Temperley–Lieb Semisimplicity and Generic Structure

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.

Write $d_t={\dim }_{\mathbb{C}}{W}_{n,t}$. By [F1] and [F3], $\dim {\mathrm{T}\mathrm{L}}_n={\sum }_t{d}_t^2$. By [F2], each simple module is exactly one $D_{n,t}$ with nonzero form quotient, and these modules are pairwise nonisomorphic. If ${\varphi }_{n,t}$ is degenerate then either it is identically zero, so $\dim D_{n,t}=0<d_t$, or its radical is nonzero, so again $\dim D_{n,t}<d_t$.

Suppose ${\mathrm{T}\mathrm{L}}_n$ is semisimple. By [F4] and [F6], each division algebra factor $D_i$ is finite-dimensional over the central copy of $\mathbb{C}$. For $x\in D_i$, left multiplication by $x$ has a complex eigenvalue $\lambda$, and a nonzero eigenvector gives $(x-\lambda )y=0$, which forces $x=\lambda$ in a division algebra. Thus $D_i=\mathbb{C}$, the right simple module for the factor $M_{m_i}(D_i)$ has dimension $m_i$, and the right regular module contains $m_i$ copies of it. Consequently $\dim {\mathrm{T}\mathrm{L}}_n$ is the sum of the squares of the dimensions of its pairwise nonisomorphic simple right modules. By [F2], these are the $D_{n,t}$. If any form were degenerate, step 1.1 would make this sum strictly smaller than ${\sum }_t{d}_t^2=\dim {\mathrm{T}\mathrm{L}}_n$, a contradiction. Thus all forms are nondegenerate.

Conversely suppose every form is nondegenerate. Then [F2] gives $D_{n,t}=W_{n,t}\mathrm{\ne }0$ for every allowed $t$, so these are the complete simple right-module list and [F3] says that the squares of their dimensions sum to $\dim {\mathrm{T}\mathrm{L}}_n$. By [F5], the quotient by the Jacobson radical is semisimple, and its simple right modules are exactly those of ${\mathrm{T}\mathrm{L}}_n$ because the radical annihilates every simple right module. In particular, this quotient is nonzero. Applying the right-module dimension formula from step 2.1 to it shows that its dimension is the sum of the squares of those same simple dimensions. Therefore the quotient has the same dimension as ${\mathrm{T}\mathrm{L}}_n$, the radical has dimension zero, and ${\mathrm{T}\mathrm{L}}_n$ is semisimple.

Case $\mathrm{\ell }=\mathrm{\infty }$. No positive quantum integer vanishes by [F3], so every factor in [F1] is nonzero and [F2] gives semisimplicity at every rank.

Case $\mathrm{\ell }=1$. Then $q=\pm 1$ and $[r{]}_q=r{q}^{r-1}\mathrm{\ne }0$ for every positive $r$ by [F3]. Every Gram determinant in [F1] is nonzero, so [F2] gives semisimplicity at every rank.

Case $3\le \mathrm{\ell }<\mathrm{\infty }$ and $n<\mathrm{\ell }$. For every allowed $p$, each denominator index $j$ and each numerator index $n-2p+1+j$ in [F1] is positive and less than $\mathrm{\ell }$. Neither vanishes by [F3], so all forms are nondegenerate and [F2] gives semisimplicity.

Case $3\le \mathrm{\ell }<\mathrm{\infty }$ and $n\ge \mathrm{\ell }$. Write $n=a\mathrm{\ell }+s$ with $a\ge 1$ and $0\le s<\mathrm{\ell }$. If $s=0$, take $p=j=1$; if $1\le s\le \mathrm{\ell }-2$, take $p=j=s+1$; if $s=\mathrm{\ell }-1$, take $p=\mathrm{\ell }-1,j=\mathrm{\ell }-2$. In every case $1\le j\le p<\mathrm{\ell }$ and $p\le \lfloor n\mathrm{/}2\rfloor$: for $s=0$, $p=1$ and $n\ge 3$; for $1\le s\le \mathrm{\ell }-2$, $2p=2s+2\le \mathrm{\ell }+s\le n$; and for $s=\mathrm{\ell }-1$, $2p=2\mathrm{\ell }-2\le 2\mathrm{\ell }-1\le n$. Direct substitution gives $n-2p+1+j=a\mathrm{\ell }$. The exponent $d_{n,p-j}$ is positive because $p-j$ is $0$ or $1$, and $[j{]}_q\mathrm{\ne }0$ because $j<\mathrm{\ell }$. Thus the determinant in [F1] is zero, one form is degenerate, and [F2] gives nonsemisimplicity.

