Express the following power series in two commuting variables $x,y$ as an infinite product, or a short sum of infinite products: $$ \frac{P(xy)^2}{(1-x)}\sum_{k=-\infty}^\infty(2k+1)x^{k^2}y^{k^2+k}. $$ Here $P(x)=\prod_{i=1}^\infty(1-x^i)^{-1}$ is the partition generating function.

Does it have any special properties e.g. automorphic form?


The $2$-residue of an integer node $(n,m)$ in the plane is $m-n$ mod $2$. So the $2$-residues alternate as $0,1$ in a checkerboard pattern. The Young diagram $[\lambda]$ of a partition $\lambda$ is a set of nodes in the plane. An addable node of $\lambda$ does not belong to $[\lambda]$, but can be adjoined to give the Young diagram of a partition (of $|\lambda|+1$).

Now define, for $i=0,1$: $c_i(\lambda)$ is the number of nodes in $[\lambda]$ with $2$-residue $i$. $a_i(\lambda)$ is the number of addable nodes of $\lambda$ with $2$-residue $i$.

Then my power series is the generating function of $$ \sum_\lambda a_0(\lambda)x^{c_0(\lambda)}y^{c_1(\lambda)} $$ Here $\lambda$ ranges over all partitions. I'll leave it as an exercise to work out the corresponding identity for the other generating function $\sum_\lambda a_1(\lambda)x^{c_0(\lambda)}y^{c_1(\lambda)}$, using the first.

The coefficient of a given monomial $x^ay^b$ is the dimension of a certain algebra, naturally associated to the symmetric group $S_{a+b}$, defined in characteristic $2$.

Other Information

Using the Jacobi Triple Product identity I can factorize the generating function for the $2$-residues of partitions as $$ \sum_{\lambda}x^{c_0(\lambda)}y^{c_1(\lambda)} =P(xy)^2\sum\limits_{k=-\infty}^\infty x^{k^2}y^{k^2+k} $$ $$ =\prod\limits_{i=1}^\infty\frac{(1+x^{2i-1}y^{2i})(1+x^{2i-1}y^{2i-2})(1+x^iy^i)}{(1-x^iy^i)}. $$ Experts on the modular representation of the symmetric group will understand the significance of the left hand side and the first equality. The identity has a combinatorial proof that can be deduced from Cilanne E. Boulet, A four-parameter partition identity, arXiv:math/0308012v1

The generating function I'm interested in can be got (almost) using partial differentiation from this.

Also if we set $x=y$ in the original, standard results give: $$ \sum_\lambda a_0(\lambda)x^{|\lambda|}=\frac{1}{2(1-x)} \frac{P(x)^4+P(x^2)^2}{P(x)^3} $$ $$ \sum_\lambda a_1(\lambda)x^{|\lambda|}=\frac{1}{2(1-x)} \frac{P(x)^4-P(x^2)^2}{P(x)^3} $$

  • $\begingroup$ $P(x)=\prod_{i=1}^\infty(1−x^i)^{−1}$ is the partition generating function. $\endgroup$
    – Bernikov
    Jan 25, 2012 at 11:04
  • $\begingroup$ why does partial differentiation fail, exactly? It looks like the operator $2y \frac{d}{dy} - 2x \frac{d}{dx} + 1$ might recover your generating function. $\endgroup$ Apr 1, 2012 at 18:12

1 Answer 1


Let us do the following transform: \begin{align} \begin{cases}e^{\pi i \tau}=q=xy\\ e^{\pi i z}=u=\sqrt{y}. \end{cases} \end{align} Then, we just need consider the function \begin{align} f(z;\tau)&=\frac{P(q)^2}{\left(1-q/u^2\right)^2}\frac{\partial}{\partial u}\left(u\sum_{k\in\mathbb{Z}}q^{k^2}u^{2k}\right)\\ &=\frac{P(q)^2}{\left(1-q/u^2\right)^2}\sum_{k\in\mathbb{Z}}q^{k^2}u^{2k}+\frac{uP(q)^2}{\left(1-q/u^2\right)^2}\frac{\partial}{\partial u}\left(\sum_{k\in\mathbb{Z}}q^{k^2}u^{2k}\right). \end{align} Let \begin{align} \vartheta(z;\tau):=\sum_{k\in\mathbb{Z}}q^{k^2}u^{2k}=\prod_{n\ge 1}(1-q^{2n})(1+u^{-2}q^{2n-1})(1+u^2q^{2n-1}) \end{align} be the Jacobi Theta function, then it is easy seen that \begin{align} f(z;\tau)=\frac{\vartheta(z;\tau)P(q)^2}{\left(1-q/u^2\right)^2}\left(1+2\sum_{n\ge 1}\left(\frac{1}{1+u^{-2}q^{2n-1}}-\frac{1}{1+u^{2}q^{2n-1}}\right)\right). \end{align} Namely, \begin{align} f(z;\tau)=\frac{\vartheta(z;\tau)\eta(\tau)^2e^{\frac{\pi i\tau}{12}}}{\left(1-e^{\pi i \tau}e^{-4\pi i z}\right)^2}\left(1+2\sum_{n\ge 1}\left(\frac{1}{1+e^{-4\pi i z}e^{(2n-1)\pi i \tau}}-\frac{1}{1+e^{4\pi i z}e^{(2n-1)\pi i \tau}}\right)\right), \end{align} where $\eta(\tau)$ is the Dedekind eta function.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.