Existence and highest weight classification of root-of-unity tiltings

The standard-costandard degree-one vanishing in [F2] extends to all positive degrees. For $j\ge 2$, the projective sequence for ${\mathrm{\Delta }}_{\eta }$ in [F1] identifies ${\mathrm{Ext}}^j({\mathrm{\Delta }}_{\eta },{\mathrm{\nabla }}_{\mu })$ with ${\mathrm{Ext}}^{j-1}(K_{\eta },{\mathrm{\nabla }}_{\mu })$; a Weyl flag of $K_{\eta }$ and induction on $j$ make the latter zero. Filtering either variable proves all positive Ext groups vanish between any Weyl-filtered and dual-Weyl-filtered pair. Likewise, if $N$ satisfies ${\mathrm{Ext}}^1(N,{\mathrm{\nabla }}_{\mu })=0$ for every label, then the injective sequences in [F1] give all positive groups zero: filter their dual-Weyl cokernels and dimension shift successively in the second variable. In particular Ext from $N$ in every positive degree vanishes against any dual-Weyl-filtered module.

Let $\eta$ be maximal in $\mathrm{\Gamma }$ and $e$ its primitive idempotent. Then $Ae=P_{\eta }={\mathrm{\Delta }}_{\eta }$ and $eAe=\mathbb{C}$. Apply the canonical maximal-Weyl-trace argument of [F2] to each projective column and sum the columns: multiplication gives an injection $Ae{\otimes }_{\mathbb{C}}eA\to A$ with image $J=AeA$. Thus $J$ is projective on both sides. The quotient $A^{\mathrm{\prime }}=A\mathrm{/}J$ models the Serre subcategory on $\mathrm{\Gamma }\setminus \{\eta \}$. Its projective columns are obtained by removing the maximal trace from the old projectives, so they retain the other Weyl sections; equivalently this is exactly the smaller root truncation in [F1]. Its injectives are dual-Weyl-filtered, with unchanged smaller-label families.

Suppose $N$ has the Ext vanishing of step 1.1. Choose a projective resolution $P_{\bullet }\to N$ with finite dimensional projective terms, and tensor it with $A^{\mathrm{\prime }}$. Each resulting term is projective over $A^{\mathrm{\prime }}$. For any injective $A^{\mathrm{\prime }}$-module $I^{\mathrm{\prime }}$, adjunction identifies ${\mathrm{Hom}}_{A^{\mathrm{\prime }}}(A^{\mathrm{\prime }}{\otimes }_A{P}_{\bullet },I^{\mathrm{\prime }})$ with ${\mathrm{Hom}}_A(P_{\bullet },I^{\mathrm{\prime }})$, whose positive cohomology is zero by step 1.1 and the dual Weyl filtration of $I^{\mathrm{\prime }}$. Exactness of Hom into an injective identifies that cohomology with Hom from the corresponding homology module. Injectives detect nonzero homology: a nonzero finite homology module has a simple submodule $L$; embed $L$ into its injective envelope $I_L$ and extend that map to the homology module by injectivity. The extension is nonzero on $L$, so Hom into $I_L$ is nonzero. No assertion that one copy of each injective envelope embeds every finite module is needed. Hence ${\mathrm{Tor}}_j^A(A^{\mathrm{\prime }},N)=0$ for all $j>0$. Tensoring $0\to J\to A\to A^{\mathrm{\prime }}\to 0$ therefore gives the actual exact sequence $0\to Ae{\otimes }_{\mathbb{C}}eN\to N\to N^{\mathrm{\prime }}\to 0$, $N^{\mathrm{\prime }}=A^{\mathrm{\prime }}{\otimes }_{A}N$.

The exact tensor resolution from step 2.1 and adjunction give ${\mathrm{Ext}}_{A^{\mathrm{\prime }}}^j(N^{\mathrm{\prime }},{\mathrm{\nabla }}_{\mu })={\mathrm{Ext}}_A^j(N,{\mathrm{\nabla }}_{\mu })=0$ for every smaller label and every $j>0$. Induct on $\mathrm{\mid }\mathrm{\Gamma }\mathrm{\mid }$ to conclude $N^{\mathrm{\prime }}$ is Weyl-filtered; the base empty category contains only zero. The kernel in step 2.1 is a direct sum of ${\mathrm{\Delta }}_{\eta }$, so $N$ is Weyl-filtered. The converse is step 1.1. Exact ordinary duality in the relabelled finite truncation ${\mathrm{\Gamma }}^{\ast }=-w_0\mathrm{\Gamma }$ proves the asserted dual-Weyl criterion. This proves the needed Ext criteria rather than importing them from category $\mathcal{O}$.

Fix $\lambda$ and take the finite saturated ideal $\mathrm{\Gamma }=\{\mu \in P^{+}:\mu \le \lambda \}$. Start with $X={\mathrm{\Delta }}_{\zeta }(\lambda )$ and process the other labels in a linear order descending in dominance. At the step for $\mu$, let $m=\dim {\mathrm{Ext}}^1({\mathrm{\Delta }}_{\mu },X)$ and choose the universal extension $0\to X\to X^{\mathrm{\prime }}\to {\mathrm{\Delta }}_{\mu }^{\oplus m}\to 0$ whose components are a basis of that Ext space. The connecting map from $\mathrm{Hom}({\mathrm{\Delta }}_{\mu },{\mathrm{\Delta }}_{\mu }^{\oplus m})$ onto this Ext space is an isomorphism because $\mathrm{End}({\mathrm{\Delta }}_{\mu })=\mathbb{C}$. Also ${\mathrm{Ext}}^1({\mathrm{\Delta }}_{\mu },{\mathrm{\Delta }}_{\mu })=0$ by the highest-vector splitting lemma, so the exact sequence of Ext makes ${\mathrm{Ext}}^1({\mathrm{\Delta }}_{\mu },X^{\mathrm{\prime }})=0$. Already processed labels $\nu$ stay killed: they are not strictly below $\mu$, and the same splitting lemma gives ${\mathrm{Ext}}^1({\mathrm{\Delta }}_{\nu },{\mathrm{\Delta }}_{\mu })=0$. The original top label is killed from the start. Every step preserves a Weyl flag, top multiplicity one and support at most $\lambda$. Finite dimensionality of the algebra in [F1] makes each Ext space finite dimensional and the process finite.

The final $X$ has all the degree-one vanishings in the costandard criterion of step 3.1, so it is dual-Weyl-filtered as well as Weyl-filtered, hence tilting. Decompose it into indecomposable tiltings using [F2]. Exactly one summand contains its one-dimensional $\lambda$ space. That summand has highest label $\lambda$, multiplicity one, and normalized bottom inclusion by the maximal trace lemma; all remaining Weyl labels are strictly smaller. Call it $T_{\zeta }(\lambda )$. Its uniqueness and the decomposition of every tilting into the objects so obtained follow from [F2]. All constructions occur in an actual finite root truncation, so these are genuine module filtrations and objects in the full category.