Temperley–Lieb Root-of-Unity and Modular Representation Theory

If $\kappa =1$, then $q=\pm 1$. By [F1], every positive quantum integer is nonzero, so each factor $[t+1+j{]}_q\mathrm{/}[j{]}_q$ in [F3] is nonzero. Therefore $\det G_{n,p}\mathrm{\ne }0$ in this case.

Suppose $\kappa \ge 2$. By [F5], there is an integer $m\ge 1$ with $t+1=m\kappa$. Fix a root $q_0=q$ and write $ϵ=q_0^{m\kappa }\in \{1,-1\}$. For any $j$ not divisible by $\kappa$, $[j{]}_{q_0}\mathrm{\ne }0$ and $[m\kappa +j{]}_{q_0}=ϵ[j{]}_{q_0}$, by substituting $q_0^{m\kappa }=q_0^{-m\kappa }=ϵ$ in the quantum-integer formula. Thus this quotient is nonzero.

If $j=s\kappa$ with $s\ge 1$, the quotient in [F3] has a removable zero-over-zero. Write nearby parameters as $q=e^u$, with $u_0$ satisfying $e^{\kappa u_0}={ϵ}_0\in \{1,-1\}$. Since $\kappa \ge 2$, $q_0\mathrm{\ne }\pm 1$ and $\sinh u_0\mathrm{\ne }0$. From $[a\kappa {]}_{e^u}=\sinh (a\kappa u)\mathrm{/}\sinh u$, the ratio has limit ${\lim }_{u\to u_0}\frac{[(m+s)\kappa {]}_{e^u}}{[s\kappa {]}_{e^u}}=\frac{m+s}{s}(-1{)}^{mv}\mathrm{\ne }0$, where $\kappa u_0=\pi iv$ for an integer $v$; this follows by taking the ratio of the first derivatives of the two vanishing numerators. Also $d(q+q^{-1})\mathrm{/}dq=1-q^{-2}\mathrm{\ne }0$ at $q_0$, so this quotient also has a removable continuation locally in $\delta$ at the chosen root. Thus every rational factor in [F3] is regular and nonzero there, and their product equals the determinant polynomial by [F3].

The exponents $d_{n,p-j}$ in [F3] are nonnegative integers, and an empty product is $1$. Steps 1.1–2.1 therefore give $\det G_{n,p}\mathrm{\ne }0$ for every critical label, so the finite-dimensional form ${\varphi }_{n,t}$ is nondegenerate.

By [F4], a nondegenerate cell form has zero radical and its quotient by that radical is simple. The quotient is $W_{n,t}$ itself, so this standard module is simple.

For a parameter variable $b$, let ${\varphi }_b$ be the link-state form on the fixed basis of $W_{2p,0}(\mathbb{C},b)$. If $a,c$ are perfect matchings, their overlay is a disjoint union of $C(a,c)\ge 1$ alternating cycles, counting a shared arc as a two-edge cycle; therefore ${\varphi }_b(a,c)=b^{C(a,c)}$. Define the bilinear form ${\varphi }^{\mathrm{\prime }}(a,c)={\frac{d}{db}{\varphi }_b(a,c)\mid }_{b=0}\mathrm{.}$ It follows that ${\varphi }^{\mathrm{\prime }}(a,c)=1$ when $C(a,c)=1$ and $0$ when $C(a,c)>1$.

Delete site $2p$ and its incident arc from a perfect matching, and declare its partner to be the single defect. This is a bijection from perfect matchings on $2p$ sites to the link-state basis of $W_{2p-1,1}$: the inverse joins the unique defect to the appended site $2p$. For a pair $a,c$, deleting the two arcs incident to site $2p$ turns exactly the overlay cycle containing that site into the open path between the defects and leaves the other $C(a,c)-1$ cycles unchanged. Hence the ordinary one-defect Gram value of the images is $b^{C(a,c)-1}$, whose value at $b=0$ equals ${\varphi }^{\mathrm{\prime }}(a,c)$. Thus the matrix of ${\varphi }^{\mathrm{\prime }}$ is exactly the Gram matrix of $W_{2p-1,1}(\mathbb{C},0)$.

