Tensor stability and dual stability of root-of-unity tilting modules

The finite tensor lattice $V_A(\lambda ){\otimes }_A{V}_A(\mu )$ has the constructed finite based global basis by [F1], with exactly this underlying integral lattice. Apply [F2] before specializing; every section is an integral highest lattice and every quotient is free. Its specialization is therefore a genuine Weyl filtration of ${\mathrm{\Delta }}_{\zeta }(\lambda )\otimes {\mathrm{\Delta }}_{\zeta }(\mu )$ with all submodule injections intact. This proof uses an integral filtration and never specializes a generic direct-sum decomposition.

Let $M,N$ have Weyl filtrations. Tensor a filtration of $M$ with $N$; exactness gives sections ${\mathrm{\Delta }}_{\zeta }(\lambda )\otimes N$. Tensor each filtration of $N$ with this Weyl factor and use step 1.1 to give each of its sections a Weyl filtration. Pull these finite filtrations back through the successive quotient maps to refine the original filtration. This produces a finite Weyl filtration of $M\otimes N$. The zero case has the empty filtration.

If $M,N$ are dual-Weyl-filtered, then $M^{\ast },N^{\ast }$ are Weyl-filtered by [F3]. Step 2.1 makes $N^{\ast }\otimes M^{\ast }$ Weyl-filtered; its dual is dual-Weyl-filtered and is isomorphic to $M\otimes N$ by the dual tensor isomorphism and the pivotal double-dual identifications. Thus dual Weyl filtrations are tensor stable as well. Ordinary duality exchanges the two filtration kinds, so it preserves their intersection. Concatenating filtrations proves finite-sum closure. The trivial one-dimensional module is both ${\mathrm{\Delta }}_{\zeta }(0)$ and ${\mathrm{\nabla }}_{\zeta }(0)$, hence is tilting. These arguments prove exactly the asserted tensor, dual, sum and unit statements.