Projective covers of complex-root Temperley–Lieb simple modules

Fix a nonzero head $L=L_{n,p}$ and its finite-dimensional projective cover $\pi :P\to L$ from [F3]. Since $L$ is cyclic, choose a surjection $\alpha :A_n\to L$. Projectivity of $P$ and of the finite free module $A_n$ gives maps $s:P\to A_n$ and $t:A_n\to P$ with $\alpha s=\pi$ and $\pi t=\alpha$. Then $\pi ts=\pi$. Consequently $P=ts(P)+\ker \pi$; superfluity gives $ts(P)=P$, so $ts$ is invertible by finite dimensionality. Replacing $s$ by $s(ts{)}^{-1}$ makes $ts=1$. Thus $P$ is a direct summand $e{A}_n$ of the right regular module for an idempotent $e$ (a right-module endomorphism of $A_n$ is left multiplication by its value at $1$). The cover is indecomposable: a nonzero decomposition would give a proper summand surjecting onto $L$, contrary to the superfluous kernel.

If $S$ is any simple right module and $h:P\to S$ is nonzero, then its kernel is maximal. Superfluity of $\ker \pi$ implies $\ker \pi \subseteq \ker h$: otherwise their sum is $P$ and the superfluity condition would force $\ker h=P$. Thus $h$ factors through $L$. On a finite-dimensional complex simple module an endomorphism $u$ has an eigenvalue $\lambda$; since $u-\lambda 1$ has a nonzero kernel, [F5] makes it zero. Thus its endomorphism space is $\mathbb{C}$. By [F1], $\dim {\mathrm{Hom}}_{A_n}(e{A}_n,S)$ is $1$ for $S=L$ and $0$ otherwise. Evaluation at $e$ identifies this Hom space with $Se$: the inverse sends $v\in Se$ to the map $ea\mapsto va$. Hence $\dim Se$ has those same values. This also proves any finite-dimensional projective module surjecting onto $L$ contains $P$ as a summand, by repeating step 1.1 with that module in place of $A_n$. Finally every indecomposable projective has a simple quotient, hence is the cover of that quotient by this summand assertion.

Left multiplication by $e$ is an exact functor on left modules: for an exact sequence, a lift of $ey$ can be replaced by $e$ times that lift. The cellular basis itself identifies each ideal layer in [F2] with $U_s\otimes V_{n,s}$: map its basis tensor $x\otimes y$ to the class of $C_{x,y}^t$. Right multiplication acts only on $y$ by the cellular triangular rule, and left multiplication acts only on $x$ by reflecting that rule; this proves the bimodule identification, including the zero-form layers. Applying the exact functor gives a filtration of $e{A}_n=P$ with layers $(e{U}_s)\otimes V_{n,s}$, hence $\dim e{U}_s$ copies of $V_{n,s}$. To compute this multiplicity, write $\sigma (a)=a^{\ast }$ for diagram reflection and convert a left module $U$ to the right module with action $u\cdot a=\sigma (a)u$. By the cellular-basis construction under reflection, this converts $U_s$ to $V_{n,s}$, so $\dim (e{U}_s)=\dim (V_{n,s}\sigma (e))$. The idempotent $e$ is primitive because $e{A}_n$ is an indecomposable projective cover of $L=L_{n,p}$. Evaluation at $e$ gives ${\mathrm{Hom}}_{A_n}(e{A}_n,L)\cong Le\cong \mathbb{C}$. For the contragredient dual $L^{\vee }={\mathrm{Hom}}_{\mathbb{C}}(L,\mathbb{C})$ with $(\phi \cdot a)(x)=\phi (x\sigma (a))$, restriction to $Le$ gives $L^{\vee }\sigma (e)\cong (Le{)}^{\ast }\mathrm{\ne }0$. Thus the indecomposable projective $\sigma (e)A_n$ has simple top $L^{\vee }\cong L$, by the self-duality in [F1], and is another projective cover of $L$. Uniqueness of covers [F3] gives $e{A}_n\cong \sigma (e)A_n$, so $e$ and $\sigma (e)$ are equivalent idempotents: an isomorphism $e{A}_n\to \sigma (e)A_n$ supplies elements $x\in \sigma (e)A_{n}e$ and $y\in e{A}_n\sigma (e)$ with $xy=\sigma (e)$ and $yx=e$. The maps $me\mapsto my$ and $mf\mapsto mx$ identify $Me$ with $Mf$ for every right module $M$, where $f=\sigma (e)$, so $\dim (V_{n,s}\sigma (e))=\dim (V_{n,s}e)$. The functor $M\mapsto Me$ is exact on right modules; $Se$ is one-dimensional for $S\cong L$ by Schur's lemma and zero for every other simple, since $e{A}_n$ covers $L$. Therefore $\dim (V_{n,s}e)=[V_{n,s}:L]$. This proves the standard filtration formula $[P_{n,p}:V_{n,s}]=[V_{n,s}:L_{n,p}]$, without assuming semisimplicity or quasiheredity. Its layer order is increasing defect number, equivalently decreasing arc count.