For $W_{2p-1,1}$ the arc count is $p-1$, the defect count is $1$, and $q^2=-1$ has order $\kappa =2$. Since $\kappa \mid (1+1)$, this is a critical label. By [F4], its Gram form is nondegenerate; step 2.1 then shows that ${\varphi }^{\mathrm{\prime }}$ is nondegenerate.

For two link states $x,y$ on $2p$ sites with two defects, define $D_{x,y}$ to be the Temperley–Lieb diagram whose bottom arcs are those of $x$, whose top arcs are those of $y$, and whose two unmatched bottom sites are joined to the two unmatched top sites in left-to-right order by propagating strands. This matching is planar and is an allowed basis diagram. Let $x^{\mathrm{\prime }}$ and $y^{\mathrm{\prime }}$ be the perfect matchings obtained by joining the two defects of $x$ and $y$, respectively. If $z$ is a basis perfect matching and $C(y^{\mathrm{\prime }},z)$ is the number of cycles in the overlay of $y^{\mathrm{\prime }}$ and $z$, stacking $z$ above $D_{x,y}$ leaves the bottom matching $x^{\mathrm{\prime }}$ and removes exactly $C(y^{\mathrm{\prime }},z)-1$ closed cycles. Therefore $z\cdot D_{x,y}={\delta }^{C(y^{\mathrm{\prime }},z)-1}{x}^{\mathrm{\prime }}$, where the coefficient at $\delta =0$ is defined as $1$ when $C(y^{\mathrm{\prime }},z)=1$ and $0$ when $C(y^{\mathrm{\prime }},z)>1$. This coefficient is exactly ${\varphi }^{\mathrm{\prime }}(y^{\mathrm{\prime }},z)$, so $z\cdot D_{x,y}={\varphi }^{\mathrm{\prime }}(y^{\mathrm{\prime }},z)x^{\mathrm{\prime }}$. The same equality holds for every $z$ by linearity.

Let $z\mathrm{\ne }0$. Nondegeneracy of ${\varphi }^{\mathrm{\prime }}$ gives a nonzero functional ${\varphi }^{\mathrm{\prime }}(-,z)$, so at least one perfect-matching basis vector $y^{\mathrm{\prime }}$ has ${\varphi }^{\mathrm{\prime }}(y^{\mathrm{\prime }},z)\mathrm{\ne }0$. For any target basis matching $x^{\mathrm{\prime }}$, cut from $x^{\mathrm{\prime }}$ and $y^{\mathrm{\prime }}$ the arcs incident to site $2p$ to obtain two-defect link states $x,y$. By step 3.2, $z\cdot D_{x,y}$ is a nonzero scalar multiple of $x^{\mathrm{\prime }}$. Thus the submodule generated by $z$ contains every basis vector and equals $W_{2p,0}$. Every nonzero submodule contains such a $z$, so the module is simple.

It suffices to perform computations for left modules: send a right module to the left module with $a\cdot v=v\cdot a^{\ast }$, where $\ast$ is diagram reflection. This is an exact equivalence preserving Hom direction, induction and restriction. Fix a square root of $q$ and put $g_i=q^{1\mathrm{/}2}1-q^{-1\mathrm{/}2}{d}_i$. Expanding with [F1] gives $g_i^{-1}=q^{-1\mathrm{/}2}1-q^{1\mathrm{/}2}{d}_i$ and $g_i{g}_{i+1}{g}_i=g_{i+1}{g}_i{g}_{i+1}$. Let $T_n:A_{n+1}\to A_n$ close the last top endpoint to the last bottom endpoint externally on the right, removing any resulting loop with factor $\delta$. This is an $A_n$-bimodule map by the matching rule. Define $M_n=g_n\cdots g_2{g}_1^2{g}_2\cdots g_n$ and $F_n=T_n(M_n)$; set $F_0=\delta$. Every summand of $M_n$ has an even total number of square-root factors, so $F_n$ has coefficients in $\mathbb{Z}[q,q^{-1}]$, independent of the chosen square root.

