Jacobi triple product

Template:Redirects here In mathematics, the Jacobi triple product is the identity:

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

for complex numbers x and y, with |x| < 1 and y ≠ 0. It was introduced by Template:Harvs 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

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

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

The Rogers–Ramanujan identities follow with $x=q^2\sqrt{q}$, $y^2=-\sqrt{q}$ and $x=q^2\sqrt{q}$, $y^2=-q\sqrt{q}$.

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

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

can be written in the form

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

Using the Jacobi triple product identity, the theta function can be written as the product

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

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:

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

where $(a;q{)}_{\mathrm{\infty }}$ 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

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

Proof

Let $f_x(y)=\prod _{m=1}^{\mathrm{\infty }}(1-x^{2m})(1+x^{2m-1}{y}^2)(1+x^{2m-1}{y}^{-2})$

Substituting Template:Mvar for Template:Mvar and multiplying the new terms out gives

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

Since $f_x$ is meromorphic for $|y|>0$, it has a Laurent series

$f_x(y)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n(x)y^{2n}$

which satisfies

$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n(x)x^{2n+1}{y}^{2n}=x{f}_x(xy)=y^{-2}{f}_x(y)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_{n+1}(x)y^{2n}$

so that

$c_{n+1}(x)=c_n(x)x^{2n+1}=\dots =c_0(x)x^{(n+1{)}^2}$

and hence

$f_x(y)=c_0(x)\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{n^2}{y}^{2n}$

Evaluating Template:Math

Showing that $c_0(x)=1$ (the polynomial of x of $y^0$ is 1) is technical. One way is to set $y=1$ and show both the numerator and the denominator of

$\frac{1}{c_0(e^{2i\pi z})}=\frac{\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{2i\pi n^{2}z}}{\prod _{m=1}^{\mathrm{\infty }}(1-e^{2i\pi mz})(1+e^{2i\pi (2m-1)z}{)}^2}$

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, Template:ISBN
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