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Jacobi triple product

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"Triple product identity" redirects here. For the ternary operation on vector spaces, see Triple product.

In mathematics, the Jacobi triple product is the identity:

∏ m = 1 ∞ ( 1 − x 2 m ) ( 1 + x 2 m − 1 y 2 ) ( 1 + x 2 m − 1 y 2 ) = ∑ n = − ∞ ∞ x n 2 y 2 n , {\displaystyle \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

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

ϕ ( q ) = ∏ m = 1 ∞ ( 1 − q m ) = ∑ n = − ∞ ∞ ( − 1 ) n q 3 n 2 − n 2 . {\displaystyle \phi (q)=\prod _{m=1}^{\infty }\left(1-q^{m}\right)=\sum _{n=-\infty }^{\infty }(-1)^{n}q^{\frac {3n^{2}-n}{2}}.}

The Rogers–Ramanujan identities follow with x = q 2 q {\displaystyle x=q^{2}{\sqrt {q}}} , y 2 = − q {\displaystyle y^{2}=-{\sqrt {q}}} and x = q 2 q {\displaystyle x=q^{2}{\sqrt {q}}} , y 2 = − q q {\displaystyle 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 π τ {\displaystyle x=e^{i\pi \tau }} and y = e i π z . {\displaystyle y=e^{i\pi z}.}

Then the Jacobi theta function

ϑ ( z ; τ ) = ∑ n = − ∞ ∞ e π i n 2 τ + 2 π i n z {\displaystyle \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

∑ n = − ∞ ∞ y 2 n x n 2 . {\displaystyle \sum _{n=-\infty }^{\infty }y^{2n}x^{n^{2}}.}

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

ϑ ( z ; τ ) = ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) [ 1 + e ( 2 m − 1 ) π i τ + 2 π i z ] [ 1 + e ( 2 m − 1 ) π i τ − 2 π i z ] . {\displaystyle \vartheta (z;\tau )=\prod _{m=1}^{\infty }\left(1-e^{2m\pi {\rm {i}}\tau }\right)\left\left.}

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:

∑ n = − ∞ ∞ q n ( n + 1 ) 2 z n = ( q ; q ) ∞ ( − 1 z ; q ) ∞ ( − z q ; q ) ∞ , {\displaystyle \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 ) ∞ {\displaystyle (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 | a b | < 1 {\displaystyle |ab|<1} it can be written as

∑ n = − ∞ ∞ a n ( n + 1 ) 2 b n ( n − 1 ) 2 = ( − a ; a b ) ∞ ( − b ; a b ) ∞ ( a b ; a b ) ∞ . {\displaystyle \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 ) = ∏ m = 1 ∞ ( 1 − x 2 m ) ( 1 + x 2 m − 1 y 2 ) ( 1 + x 2 m − 1 y − 2 ) {\displaystyle 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 ( x y ) = 1 + x − 1 y − 2 1 + x y 2 f x ( y ) = x − 1 y − 2 f x ( y ) {\displaystyle 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 {\displaystyle f_{x}} is meromorphic for | y | > 0 {\displaystyle |y|>0} , it has a Laurent series

f x ( y ) = ∑ n = − ∞ ∞ c n ( x ) y 2 n {\displaystyle f_{x}(y)=\sum _{n=-\infty }^{\infty }c_{n}(x)y^{2n}}

which satisfies

∑ n = − ∞ ∞ c n ( x ) x 2 n + 1 y 2 n = x f x ( x y ) = y − 2 f x ( y ) = ∑ n = − ∞ ∞ c n + 1 ( x ) y 2 n {\displaystyle \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 2 n + 1 = ⋯ = c 0 ( x ) x ( n + 1 ) 2 {\displaystyle 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 ) ∑ n = − ∞ ∞ x n 2 y 2 n {\displaystyle f_{x}(y)=c_{0}(x)\sum _{n=-\infty }^{\infty }x^{n^{2}}y^{2n}}

Evaluating c0(x)

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

1 c 0 ( e 2 i π z ) = ∑ n = − ∞ ∞ e 2 i π n 2 z ∏ m = 1 ∞ ( 1 − e 2 i π m z ) ( 1 + e 2 i π ( 2 m − 1 ) z ) 2 {\displaystyle {\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 ↦ − 1 4 z {\displaystyle 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 {\displaystyle 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

  1. Andrews, George E. (1965-02-01). "A simple proof of Jacobi's triple product identity". Proceedings of the American Mathematical Society. 16 (2): 333. doi:10.1090/S0002-9939-1965-0171725-X. ISSN 0002-9939.
  2. Chapter 14, theorem 14.6 of Apostol, Tom M. (1976), Introduction to analytic number theory, Undergraduate Texts in Mathematics, New York-Heidelberg: Springer-Verlag, ISBN 978-0-387-90163-3, MR 0434929, Zbl 0335.10001
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