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I take a random but practical example direct from "R-matrices and the magic square" by Bruce Westbury: The adjoint irrep $A$ of the $E_7$ family has quantum dimension
$qi[2*m+3]*qi[3*m/2+2]*qi[3*m/2]/qi[m/2]/qi[m/2+2]$
where $qi[m]$ (at the value $q$) denotes the quantum integer function and $m$ is "mostly" an integer (but not throughout, e.g. $G_2$ has $m=-5/2$). At $q=1$, we get $(m+4)|120$. Let's stay at integer $m$ for simplicity and take, say, $m=116: Dim(A)=2068$. OK, now a different value, say, $q=i: Dim(A)=-44/5$. Owch.
I conclude that $qi[58]*qi[60]|qi[174]*qi[176]*qi[235]$ (in the Laurent polynomial sense, see below) fails to hold, although $58*60|174*176*235$.
Can you give an condition for $qi[m]|qi[n]$ (or even better, for arbitrary products of quantum integers in numerator and denominator allowed) that is sharper than $m|n$? $|$ shall mean that the denominator of the fully cancelled form for $qi[m]|qi[n]$ only may contain some $q^k$ (i.e. not $q-1/q$ or something like that). (Random Example: $qi[2]*qi[4]|qi[8]$ is false because the denominator contains $1+q^8$.)

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    $\begingroup$ The a better latexism for the divisibility relation is \mid rather than the more obvious |, because it gets the spacing right: $a\mid b$ versus $a|b$. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 15, 2012 at 19:58
  • $\begingroup$ What exactly is a quantum integer? $\endgroup$
    – Gerry Myerson
    Commented Feb 15, 2012 at 23:24

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Factor a fraction in the $q$-integers as a product of cyclotomic polynomials $\Phi_d$, using that $q^n-1 = \prod_{d|n} \Phi_d$, and look at the multiplicity of each polynomial $\Phi_d$.

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  • $\begingroup$ That answers the question precisely and constructively... but not in the sense I hoped :-) Is the failure of the 58∗60|174∗176∗235 or the simpler 2*4|8 example due to the following (drawn from the hat) Conjecture: qi[p1*p2] does not divide q[p1*p3]*q[p2*p4], you need some qi[p1*p2*p5] the numerator. (Loosely speaking.) P.S. @Gerry: qi[m,q]=Sum[q^i,{i,-k+1,k-1,2}]. qi[m,1]=m. P.P.S. @Mariano: I live, I learn. Maybe even TeX. :-) $\endgroup$
    – Hauke Reddmann
    Commented Feb 16, 2012 at 12:19
  • $\begingroup$ Thanks, but, what does Sum$[q^i,\lbrace i,-k+1,k-1,2\rbrace]$ mean?Is that $q^{-k+1}+q^{-k+3}+q^{-k+5}+\cdots+q^{k-1}$? $\endgroup$
    – Gerry Myerson
    Commented Feb 16, 2012 at 22:44

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