For clarity give a basis proof of the induction sequence. In left-module notation put $h_r=d_r{d}_{r+1}\cdots d_n$ for $1\le r\le n$, and $h_{n+1}=1$. A state $d$ is $r$-admissible when it has no adjacent arc at an index $i\ge r$. The tensors $h_r\otimes d$ with $d$ $r$-admissible, including all $1\otimes d$, span $A_{n+1}{\otimes }_{A_n}{V}_{n,p}$. First, a diagram word is a sum of $h_{r}u$ with $u\in A_n$: reverse a spanning Jones normal form from [F1]. Its initial ascending block is either $h_r$ or contains no $d_n$, and every later block lies in $A_n$. The surjection from the presented algebra to the diagram algebra is supplied by the local relations, so this spanning assertion applies to diagram words. Next, if $d$ has its largest adjacent arc at $s\ge r$, write $d=d_{s}f$. For $n\ge 3$ this can be done without division: move the endpoint immediately outside that adjacent arc through the arc, so applying $d_s$ restores $d$ and creates no loop. If $s=1$ use site $3$; otherwise use site $s-1$. The outside site is either a defect or an endpoint of another arc, and reconnecting its former partner supplies the inverse state $f$. Now the relations give $h_r\otimes d=h_{s+2}\otimes d_r\cdots d_{s-1}d$. The latter state is $(s+2)$-admissible: the adjacent arc at $s$ has moved to $r$, all intervening endpoints shift by two, and the tail beyond $s+1$ still has no adjacent arc. Repetition, increasing $r$, proves spanning. At $n=1,2$ the same conclusion is checked on the at most two states, except $n=2,p=1,\delta =0$, when the needed factorization $d=d_{1}f$ fails.

The braid relations give $g_i{M}_n=M_n{g}_i$ for $i<n$. To see this directly, commute $g_i$ past all $g_j$ with $\mathrm{\mid }j-i\mathrm{\mid }>1$ in the descending and ascending strings; the remaining equality is $g_i{g}_{i+1}{g}_i=g_{i+1}{g}_i{g}_{i+1}$ on each side of the middle string. This proves the equality inductively in the length of the strings, starting with $g_1{g}_2{g}_1^2{g}_2=g_2{g}_1^2{g}_2{g}_1$. Bimodule linearity of $T_n$ gives $g_i{F}_n=F_n{g}_i$, hence $d_i{F}_n=F_n{d}_i$. Thus $F_n$ is central. Reversing the palindromic word $M_n$ and reflecting its closure leaves it unchanged, so $F_n^{\ast }=F_n$.

Compute $F_n$ on the state $z_p$ with adjacent arcs $(1,2),(3,4),\dots ,(2p-1,2p)$ and all remaining sites defects. At a cup, expansion of the two adjacent crossings uses the identities $d_i{g}_{i+1}{g}_i=-d_i{d}_{i+1}$ and $g_i{g}_{i+1}{d}_i=-d_{i+1}{d}_i$; for example the first is $q{d}_i-d_i{d}_{i+1}-\delta d_i+q^{-1}{d}_i{d}_{i+1}{d}_i=-d_i{d}_{i+1}$. The two minus signs on the two sides of each cup cancel. Thus each initial cup can be removed without changing the coefficient of $z_p$. For the $t=n-2p$ defect strands, expand the auxiliary strand successively from its first crossing. A smoothing connecting two defects vanishes in the standard quotient. There are two remaining continuations: following the auxiliary strand on its two sides gives respectively $q^t\delta -q^{t-1}=q^{t+1}$ and $q^{-t}\delta -q^{-t+1}=q^{-(t+1)}$. The factors arise by choosing the identity smoothing at each subsequent crossing until the final cap; that cap either closes a loop or chooses the second smoothing, giving the displayed two terms on each side. Hence the coefficient of $z_p$ is $f_{n,p}$. For generic $q$, [F3] makes the form nondegenerate. Capping $z$ against $y$ and then attaching $x$ gives the rank-one diagram operator $z\mapsto \varphi (y,z)x$, because lower-propagating terms vanish. Thus every nonzero vector generates the standard, which is simple. Centrality now makes $F_n$ scalar on the entire standard: over $\mathbb{C}$ a central operator has an eigenvalue, whose nonzero eigenspace is the simple module. Since the action matrices and $f_{n,p}$ are Laurent polynomial in $q$, this scalar identity extends to every $q\mathrm{\ne }0$.

