# What is the ring structure on Lusztig's integral form of quantum $\mathfrak{sl}(2)$?

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?

• Should there be a $c$ in the right side of your definition? May 13, 2023 at 3:39
• @WillSawin thanks! fixed now May 13, 2023 at 3:57
• 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. May 13, 2023 at 4:16
• @LSpice Good to know! Thanks for the edits May 13, 2023 at 5:19

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 $$b>a$$)