30
$\begingroup$

I'm taking a short number theory course this summer. The first topic we covered was Jacobi's triple product identity. I still have no sense of why this is important, how it arises, how it might have been discovered, etc. The proof we studied is a bit on the clever side for my taste, giving me no sense of how said proof might have been discovered. Can anyone help?

$\endgroup$

6 Answers 6

31
+50
$\begingroup$

The full appreciation of Jacobi's triple product identity can not be done without some understanding of the elliptic functions. However, it is possible to develop some parts of the theory of elliptic functions without any complex analysis.

Anyways back to the triple product identity, it says that

$$ \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} = \prod_{n = 1}^{\infty}(1 - q^{2n})(1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1})$$

This is a highly non obvious identity which represents equality of a series with a product and to a beginner it might be very difficult to establish. One of the reasons I feel it is important is that it can be used to prove many different identities which have very nice and surprising applications. I will give a famous example here.

If we replace $q$ by $q^{3/2}$ and $z$ by $-q^{1/2}$ then we get

$$ \prod_{n = 1}^{\infty}(1 - q^{3n})(1 - q^{3n - 1})(1 - q^{3n - 2}) = \prod_{n = 1}^{\infty}(1 - q^{n}) = \sum_{n = -\infty}^{\infty}(-1)^{n}q^{(3n^{2} + n)/2}$$

This is Euler's famous Pentagonal theorem which he obtained by multiplying the product by hand to obtain first few terms of the series and then he guessed the pattern of exponents in the series as $(3n^{2} + n)/2$. But it took him some years to prove the identity.

This can be used to evaluate partitions of a positive integer. If $n$ is a positive integer and $p(n)$ denotes the number of partitions of $n$ (i.e. number of ways in which $n$ can be expressed as sum of positive integers without taking into account order of summands) then it can be easily established that

$$1 + \sum_{n = 1}^{\infty}p(n)q^{n} = \frac{1}{{\displaystyle \prod_{n = 1}^{\infty}(1 - q^{n})}} = \frac{1}{{\displaystyle \sum_{n = -\infty}^{\infty}(-1)^{n}q^{(3n^{2} + n)/2}}}$$

so that

$$ \left(1 + \sum_{n = 1}^{\infty}p(n)q^{n}\right)\sum_{n = -\infty}^{\infty}(-1)^{n}q^{(3n^{2} + n)/2} = 1$$

Equating coefficients of $q^{n}$ we get the recursive formula for $p(n)$ as

$$ p(n) = p(n - 1) + p(n - 2) - p(n - 5) - p(n - 7) + p(n - 12) + p(n - 15) - \cdots$$

which is the simplest possible way to calculate the number of partitions of a positive integer. This is one of the applications of Jacobi's triple product identity which can be understood without any reference to complex analysis or elliptic function theory.

$\endgroup$
2
  • 1
    $\begingroup$ To dfeuer: I am not sure what proof you have studied for the Jocobi's triple product identity, but I would advise you to have a look at the proof by Jacobi which is based on simple algebraical manipulation (but highly non-obvious). I have described the same proof by Jacobi in my blog post paramanands.blogspot.com/2011/02/… $\endgroup$
    – Paramanand Singh
    Jul 13, 2013 at 19:18
  • $\begingroup$ This is a very clean and elegant answer $\endgroup$ Jan 9, 2021 at 22:29
30
$\begingroup$

I will try to give a rather elementary motivation which I believe to be close to how this identity was actually discovered.


Let us consider the sine function $f(z)=\sin z$.

  • As a function of complex argument, it has one period: $f(z+2\pi)=f(z)$. Further, it is holomorphic in the whole complex plane with only simple zeroes given by $\pi\mathbb{Z}$.

  • It has a (finite) exponential series representation $$f(z)=\frac{1}{2i}\left(e^{iz}-e^{-iz}\right).\tag{1}$$

  • It has an infinite product representation $$f(z)=z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{\pi^2 n^2}\right).\tag{2}$$

Note that the representation (2) almost follows from the analytic properties of $f(z)$ by Liouville theorem-type arguments ("almost" is due to the need to control the infinite point).