Case $\mathrm{\ell }=2$. At $n=0$, ${\mathrm{T}\mathrm{L}}_0=\mathbb{C}$ is semisimple. For even $n\ge 2$, $p=j=1$ gives $\det G_{n,1}=[n{]}_q\mathrm{/}[1{]}_q=0$, so [F2] gives nonsemisimplicity. If $n$ is odd, then $n-2p+1=2m>0$ for every allowed $p$. For odd $j$, both quantum integers in each factor of [F1] are nonzero; for $j=2k$, numerator and denominator have simple zeros at $q^2=-1$, and their ratio tends to $(-1{)}^m(m+k)\mathrm{/}k\mathrm{\ne }0$. Hence all continued determinants are nonzero and [F2] gives semisimplicity for odd $n$.

The cases $\mathrm{\ell }=\mathrm{\infty }$, $\mathrm{\ell }=1$, $3\le \mathrm{\ell }<\mathrm{\infty }$ with $n<\mathrm{\ell }$ or $n\ge \mathrm{\ell }$, and $\mathrm{\ell }=2$ exhaust all possibilities. Steps 1.1–1.5 prove exactly the branches in the Statement, including the rank-zero order-two endpoint.

For basis diagrams $D,E:n\to n$, the closed graph used to evaluate ${\tau }_n(DE)$ is obtained by placing the two layers in a cyclic order and joining the exterior endpoints. Cutting this cyclic graph between the $D$ and $E$ layers gives the graph for ${\tau }_n(ED)$ with the same components and the same number of closed loops. The normalizing power ${\delta }^{-n}$ is the same, so ${\tau }_n(DE)={\tau }_n(ED)$. Bilinearity extends the identity to all $D,E$.

In the closure of $x\otimes 1$, the added vertical strand closes to one additional circle and is disjoint from the closed graph for $x$. Thus its loop count is $c(x)+1$, while the normalizing exponent changes from $-n$ to $-(n+1)$; the two powers of $\delta$ cancel and ${\tau }_{n+1}(x\otimes 1)={\tau }_n(x)$.

In the closure of $(x\otimes 1)d_n$, the last cap and cup connect the new boundary strand to the preceding strand and create no additional closed component. The normalization still gains one factor ${\delta }^{-1}$ from rank $n$ to rank $n+1$. Hence ${\tau }_{n+1}((x\otimes 1)d_n)={\delta }^{-1}{\tau }_n(x)$ on basis diagrams, and bilinearity gives it for every $x$.

For ranks zero and one, $p_0=p_1=1$ has all asserted properties, ${\epsilon }_s(p_s)=1$, and trace $1=[1]\mathrm{/}{\delta }^0=[2]\mathrm{/}\delta$ where applicable. At rank two, $p_2=1-{\delta }^{-1}{d}_1$; $d_1^2=\delta d_1$ gives $p_2^2=p_2$ and $d_1{p}_2=p_2{d}_1=0$, ${\epsilon }_2(p_2)=1$, its corner is $\mathbb{C}{p}_2$ since $1,d_1$ span ${\mathrm{T}\mathrm{L}}_2$, and ${\tau }_2(p_2)=1-{\delta }^{-2}=[3]\mathrm{/}{\delta }^2$.

Fix $r\ge 2$ and assume inductively that through rank $r$ the recursively defined elements are nonzero idempotents, satisfy ${\epsilon }_s(p_s)=1$, are annihilated on both sides by every generator, have corner $\mathbb{C}{p}_s$, and satisfy ${\tau }_s(p_s)=[s+1]\mathrm{/}{\delta }^s$.

Put $P=p_r\otimes 1$, $Q=p_{r-1}\otimes 1\otimes 1$, $a=d_{r-1}$, $b=d_r$, and $\mu =[r-1]\mathrm{/}[r]$. The recursion at rank $r$ gives $P=Q-\mu QaQ$; $Q$ is idempotent, commutes with $b$, and $bab=b$ by [F1]. Therefore $bPb=\delta Qb-\mu QbabQ=(\delta -\mu )Qb=([r+1]\mathrm{/}[r])Qb$ by [F3].