On the submodule and quotient in [F2], $F_{n-1}$ acts by $f_{n-1,p}$ and $f_{n-1,p-1}$. Their difference is $-(q-q^{-1})(q^{n-2p+1}-q^{-(n-2p+1)})$. Under the noncritical and $q\mathrm{\ne }\pm 1$ hypotheses it is nonzero. The operator on the restriction satisfies $(F_{n-1}-f_{n-1,p})(F_{n-1}-f_{n-1,p-1})=0$, because each factor kills its respective subquotient. Its two spectral projections therefore give an invariant direct sum splitting the restriction sequence.

Append an adjacent arc at sites $n+1,n+2$ to $d$, obtaining $\bar{d}\in V_{n+2,p+1}$. The map $\mathrm{\Phi }(u\otimes d)=u\bar{d}$ into $V_{n+2,p+1}\downarrow$ is balanced and $A_{n+1}$-linear: elements of $A_n$ act only on the first $n$ sites. Its image of $h_r\otimes d$ has its rightmost adjacent arc at $r$; deleting that arc recovers $d$. Conversely every $(n+2,p+1)$-state has a rightmost adjacent arc; delete it at position $r$, and the remaining state is $r$-admissible. These two operations are inverse, so the spanning tensors in step 1.2 map bijectively to the link-state basis. They are therefore a basis and $\mathrm{\Phi }$ is an isomorphism, except in the stated exceptional case. Applying the restriction sequence [F2] to $V_{n+2,p+1}$ now gives the asserted induction sequence. Applying step 4.1 at rank $n+2$ splits it whenever $(n,p)$ is noncritical, since $n+2-2(p+1)+1=n-2p+1$.

Suppose $(n,p)$ is critical, $q\mathrm{\ne }\pm 1$, and $z_p$ is as in step 3.1. In the induced basis the coefficient of $h_{2p+1}\otimes z_p$ in $F_{n+1}(1\otimes z_p)$ is $(-1{)}^n({\delta }^2-4)\mathrm{\ne }0.$ Here is the local expansion computing it. First take $p=0$. To leave $h_1=d_1\cdots d_n$ after reduction, the first pair of crossings of the closed auxiliary strand has three contributing smoothings, with weights $-1,-1,\delta q^{-1}$; the fourth, of weight $q$, closes off the needed through path. This gives $-2+\delta q^{-1}$. At each of the next $n-1$ pairs exactly one continuation retains the required external matching, of weight $-1$: the other continuation either caps the auxiliary path prematurely or makes an additional bottom arc. At the final pair the corresponding three weights are $-1,-1,\delta q$, giving $-2+\delta q$. Their product is $(-1{)}^{n-1}(-2+\delta q^{-1})(-2+\delta q)=(-1{)}^n({\delta }^2-4)$. The required exterior matching of $h_1$ has top pair $(1,2)$, bottom pair $(n,n+1)$, and through paths from top $j+2$ to bottom $j$ for $1\le j<n$. Follow the auxiliary strand subject to these connections: a premature cap at an intermediate row would create another exterior arc instead of the required through path, so its continuation is excluded. For $p>0$, remove the first $p$ cups using the two local identities in step 3.1, on both sides of the auxiliary path. This leaves the same computation on $n-2p$ sites; $n-2p>0$ by criticality and $\kappa \ge 2$, and its parity equals that of $n$. The target basis tensor is $h_{2p+1}\otimes z_p=h_1\otimes z_p$, by successive $d_i{d}_{i+1}{d}_i=d_i$ reductions. One may do the cup removal first for $\delta \mathrm{\ne }0$ by inserting ${\delta }^{-1}{d}_{2j-1}$ at each initial cup. The resulting coefficient identity extends to $\delta =0$: the induced basis reduction in step 1.2 uses no division, so its matrices, and this coefficient, are Laurent polynomials in $q$. Finally ${\delta }^2-4=(q-q^{-1}{)}^2\mathrm{\ne }0$.

In the induction sequence at a critical $(n,p)$ the two standard constituents have the same eigenvalue $f=f_{n+1,p}=f_{n+1,p+1}$. The central endomorphism $F_{n+1}-f$ kills both constituents but is nonzero on the induction by step 6.1. Consequently it factors through the quotient $V_{n+1,p}$ into the submodule $V_{n+1,p+1}$, giving a nonzero map between these adjacent reflected standards. Explicitly, lift a quotient vector to the induction, apply $F_{n+1}-f$, and use the inverse of the submodule injection; the result is independent of the lift because this operator kills the submodule.

