Jacobi's triple product formula

Statement

\prod_{n \geq 1} ((1 - x^{2 n}) (1 + x^{2 n - 1} y^{2}) (1 + x^{2 n - 1} y^{- 2})) = \sum_{k \in Z} x^{k^{2}} y^{2 k}

Notable specializations

To get Euler's pentagonal number theorem, we can substitute $x=q^{3/2}, y {#2} = -q^{1/2}$ .

Two results by Gauss:

\prod_{n \geq 1} (1 - x^{n})^{3} = \sum_{k \in Z} (- 1)^{k} k x^{k (k + 1) / 2}

and

\prod_{n \geq 1} \frac{1 - x^{n}}{1 + x^{n}} = \sum_{k \in Z} (- 1)^{k} x^{k^{2}}

Proof

First, let us perform a change of variables, setting $x^{2} = q, y^{2} = z - q^{1 / 2}$ :

\prod_{n \geq 1} (1 - q^{n}) (1 + z q^{n}) (1 + z^{- 1} q^{n - 1}) = \sum_{k \in Z} q^{k (k + 1) / 2}

Let us divide by the first term in the product on both sides, as we notice that it is the generating function for size of young diagrams

\prod_{n \geq 1} (1 + z q^{n}) (1 + z^{- 1} q^{n - 1}) = (\sum_{k \in Z} q^{k (k + 1) / 2}) (\prod_{n \geq 1} \frac{1}{1 - q^{n}})

Some observations:

$\prod_{n \geq 1} (1 + z q^{n}) = \sum_{a \geq 0} z^{a} (\sum_{μ} q^{| μ |})$ where $μ$ is a partition with $a$ distinct parts (this comes from the formula of distinct parts here)
$\prod_{n \geq 1} (1 + z^{- 1} q^{n - 1}) = \sum_{b \geq 0} z^{- b} (\sum_{ν} q^{| ν | - b})$ where $ν$ is a partition with $b$ distinct parts similarly
$\prod_{n \geq 1} \frac{1}{1 - q^{n}} = \sum_{λ} q^{| λ |}$ where $λ$ is any partition.

So it suffices to create a bijection between the following two sets:

$(a, b, μ, ν)$ with $a, b \in Z_{\geq 0}$ , $μ ⊢ a, ν ⊢ b$ both with distinct parts
$(k, λ)$ $k \in Z$ and $λ$ a partition
Satisfying
$a - b = k$
$| μ | + | ν | - b = | λ | + \frac{k (k + 1)}{2}$

We shall construct it as follows:

Given the tuple $(a, b, μ, ν)$ , we shift $μ, ν$ as follows:
- For each row $i$ (starting at 0), shift the entire row $i$ to the right
- It should look like a staircase on the left
- Denote the shifted diagram as $\tilde{μ}$ and $\tilde{ν}$
Put dots on the first of each row in both $\tilde{μ}$ and $\tilde{ν}$
Flip $\tilde{ν}$ across the diagonal to produce $\tilde{\tilde{ν}}$
Glue the last $min (a, b)$ dots on the diagonal of $\tilde{μ}$ and $\tilde{\tilde{ν}}$ together.
Remove the little staircase to the left / above so that what we get at the end is a Young diagram
We let $λ$ be the resulting Young diagram and we cut out $k$ rows (if we cut on the left) or $- k - 1$ rows (if we cut on top)

In that case that $a \geq b$ , we note that

$k = a - b$
$| λ | + \frac{k (k + 1)}{2} = | μ | + | ν | - b$ which you can interpret as the size of the construction that we got (without removing the staircase) is equal to the sum of the two sizes minus the diagonal that we cut out, which is $b$
In the case that $b > a$ , we note that
we delete $b - a - 1$ rows (note we subtract one because we lose one from the gluing), so $k = a - b$
$| λ | + \frac{(b - a - 1) (b - a)}{2} = | λ | + \frac{k (k + 1)}{2} = | μ | + | ν | - b$

Now the inverse construction is given as follows (we are essentially reversing all the steps):

If $k \geq 0$ , we take $λ$ and attach a $k$ -staircase to the upper left. We then extend the staircase to chop into $\tilde{μ}$ and $\tilde{\tilde{ν}}$
Otherwise, we attack a $- k - 1$ staircase on top and do the same thing

Example of bijection:

Example 1: Let $μ = (7, 6, 4, 3, 1)$ and $ν = (6, 5, 3)$ .

The result of step 1 on $μ$ looks like
20260318_132718.jpg|300
The result of steps 1-3 on $ν$ is (note that the top of the diagonal is dotted, because this will be glued on top of the diagonal of $μ$ )
20260318_133019.jpg|150
Steps 4, 5:

Example 2: Let $μ = (5, 4, 2), ν = (9, 8, 7, 4, 3, 1)$ . Then, the result will look like

Now for the other direction of the bijection: