Rank-one Weyl strings, tensor filtrations, and tilting classification

The generic string lattice is stable under every divided power by [F2,F3], and $T_{c,t}{v}_j={\left[\begin{array}{c}a-2j+c\\ t\end{array}\right]}_q{v}_j$ by the toral product. Hence it specializes to the stated full-weight module. More generally, if a full-weight module is supported on $[-a,a]$ and $w$ has weight $a$, the assignment $v_j\mapsto F^{(j)}w$ defines its unique map from $\mathrm{\Delta }(a)$. Indeed the support kills every $E^{(r)}w$ for $r>0$ and every $F^{(j)}w$ for $j>a$. Divided-power straightening gives $E^{(r)}{F}^{(j)}w={\left[\begin{array}{c}a-j+r\\ r\end{array}\right]}_q{F}^{(j-r)}w$, while $F^{(r)}{F}^{(j)}={\left[\begin{array}{c}r+j\\ r\end{array}\right]}_q{F}^{(r+j)}$. These are Laurent identities, obtained by commuting $E$ past $F$ and applying Pascal before specialization. They verify all divided actions; the full toral characters verify the remaining generators.

The contragredient action $(x\phi )(u)=\phi (S(x)u)$ defines $\mathrm{\nabla }(a)$; its weights are the negatives of the original weights and its toral character at weight $m$ is the full character at $m$, by substituting $K^{-1}$ in the generic toral product. Double duality is identified by $K$, since $S^2=\mathrm{Ad}K$. A map $\mathrm{\Delta }(a)\to \mathrm{\nabla }(b)$ vanishes for $a>b$ because the generating weight-$a$ vector has nowhere to go. For $a<b$, dualizing gives a map $\mathrm{\Delta }(b)\to \mathrm{\nabla }(a)$, also zero. For $a=b$, step 1.1 maps onto the one-dimensional highest line of $\mathrm{\nabla }(a)$, and every map is determined by that line. This proves the Hom formula.

Work integrally over $A=\mathbb{Z}[v,v^{-1}]$ in the quantum plane $P=A⟨x,y⟩\mathrm{/}(yx-vxy)$. Let $x,y$ have weights $1,-1$, $Ex=0$, $Ey=x$, $Fx=y$, $Fy=0$. The coproduct of [F1] preserves the relation: $E(yx-vxy)=v{x}^2-v{x}^2=0$ and $F(yx-vxy)=v{y}^2-v{y}^2=0$. On $P_n$, with basis $X_j=x^{n-j}{y}^j$, direct induction gives $E^{(r)}{X}_j={\left[\begin{array}{c}j\\ r\end{array}\right]}_v{X}_{j-r}$ and $F^{(r)}{X}_j={\left[\begin{array}{c}n-j\\ r\end{array}\right]}_v{X}_{j+r}$. These matrices and the full toral characters are Laurent integral. This is the dual Weyl lattice: if $w_j=(v_{n-j}{)}^{\ast }$ in $\mathrm{\Delta }(n{)}^{\vee }$, then $E{w}_j=-v^{n-2j+2}[j]w_{j-1}$ and $F{w}_j=-v^{-n+2j}[n-j]w_{j+1}$. Rescale by $c_0=1$, $c_j\mathrm{/}{c}_{j-1}=-v^{-n+2j-2}$; the resulting coefficients are $[j]$ and $[n-j]$. The same calculation for divided powers proves the asserted integral identification.

For $a,b\ge 1$, multiplication and insertion of the invariant tensor $x\otimes y-v^{-1}y\otimes x$ give a sequence $0⟶P_{a-1}\otimes P_{b-1}\stackrel{i}{\to }{P}_a\otimes P_b\stackrel{\mu }{\to }{P}_{a+b}⟶0,$ where $i(f\otimes g)=fx\otimes yg-v^{-1}fy\otimes xg$. Both maps are module maps by the module-algebra identities of step 3.1; invariance for higher divided powers follows from their integral matrices, or directly from their weight shifts on the two-factor tensor. Also $\mu i=0$ by $yx=vxy$. Here is exactness over $A$, before any root specialization. For $f_i=x^{a-1-i}{y}^i$ and $g_j=x^{b-1-j}{y}^j$, $i(f_i\otimes g_j)=v^{i+b-1-j}{X}_i\otimes Y_{j+1}-v^{-1}{X}_{i+1}\otimes Y_j\mathrm{.}$ For each fixed $i+j$ these are adjacent differences along the consecutive path of target basis pairs, with unit coefficients. Successively eliminating from its first vertex proves independence and a free rank-one quotient. Multiplication sends each target vertex to a unit multiple of the same monomial, so it identifies this quotient with that monomial line. This proves exactness and freeness in every degree. It therefore remains exact at every specialization. When $a=0$ or $b=0$, $P_0=A$ is the tensor unit and the assertion is immediate. Iterating gives the dual-Weyl factors $a+b,a+b-2,\dots ,\mathrm{\mid }a-b\mathrm{\mid }$; dualizing reverses tensor order and gives the Weyl factors. Tensoring finite filtrations is exact over $\mathbb{C}$, proving tensor closure of both filtration classes and hence of tiltings.

