Bocherds Product for j-function

Consider the $j$ function

$\displaystyle j(\tau)=E_{4}^{3}(\tau) / \Delta(\tau)=q^{-1}+744+196884 q+21493760 q^{2}+\ldots =\sum_{n} c(n) q^n$

We have

$\displaystyle j(p)-j(q)=\left(\frac{1}{p}-\frac{1}{q}\right) \prod_{m, n=1}^{\infty}\left(1-p^{m} q^{n}\right)^{c(m n)}$

This identity is equivalent to some non-trivial polynomials relations in the coefficients of the $j$ function. For example $c(4)=c(3)+\frac{c(1)^{2}-c(1)}{2}$

Proof:

Apply Hecke operator $T_m$ on $j$ to get

$\displaystyle j_m (\tau) =\sum_{a d=m,~ 0 \leq b<d} j\left(\frac{a \tau+b}{d}\right)$

Let us say $j_{m}(\tau)=\sum_{n \in \mathbb{Z}} c_{m}(n) q^{n}$ is the Fourier expansion for $j_m(\tau)$

We can compute and see that

$\displaystyle c_m(n) = \sum_{d|(m,n)} \frac{n}{d} c\left(\frac{mn}{d^2}\right)$

So we can observe the only negative coefficients that appear in $j_m(\tau)$ is the $-m$ coefficient, that is $c_m(-m).$

Hence we can eliminate that coefficient by subtracting $j^m(\tau)$ , to get coefficients all bigger than $-m$ . Repeating this by subtracting with powers of $j^k(\tau)$ , we can remove all the non-positive coefficients so that for a polynomial $P_m(j)$

$\displaystyle j_m(\tau) -P_m (j) =O(q)$

This is a modular function of weight $0$ and hence has to be a constant. At $q=0$ , the expression vanishes, therefore we get

$\displaystyle j_m(\tau) =P_m (j)$

Now from this equation, we can derive a lot identities, that essentially reflect polynomial relations between the coefficients $c(n) .$

$j_m(\tau) = q^{-m}+c_{m}(1) q+\cdots .$

because there are the only negative coefficient as we said has to be $c_m(-m) =c(-1) =1 .$

$j_m(\tau) =P_m (j)$ gives that $c_m(1) = mc(m)$ so that

$j_m(\tau) = q^{-m}+mc(m) q+\cdots .$

$\displaystyle j \cdot P_{m}(j)=q^{-m-1}+\sum_{l=0}^{m} c(l) P_{m-l}(j)+m c(m) +\cdots$

$\displaystyle j \cdot P_{m}(j)=P_{m+1}(j)+\sum_{l=0}^{m} c(l) P_{m-l}(j)+m c(m) + O(q)$

By similar reasoning as above, we get the equality

$\displaystyle j \cdot P_{m}(j)=P_{m+1}(j)+\sum_{l=0}^{m} c(l) P_{m-l}(j)+m c(m)$

$\displaystyle m c(m) =j \cdot P_{m}(j)-P_{m+1}(j)-\sum_{l=0}^{m} c(l) P_{m-l}(j)$

This expression says that the coefficients $c(m)$ can be retrieved as certain polynomials of the $j$ function.

We have $\displaystyle \sum_{m=1}^{\infty} m c(m)p^{m} =p \frac{d (j(p) -p^{-1}}{dp} = p\frac{d j(p)}{dp} +p^{-1}$

Susbstituting the right hand side expression for $mc(m)$ in the above expression, we get

$\displaystyle j \sum_{m=1}^{\infty} P_{m}(j) p^{m}-p^{-1} \sum_{m=2}^{\infty} p^{m} -\left(j(p)-p^{-1}\right)\left(1+\sum_{m=1}^{\infty} P_{m}(j) p^{m}\right)+c(0)$

(The last two terms come from the observation that right hand side expression is a convolution of $c(m)$ and $P_m(j)$ )

This can be simplified to

$\displaystyle (j-j(p))\left(1+\sum_{m=1}^{\infty} P_{m}(j) p^{m}\right)+p^{-1}$

Therefore

$\displaystyle p \frac{d}{d p} j(p)=(j-j(p))\left(1+\sum_{m=1}^{\infty} P_{m}(j) p^{m}\right)$

$\displaystyle p \frac{d}{d p} j(p)=(j-j(p))\left(1+\sum_{m=1}^{\infty} j_{m}(\tau) p^{m}\right)$

$\displaystyle p \frac{d}{d p} j(p)=(j-j(p))\left(1+\sum_{m=1}^{\infty} \sum_{n \in \mathbb{Z}} c_{m}(n) q^{n} p^{m}\right)$

