# Curves on K3 and modular forms

The paper of Bryan and Leung "The enumerative geometry of $K3$ surfaces and modular forms" provides the following formula. Let $S$ be a $K3$ surface and $C$ be a holomorphic curve in $S$ representing a primitive homology class. If $N_g(n)$ is the number of curves of geometric genus $g$ with $n$ nodes passing through $g$ generic points in $S$ in the linear system $|C|$ with any $g$ and $n$ satisfying $C^2=2(g+n)+2$ then $$\sum\limits_{n=0}^{\infty}N_{g}(n)q^n=\frac{q}{\Delta(q)}\left(q\frac{d}{dq}G_2(q)\right)^g$$ where $\Delta(q)$ and $G_2(q)$ are the modular forms: $$\Delta(q)=q\prod\limits_{n=1}^{\infty} (1-q^n)^{24},\ G_2(q)=-\frac{1}{24}+\sum\limits_{n=1}^{\infty}\sum\limits_{k|n}kq^n.$$ A relation between these two forms is well known in Number Theory. Unfortunately, I did not manage to find the relation and the first question is "What is the relation?". The main question is

How can one explain the relation between $\Delta(q)$ and $G_2(q)$ in the context of counting curves?

Some ideas are provided by the original paper of Bryan and Leung: these two modular forms correspond to numbers of (multiple) covers of a nodal rational curve and of an elliptic curve.

• You have a couple typos in your formulas above: $\frac{d}{dq}$ should be replaced by $q\frac{d}{dq}$, the exponent of $\Delta$ should be +24 not -24 and there should be a factor of $q$ in front (see my answer below). Also, to adhere to common convention, you should add the constant $-\frac{1}{24}$ to $G_2$. – Jim Bryan Jun 10 '14 at 23:00

## 1 Answer

The answer to your first question: "What is the relationship between $G_2$ and $\Delta$?" is

$$q\frac{d}{dq} \log \Delta = -24G_2$$

where

$$\Delta = q\prod_{m=1}^\infty (1-q^m)^{24}$$ and $$G_2 = -\frac{1}{24} +\sum_{d=1}^\infty \sum_{k|d}k q^d$$ (note that I've included the constant -1/24 which is in usual definition of $G_2$ which you admitted). The above relation is easy to prove directly: take the log of the definition of $\Delta$, use the taylor series of log, rearrange the sum, differentiate, and out pops $-24G_2$.

To answer your second question of explaining this relationship in terms of counting curves, this can be done in the context of Gromov-Witten theory of the elliptic curve. Suppose that we wish to count the number of unramified covers of degree d of an elliptic curve $E$. We can either count connected covers, or possibly disconnected covers (and either way we should count each cover by the reciprocal of the number of automorphisms of the cover). Let $N_{conn}(d)$ and $N_{disc}(d)$ be the number of connected and possibly disconnected degree $d$ covers respectively. Let $$Z = \sum_{d>0}N_{disc}(d)q^d \quad\text{ and }\quad F=\sum_{d>0}N_{conn}(d)q^d$$ be the associated generating functions. Then the combinatorial relationship between possibly disconnected and connected covers yields $$Z=\exp{F}.$$

On the other hand, the number of connected covers is determined by covering space theory to be $$N_{conn}(d)=\frac{1}{d}\cdot\#\{\text{index d subgroups of \mathbb{Z}\oplus \mathbb{Z}}\}$$

which one easily can prove is given by $\frac{1}{d}\sum_{k|d}k$ by thinking about index $d$ sublattices or equivalently 2 by 2 integer matrices of determinant $d$. Thus we see that $$q\frac{d}{dq}F= G_2 +1/24.$$

On the other hand, possibly disconnected covers are determined by their monodromy and this yields $$N_{disc}(d) = \frac{1}{d!}\cdot |\operatorname{Hom}(\mathbb{Z}\oplus \mathbb{Z},S_d)|$$ where $S_d$ is the symmetric group and the $1/d!$ prefactor is to remove the choice of a labelling of one fiber (necessary to get monodromy). The right hand side of the above equation can be computed with elementary group theory and one gets that $N_{disc}(d)$ is given by the number of conjugacy classes in $S_d$ which is in turn given by $p(d)$ the number of partitions of $d$. Thus $Z$ is the well known generating function for $p(d)$: $$Z=\prod_{m>0}(1-q^m)^{-1} = \left(\frac{q}{\Delta}\right)^{\frac{1}{24}}$$.

Putting this whole discussion together, we see that the relationship between possibly disconnected and connected covers of the elliptic curve, i.e. $Z=\exp (F)$ becomes

$$q\frac{d}{dq}\log \left(\left(\frac{q}{\Delta}\right)^{\frac{1}{24}}\right) = G_2 +1/24$$

which is equivalent to the original relationship.

• This answer was pretty long, so I put this as a comment. The above has nothing to do with the K3. The appearance of modular forms in the generating function for the number of curves on K3 is still somewhat mysterious, although there reasons from physics to expect that the generating function for the number of BPS states has modular properties. – Jim Bryan Jun 10 '14 at 18:38
• Interesting! I think it's worth mentioning how this argument changes if $\mathbb{Z} \oplus \mathbb{Z}$ is replaced by an arbitrary finitely generated group $G$. In this case the subgroup counting gets more complicated: one must take a sum over index-$d$ subgroups $H$ of the coefficients $\frac{1}{|\text{Aut}(G/H)|}$. If $H$ is normal, and in particular if $G$ is abelian, then this automorphism group is isomorphic to $G/H$ and hence the coefficient is $\frac{1}{d}$, but in general this automorphism group is isomorphic to $N_G(H)/H$ which is smaller. – Qiaochu Yuan Jun 11 '14 at 5:02
• @Qiaochu Yuan: the only other case I've thought about along these lines is covers of $\mathbb{RP}^2$ which is amusing: since the only connected covers are the degree 1 cover and the universal cover in degree two, $F=q+\frac{q^2}{2}$ and so we learn that $\exp (q+\frac{q^2}{2})$ is the generating series which counts order 2 elements in the symmetric group. (It is a fun exercise to derive this combinatorially). – Jim Bryan Jun 13 '14 at 14:56