Every symmetric pair has the form $(n+j,p),(n+j,p+j)$ for a critical $(n,p)$ and $1\le j<\kappa$. For $j\ge 2$, induction and restriction at the noncritical labels give $\mathrm{Hom}(V_{n+j,p},V_{n+j,p+j})\cong \mathrm{Hom}(V_{n+j-1,p}\uparrow ,V_{n+j,p+j})\cong \mathrm{Hom}(V_{n+j-1,p},V_{n+j,p+j}\downarrow )\cong \mathrm{Hom}(V_{n+j-1,p},V_{n+j-1,p+j-1})\mathrm{.}$ In the first and last isomorphisms the extra standard summand has a different $F$-eigenvalue: substitute $q^{2(n-2p+1)}=1$ in step 3.1, and equality would force $j=1$ or $j=\kappa$. Thus its Hom is zero. The middle adjunction is explicit: $f\mapsto (v\mapsto f(1\otimes v))$, with inverse $g\mapsto (a\otimes v\mapsto ag(v))$. This works also when an extraneous summand is out of range. Reducing $j$ to $1$ and applying step 7.1 proves nonvanishing for every reflected pair. Returning along step 1.1 gives all statements for the repository's right modules.

First work at generic $q$. The rightmost-defect submodule has nondegenerate form by [F2], so its orthogonal complement is invariant and maps isomorphically to the restriction quotient. The smaller standards are simple by their rank-one diagram operators; an invariant form on such a standard is a scalar multiple of its nondegenerate standard form, because their associated maps into the dual differ by an endomorphism, and an endomorphism of a finite-dimensional complex simple module is scalar (take any eigenvalue). Subtracting the submodule projection from quotient basis lifts gives a unitriangular Gram splitting $G_{n,p}=U^{\mathsf{T}}(G_{n-1,p}\oplus {\alpha }_{n,p}{G}_{n-1,p-1})U$. Compute its scalar as follows. Group the link states into the four types: $n,n-1$ both defects; $n$ a defect and $n-1$ on an earlier arc; $(n-1,n)$ an arc; and $n$ on an arc ending earlier than $n-1$. Put $A=G_{n-2,p}$, $B=G_{n-2,p-1}$ and write the smaller-rank matrix as $G_{n-1,p}=\left(\begin{array}{cc}A& A{V}_0\\ V_0^{\mathsf{T}}A& V_0^{\mathsf{T}}A{V}_0+{\alpha }_{n-1,p}B\end{array}\right)$. In the current off-diagonal Gram block the first column block is $(0,B{)}^{\mathsf{T}}$: types one and three have unmatched defects, while cutting the terminal arc in types two and three removes no closed component. The type-three diagonal block is $\delta B$, since the shared terminal arc closes one loop. If $(X,Y{)}^{\mathsf{T}}$ is the first column block of the projection coefficient matrix, multiplication by $G_{n-1,p}$ gives $AX+A{V}_{0}Y=0$ and $V_0^{\mathsf{T}}AX+(V_0^{\mathsf{T}}A{V}_0+{\alpha }_{n-1,p}B)Y=B$. Thus $X=-V_{0}Y$ and $Y={\alpha }_{n-1,p}^{-1}I$. Its quadratic contribution is ${\alpha }_{n-1,p}^{-1}B$ after the four $V_0^{\mathsf{T}}A{V}_0$ terms cancel with coefficients $1-1-1+1$. Comparing the type-three block gives ${\alpha }_{n,p}=\delta -{\alpha }_{n-1,p}^{-1}$. At $n=2p$, deleting the terminal site closes exactly one additional component, so $G_{2p,p}=\delta G_{2p-1,p-1}$ and ${\alpha }_{2p,p}=\delta$. The recurrence solves to ${\alpha }_{n,p}=[n-2p+2{]}_q\mathrm{/}[n-2p+1{]}_q$ by the quantum-integer recurrence. If the $A$ block is empty it is omitted; the same cancellation still holds. Now at a noncritical root, the two eigenvalues used for [F1] differ by $-(q-q^{-1})(q^{n-2p+1}-q^{-(n-2p+1)})\mathrm{\ne }0$. The spectral projector onto the rightmost-defect summand is $(F_{n-1}-f_{n-1,p-1})\mathrm{/}(f_{n-1,p}-f_{n-1,p-1})$. Its entries are rational functions regular at the root; subtracting this projection from each standard quotient lift gives a unitriangular splitting matrix $U$, also regular there. Since $F_{n-1}^{\ast }=F_{n-1}$ and the eigenvalues differ, the two summands are orthogonal. For generic $q$ this is exactly the orthogonal splitting just computed, since the submodule form is nondegenerate and its orthogonal complement is unique. Thus this Gram matrix identity is an identity of rational matrices regular at every noncritical root, and specializes there. This continuity argument applies to the entire matrix identity, rather than inferring a rank from a specialized determinant. If $n=2p$, there is only the quotient summand and direct capping gives $G_{2p,p}=\delta G_{2p-1,p-1}$, the same formula with $\alpha =[2]\mathrm{/}[1]=\delta$.