Substituting the Hecke relation we saw above

$\displaystyle c_m(n) = \sum_{d|(m,n)} \frac{n}{d} c\left(\frac{mn}{d^2}\right)$

we get

$\displaystyle p \frac{d}{d p} j(p)=(j-j(p))\left(1+\sum_{m=1}^{\infty} \sum_{n \in \mathbb{Z}} \sum_{d|(m,n)} \frac{n}{d} c\left(\frac{mn}{d^2}\right) q^{n} p^{m}\right)$

$\displaystyle \frac{d \log (j(p) -j(q)) }{d p} +p^{-1}=-\frac{d}{dp}\left(1+\sum_{m=1}^{\infty} \sum_{n \in \mathbb{Z}} \sum_{d|(m,n)} \frac{1}{d} c\left(\frac{mn}{d^2}\right) q^{n} p^{m}\right)$

$\displaystyle \frac{d \log (j(p) -j(q)) }{d p} +p^{-1}=-\frac{d}{dp}\left(1+\sum_{m=1}^{\infty} \sum_{n \in \mathbb{Z}} \sum_{d=1}^{\infty}\frac{1}{d} c_{1}(m n) p^{m d} q^{n d}\right)$

$\displaystyle \frac{d \log (j(p) -j(q))+ \log p }{d p}=-\frac{d}{dp}\left(1+\sum_{m=1}^{\infty} \sum_{n =-1}^{\infty} \sum_{d=1}^{\infty} c_{1}(m n)\log \left(1-p^mq^n\right)\right)$

Hence

$\displaystyle j(p)-j(q)=p^{-1} \prod_{m \geq 1, n \geq-1}\left(1-p^{m} q^{n}\right)^{c_{1}(m n)}$

$\displaystyle j(p)-j(q)= \left(p^{-1}-q^{-1}\right) \prod_{m, n \geq 1}^{\infty}\left(1-p^{m} q^{n}\right)^{c(m n)}$

Dividing by $p-q$ and taking the limit $p=q$ , we get

$\displaystyle \frac{d}{d q} j(q) =-q^{-2} \prod_{m, n \geq 1}\left(1-q^{m+n}\right)^{c(m n)}$

$\displaystyle -q\frac{d}{d q} j(q) =q^{-1} \prod_{k \geq 1}\left(1-q^{k}\right)^{b(k)}$

where

$\displaystyle b(k) =\sum_{m+n =k} c(mn)$

Finally $\displaystyle -q\frac{d}{d q} j(q) =\frac{E_{14}(\tau)}{2 \zeta(14) \Delta(\tau)}$ because the difference $-q\frac{d}{d q} j(q)-\frac{E_{14}(\tau)}{2 \zeta(14) \Delta(\tau)}$ is a modular form of weight 2.

$\displaystyle \frac{E_{14}(\tau)}{2 \zeta(14) \Delta(\tau)} = q^{-1} \prod_{k \geq 1}\left(1-q^{k}\right)^{b(k)}$

Consider the form

$\displaystyle \sum_{k=--4} a(k) q^{k} := (j(q^4)-744) \sum_{n \in \mathbb Z} q^{n^2}$

We see that $a(l^2) =b(l)$ because

$a(l^2) =\sum_{4m+n^2 =l^2} c(m) = \sum_{n} c_{1}\left(\frac{l+n}{2}\frac{ l-n}{2}\right)=\sum_{a+b=l} c(ab)=b(l)$

Therefore we have

$\displaystyle \frac{E_{14}(\tau)}{2 \zeta(14) } = q^{-1}\Delta(\tau) \prod_{k \geq 1}\left(1-q^{k}\right)^{a(k^2)}$

Using $\Delta(\tau) =q\prod_{k \ge 1}\left(1-q^{k}\right)^24$

we get the Bocherds product formula

$\displaystyle \frac{E_{14}(\tau)}{2 \zeta(14) } = \prod_{k \geq 1}\left(1-q^{k}\right)^{a(k^2)+24}$

Bocherds Correspodence: In general for Bocherds gives similar conceptual explanations for products.
For example in
$\Delta(\tau) =q\prod_{k \ge 1}\left(1-q^{k}\right)^24$

The exponents 24 can all be seen to be fourier coefficients of a weight 1/2 modular form $12\sum_{n \in \mathbb Z} q^{n^2}$ )

The correspondence in general roughly says that products with exponents coming from weight 1/2 modular forms are also modular forms.

Share this:

Related

Leave a comment Cancel reply