Combinatorial interpretation of this identity of Gauss?

Question

Gauss came up with some bizarre identities, namely $$ \sum_{n\in\mathbb{Z}}(-1)^nq^{n^2}=\prod_{k\geq 1}\frac{1-q^k}{1+q^k}. $$

How can this be interpreted combinatorially? It strikes me as being similar to many partition identities. Thanks.

Answer 1

Here's a combinatorial interpretation, but I have no idea how to turn it into a combinatorial proof.

$\prod(1+q^k)$ is the generating function that counts the number of partitions into distinct parts. $\prod(1-q^k)$ is the generating function that counts the excess of partitions into an even number of distinct parts over the partitions into an odd number of distinct parts. Thus

$$\sum_{n\in\mathbb{Z}}(-1)^nq^{n^2}\prod_{k\geq 1}\left(1+q^k\right)=\prod_{k\geq 1}\left(1-q^k\right)$$

considers partitions into a square and distinct parts and states that the excess of such partitions with an odd square over such partitions with an even square (where each non-zero square occurs in two colours, positive and negative) is equal to the excess of partitions into an even number of distinct parts over the partitions into an odd number of distinct parts.

Answer 2

There is a classical proof by Andrews which you can find in my survey here (section 5.5). There is also a bijective proof of a more general identity I gave in this paper (section 2.2). Enjoy!