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Take the graded algebra $A=\mathbb C[a_1,a_2,a_3,\dots]$ with grading given by $\deg a_{i}=2i$. Let $a(t)$ be the generating function for generators ($a(t)=a_1t+a_2t^2+\cdots\in A[[t]]$). What is the quotient of $A$ by relations, generating function of which is given by $a(t)^2$ — i.e. is there some nice description of the graded algebra $$ \mathbb C[a_1,a_2,a_3,\dots]/(a(t)^2=0):=\mathbb C[a_1,a_2,a_3,\dots]/(a_1^2,2a_1a_2,2a_1a_3+a_2^2,\dots,\sum_{i+j=k}a_ia_j,\dots)? $$ What about Hilbert series, at least?

Upd (from off-site communications). Feigin-Stoyanovsky looks relevant: the algebra in question appears in theorem 2.2.1 as some representation of $\widehat{sl}_2$, and formula 2.3.3 tells that Hilbert series is exactly $G(q)$ of Rogers-Ramanujan. Mysterious but interesting.

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What kind of object is $a_i$? A polynomial? If so, in what variable(s)? –  Gerry Myerson Jun 29 '11 at 12:58
    
@Grigory, I love generating functions, I just don't know what $\deg a_i=2i$ means if $a_i$ is a variable. –  Gerry Myerson Jun 30 '11 at 4:44
    
@Gerry Ah. It's just a formal grading -- I want it to be a graded algebra (note that with this grading all relations are homogeneous). –  Grigory M Jun 30 '11 at 5:22
    
I'm confused: how does $a(t)$ live in $\mathbb{C} \langle a_1,a_2,\ldots \rangle$? –  Pete L. Clark Jun 30 '11 at 9:58
    
@Grigory: okay, thanks. (I've never seen that notation before, and I only just now noticed the $:=$.) –  Pete L. Clark Jun 30 '11 at 10:11

1 Answer 1

Any coefficient of $a(t)^2$ has a «middle term» of the form either $a_i^2$ or $a_ia_{i+1}$. So in our algebra $a_1^2=0$, $a_1a_2=0$, $a_2^2=-2a_1a_3$, $a_2a_3=-a_1a_4$ and so on. And it's not hard to see that the algebra is generated (additively) by monomials of the form $a_{n_1}a_{n_2}...a_{n_k}$ s.t. $n_i+1<n_{i+1}$. Thm: these monomials form an additive basis (i.e. they are lineary independent).

The lowest degree generator of «length $l$» (that is, consisting of $l$ variables) is $a_1a_3...a_{2l-1}$ and it has degree $l^2$. Now it's not hard to see that the component of length $l$ has Hilbert series $\frac{q^{l^2}}{(1-q)(1-q^2)...(1-q^l)}$ (term $1-q^i$ in denominator corresponds to a «shift» $a_{n_1}a_{n_2}...a_{n_k}\mapsto (a_{n_1}a_{n_2}...a_{n_{k-i}})(a_{n_{k-i+1}+1}...a_{n_k+1})$, if you will). Summation by $l$ yields Hilbert-Poincare series equal to Rodgers-Ramanujan.

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