Set $X=PbP$ and $c=[r]\mathrm{/}[r+1]$. Since $PQ=QP=P$, step 3.1 gives $X^2=P(bPb)P=([r+1]\mathrm{/}[r])X$, so $(P-cX{)}^2=P-cX=p_{r+1}$. Also $b{p}_{r+1}=bP-cbPbP=0$ by step 3.1 and $QbP=bP$; and $p_{r+1}b=Pb-cPbPb=0$ because $PbPb=P(bPb)=([r+1]\mathrm{/}[r])PQb=([r+1]\mathrm{/}[r])Pb$. For $i<r$, both annihilation identities follow from the induction hypothesis. Juxtaposing a vertical strand preserves the identity coefficient, so ${\epsilon }_{r+1}(P)={\epsilon }_r(p_r)=1$. The correction term $PbP$ has fewer than $r+1$ through strands, since every diagram term passes through $b=d_r$, which itself has fewer than $r+1$, and stacking cannot increase that count by [F6]. Hence ${\epsilon }_{r+1}(PbP)=0$, so ${\epsilon }_{r+1}(p_{r+1})=1$ and $p_{r+1}\mathrm{\ne }0$.

Every nonempty generator word has a rightmost letter which kills $p_{r+1}$ by step 4.1. Since generator words span ${\mathrm{T}\mathrm{L}}_{r+1}$ by [F5], $p_{r+1}{\mathrm{T}\mathrm{L}}_{r+1}{p}_{r+1}=\mathbb{C}{p}_{r+1}$.

By [F4], ${\tau }_{r+1}(P)={\tau }_r(p_r)$ and, using cyclicity, ${\tau }_{r+1}(X)={\tau }_{r+1}(PbP)={\tau }_{r+1}(b{P}^2)={\tau }_{r+1}(Pb)={\delta }^{-1}{\tau }_r(p_r)$. Hence ${\tau }_{r+1}(p_{r+1})={\tau }_r(p_r)(1-[r]\mathrm{/}(\delta [r+1]))=[r+2]\mathrm{/}{\delta }^{r+1}$ by [F3] and the induction hypothesis. This proves the trace recurrence and closes the induction.

Since $\delta \mathrm{\ne }0$ and $[r{]}_q\mathrm{\ne }0$ for $1\le r\le n=\mathrm{\ell }-1$, [F2] supplies the nonzero primitive idempotent $p_n$ and gives ${\tau }_n(p_n)=[\mathrm{\ell }{]}_q\mathrm{/}{\delta }^n=0$ by [F4].

For any $x\in {\mathrm{T}\mathrm{L}}_n$, cyclicity [F3] and $p_n^2=p_n$ imply ${\tau }_n(p_{n}x(1-p_n))={\tau }_n((1-p_n)p_{n}x)=0$, so ${\tau }_n(p_{n}x)={\tau }_n(p_{n}x{p}_n)$. The one-dimensional corner in [F2] gives $p_{n}x{p}_n={\lambda }_x{p}_n$ for some ${\lambda }_x\in \mathbb{C}$. Hence ${\tau }_n(p_{n}x)={\lambda }_x{\tau }_n(p_n)=0$ for every $x$, while $p_n\mathrm{\ne }0$. Thus $p_n$ is a nonzero vector in the radical of the trace form, so the form is degenerate.

Semisimplicity is the independent assertion in [F1]; combining it with step 2.1 proves that trace degeneracy occurs at rank $\mathrm{\ell }-1$ while the algebra is still semisimple.

A basis diagram has $n$ propagating strands exactly when it is the identity. If either factor in a product has fewer than $n$ propagating strands, every exterior through path in the product must pass through a distinct propagating path in that factor; hence the product also has fewer than $n$. Thus the span of nonidentity diagrams is a two-sided ideal and $\epsilon$ is an algebra homomorphism. Its kernel is spanned by nonempty generator words by [F1].

If $K_{\epsilon }$ is projective, the surjection $\epsilon :A\to K_{\epsilon }$ splits as a left $A$-module map $s$. Put $e=s(1)$. Then $\epsilon (e)=1$ and, for every $a\in A$, $ae=s(\epsilon (a))=\epsilon (a)e$. Taking $a=e$ gives $e^2=e$, and taking $a=d_i$ gives $d_{i}e=0$.