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The question:

I have been trying to prove the following $q$-series identity for quite some time now: $$ \sum_{k \geq 0} \frac{q^{2k^2}}{(q)_{2k}} = \sum_{m,k \geq 0} \frac{q^{m^2 + 3k m + 4k^2}}{(q)_k(q)_m} \left( 1 - q^{2m + 4k +1} \right) \tag{*} $$ Where for a non-negative number $k$ we let $(q)_k = (1-q)...(1-q^k)$. I have verified this conjecture with SageMath to order $q^{2000}$.

I am no expert on the topic of $q$-series and partition identities so I have no intuition to decide if this should be an easy consequence of some well known $q$-hypergeometric summation formula or not. I am hoping that if that is the case, some expert here will point me in the right direction or share a proof.

I can prove that for every $n \geq 0$ the coefficient of $q^n$ on the RHS of (*) is bigger or equal to the same coefficient on the LHS, so only the other inequality is required.

Some background:

For a history of this identity please refer to the article we just posted with this conjecture [7]. I will collect below many things I know about this identity.

Here is a reason (besides pride and the list of techniques below that I have tried) that led me to believe that it may not be a trivial identity:

Sums like either of the two terms of the RHS of (*) are known as Nahm sums. In this particular case this is a Nahm sum associated to the matrix $(\begin{smallmatrix}8 & 3 \\ 3 & 2\end{smallmatrix})$. There is a famous conjecture [4] regarding modularity of these sums. This particular matrix is known to have the right asymptotic at $q \rightarrow 1$ for it to be modular. The LHS of (*) is the character of the Ising model vertex algebra which is known to be modular. Zagier studied this particular type of Nahm sums in [5] and did not find a modular sum with only one term and the matrix as in the RHS of (*). What this identity is saying is that by allowing linear combinations of Nahm sums of a fixed matrix that satisfies the right assymptotics, one may find modular sums.

Here are some techniques that I have tried so far:

  • The theory of Bailey pairs and Bailey chains. None of the Bailey pairs with parameters found in Sills recent article [1] works.
  • Polynomial approximations. Most of the RR type identities admit proofs along the lines of finding a polynomial identity that in certain limit gives the wanted identity. There are well known polynomial approximations to the LHS of this identity. For example one can look at the Santos Polynomials [2]. The RHS admits several polynomial approximations, for example one could take $$ P_L(q) = \sum_{m=0}^L q^{m^2} \binom{L}{m}_q \sum_{k = 0}^m q^{2k^2 +km} \binom{m}{k}_q $$ where the $q$-binomials in this setting is defined as $\binom{m}{n}_q = \frac{(m)_q}{(n)_q(m-n)_q}$.
  • Two variable generalizations. Most of the identities in Slater's list [3] admit two variable generalizations. In Sills list [1] mentioned above for example, there is a two variable generalization of Slater's identity (39) (which goes by the name in the title of this post), which has the LHS of (*) as one of the members. The RHS of (*) also comes with a natural two-variable generalization (see please [7] for this generating function). However I could not find a natural way of matching this two-variable generalization to any of the known expressions equal to the LHS of (*).
  • Functional equations: the generating series (or rather its two variable generalization alluded above) satisfies a functional equation of order $11$ so in principle if one were to guess the two variable version of the LHS a computer could decide if it satisfies this functional equation. This equation can be written as a system of $5$ (or even 2) variables and of order $2$ as written in our article.
  • Combinatorial description: The LHS of (*) counts naturally partitions satisfying some difference conditions due to a theorem of Hirschorn. There are many more families of partitions that are counted by the same generating function, a list below to the equivalent forms of this side should give you an idea. The RHS of (*) also is the generating function of a family of partitions. It arises by studying a Gröbner basis for an ideal sheaf in an affine arc space. As is typically the case in these Gröbner bases computations, the partitions involved are just horrible. It is somewhat of a miracle that we could find the generating series.

Here are some things that I have not tried:

  • The character of the Ising model satisfies an explicit modular differential equation. I do not know how to check that the RHS of (*) satisfies this equation or not. I do not even know how to compute $q \frac{d}{dq}$ of it.
  • Relations with other similar conjectures: if you look into our paper you will find similar conjectural expressions for the characters of other modules for the Ising vertex algebra. Since these are known to satisfy certain algebraic relations, perhaps these can be exploited to prove all of these identities at once.