Consider $0\to \mathrm{\nabla }(b)\to M\to \mathrm{\Delta }(a)\to 0$. If $a\ge b$, its support is $[-a,a]$. A weight-$a$ lift of the quotient generator is highest by support; step 1.1 supplies a map $\mathrm{\Delta }(a)\to M$ whose composite with the quotient fixes that generator and is the identity. This splits the sequence. If $a<b$, dualizing gives the already proved case with $a,b$ interchanged. This proves Ext vanishing. Induction along two finite filtrations gives the same splitting/lifting property with a Weyl-filtered first end and dual-Weyl-filtered second end: at each filtration step lift across the preceding short exact sequence, whose obstruction is an extension of the displayed type; its splitting removes that obstruction. Equivalently, applying the elementary Hom-extension exact sequence at each step gives ${\mathrm{Ext}}^1(D,N)=0$ for every such pair $D,N$.

Let $M$ be Weyl-filtered with maximal label $a$. Its support is $[-a,a]$. An occurrence of a quotient $\mathrm{\Delta }(a)$ can be moved to the bottom of the filtration: lift its weight-$a$ generator across the preceding submodule and apply step 1.1, splitting that extension. Repeating for all occurrences gives the injective evaluation map $\mathrm{\Delta }(a){\otimes }_{\mathbb{C}}{M}_a\to M$, whose image is the sum of all maps from $\mathrm{\Delta }(a)$ and whose quotient is Weyl-filtered with labels below $a$. It is injective because after these splittings it is precisely the direct sum of the top standard factors, each with a one-dimensional highest line. Any idempotent $e$ on $M$ preserves this image and acts there as $1\otimes e_a$. Its image and kernel therefore have top submodules that are sums of $\mathrm{\Delta }(a)$'s, and their quotients are summands of the lower-label filtered quotient. Induction on the maximal label shows both are Weyl-filtered. Duality gives the corresponding assertion for dual-Weyl filtrations. Thus every summand of a tilting is tilting.

The module $V=\mathrm{\Delta }(1)$ is its own dual Weyl, as the rescaling of step 3.1 shows at $n=1$. Its powers are tilting by step 4.1. Recursively splitting a finite-dimensional module into nontrivial direct sums terminates by dimension, giving an indecomposable decomposition. Summand closure makes these tiltings. Exactly one summand of $V^{\otimes a}$ has weight $a$, whose whole space is one-dimensional; denote it by $T(a)$. For a finite-dimensional indecomposable module, an endomorphism $h$ is either invertible or nilpotent: for large $N$, stabilized kernel and image give $M=\ker h^N\oplus \mathrm{im}{h}^N$. Consequently an endomorphism nonzero on a preserved one-dimensional highest line is invertible.

Let $M$ be any indecomposable finite tilting, of maximal label $a$. Step 6.1 embeds $\mathrm{\Delta }(a)\otimes M_a$ into $M$, with lower-label Weyl-filtered quotient. Choose a vector and linear functional on $M_a$ pairing to $1$. They define maps between the standard submodules of $T(a)$ and $M$. Step 5.1 extends them to maps $f:T(a)\to M$ and $g:M\to T(a)$, because the source quotients are Weyl-filtered and the targets dual-Weyl-filtered. The composite $gf$ fixes the highest line, hence is invertible by step 7.1. Thus $T(a)$ is a nonzero summand of $M$; indecomposability forces an isomorphism. It also forces $\dim M_a=1$, proves uniqueness for each $a$, and distinguishes different $a$ by their maximal integral weights. Decomposing any finite tilting proves all the asserted classification and generation claims. The choices here concern individual finite-dimensional vector spaces and terminating finite decompositions, not a choice axiom.