# Extension of the Jacobi triple product identity

The Jacobi triple product and the mathematical identity of it is: $$\prod\limits_{n=1}^{ \infty }(1-q^{2n})(1+zq^{2n-1})(1+z^{-1}q^{2n-1})=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2}$$

I would like to extend the idea for $\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$

My idea is below for extension:

Let's assume we define $G(z,q,h)$ as

$$G(z,q,h)\prod\limits_{n=1}^{ \infty }(1+zq^{2n-1}h^{3n^2-3n+1})(1+z^{-1}q^{2n-1}h^{-3n^2+3n-1})=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$$

$z=ZQ^{2}h^{3}$

$q=Qh^{3}$

$$G(ZQ^{2}h^{3},Qh^{3},h)\prod\limits_{n=1}^{ \infty }(1+ZQ^{2}h^{3}(Qh^{3})^{2n-1}h^{3n^2-3n+1})(1+(ZQ^2h^3)^{-1}(Qh^3)^{2n-1}h^{-3n^2+3n-1})=\sum\limits_{n = - \infty }^ \infty Z^n Q^{2n}h^{3n} Q^{n^2} h^{3n^2} h^{n^3}$$

$$G(ZQ^{2}h^{3},Qh^{3},h)\prod\limits_{n=1}^{ \infty }(1+ZQ^{2n+1}h^{3n^2+3n+1})(1+Z^{-1}Q^{2n-3}h^{-3n^2+9n-7})=\sum\limits_{n = - \infty }^ \infty Z^n Q^{2n+n^2}h^{3n+3n^2+n^3}$$

$$G(ZQ^{2}h^{3},Qh^{3},h)\frac{(1+Z^{-1}Q^{-1}h^{-1})}{(1+ZQh)}\prod\limits_{n=1}^{ \infty }(1+ZQ^{2n-1}h^{3n^2-3n+1})(1+Z^{-1}Q^{2n-1}h^{-3n^2+3n-1})=\sum\limits_{n = - \infty }^ \infty Z^n Q^{2n+n^2}h^{3n+3n^2+n^3}$$

$$G(ZQ^{2}h^{3},Qh^{3},h)\frac{(1+Z^{-1}Q^{-1}h^{-1})}{(1+ZQh)} \frac{\sum\limits_{n = - \infty }^ \infty Z^n Q^{n^2} h^{n^3}}{G(Z,Q,h)}=\sum\limits_{n = - \infty }^ \infty Z^n Q^{2n+n^2}h^{3n+3n^2+n^3}$$

$$G(ZQ^{2}h^{3},Qh^{3},h)\sum\limits_{n = - \infty }^ \infty Z^n Q^{n^2} h^{n^3}=G(Z,Q,h)ZQh\sum\limits_{n = - \infty }^ \infty Z^n Q^{2n+n^2}h^{3n+3n^2+n^3}$$

$$G(ZQ^{2}h^{3},Qh^{3},h)\sum\limits_{n = - \infty }^ \infty Z^n Q^{n^2} h^{n^3}=G(Z,Q,h)\sum\limits_{n = - \infty }^ \infty Z^{n+1} Q^{1+2n+n^2}h^{1+3n+3n^2+n^3}$$

$$G(ZQ^{2}h^{3},Qh^{3},h)\sum\limits_{n = - \infty }^ \infty Z^n Q^{n^2} h^{n^3}=G(Z,Q,h)\sum\limits_{n = - \infty }^ \infty Z^{n+1} Q^{(n+1)^2}h^{(n+1)^3}$$

$$G(ZQ^{2}h^{3},Qh^{3},h)\sum\limits_{n = - \infty }^ \infty Z^n Q^{n^2} h^{n^3}=G(Z,Q,h)\sum\limits_{n = - \infty }^ \infty Z^n Q^{n^2} h^{n^3}$$

$$G(ZQ^{2}h^{3},Qh^{3},h)=G(Z,Q,h) \tag 1$$

If we continue the rule, we can get $$G(ZQ^{2}h^{3},Qh^{3},h)=G(ZQ^{2}h^{3}Q^2h^{6}h^{3},Qh^{3}h^{3},h)=G(ZQ^{4}h^{12},Qh^{6},h)$$

If $h=1$ then $G(z,q,1)=\prod\limits_{n=1}^{ \infty }(1-q^{2n})$ can be gotten from Jacobi_triple_product.

I wonder how I can find the function $G(z,q,h)$. Please help me which Technics can be applied to find it. Also If you know there is other works about this subject, please share links and references.

