The Temperley–Lieb link-state Gram determinant

For $n=0$ and $n=1$, the only allowed arc count is $p=0$; the all-defect link state has Gram matrix $(1)$, agreeing with the empty product.

For any $n$, the $p=0$ module is spanned by the all-defect state and its form value is $1$. Now fix $n\ge 2$ and assume inductively that the determinant formula holds at all smaller ranks; because no positive quantum integer vanishes when $q^2$ is not a root of unity, those smaller Gram forms are nondegenerate.

Suppose $n>2p>0$. By [F2], restriction has the rightmost-defect submodule $U\cong W_{n-1,n-1-2p}$ and quotient $W_{n-1,n-1-2(p-1)}$. Deleting the rightmost defect preserves the reflected-and-capped matching graph and its loop count, so the restricted form on $U$ is exactly the standard form with matrix $G_{n-1,p}$; it is nondegenerate by induction. Invariance [F1] makes $U^{\perp }$ invariant, and nondegeneracy on $U$ gives $W_{n,n-2p}=U\oplus U^{\perp }$. The quotient map identifies $U^{\perp }$ with the quotient module. The quotient's standard form is nondegenerate by induction, so [F3] makes it absolutely simple and the form identifies it with its dual. The form induced from $U^{\perp }$ is invariant; by [F4, F5], its map to the dual is a scalar multiple of the standard form's map, say ${\alpha }_{n,p}$ times that map. Subtracting the submodule projection from each quotient lift is a block-triangular basis change with identity diagonal, so the Gram matrix in this basis is $G_{n-1,p}\oplus {\alpha }_{n,p}{G}_{n-1,p-1}$ and the change-of-basis determinant is $1$.

If $n=2p$ with $p>0$, deleting the rightmost endpoint and turning its partner into a defect bijects the link-state basis with that of $W_{2p-1,p-1}$. In the glued matching graph, the reduced pairing has one open path between the new defect ends; restoring the two terminal arcs closes exactly that path into one additional component and leaves every other component unchanged. Hence $G_{2p,p}=\delta G_{2p-1,p-1}$, which is the boundary scalar ${\alpha }_{2p,p}=\delta$.

For $n>2p$, refine the ordered link states into four types: type 1 has $n$ and $n-1$ both defects; type 2 has $n$ a defect and $n-1$ on an earlier arc; type 3 joins $n-1$ to $n$; and type 4 has $n$ on an arc whose partner is earlier than $n-1$. The gluing rule gives $G_{n,p}^{12}=\left(\begin{array}{cc}0& \ast \\ G_{n-2,p-1}& \ast \end{array}\right)$ and $G_{n,p}^{22}=\left(\begin{array}{cc}\delta G_{n-2,p-1}& \ast \\ \ast & \ast \end{array}\right)$: the type-1/type-3 pairing has unmatched defects and is zero, cutting a type-2/type-3 terminal arc leaves the same exterior matching, and a type-3/type-3 pairing creates one loop. Let $A=G_{n-2,p}$, $B=G_{n-2,p-1}$, and write the rank-$(n-1)$ form as $G_{n-1,p}=\left(\begin{array}{cc}A& A{V}_0\\ V_0^{\mathsf{T}}A& V_0^{\mathsf{T}}A{V}_0+{\alpha }_{n-1,p}B\end{array}\right)$; at $n=2p+1$ the $A,V_0$ blocks are empty and their products are omitted. If $V=\left(\begin{array}{cc}{V}^{11}& \ast \\ V^{21}& \ast \end{array}\right)$ is the off-diagonal block in the orthogonalizing basis change of step 3.1, then $G_{n,p}^{12}=G_{n-1,p}V$ gives $0=A{V}^{11}+A{V}_0{V}^{21}$ and $B=V_0^{\mathsf{T}}A{V}^{11}+(V_0^{\mathsf{T}}A{V}_0+{\alpha }_{n-1,p}B)V^{21}$. The induction hypothesis makes $A$ invertible when nonempty, so the first equation gives $V^{11}=-V_0{V}^{21}$; at $n=2p+1$ these blocks are omitted. Substitution in the second equation cancels the $A$-terms and yields $B={\alpha }_{n-1,p}B{V}^{21}$. The induction hypothesis makes $B$ invertible for generic $q$, so multiplication by $B^{-1}$ gives $I={\alpha }_{n-1,p}{V}^{21}$ and hence ${\alpha }_{n-1,p}\mathrm{\ne }0$ and $V^{21}={\alpha }_{n-1,p}^{-1}I$.

