Critical Temperley–Lieb Gram determinants in a digit unit interval

Each Gram entry is a monomial in the loop parameter with integer coefficient, so its determinant is a polynomial $D(T)\in \mathbb{Z}[T]$. Clear the quantum-integer denominators in [F1], obtaining an equality between integer polynomials for every complex loop parameter. Every complex number has the form $q+q^{-1}$ for a nonzero $q$, and a polynomial with infinitely many distinct complex zeros is zero. Thus this is an integer polynomial identity, which reduces to an identity over every field. No division after specialization is involved in this conclusion.

Adjoin $q_0\mathrm{\ne }0$ with $q_0+q_0^{-1}=\delta$ using [F3], and work in formal Laurent series over that extension. Set $q(t)=q_0(1+t)$ and $\delta (t)=q(t)+q(t{)}^{-1}$. As a rational function of $t$, $\delta (t)$ has a pole at $t=-1$, so it is nonconstant. For every nonzero polynomial $P\in K[T]$, $P(\delta (t))$ is a nonzero rational function: if $\mathrm{deg}P=d>0$, it has a pole of order $d$ at $t=-1$, and if $d=0$ it is a nonzero constant. The Laurent expansion at $t=0$ of a nonzero rational function is nonzero. Thus substitution $P(T)\mapsto P(\delta (t))$ is injective, and every nonzero polynomial remains a nonzero series. Every quantum integer $[j{]}_{\delta (t)}$ is nonzero by its Laurent expression $q(t{)}^{1-j}(1-q(t{)}^{2j})\mathrm{/}(1-q(t{)}^2)$, interpreted as a rational identity. We may consequently use the product in [F1] in the Laurent-series field. Its constant coefficient is $D(\delta )$.

Suppose the characteristic is $p>0$ and put $c=1$ for odd $p$ and $c=2$ for $p=2$. For every positive integer $j$, the order of $1-(1+t{)}^{2j}$ is $c{p}^{{\nu }_p(j)}$: factor $2j=p^{h}u$ with $p\nmid u$, use $(1+t{)}^{p^h}=1+t^{p^h}$, and the coefficient of $t^{p^h}$ is the nonzero residue of $u$. If $q_0^2\mathrm{\ne }1$, finite $\mathrm{\ell }$ is the order of $q_0^2$, and it is prime to $p$ because a field has no nonidentity element of order $p$. Thus $[j{]}_{\delta (t)}$ has order zero for $\mathrm{\ell }\nmid j$ and order $c{p}^{{\nu }_p(j\mathrm{/}\mathrm{\ell })}$ for $\mathrm{\ell }\mid j$. If $q_0^2=1$, the recurrence gives $[j{]}_{\delta }=j{q}_0^{j-1}$ and $\mathrm{\ell }=p$; the denominator $1-q(t{)}^2$ has order $c$, so the order of $[j{]}_{\delta (t)}$ is $c(p^{{\nu }_p(j)}-1)$. These statements hold also in characteristic two.

For a critical label $m+1=B\mathrm{\ell }$, the numerator and denominator of a factor in [F1] both have order zero when $\mathrm{\ell }\nmid j$. If $j=j^{\mathrm{\prime }}\mathrm{\ell }$, step 3.1 shows that their orders agree whenever ${\nu }_p(B+j^{\mathrm{\prime }})={\nu }_p(j^{\mathrm{\prime }})$, in both of its cases, including $\mathrm{\ell }=p$. Write $B=p^v{b}^{\mathrm{\prime }}$ as in the Statement and suppose $0<j^{\mathrm{\prime }}<p^v(p-b)$. If $p^v\nmid j^{\mathrm{\prime }}$, then ${\nu }_p(j^{\mathrm{\prime }})<v$ and adding $B$ preserves that valuation. Otherwise $j^{\mathrm{\prime }}=p^{v}h$ with $1\le h<p-b<p$, so both $h$ and $b^{\mathrm{\prime }}+h$ are nonzero modulo $p$ and both valuations equal $v$. Thus the equality holds for every $j^{\mathrm{\prime }}\le \lfloor (A-B)\mathrm{/}2\rfloor$ under the stated bound. Every factor in the determinant product is consequently a Laurent-series unit with nonzero constant term. Their product has nonzero constant term, proving $D(\delta )\mathrm{\ne }0$. This nonvanishing descends to the original field, so [F2] proves simplicity.

In characteristic zero with finite $\mathrm{\ell }$, one has $q_0^2\mathrm{\ne }1$. Each quantum integer whose index is a multiple of $\mathrm{\ell }$ has a simple zero along $q(t)$, because differentiating $1-q(t{)}^{2j}$ gives nonzero coefficient $-2j$ at $t=0$. All other indices give units. Therefore each numerator and denominator in the critical determinant product has the same order, either zero or one, and its quotient has nonzero constant term. The same argument as in step 4.1 proves simplicity. If $\mathrm{\ell }=\mathrm{\infty }$, every specialized numerator and denominator in [F1] is nonzero, so the determinant is nonzero directly. Finally, when $m=n-2$, the product consists solely of $[n{]}_{\delta }$ to exponent one. Thus $[n{]}_{\delta }\mathrm{\ne }0$ implies the last simplicity assertion.