Now a natural generalization of (2) comes out when we try to construct doubly-periodic functions. Liouville theorem forbids to such functions being holomorphic, therefore one can go in two directions:

Let us choose the 2nd direction. The simplest generalization of (2), obtained by further periodization, is $$\vartheta_A(z)=\sin z\prod_{n=1}^{\infty}\left(\cos2\pi n\tau-\cos2z\right).\tag{3}$$ Naively, the function $\vartheta_A(z)$ is doubly-periodic and has simple zeros at $\pi\mathbb{Z}+\pi\tau\mathbb{Z}$. However, there is a problem - the product in (3) is very ill-defined. This can be cured by multiplying it by another ill-defined product independent of $z$ to get a well-defined quantity $$\vartheta_{B}(z)=\sin z\prod_{n=1}^{\infty}\left(1-2q^{2n}\cos2z+q^{4n}\right),\tag{4}$$ with $q=e^{i\pi\tau},\tau\in \mathbb{H}$. The price to pay for well-definedness is the loss of double-periodicity: though we still have $\vartheta_{B}(z+\pi)=-\vartheta_{B}(z)$, the second period is no longer a true one: $$\vartheta_{B}(z+\pi\tau)=-q^{-1}e^{-2iz}\vartheta_{B}(z).\tag{5}$$

So, once again, (4) defines a holomorphic almost doubly periodic function having simple zeros at $\pi\mathbb{Z}+\pi\tau\mathbb{Z}$. This is an analog of (2). Now the question is: what is the corresponding analog of (1)?

The answer is obtained relatively easily. Consider the function $$\vartheta_{C}(z)=\sum_{n\in\mathbb{Z}}(-1)^n q^{(n+1/2)^2}e^{i(2n+1)z}.\tag{6}$$ It is straightforward to verify that this function has the same periodicity properties as $\vartheta_{B}(z)$: $$\vartheta_{C}(z+\pi)=-\vartheta_{C}(z),\qquad \vartheta_{C}(z+\pi \tau)=-q^{-1}e^{-2iz}\vartheta_{C}(z).\tag{7}$$ Both functions are odd and in particular vanish at $z=0$. Further, $\vartheta_{C}(z)$ has exactly one zero (just as $\vartheta_{B}(z)$) inside the fundamental parallelogram. The last property is obtained by integrating $\vartheta_{C}'(z)/\vartheta_{C}(z)$ over its boundary and using quasiperiodicity (for holomorphic functions, the result should be equal to the number of zeros inside the fundamental parallelogram counted with their multiplicities).

Now, since we have two functions with the same zeros and the same quasiperiodicity properties, they can only be proportional because of Liouville theorem: $$\vartheta_{B}(z)=c(q)\cdot \vartheta_{C}(z).\tag{9}$$ But, if we recall (4) and (6), this is nothing but the Jacobi triple product identity - it remains only to fix the $z$-independent coefficient $c(q)$. In other words, this identity arises by comparing two natural ways (analogous to (1) and (2)) of writing a doubly-(quasi)periodic function. Actually, $\vartheta_{B,C}(z)$ are proportional to the Jacobi theta function $\vartheta_1(z,q)$.

$\endgroup$
1
  • $\begingroup$ This looks interesting, but unfortunately most of it goes over my head because I have not yet studied complex analysis. Please do not delete it, however. $\endgroup$
    – dfeuer
    Jul 8, 2013 at 0:08
7
$\begingroup$

I would like to offer very elementary proof for Jacobi Triple Product Identity. It is very direct way proof and elementary proof.

Let's define $$ F(z)= \prod_{n = 1}^{\infty}(1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1})$$

$$ F(zq^2)= \prod_{n = 1}^{\infty}(1 + zq^2q^{2n - 1})(1 + z^{-1}q^{-2}q^{2n - 1})$$

$$ F(zq^2)= \prod_{n = 1}^{\infty}(1 + zq^{2n + 1})(1 + z^{-1}q^{2n - 3})$$

$$ F(zq^2)= (1 + z^{-1}q^{ -1})\prod_{n = 1}^{\infty}(1 + zq^{2n + 1})(1 + z^{-1}q^{2n - 1})$$

