what else is in $\prod_{j=1}^n(1+q^j)$?

Question

From time to time, I run into the finite product $\prod_{j=1}^n(1+q^j)$. And, the more it happens, the more fascinated I've become. So, herein, I wish to get help in collecting such results. To give some perspective into what I look for, check out the below examples. First, some nomenclature: $(q)_k=(1-q)(1-q^2)\cdots(1-q^k)$ and $\binom{n}k_q=\frac{(q)_n}{(q)_k(q)_{n-k}}$.

(0) It's almost silly, but the sum of elementary functions of the specialization $\pmb{q}=(q,q^2,\dots,q^n)$: $$e_0(\pmb{q})+e_1(\pmb{q})+\cdots+e_n(\pmb{q})=\prod_{j=1}^n(1+q^j).$$

(1) The classical $q$-binomial theorem, which results from counting restricted distinct partitions or number of weighted tilings: $$\sum_{k=0}^nq^{\binom{k+1}2}\binom{n}k_q=\prod_{j=1}^n(1+q^j).$$

(2) I can't remember where I saw this (do you?) but $$\sum_{k=0}^nq^k\binom{n}k_{q^2}=\prod_{j=1}^n(1+q^j).$$

(3) The $H$-polynomial of a symplectic monoid $MSp_n$ (see this paper, page 13): $$\sum_{k=0}^n(-1)^kq^{k^2}\binom{n}k_{q^2}^2\prod_{i=1}^k(1-q^{2i})\prod_{j=1}^{n-k}(1+q^j)^2=\prod_{j=1}^{2n}(1+q^j),$$ although the authors did not seem to be aware of the RHS.

QUESTION. Can you provide such formulas (in any field) with the same RHS (always a finite product) as in above, together with resources or references?

Thank you.

Answer 1

Up to scaling, $\prod_{j=1}^n(1+q^j)$ is the character of the principal specialization of the spinor representation of $\mathfrak{so}(2n+1)$. This was first explicitly stated by J. W. B. Hughes, Lie algebraic proofs of some theorems on partitions, in Number Theory and Algebra, Academic Press, 1977, pp. 135--155.

Answer 2

In a $2n$ dimensional vector space equipped with a symplectic form, you can look at all the Lagrangian subspaces, and this forms a subvariety of the Grassmannian $G(n,2n)$ called the Lagrangian Grassmannian, $\Lambda(n)$. It is a smooth projective variety which inherits a schubert cell decomposition from the Grassmannian, and lots of other Schubert calculus goodness. It's cells are indexed by strict partitions with largest part at most $n$. The number of points of $\Lambda(n)$ over finite fields is given by $\prod_{i=1}^n(1+q^i)$, so unimodality of the coefficients can be obtained from the Hard Lefschetz theorem. (Not like this is a simpler way of looking at it, but perhaps a geometric point of view can be useful for certain applications, for example the classical q-binomial theorem follows from a certain stratification of $\Lambda(n)$.)

A reference for an explicit presentation of the cohomology ring using Schur Q-polynomials can be found in "Algebro—Geometric applications of Schur S- and Q-polynomials" by Piotr Pragacz, Topics in Invariant Theory ,M. P. Malliavin Ed., Springer Lecture Notes in Math., 1478, 1991.

Answer 3

Let $exp(z,q)=\sum_{k=0}^{\infty}z^k/[k]_q!$ be the usual $q-$analogue of the exponential function $e^x.$
The identity $\sum_{k=0}^n{q^k}{\binom{n}{k}_{q^2}}=\prod_{j=1}^n(1+q^j)$ can be obtained by comparing coefficients in $$exp(\frac{z}{1+q},q^2)exp(\frac{qz}{1+q},q^2)=exp(z,q),$$ which is a natural $q-$analogue of $e^\frac{x}{2} e^\frac{x}{2}=e^x.$ This identity occurs in Séminaire Lotharingien de Combinatoire, B05a (1981), but is perhaps older.