Since $\kappa \ge 2$, $q\mathrm{\ne }\pm 1$; hence $[a{]}_q=0$ precisely when $\kappa \mid a$. At a noncritical label the denominator of $\alpha$ is nonzero and its numerator vanishes precisely when $r=\kappa -1$. The kernel dimension of the specialized block form in step 1.1 is therefore $\dim R_{n-1,p}+\dim V_{n-1,p-1}$ in that case, and $\dim R_{n-1,p}+\dim R_{n-1,p-1}$ otherwise. Critical labels give zero by [F3]; the all-defect form is $(1)$, giving $R_{n,0}=0$. For $n\ge 1$, subtraction from [F4] gives the companion recursion, with initial $\dim L_{0,0}=1$: $\dim L_{n,p}=\dim V_{n,p}$ at critical labels; at noncritical labels it is $\dim L_{n-1,p}$ if $r=\kappa -1$, and $\dim L_{n-1,p}+\dim L_{n-1,p-1}$ otherwise.

Prove the reflected-dimension formula by induction on $n$, with out-of-range labels assigned dimension zero. Write $a=p+r-\kappa$. If $r\mathrm{\notin }\{1,\kappa -1\}$, neither predecessor is critical. Their reflection indices are respectively $a-1$ and $a$, so induction and step 2.1 give $\dim R_{n,p}=\dim L_{n-1,a-1}+\dim L_{n-1,a}$. The reflected label $a$ has remainder $\kappa -r$, which is also neither $1$ nor $\kappa -1$; its companion recursion gives exactly $\dim L_{n,a}$.

When $\kappa =2$, every noncritical label has $r=1=\kappa -1$ and both predecessors are critical. The radical recursion gives $\dim R_{n,p}=\dim V_{n-1,p-1}=\dim L_{n-1,p-1}$. Since $a=p-1$ and the companion recursion at $(n,a)$ also has remainder $1$, this is $\dim L_{n,a}$. The same formula holds for $n=2p$: it gives $\dim R_{2p,p}=\dim V_{2p-1,p-1}=\dim V_{2p,p}$, as required by the identically zero form.

If $r=\kappa -1$ and $\kappa >2$, the predecessor $(n-1,p-1)$ is critical, while $(n-1,p)$ has reflected index $a-1$. Consequently $\dim R_{n,p}=\dim L_{n-1,a-1}+\dim V_{n-1,p-1}$. Here $a=p-1$, and criticality gives $V_{n-1,p-1}=L_{n-1,a}$. The companion recursion at $(n,a)$ has remainder $1\mathrm{\ne }\kappa -1$, so this sum is $\dim L_{n,a}$. If $r=1$ and $\kappa >2$, $(n-1,p)$ is critical and its radical is zero; the reflection of $(n-1,p-1)$ has index $a$. Thus $\dim R_{n,p}=\dim L_{n-1,a}$. The reflected label at rank $n$ has remainder $\kappa -1$, whose companion recursion is precisely $\dim L_{n,a}=\dim L_{n-1,a}$.