Here are some equivalent forms for both sides of the identity:

The LHS of (*) is very well known and admits the following well known forms: $$ \begin{aligned} % \sum_{k \geq 0} \frac{q^{2k^2}}{(q)_{2k}} &= \prod_{n=1}^\infty \frac{1}{1-q^n} \sum_{m \in \mathbb{Z}} \left( q^{12m^2+m} - q^{12m^2+7m+1} \right) \\ % &= \frac{1}{2} \left( \prod_{m=1}^\infty \left( 1+q^{m-1/2} \right) + \prod_{m=1}^\infty \left( 1-q^{m-1/2} \right) \right) \\ % &= \prod_{k=1}^\infty \frac{(1+q^{8k - 5})(1+q^{8k-3})(1-q^{8k})}{1-q^{2k}} \\ &= \sum_{k = (k_1, k_2, \ldots, k_8) \in \mathbb{Z}^8_{\geq 0}} \frac{q^{k^T C_{E_8}^{-1} k}}{(q)_{k_1} \dots (q)_{k_8}} \end{aligned} $$ This last form is striking, here $C_{E_8}$ is the Cartan matrix for the simple Lie algebra $E_8$.

The RHS can be easily written as $$ \sum_{m \geq 0} \frac{q^{m^2}}{(q)_m} \sum_{k =0}^m q^{2k^2 + km} \binom{m}{k}_q (1 - q^k + q^m) $$ Perhaps a better approach to try to use the trick of Section 5 of [6], the RHS (*) is also equivalent to

$$ \sum_{m \geq 0} \frac{q^{m^2}}{(q)_m} \left( 1 - q^{2m +1} \right) \sum_{0 \leq 2k \leq m} \frac{(q^{-m}, q^{1-m}; q^2)_k}{(q;q)_k} q^{(m+1)k} $$

Finally, to show a not-so-straightforward equivalent form, by using Ismail-Garret-Statton generalization of the Rogers-Ramanujan identity one can sum first $m$ in (*) to obtain the equivalent form:

$$ \sum_{k \geq 0} (-1)^k \frac{q^{\frac{k(k+1)}{2}}}{(q)_k} \left( \frac{q^{2k}a_{3k} - a_{3k+2}}{(q,q^4;q^5)_{\infty}} - \frac{q^{2k} b_{3k} -b_{3k+2}}{(q^2,q^3;q^5)_{\infty}}\right) $$ where $a_k$, $b_k$ are the Schur polynomials defined by the recursion $x_{k+2} = x_{k+1} + q^k x_k$ and the initial conditions $$ a_0 = b_1 = 1, \qquad a_1 = b_0 = 0. $$

References:

[1] A. Sills Identities of the Rogers-Ramanujan-Slater Type. International Journal of Number Theory. Vol. 03, No. 02, pp. 293-323 (2007) ]

[2] G. E. Andrews and J. P. Santos. Rogers-Ramanujan type identities for partitions with attached odd parts. The Ramanujan Journal, 1:91–99, 1997.

[3] L. J. Slater. Further identities of the rogers-ramanujan type. Proc. London Math. Soc., 54(2):147–167, 1952.

[4] W. Nahm. Conformal field theory and torsion elements of the Bloch group. In Frontiers in number theory, physics, and geometry. II, pages 67–132. Springer, Berlin, 2007.

[5] D. Zagier. The dilogarithm function. In Frontiers in number theory, physics, and geometry. II, pages 3–65. Springer, Berlin, 2007.

[6] G. Andrews, K. Bringmann, K. Mahlburg Double series representations for Schur's partition function and related identities Journal of Combinatorial Theory, Series A Volume 132, May 2015, Pages 102-119

[7] J. van Ekeren, R. Heluani The singular support of the Ising model

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1 Answer 1

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The following proof is due to George E. Andrews (who is now a coauthor on the mentioned paper).

Consider the following families of polynomials indexed by $n$

$$ S_n = \sum_{k \geq 0} q^{2k^2} \binom{n-k}{2k}_q $$

$$ T_n = \sum_{m,k\geq 0} q^{m^2+3k m+4k^2} \left( \binom{n-3k-m}{k}_q \binom{n-4k-m}{m}_q - q^k \binom{n-3k-m-1}{k}_q\binom{n-4k-m-1}{m-1}_q \right). $$

The limit as $n \rightarrow \infty$ of $S_n$ is the LHS of the desired identity while the limit as $n \rightarrow \infty$ of $T_n$ is the RHS of the desired identity.

For every $n \geq 0$ one can check that $S_n = T_n$. We used the Mathematica package qMultiSum which was provided by RISC to check that both sides satisfy the order 8 recursion $$ q^{4n+15}S_n + q^{2n+11}(1+q)(S_{n+3} - S_{n+4}) - q^3 S_{n+5} + (1+q+q^2)(qS_{n+6} - S_{n+7}) + S_{n+8} = 0. $$

In fact the same technique works for the three conjectures of our paper so now we have a PBW basis for the Ising model and three new quasi-particle sum expressions for the three characters of the minimal model at central charge $c=1/2$

I'll leave this question open since it still uses a computer to check the recursion. Perhaps someone can find a proof by pen and paper.

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