generating function for size of young diagrams

Statement

\sum_{λ} q^{| λ |} = \frac{1}{(1 - q) (1 - q^{2}) \dots}

where $λ$ is a Young diagram.

Proof 1 (with q-binomial coefficients)

By this, we see that we may write

\sum_{λ \subseteq k \times (n - k)} q^{| λ |} = {[\binom{n}{k}]}_{q} = \frac{(q^{n} - 1) (q^{n - 1} - 1) \dots (q^{n - k + 1} - 1)}{(q^{k} - 1) (q^{k - 1} - 1) \dots (q - 1)}

Take the limit as $n \to \infty$ , then

\sum_{λ with at most k rows} q^{| λ |} = \sum_{λ with at most k columns} q^{| λ |} = \frac{1}{(1 - q) (1 - q^{2}) \dots (1 - q^{k})}

and taking the limit as $k \to \infty$ , we have our result.

Note that the limits here are defined as follows: let $f_{n} (x) = a_{0, n} + a_{1, n} x + a_{2, n} x^{2} + \dots$ be a formal power series in $x$ . Then, $lim_{n \to \infty} f_{n} (x)$ is the formal power series with coefficients $a_{i} = lim_{n \to \infty} a_{i, n}$ . Notably, we require that each of the coefficients must converge. It is true here as the coefficient of $q^{i}$ is only affected by the products of $q^{j}$ for $j < i$ , so it will be a finite product.

Proof 2 (counting multiplicities)

Write $λ = (m_{1}, m_{2}, \dots)$ where each $m_{i}$ is the multiplicity of $i$ . Note that saying that $λ$ has $\leq k$ columns is the same as saying that $m_{i} = 0$ for all $i > k$ . Then, $| λ | = \sum_{i} i m_{i}$ . Plugging this in, we have

\sum_{λ with at most k rows} q^{| λ |} = \sum_{(m_{1}, m_{2}, \dots, m_{k}), m_{i} \geq 0} q^{m_{1} + 2 m_{2} + 3 m_{3} + \dots + k m_{k}} = \prod_{i = 1}^{k} \sum_{m_{i} \geq 0} q^{i m_{i}} = \prod_{i = 1}^{k} \frac{1}{1 - q^{i}}

and we can take the limit as $k \to \infty$ to conclude.