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 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 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 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, 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 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.

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.

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 $$G_2 = -\frac{1}{24} +\sum_{k=1}^\infty \sum_{d|k}k q^k$$ (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 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 disconnected degree $d$ covers respectively. Let $$Z = \sum_{d>0}N_{disc}q^d \quad\text{ and }\quad F=\sum_{d>0}N_{conn}q^d $$ be the associated generating functions. Then the combinatorial relationship between 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, 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 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.

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 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 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 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, 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 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.