thomas theory is conjectured to be the (inverse of) the Igusa cusp form $\chi_{10}(q,y,p)$. The variables $q,y,p$ aren't conventional, I just want to emphasize that it is a three-variable automorphic form. So we have

$$Z^{\text{DT}}_{K3\times E} = \frac{1}{\chi_{10}(q,y,p)}.$$

I've been working with a compact, smooth Calabi-Yau threefold $X$ who has three Kahler classes $d_{1}, d_{2}, d_{3}$. One can compute an equality of the form

$$\frac{1}{2} \log\bigg( \frac{1}{\chi_{10}(q,y,p)}\bigg) = F^{\text{GW}, 1}_{X}(d_{1}, d_{2}, d_{3})$$

where $F^{\text{GW}, 1}_{X}$ is actually the genus one Gromov-Witten potential. There is a non-trivial change of variables between the three parameters on each side.

Now, this could be an accidental thing. However, I know string theorists use duality to convert a "hard" problem into an "easy" one. Well the lefthand side of the above equation is a hard computation; it's the full partition function on K3$\times E$. The righthand side though is simply a "one-loop perturbative computation" in Gromov-Witten theory, as a physicist might say. This is (relatively) easy.

So my question is: is there possibly a string duality lurking here? If so, are there any more details which jump to anyone's mind? I'm a little hesitant, because I know GW and DT theories rightly belong in topological string theory and the web of dualities seems to correspond to the full, physical string theories. But it looks very, very suggestive to me.

$\endgroup$
  • $\begingroup$ What do you mean by "three Kähler classes"? Do you mean that your threefold is hyper-Kähler? $\endgroup$ – Danu Aug 19 '17 at 13:47
  • $\begingroup$ @Danu No, what I meant is that I'm only "counting curves" in three distinct Kahler classes in $X$. In some geometry with $h^{1,1}$ at least three, you can introduce formal variables $d_{1}, d_{2}, d_{3}$ which track the degree along each of the three curves of interest. Of course, there could be more curve classes, but you might choose to have "degree zero" on them, which amounts to essentially setting their formal variables to zero. $\endgroup$ – Benighted Aug 19 '17 at 14:31
  • $\begingroup$ Thanks for the answer. I must admit that I'm not really acquainted with questions of enumerative geometry and the physical approaches to them, but if you'd like to explain some of it to me you can find me in Physics Chat. $\endgroup$ – Danu Aug 19 '17 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.