At a critical label, [F1] says the only standard containing $L_{n,p}$ is $V_{n,p}$, once. Step 3.1 gives $P_{n,p}\cong V_{n,p}$. The same argument applies to every label at $\kappa =1,\mathrm{\infty }$. At a finite-order noncritical nonzero head, the standards containing it are $V_{n,p}$ and, when present, its neighbor on the lower-defect side, $V_{n,b}$ with $b=p+r$. This follows by solving the adjacent-reflection relation in [F1] in the reverse direction. If $k=0$, $n-2b+1=-r<0$ and $b$ is absent. If $k>0$, $n-2b=k\kappa -r-1\ge 0$ and it is present. Each occurrence has multiplicity one. At $\kappa =2$, even rank, the possible $b=n\mathrm{/}2$ is included because the zero-form standard is the actual simple $L_{n,n\mathrm{/}2-1}$ by [F1]. Thus step 3.1 gives either $P_{n,p}=V_{n,p}$ or exactly the asserted two-standard filtration. Since $b>p$, its $V_{n,b}$ layer is a submodule and its $V_{n,p}$ layer the quotient, yielding the exact sequence. It is nonsplit: both standards are nonzero, whereas the cover in step 1.1 is indecomposable.

For $k>0$, the starting label $(n-r,p)$ has defect number $k\kappa -1$ and is critical; its standard is projective by step 4.1. Induction preserves projectivity: the adjunction in [F4] identifies maps from an induced module with maps from the original module into a restriction, and restriction preserves surjections, so the defining lifting property is preserved. The $r$-fold induced module $Q$ is therefore projective. Repeatedly applying the standard induction sequences gives a filtration of $Q$ whose layers are quotients of the standards $V_{n,p+j}$, $0\le j\le r$, with multiplicities at most $\left(\genfrac{}{}{0px}{}{r}{j}\right)$. To justify this even though induction is only right exact, induct the existing filtration: the images of its submodules still form a filtration, and each new layer is a quotient of the induction of the old layer; the two-term sequence in [F4] refines these layers. No left-exactness is assumed. The exceptional induction input $(2,1,0)$ is never encountered: every intermediate defect is at least $k\kappa -1-(r-1)>0$.

On the standard indexed by $p+j$ the central eigenvalue is $q^{k\kappa +r-2j}+q^{-(k\kappa +r-2j)}$. It equals $f_{n,p}$ exactly when $j\equiv 0$ or $j\equiv r(\mathrm{m}\mathrm{o}\mathrm{d}\kappa )$: use $x+x^{-1}=y+y^{-1}$ if and only if $(x-y)(x-y^{-1})=0$, and $q^{2k\kappa }=1$. Since $0\le j\le r<\kappa$, only $j=0,r$ qualify. Generalized eigenspaces of a central operator are submodules; the associated polynomial spectral projection is exact on finite-dimensional modules, as is seen by applying the same projection to a lift. Thus the $f_{n,p}$-space $Q_f$ is projective and step 5.1 bounds its dimension by $\dim V_{n,p}+\dim V_{n,p+r}$. Successive standard quotient maps give a surjection $Q\to V_{n,p}\to L_{n,p}$; its restriction to $Q_f$ remains onto, because the other generalized eigenspaces have distinct eigenvalues. Step 2.1 makes $P_{n,p}$ a direct summand of $Q_f$. Step 4.1 already gives $\dim P_{n,p}=\dim V_{n,p}+\dim V_{n,p+r}$, so the bound is saturated and $Q_f\cong P_{n,p}$, proving the stated realization.

At $\delta =0$, $\kappa =2$. Odd ranks have only critical labels and are covered by step 4.1. At rank $2m$, the simple labels are $0\le p<m$ by [F1]; each has $r=1$ and $k>0$. The repeated-induction realization in step 6.1 therefore has just one induction, from the odd-rank critical standard $V_{2m-1,p}$. Its two standard constituents both have eigenvalue $0$, so the entire induction is the selected generalized eigenspace. This proves the final displayed formulas and excludes a cover for the zero quotient label. At rank zero the algebra is $\mathbb{C}$, whose single simple regular module is projective. All covers and all exceptional cases have now been included.