The top-left block of $G_{n,p}^{22}=V^{\mathsf{T}}{G}_{n-1,p}V+{\alpha }_{n,p}{G}_{n-1,p-1}$ is $\delta B$. Its first term is $(V^{11}{)}^{\mathsf{T}}A{V}^{11}+(V^{11}{)}^{\mathsf{T}}A{V}_0{V}^{21}+(V^{21}{)}^{\mathsf{T}}{V}_0^{\mathsf{T}}A{V}^{11}+(V^{21}{)}^{\mathsf{T}}(V_0^{\mathsf{T}}A{V}_0+{\alpha }_{n-1,p}B)V^{21}$. Substituting $V^{11}=-V_0{V}^{21}$ makes the coefficients of $(V^{21}{)}^{\mathsf{T}}{V}_0^{\mathsf{T}}A{V}_0{V}^{21}$ equal $1-1-1+1=0$ and leaves ${\alpha }_{n-1,p}(V^{21}{)}^{\mathsf{T}}B{V}^{21}={\alpha }_{n-1,p}^{-1}B$. The first block of $G_{n-1,p-1}$ is $B$, so cancelling invertible $B$ gives ${\alpha }_{n,p}=\delta -1\mathrm{/}{\alpha }_{n-1,p}$.

The quantum-integer recurrence [F6] solves this scalar recurrence as ${\alpha }_{n,p}=[n-2p+2{]}_q\mathrm{/}[n-2p+1{]}_q$: the boundary value at $n=2p$ is $[2{]}_q\mathrm{/}[1{]}_q=\delta$, and if ${\alpha }_{n-1,p}=[n-2p+1{]}_q\mathrm{/}[n-2p{]}_q$, then $\delta -1\mathrm{/}{\alpha }_{n-1,p}=\delta -[n-2p{]}_q\mathrm{/}[n-2p+1{]}_q=[n-2p+2{]}_q\mathrm{/}[n-2p+1{]}_q$.

The determinant of the orthogonal block form in step 3.1 and its unit-determinant basis change gives $\det G_{n,p}=([n-2p+2{]}_q\mathrm{/}[n-2p+1{]}_q{)}^{d_{n-1,p-1}}\det G_{n-1,p}\det G_{n-1,p-1}$ for $n>2p$, while step 3.2 gives $\det G_{2p,p}={\delta }^{d_{2p-1,p-1}}\det G_{2p-1,p-1}$; also $\det G_{n,0}=1$ by step 2.1.

Let $P_{n,p}={\prod }_{j=1}^p([n-2p+1+j{]}_q\mathrm{/}[j{]}_q{)}^{d_{n,p-j}}$. For $n>2p$, put $h=n-2p+1$. If $p=1$, $P_{n,1}=([h+1]\mathrm{/}[h])P_{n-1,1}$. If $p\ge 2$, the numerator $[h+j]$ for $2\le j<p$ receives exponents $d_{n-1,p-j-1}$ and $d_{n-1,p-j}$ from the two lower-rank products, summing to $d_{n,p-j}$; at $j=1$ the missing exponent is supplied by $([h+1]\mathrm{/}[h]{)}^{d_{n-1,p-1}}$, whose denominator cancels the extra $[h]$, and at $j=p$ the second lower-rank product supplies exponent $d_{n,0}=1$. The denominator exponents add as $d_{n-1,p-j}+d_{n-1,p-j-1}=d_{n,p-j}$ (with the endpoint $j=p$ supplied by the first lower-rank product). Thus $P_{n,p}$ satisfies the recurrence in step 7.1 by Pascal's identity for $d_{n,r}$.

At $n=2p$, $P_{2,1}=[2]=\delta$ for $p=1$. For $p\ge 2$, compare net exponents of each quantum integer in $P_{2p,p}$ and $[2{]}^{d_{2p-1,p-1}}{P}_{2p-1,p-1}$: for $2\le m<p$ the exponent is $d_{2p,p-m+1}-d_{2p,p-m}=d_{2p-1,p-m+1}-d_{2p-1,p-m-1}$ by Pascal's identity, exactly the exponent supplied by the latter product. At $m=p$, both exponents are $d_{2p,1}-d_{2p,0}=2p-2=d_{2p-1,1}$ (for $p=2$, the latter comes from the prefactor $[2{]}^{d_{3,1}}$); for $m=p+1$ both exponents are $d_{2p,0}=d_{2p-1,0}=1$, and $[1]=1$. Hence $P_{2p,p}={\delta }^{d_{2p-1,p-1}}{P}_{2p-1,p-1}$, the boundary recurrence in step 7.1.

Induction using steps 7.1–8.2 now gives $\det G_{n,p}=P_{n,p}$ for every $q^2$ not a root of unity. Every Gram entry is polynomial in $\delta$, so its determinant is polynomial; the displayed product is rational in $q$ and agrees with that determinant on the infinite set where $q^2$ is not a root of unity. After putting the difference over a common denominator, its polynomial numerator has infinitely many zeros and is identically zero. Hence the entire product is the polynomial in $\delta$ represented by this rational-function identity, and the formula holds for every $q\in {\mathbb{C}}^{\times }$, including parameter values where separate factors are singular.