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Consider the quantum group $U_q(\mathfrak{sl}_2)$, with generators $E,F,K$ such that $[E,F]=\frac{K-K^{-1}}{q-q^{-1}}$. Write $[n]=\frac{q^n-q^{-n}}{q-q^{-1}}$, and $[n]!=[n][n-1]\dotsm[1]$.

In Quantum deformations of certain simple modules over enveloping algebras, Lusztig defined the following elements inside his integral form for $U_q(\mathfrak{sl}_2)$ (I am replacing $q$ by its square root in his notation): $$ \newcommand\qbinom{\genfrac[]0{}}\qbinom{K;\ c}t=\frac{1}{[t]!}\prod_{s=0}^{t-1}\frac{(q^{c-s}K-q^{s-c}K^{-1})}{(q-q^{-1})}.$$ These are $q$-deformations of the element $\binom{H+c}{t}$ in Kostant's $\mathbb{Z}$-form for $U(\mathfrak{sl}_2)$.

In Kostant's $\mathbb{Z}$-form, we have identities such as $\binom{H}{1}^2=2\binom{H}{2}+\binom{H}{1}$. However, if one tries to quantize both sides, the result is $$ \qbinom{K;\ 0}1^2=q^2[2]\qbinom{K;\ 0}2+qK^{-1}\qbinom{K;\ 0}1.$$

For various reasons, I am interested in identities whose coefficients do not involve powers of $K$. For instance, it turns out $$ \qbinom{K;\ 0}1^2=\qbinom{K;\ 1}2+\qbinom{K;\ 0}2.$$

In general, I believe one may be able to apply iteratively a relation in Lusztig's paper, to prove that products between such elements are always $\mathbb{Z}(q)$-linear combinations of $\qbinom{K;c}t$'s. A possible caveat, is that these elements are not linearly independent over $\mathbb{Z}(q)$. For instance, one has the relation $$ \qbinom{K;\ 0}2-[3]\qbinom{K;\ 1}2+[3]\qbinom{K;\ 2}2-\qbinom{K;\ 3}2=0.$$

[EDIT: Assuming a certain PBW-like result, I think one may be able to prove that Lusztig’s elements for $c\le t$ form a basis.]

This is such a well studied object, so I was hoping this ring structure would be worked out somewhere. In short:

What is the product $\qbinom{K;\ a}t\qbinom{K;\ b}s$ in terms of Lusztig's elements?

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    $\begingroup$ Should there be a $c$ in the right side of your definition? $\endgroup$
    – Will Sawin
    May 13, 2023 at 3:39
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    $\begingroup$ @WillSawin thanks! fixed now $\endgroup$ May 13, 2023 at 3:57
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    $\begingroup$ Please don't use displaymath in titles. TeX note: The ferociously general command \genfrac is meant for situations like yours. For example, $\genfrac()0{}{H + c}t$ \genfrac()0{}{H + c}t produces the un-quantised version, $\genfrac[]0{}{K; c}t$ \genfrac[]0{}{K; c}t produces the quantised version, and, just to illustrate, $\genfrac(){}{}a q$ is a Legendre symbol \genfrac(){}{}a q. I always have to Google the syntax, but the first two arguments are the delimiters, and the last two are the "numerator" and "denominator" of a generalised fraction. Anyway, I edited accordingly. $\endgroup$
    – LSpice
    May 13, 2023 at 4:16
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    $\begingroup$ @LSpice Good to know! Thanks for the edits $\endgroup$ May 13, 2023 at 5:19

1 Answer 1

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I figured it out, but I would really appreciate a reference!

The formula is quite nice. For $c,d\ge0$ such that $c\le t $ and $d\le s$, the following holds: $$ \newcommand\qbinom{\genfrac[]0{}}\qbinom{K;\ c}t\qbinom{K;\ d}s=\sum_{i\ge 0}\qbinom{t-c+d}{i-c}\qbinom{s-d+c}{i-d}\qbinom{K;\ i}{t+s}.$$

(Here, the binomial $\qbinom{a}{b}$ is set to zero whenever $b<0$ or $b>a$)

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