Multiple both side by $zq$

$$ zqF(zq^2)= zq(1 + z^{-1}q^{ -1})\prod_{n = 1}^{\infty}(1 + zq^{2n + 1})(1 + z^{-1}q^{2n - 1})$$

$$ zqF(zq^2)= (1 + zq)\prod_{n = 1}^{\infty}(1 + zq^{2n + 1})(1 + z^{-1}q^{2n - 1})$$ $$ zqF(zq^2)= \prod_{n = 1}^{\infty} (1 + zq)(1 + zq^{2n + 1})(1 + z^{-1}q^{2n - 1})$$

$$ zqF(zq^2)= \prod_{n = 1}^{\infty} (1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1})$$ $$ zqF(zq^2)= F(z)$$

Let's define $G(z)$

$$G(z)= \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} $$

$$G(zq^2)= \sum_{n = -\infty}^{\infty}z^{n}q^{2n}q^{n^{2}} $$

$$G(zq^2)= \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}+2n} $$ Multiple both side by $zq$ $$zqG(zq^2)= zq\sum_{n = -\infty}^{\infty} z^{n}q^{n^{2}+2n} $$ $$zqG(zq^2)= \sum_{n = -\infty}^{\infty} z^{n+1}q^{n^{2}+2n+1} $$ $$zqG(zq^2)= \sum_{n = -\infty}^{\infty} z^{n+1}q^{(n+1)^2} = \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}}$$

$$G(z)=zqG(zq^2)$$

Because $F(z)$ and $G(z)$ satisfy same function relation, we can write
$$G(z)=A(q)F(z)$$

where A(q) is only depends on $q$

We got the relation $$ \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} =A(q) \prod_{n = 1}^{\infty}(1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1}) \tag{1}$$

We only need to find $A(q)$ .

$z-->-z$ in Equation $(1)$

$$ \sum_{n = -\infty}^{\infty}(-1)^n z^{n}q^{n^{2}} =A(q) \prod_{n = 1}^{\infty}(1 - zq^{2n - 1})(1- z^{-1}q^{2n - 1})$$

$$ \sum_{n = -\infty}^{\infty}(-1)^n z^{n}q^{n^{2}} \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} =A^2(q) \prod_{n = 1}^{\infty}(1 - zq^{2n - 1})(1- z^{-1}q^{2n - 1})(1 + zq^{2n - 1})(1+ z^{-1}q^{2n - 1})$$

$$ \sum_{n = -\infty}^{\infty}(-1)^n z^{n}q^{n^{2}} \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} =A^2(q) \prod_{n = 1}^{\infty}(1 - z^2q^{2(2n - 1)})(1- z^{-2}q^{2(2n - 1)})$$

$z-->-z^2$ $q-->q^2$

in Equation $(1)$

$$ \sum_{n = -\infty}^{\infty}(-1)^nz^{2n}q^{2n^{2}} =A(q^2) \prod_{n = 1}^{\infty}(1 - z^2q^{2(2n - 1)})(1 - z^{-2}q^{2(2n - 1)})$$

$$ \sum_{n = -\infty}^{\infty}(-1)^n z^{n}q^{n^{2}} \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} =\frac{A^2(q)}{A(q^2)} \sum_{n = -\infty}^{\infty}(-1)^nz^{2n}q^{2n^{2}}$$

if we focus $z^0$ term and we multiply in left side. We get that $$\sum_{n = -\infty}^{\infty}(-1)^{n}q^{2n^{2}} =\frac{A^2(q)}{A(q^2)}$$

$z-->-1$ $q-->q^2$ in Equation $(1)$ $$ \sum_{n = -\infty}^{\infty}(-1)^{n}q^{2n^{2}} =A(q^2) \prod_{n = 1}^{\infty}(1 -q^{2(2n - 1)})(1 -q^{2(2n - 1)})$$

$$\frac{A^2(q)}{A(q^2)}= A(q^2) \prod_{n = 1}^{\infty}(1 -q^{2(2n - 1)})(1 -q^{2(2n - 1)})$$

