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8 votes
1 answer
497 views

q-Integer-valued polynomials

For $n \in \mathbb{Z}_{\geq 0}$, let $[n]_q := (1-q^n)/(1-q) = (1+q+...+q^{n-1})$ as is customary, with $[0]_q=0$. Let $R$ be the subring of $\mathbb{Q}(q)[x]$ consisting of all $f$ such that $f([n]...
Sam Hopkins's user avatar
  • 24.2k
5 votes
1 answer
178 views

A $q$-analogue of a characterization of polynomials by binomial coefficients

Considering the binomial coefficient $\binom{x}{m}$ as a polynomial in $x$, the span of $\binom{x}{0}, \binom{x}{1}, \ldots, \binom{x}{d}$ is exactly the polynomials of degree $\le d$. A closely ...
Mark Wildon's user avatar
  • 11.2k
4 votes
0 answers
108 views

Quantum version of Kostant's basis of ℤ-form of U(𝔤)

Kostant showed that the subring of $\mathcal U(\mathfrak{sl}_2)$ generated by the divided powers $e^c/c!$ and $f^a/a!$ has a $\mathbb Z$-basis given by the elements $\frac{f^a}{a!}\binom hb \frac{e^c}{...
Linus S's user avatar
  • 71