Thanks a lot for responses.

Note: If $z=x^3$,$q=x^3$,$h=x$

$$G(x^3,x^3,x)\prod\limits_{n=1}^{ \infty }(1+x^3x^{6n-3}x^{3n^2-3n+1})(1+x^{-3}x^{6n-3}x^{-3n^2+3n-1})=\sum\limits_{n = - \infty }^ \infty x^{3n} x^{3n^2} x^{n^3}$$

$$xG(x^3,x^3,x)\prod\limits_{n=1}^{ \infty }(1+x^{3n^2+3n+1})(1+x^{-3n^2+9n-7})=\sum\limits_{n = - \infty }^ \infty x^{1+3n+3n^2+n^3}$$

$$xG(x^3,x^3,x)\frac{1}{(1+x)}\prod\limits_{n=1}^{ \infty }(1+x)(1+x^{3n^2+3n+1})(1+x^{-1})(1+x^{-3n^2+3n-1})=\sum\limits_{n = - \infty }^ \infty x^{(n+1)^3}$$

$$xG(x^3,x^3,x)\frac{(1+x^{-1})}{(1+x)}\prod\limits_{n=1}^{ \infty }(1+x^{3n^2-3n+1})(1+x^{-3n^2+3n-1})=\sum\limits_{n = - \infty }^ \infty x^{n^3}$$

$$G(x^3,x^3,x)\prod\limits_{n=1}^{ \infty }(1+x^{3n^2-3n+1})(1+x^{-3n^2+3n-1})=\sum\limits_{n = - \infty }^ \infty x^{n^3}$$

The relation is below for $G()$ from Equation 1 If $z=x^3$,$q=x^3$,$h=x$

$$G(x^{12},x^6,x)=G(x^3,x^3,x)=G(1,1,x) \tag 2$$

I thought If I can find a few term of $G(z,q,h)$ by hand and maybe it can be seen what the pattern of $G(z,q,h)$ has. I wonder if we can find $G(z,q,h)$ in product terms as Jacobi did in his product formula.

$$G(z,q,h)\prod\limits_{n=1}^{ \infty }(1+zq^{2n-1}h^{3n^2-3n+1})(1+z^{-1}q^{2n-1}h^{-3n^2+3n-1})=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$$

$$G(z,q,h)(1+zqh)(1+z^{-1}q^{1}h^{-1})(1+zq^3h^7)(1+z^{-1}q^{3}h^{-7})....=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$$

$$G(z,q,h)(1+q^2+q(zh+z^{-1}h^{-1}))(1+q^6+q^3(zh^7+z^{-1}h^{-7}))...=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$$

$$G(z,q,h)(1+q^2+q^6+q^8+q(zh+z^{-1}h^{-1})+q^7(zh+z^{-1}h^{-1})+q^3(zh^7+z^{-1}h^{-7})+q^5(zh^7+z^{-1}h^{-7})+q^4(zh+z^{-1}h^{-1})(zh^7+z^{-1}h^{-7}))...=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$$

$$G(z,q,h)( 1+q^2+q^6+q^8+q(zh+z^{-1}h^{-1})+q^7(zh+z^{-1}h^{-1})+q^3(zh^7+z^{-1}h^{-7})+q^5(zh^7+z^{-1}h^{-7})+q^4(z^2h^8+z^{-2}h^{-8})+q^4(h^{6}+h^{-6}))...=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$$

$$G(z,q,h)( 1+q^2+q^6+q^8+q(zh+z^{-1}h^{-1})+q^7(zh+z^{-1}h^{-1})+q^3(zh^7+z^{-1}h^{-7})+q^5(zh^7+z^{-1}h^{-7})+q^4(z^2h^8+z^{-2}h^{-8})+q^4(h^{6}+h^{-6}))...= 1+q (zh+z^{-1}h^{-1})+q^4 (z^2h^8+z^{-2}h^{-8})+....$$

EDIT: (Updated on 15th April)