$$\frac{A^2(q)}{A^2(q^2)}= \prod_{n = 1}^{\infty}(1 -q^{2(2n - 1)})^2$$

$$\frac{A(q)}{A(q^2)}= \prod_{n = 1}^{\infty}(1 -q^{2(2n - 1)})$$ $$\frac{A(q)}{A(q^2)}= \prod_{n = 1}^{\infty} \frac{ (1 -q^{2(2n)}) (1 -q^{2(2n - 1)})}{(1 -q^{2(2n)})}$$ $$\frac{A(q)}{A(q^2)}= \prod_{n = 1}^{\infty} \frac{ (1 -q^{2n}) }{(1 -q^{4n})}$$

$$A(q)=C\prod_{n = 1}^{\infty}(1 -q^{2n})$$ where $C$ is constant $$ \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} =A(q) \prod_{n = 1}^{\infty}(1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1})$$

$$ \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} =C \prod_{n = 1}^{\infty}(1 -q^{2n}) \prod_{n = 1}^{\infty}(1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1})$$

if $q=0$ then C can be found as $C=1$

$$ \sum_{n = -\infty}^{\infty}z^{n}q^{n^{2}} = \prod_{n = 1}^{\infty}(1 - q^{2n})(1 + zq^{2n - 1})(1 + z^{-1}q^{2n - 1})$$

$\endgroup$
3
  • $\begingroup$ I don't quite understand, why can we infer $A(q)=C\prod_{n = 1}^{\infty}(1 -q^{2n})$ from the fact that $\frac{A(q)}{A(q^2)}= \prod_{n = 1}^{\infty} \frac{ (1 -q^{2n}) }{(1 -q^{4n})}$? $\endgroup$
    – user1058910
    Nov 27, 2022 at 17:18
  • 1
    $\begingroup$ Hello, $$A(q)= A(q^2) \prod_{n = 1}^{\infty} \frac{ (1 -q^{2n}) }{(1 -q^{4n})}$$ $$A(q^2)= A(q^4) \prod_{n = 1}^{\infty} \frac{ (1 -q^{4n}) }{(1 -q^{8n})}$$ $$A(q)= A(q^4) \prod_{n = 1}^{\infty} \frac{ (1 -q^{4n}) }{(1 -q^{8n})} \prod_{n = 1}^{\infty} \frac{ (1 -q^{2n}) }{(1 -q^{4n})}$$ $$A(q)= A(q^4) \prod_{n = 1}^{\infty} \frac{ (1 -q^{2n}) }{(1 -q^{8n})}$$ , Please continue this and in limit you will find $$A(q)=\lim\limits_{ k\to \infty } A(q^{2^k}) \prod_{n = 1}^{\infty} \frac{ (1 -q^{2n}) }{(1 -q^{n2^{k+1}})}$$ $\endgroup$
    – Mathlover
    Nov 27, 2022 at 19:58
  • 1
    $\begingroup$ $$A(q)= A(0) \prod_{n = 1}^{\infty} (1 -q^{2n}) $$ $$A(q)= C \prod_{n = 1}^{\infty} (1 -q^{2n}) $$ . Please remember $|q|<1$ and $ \lim\limits_{ k\to \infty } q^{2^k} =0$; I hope it will help you understand the detail $\endgroup$
    – Mathlover
    Nov 27, 2022 at 19:59
3
$\begingroup$

This answer is dedicated to give a motivated proof of Jacobi's triple product identity that explains the origin of the $\prod_{n\ge1}(1-q^{2n})$ factor.

Let $F_N(w,q)$ denote

$$ F_N(w,q)=\prod_{1\le n\le N}(1+q^{2n-1}w)(1+q^{2n-1}w^{-1})=\sum_{m\in\mathbb Z}a_{m,N}w^m.\tag1 $$

By the symmetry of the product, we have immediately that $a_{m,N}=a_{-m,N}$. By the properties of the factors, we also have $a_{m,N}=0$ for all $m>N$ and

$$ a_{N,N}=q^{1+3+5+\dots+(2N-3)+(2N-1)}=q^{N^2}.\tag2 $$

Replacing $w$ with $q^2w$, we also have the following functional equation

$$ \begin{aligned} F_N(q^2w,q) &=\prod_{1\le n\le N}(1+q^{2n+1}w)(1+q^{2n-3}w^{-1}) \\ &=q^{-1}w^{-1}\cdot{1+q^{2N+1}w\over1+q^{2N-1}w^{-1}}F_N(w,q). \end{aligned}\tag3 $$

