Jacobi triple product

In mathematics, the Jacobi triple product is the mathematical identity:

$\prod _{m=1}^{\infty }\left(1-x^{2m}\right)\left(1+x^{2m-1}y^{2}\right)\left(1+{\frac {x^{2m-1}}{y^{2}}}\right)=\sum _{n=-\infty }^{\infty }x^{n^{2}}y^{2n},$

for complex numbers x and y, with |x| < 1 and y ≠ 0.

It was introduced by Jacobi (1829) in his work Fundamenta Nova Theoriae Functionum Ellipticarum.

The Jacobi triple product identity is the Macdonald identity for the affine root system of type A₁, and is the Weyl denominator formula for the corresponding affine Kac–Moody algebra.

Properties

The basis of Jacobi's proof relies on Euler's pentagonal number theorem, which is itself a specific case of the Jacobi Triple Product Identity.

Let $x=q{\sqrt {q}}$ and $y^{2}=-{\sqrt {q}}$. Then we have

$\phi (q)=\prod _{m=1}^{\infty }\left(1-q^{m}\right)=\sum _{n=-\infty }^{\infty }(-1)^{n}q^{\frac {3n^{2}-n}{2}}.$

The Jacobi Triple Product also allows the Jacobi theta function to be written as an infinite product as follows:

Let $x=e^{i\pi \tau }$ and $y=e^{i\pi z}.$

Then the Jacobi theta function

$\vartheta (z;\tau )=\sum _{n=-\infty }^{\infty }e^{\pi {\rm {i}}n^{2}\tau +2\pi {\rm {i}}nz}$

can be written in the form

$\sum _{n=-\infty }^{\infty }y^{2n}x^{n^{2}}.$

Using the Jacobi Triple Product Identity we can then write the theta function as the product

$\vartheta (z;\tau )=\prod _{m=1}^{\infty }\left(1-e^{2m\pi {\rm {i}}\tau }\right)\left[1+e^{(2m-1)\pi {\rm {i}}\tau +2\pi {\rm {i}}z}\right]\left[1+e^{(2m-1)\pi {\rm {i}}\tau -2\pi {\rm {i}}z}\right].$

There are many different notations used to express the Jacobi triple product. It takes on a concise form when expressed in terms of q-Pochhammer symbols:

$\sum _{n=-\infty }^{\infty }q^{\frac {n(n+1)}{2}}z^{n}=(q;q)_{\infty }\;\left(-{\tfrac {1}{z}};q\right)_{\infty }\;(-zq;q)_{\infty },$

where $(a;q)_{\infty }$ is the infinite q-Pochhammer symbol.

It enjoys a particularly elegant form when expressed in terms of the Ramanujan theta function. For $|ab|<1$ it can be written as

$\sum _{n=-\infty }^{\infty }a^{\frac {n(n+1)}{2}}\;b^{\frac {n(n-1)}{2}}=(-a;ab)_{\infty }\;(-b;ab)_{\infty }\;(ab;ab)_{\infty }.$

Proof

Let $f_{x}(y)=\prod _{m=1}^{\infty }\left(1-x^{2m}\right)\left(1+x^{2m-1}y^{2}\right)\left(1+x^{2m-1}y^{-2}\right)$

Substituting xy for y and multiplying the new terms out gives

$f_{x}(xy)={\frac {1+x^{-1}y^{-2}}{1+xy^{2}}}f_{x}(y)=x^{-1}y^{-2}f_{x}(y)$

Since $f_{x}$ is meromorphic for $|y|>0$, it has a Laurent series

$f_{x}(y)=\sum _{n=-\infty }^{\infty }c_{n}(x)y^{2n}$

which satisfies

$\sum _{n=-\infty }^{\infty }c_{n}(x)x^{2n+1}y^{2n}=xf_{x}(xy)=y^{-2}f_{x}(y)=\sum _{n=-\infty }^{\infty }c_{n+1}(x)y^{2n}$

so that

$c_{n+1}(x)=c_{n}(x)x^{2n+1}=\dots =c_{0}(x)x^{(n+1)^{2}}$

and hence

$f_{x}(y)=c_{0}(x)\sum _{n=-\infty }^{\infty }x^{n^{2}}y^{2n}$

Evaluating c₀(x)

Showing that $c_{0}(x)=1$ is technical. One way is to set $y=1$ and show both the numerator and the denominator of

${\frac {1}{c_{0}(e^{2i\pi z})}}={\frac {\sum \limits _{n=-\infty }^{\infty }e^{2i\pi n^{2}z}}{\prod \limits _{m=1}^{\infty }(1-e^{2i\pi mz})(1+e^{2i\pi (2m-1)z})^{2}}}$

are weight 1/2 modular under $z\mapsto -{\frac {1}{4z}}$, since they are also 1-periodic and bounded on the upper half plane the quotient has to be constant so that $c_{0}(x)=c_{0}(0)=1$.

Other proofs

A different proof is given by G. E. Andrews based on two identities of Euler.

For the analytic case, see Apostol.

References

• Peter J. Cameron, Combinatorics: Topics, Techniques, Algorithms, (1994) Cambridge University Press, ISBN 0-521-45761-0
• Jacobi, C. G. J. (1829), Fundamenta nova theoriae functionum ellipticarum (in Latin), Königsberg: Borntraeger, ISBN 978-1-108-05200-9, Reprinted by Cambridge University Press 2012
• Carlitz, L (1962), A note on the Jacobi theta formula, American Mathematical Society
• Wright, E. M. (1965), "An Enumerative Proof of An Identity of Jacobi", Journal of the London Mathematical Society, London Mathematical Society: 55–57, doi:10.1112/jlms/s1-40.1.55