We can see first 3 term of $G(z,q,h)$ easily.To find 4th term: $$G(z,q,h)=1-q^2+q^3\left( (zh+z^{-1}h^{-1}) - (zh^7+z^{-1}h^{-7}) \right) +a_4 q^4 +....$$

$$G(z,q,h)(1+q^2+q^6+q^8+q(zh+z^{-1}h^{-1})+q^7(zh+z^{-1}h^{-1})+q^3(zh^7+z^{-1}h^{-7})+q^5(zh^7+z^{-1}h^{-7})+q^4(z^2h^8+z^{-2}h^{-8})+q^4(h^{6}+h^{-6}))...= 1+q (zh+z^{-1}h^{-1})+q^4 (z^2h^8+z^{-2}h^{-8})+....$$

$$(1-q^2+q^3\left( (zh+z^{-1}h^{-1}) - (zh^7+z^{-1}h^{-7}) \right) +a_4 q^4 +.... ) (1+q^2+q^6+q^8+q(zh+z^{-1}h^{-1})+q^7(zh+z^{-1}h^{-1})+q^3(zh^7+z^{-1}h^{-7})+q^5(zh^7+z^{-1}h^{-7})+q^4(z^2h^8+z^{-2}h^{-8})+q^4(h^{6}+h^{-6}))...= 1+q (zh+z^{-1}h^{-1})+q^4 (z^2h^8+z^{-2}h^{-8})+....$$

$a_4=-1+(z^2h^8+z^{-2}h^{-8})-(z^2h^2+z^{-2}h^{-2})$

Thus first 4 terms of $G(z,q,h)$ are:

$$G(z,q,h)=1-q^2+q^3\left( (zh+z^{-1}h^{-1}) - (zh^7+z^{-1}h^{-7}) \right) + q^4 \left(-1+(z^2h^8+z^{-2}h^{-8})-(z^2h^2+z^{-2}h^{-2}) \right) +....$$

If we order it little bit .

$$G(z,q,h)=1-q^2-q^4 +q^3\left( zh(1-h^6) + z^{-1}h^{-1}(1-h^{-6}) \right) + q^4 \left(z^2h^2(h^6-1)+z^{-2}h^{-2}(h^{-6}-1) \right) +....$$

I will update If I find more terms of $G(z,q,h)$

EDIT: (Updated on 17th April)

I have found 5th term . $a_5= (zh+z^{-1}h^{-1}) - (z h^{19}+z^{-1}h^{-19}) +(z^3 h^3+z^{-3}h^{-3}) - (z^3 h^9+z^{-3}h^{-9})$

$$G(z,q,h)=1-q^2+q^3\left( (zh+z^{-1}h^{-1}) - (zh^7+z^{-1}h^{-7}) \right) + q^4 \left(-1+(z^2h^8+z^{-2}h^{-8})-(z^2h^2+z^{-2}h^{-2}) \right) + q^5\left( (zh+z^{-1}h^{-1}) - (z h^{19}+z^{-1}h^{-19}) +(z^3 h^3+z^{-3}h^{-3}) - (z^3 h^9+z^{-3}h^{-9}) \right)+ ....$$

$$G(z,q,h)=1-q^2-q^4 +q^3\left( zh(1-h^6) + z^{-1}h^{-1}(1-h^{-6}) \right) + q^4 \left(z^2h^2(h^6-1)+z^{-2}h^{-2}(h^{-6}-1) \right) + q^5 \left( zh(1-h^{18}) + z^{-1}h^{-1}(1-h^{-18})+z^3h^3(1-h^6)+z^{-3}h^{-3}(1-h^{-6}) \right)+....$$

I have not seen a general pattern of the terms yet but I believe there is very beautiful pattern in it. If you help me to find more terms , I will be very appreciated. Maybe the pattern of terms of $G(z,q,h)$ can be seen more . Thanks.

-
Try expanding $G$ as a series in positive powers of $q$, with coefficients that are Laurent polynomials in $z$ and $h$. – S. Carnahan Apr 12 '13 at 11:00
if you typed all this up yourself i can see you do love math. – John Jiang Apr 13 '13 at 5:23
@John Jiang : I love mathematics very much. I wrote my works above. I wanted to share it. I thought maybe someone can give a hand to solve it or can give a link about the subject. – Mathlover Apr 15 '13 at 6:23

## 1 Answer

This being community-wiki, I'll add a couple of comments which probably won't be of direct help in your pursuit. But both of them indicate the value of widening the framework for questions involving these classical identities.

1) There is an old "simple" proof of the Jacobi triple product identity by George Andrews here. This already shows the value of putting the identity into a wider context.

2) For a conceptual setting, consider the work of Victor Kac (and later Howard Garland and Jim Lepowsky) which explained in a nice way the formal derivaiton by Ian G. Macdonald of identities involving Dedekind's $\eta$-function: these arise naturally in the representation theory of affine Lie algebras. In particular, Jacobi's triple product follows from the study of the affine Lie algebra built on a three dimensional simple Lie algebra.

-