Transforming (3) gives us

$$ (qw+q^{2N})F_N(q^2w,q)=(1+q^{2N+1}w)F_N(w,q).\tag4 $$

By examining the coefficients, we obtain from (4) that

$$ q^{2m-1}a_{m-1,N}+q^{2N+2m}a_{m,N}=a_{m,N}+q^{2N+1}a_{m-1,N}.\tag5 $$

Simplifying gives

$$ \begin{aligned} a_{m,N} &=q^{2m-1}\cdot{1-q^{2N-2m+2}\over1-q^{2N+2m}}a_{m-1,N} \\ &=q^{m^2}\cdot{(1-q^{2N-2m+2})(1-q^{2N-2m+4})\cdots(1-q^{2N})\over(1-q^{2N+2})(1-q^{2N+4})\cdots (1-q^{2N+2m})}a_{0,N}. \end{aligned}\tag6 $$

Plugging (2) into this formula, we have

$$ a_{0,N}={(1-q^{2N+2})(1-q^{2N+4})\cdots(1-q^{4N})\over(1-q^2)(1-q^4)\cdots(1-q^{2N})},\tag7 $$

which indicates that for all $0\le m\le N$, there is

$$ a_{m,N}=q^{m^2}{(1-q^{2N-2m+2})(1-q^{2N-2m+4})\cdots(1-q^{4N})\over(1-q^2)(1-q^4)\cdots(1-q^{2N+2m-2})(1-q^{2N+2m})}.\tag8 $$

For any given $m$, we have

$$ \lim_{N\to\infty}a_{m,N}=q^{m^2}\prod_{n\ge1}(1-q^{2n})^{-1}.\tag9 $$

By dominated convergence theorem, it can be justified that

$$ \lim_{N\to\infty}F_N(w,q)=\sum_{m\in\mathbb Z}\lim_{N\to\infty}a_{m,N}w^m.\tag{10} $$

Hence, combining (1), (9), and (10), we deduce the triple product identity:

$$ \prod_{n\ge1}(1-q^{2n})(1+q^{2n-1}w)(1+q^{2n-1}w^{-1})=\sum_{m\in\mathbb Z}q^{m^2}w^m.\tag{11} $$

I have recently written a blog post on the foundation of Jacobian theta function, which may be helpful for the OP to understand why Jacobi came up with the idea of evaluating (1).

$\endgroup$
1
$\begingroup$

We have just bumped into Jacobi's triple product from a completely different direction, looking at interacting particle systems of probability theory. I've never worked with it before, and was quite surprised when it just came out of our arguments! http://arxiv.org/abs/1606.00639

$\endgroup$
2
  • 4
    $\begingroup$ Can you maybe summarize the stuff in your paper for this answer? It would certainly attract people to read your paper if you do so. $\endgroup$ Jun 3, 2016 at 11:10
  • $\begingroup$ We compare two stochastic particle systems (Markov processes of particles jumping on the integer line Z) which both possess independent but spatially inhomogeneous stationary distributions for their particle numbers, on the other hand there is a simple map from one of these processes to the other. The first one is called asymmetric simple exclusion and has 0 or 1 particles at a site of Z while the second one is the zero range process and can have any non-negative number of particles at a site. The mapping identifies their resp. stationary distributions, and Jacobi 3 product pops up from here. $\endgroup$ Jun 4, 2016 at 22:31
1
$\begingroup$

The boson fermion correspondence says that two vertex algebras (the lattice vertex algebra attached to $\mathbf{Z}$ and the Clifford vertex algebra) are isomorphic. Taking their $q$-characters (which are obviously the left and right sides of the Jacobi formula) gives the Jacobi triple product formula.

Source - see Frenkel and Ben-Zvi's "Vertex algebras on algebraic curves" and, for more on $q$-characters, Carter "Lie algebras of finite and affine type".

The interesting thing is that (setting $q=1$) the characters are modular forms (relating this answer to the other answers) - there are a lot of results saying that the characters of vertex algebras are modular forms, but I don't know the general theory/I don't think the relationship between vertex algebras and modular forms is 100% understood yet.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .