Recall that for $t\geq2$, a partition is a $t$-core if none of its hooklengths is divisible by $t$. It is known that the $t$-cores are parametrized by ${\mathbb Z}^{t-1}$. More precisely, let $(n_0,\dots,n_{t-2})$ be a $(t-1)$-tuple of integers and set $n_{t-1}:=-\sum\limits_{i=0}^{t-2}n_i$. Define $r_i=\sum\limits_{k=0}^{t-1}(n_k^2+\epsilon(i,k)n_k)/2$ for all $t$-residues $i$, where $\epsilon(i,k)=\left\{\begin{array}{rr}1,&\,\,\mbox{if $i\geq k$}\\-1,&\,\,\mbox{if $i<k$}\end{array}\right.$. Then there is a unique partition which is a $t$-core and which has $r_i$ nodes of $t$-residue $i$ [MR1055707 Reviewed Garvan, Frank; Kim, Dongsu; Stanton, Dennis Cranks and $t$-cores. Invent. Math. 101 (1990), no. 1, 1–17]. Notice that $n_i=r_i-r_{i+1}$ for each residue $i$.

Let $r(n):=(r_0,r_1,\dots,r_{t-1})\in{\mathbb Z}^t$ depend on $n:=(n_0,n_1,\dots,n_{t-2})$ as above and set $x^{r(n)}:=x_0^{r_0}x_1^{r_1}\dots x_{t-1}^{r_{t-1}}$ for indeterminates $x_0,\dots,x_{t-1}$. So the generating function for the $t$-residues of $t$-cores is $$ K_t(x):=\sum_{n\in{\mathbb Z}^{t-1}}x^{r(n)}. $$ Let $P(y)=\prod_n(1-y^n)^{-1}$ be the partition generating function in the indeterminate $y$ and let $J=(1,\dots,1)$ be the all-$1$'s vector in ${\mathbb Z}^t$. Using the representation of partitions on an abacus with $t$ runners, the generating function for the $t$-residues of all partitions is $$ P_t(x)=P(x^J)^tK_t(x). $$ According to Ben Elias, this is the generating function of the Fock space representation of $\hat{sl_t}$. So it should have a product expansion using Heisenberg and $\hat{sl_t}$ representation theory. For example, in a previous question [Generating function related to 2-residues of partitions I noted that the generating function for the $2$-residues of partitions factorizes as $$ \begin{aligned} P_2(x) &=P(x_0x_1)^2\sum_{n\in{\mathbb Z}}x_0^{n^2}x_1^{n^2+n}\\ &=\prod\limits_{n=1}^\infty\frac{(1+x_0^{2n-1}x_1^{2n})(1+x_0^{2n-1}x_1^{2n-2})(1+x_0^nx_1^n)}{(1-x_0^nx_1^n)}. \end{aligned} $$ This is a consequence of the Jacobi Triple Product Identity.

Question: can anyone explicitly factorize $P_t(x)$ and provide a self-contained explanation of the relevant representation theory?


Denote by $\tilde{\mathcal F}$ the Fock space representation of $\widehat{sl}_t$. In fact it is even a $\widehat{gl}_t$-module, and as such irreducible. Now \begin{equation}\widehat{gl}_t= \widehat{sl}_t\oplus Heis \Big/ (Z_{\widehat{sl}_t}-Z_{Heis}),\end{equation} where $Heis$ is the infinite dimensional Heisenberg, and the notation is meant to indicate that in the direct sum, one identifies the centres. This $Heis$-action on $\tilde{\mathcal F}$ is not the standard one, but is by a smaller subalgebra of the standard one (inside the standard Heisenberg acting on $\tilde{\mathcal F}$, one essentially takes only the generators with indices divisible by $t$). In terms of this decomposition, one has \begin{equation}\tilde{\mathcal F} = V(\omega_0)\otimes {\mathcal F},\end{equation} where $V(\omega_0)$ is the basic representation of $\widehat{sl}_t$, whereas ${\mathcal F}$ is the standard Fock space representation of $Heis$. Note they both have central charge one, so the action factors through the quotient taken above. Everything so far is explained in much greater detail in Tingley's Notes on Fock space (Section 3E) or Leclerc's Fock space representations... (2.2.8).

Your $P_t(x)$ is the character of $\tilde{\mathcal F}$. In terms of the decomposition above, we understand it as one factor of $P(x^J)$, and the rest.

Now we have to understand how the rest becomes the character of the basic representation $V(\omega_0)$ of $\widehat{sl}_t$. There are several known constructions of the basic rep; the relevant one here is the Frenkel-Kac construction. That says that, as a vector space at least, \begin{equation}V(\omega_0) \cong {\mathcal F}^{t-1}\otimes {\mathbb C}(Q),\end{equation} where $\mathcal F$ is good old Fock space once again, and $Q$ is the finite root lattice of type $A_{t-1}$. This basically gives the rest of the formula, once the relevant weights are computed explicitly: you have $t-1$ more copies of $P(x^J)$, and $K_t(x)$ comes from the root lattice. The character of the basic representation is written in the form needed here for example in Theorem 20.23(d) of Carter's book Lie algebras of finite and affine type.


I believe $K_t(x)$ itself admits no simple product expansion for $t>2$. Set $t=3$; it is easy to see that an appropriate specialisation of $K_t(x)$ is the one-variable function \begin{equation} L(q) = \sum_{m,n\in\mathbb Z} q^{m^2+mn+n^2}. \end{equation} As I learned from Garvan some years ago, this function $L(q)$ of a single variable $q$ has a zero inside the unit circle, at $q_0=-\exp(-\pi/\sqrt{3})$, by a remark hidden on top of p.697 in this paper (which uses the same notation, except that it also introduces $a(q)$ for $L(q)$). This prevents $L(q)$ from admitting a product expansion in terms of $(1-q^n)$'s raised to a polynomial exponent in $n$, since functions admitting such expansions have no zeroes inside the unique circle.

  • $\begingroup$ Many thanks. The references look particularly useful. It is still not clear to me if I can write down a factorization for $P_t(x)$ as explicit as for the t=2 case. $\endgroup$ – John Murray Apr 15 '15 at 8:12
  • $\begingroup$ Dear Balazs, I'm confused by your added comments. The generating function for $3$-cores factorizes as $$ K_3(x)=\sum_{a,b\in{\mathbb Z}}x^{3(a^2+ab+b^2)-2a-b}=\prod_{n=1}^\infty\frac{(1-x^{3n})^3}{(1-x^n)} $$ $\endgroup$ – John Murray Apr 20 '15 at 12:05
  • $\begingroup$ In your original writeup, $x=(x_0, x_1, \ldots, x_{t-1})$ was a multi-index. I am continuing to use that notation. What I am saying is that a certain way of specialising the 3-variable series for $t=3$ gives $L(q)$ as above. $\endgroup$ – Balazs Apr 22 '15 at 15:53
  • $\begingroup$ My specialization is $x=x_0=x_1=x_2$. What is yours? $\endgroup$ – John Murray Apr 23 '15 at 8:55
  • $\begingroup$ Calculating from what you have given, one has \begin{equation} K_3(x_0, x_1,x_2) = \sum_{m,n\in {\mathbb Z}} x_0^{m^2 + m n + n^2 + m + n} x_1^{ m^2 + m n + n^2 + n} x_2^{m^2 + m n + n^2}\end{equation} so my specialization would be $x_0=x_1=1$. Your objection to this will be that there may still be an explicit "infinite product" factorization for $K_3$ which has poles at this substitution. But I believe such a factorization does not exist. $\endgroup$ – Balazs Apr 27 '15 at 14:59

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