Selfadjoint weight Gram valuations detect Jantzen classes

The decomposition is direct and its summands are free. The map $M\to {⨁}_{\nu }{M}_{\nu }$, $x\mapsto (e_{\nu }x{)}_{\nu }$, is an $O$-linear isomorphism with inverse $(x_{\nu }{)}_{\nu }\mapsto {\sum }_{\nu }{x}_{\nu }$ (the $e_{\nu }$ are orthogonal idempotents summing to $1$), so $M={⨁}_{\nu }{M}_{\nu }$. To prove freeness, induct on $r$ for an arbitrary submodule $N\subseteq O^r$. The assertion is immediate for $r=0$. The first-coordinate image is an ideal $J$; if nonzero, the minimum valuation among its nonzero elements is attained by some $a\in J$, and every element of $J$ is divisible by $a$, so $J=aO$. Lift $a$ to $u\in N$. Subtracting multiples of $u$ gives $N=Ou\oplus \ker ({\mathrm{pr}}_1{\mathrm{\mid }}_N)$, with the kernel a submodule of $O^{r-1}$; if $J=0$, all of $N$ lies in $O^{r-1}$. The induction proves finite freeness and rank at most $r$. Apply it to each $M_{\nu }\subseteq M$.

The summands are pairwise orthogonal and the restricted forms are nondegenerate. For $\nu \mathrm{\ne }\rho$ and $x,y\in M$, $b(e_{\nu }x,e_{\rho }y)=b(x,e_{\nu }^{\ast }{e}_{\rho }y)=b(x,e_{\nu }{e}_{\rho }y)=0$ by $e_{\nu }^{\ast }=e_{\nu }$ and orthogonality, so $M_{\nu }$ and $M_{\rho }$ are orthogonal for $b$. If $x\in M_{\nu }$ is orthogonal to $M_{\nu }$, then it is orthogonal to every $M_{\rho }$ (orthogonality) and hence to $M$, so $x=0$ by generic nondegeneracy of $b$; the identical orthogonality argument after scalar extension handles every generic vector, and thus the restriction of $b$ to $M_{\nu }$ is nondegenerate over $\mathrm{Frac}(O)$.

Exactness and the character map. The functor $N\mapsto e_{\nu }N$ on $A$-modules is exact: a morphism commutes with $e_{\nu }$, so its kernel on $e_{\nu }N$ is $e_{\nu }$ times its kernel on $N$; if $y=e_{\nu }y$ lies in the image of a quotient map, any preimage $x$ may be replaced by $e_{\nu }x$ with the same image. Hence $\mathrm{ch}$ is an additive map on the Grothendieck group $K_0$ by [F3].

Injectivity of the weight character. Write a class in $K_0$ as ${\sum }_{\mu }{a}_{\mu }[\mathrm{\Delta }(\mu )]$ with $a_{\mu }\in \mathbb{Z}$. Its coordinates on the partition weights are the row vector $a=(a_{\mu })$ multiplied by the matrix $K=(K_{\mu \nu })$, $K_{\mu \nu }=\dim e_{\nu }\mathrm{\Delta }(\mu )$. Restricting only the columns to the selected partition weights retains every coefficient $a_{\mu }$; the resulting square unitriangular matrix is invertible over $\mathbb{Z}$ by finite triangular elimination (all diagonal entries are $1$); if $\mathrm{ch}(a)=0$, then the coordinates on that submatrix vanish and the triangular system forces $a_{\mu }=0$ for every $\mu$. Hence $\mathrm{ch}$ is injective.

The Jantzen filtration decomposes. For each $s$ the orthogonality of the summands gives $M(s)={⨁}_{\nu }{M}_{\nu }(s)$ with $M_{\nu }(s)=\{x\in M_{\nu }:b(x,M_{\nu })\subseteq {\pi }^{s}O\}$, because the components of $x$ in the other summands pair trivially; consequently $M_F^s={⨁}_{\nu }(M_{\nu }(s)+\pi M_{\nu })\mathrm{/}\pi M_{\nu }={⨁}_{\nu }{e}_{\nu }{M}_F^s$. For a zero-rank summand $M_{\nu }=0$, each $e_{\nu }{M}_F^s=0$, its empty Gram determinant is $1$ as specified by the DVR leaf, and both sides of the identity are $0$. Applying the DVR leaf [F1] to the free module $M_{\nu }$ with its restricted form and Gram matrix $G_{\nu }$ gives the displayed equality ${\sum }_{s>0}{\dim }_F{e}_{\nu }{M}_F^s={\nu }_{\pi }(\det G_{\nu })$.

Conclusion. By step 2.1, the multiplicities appearing in a Jantzen class equality are exactly the weight Gram valuations ${\nu }_{\pi }(\det G_{\nu })$; by step 1.4 the weight character detects class equalities, so a class equality involving Jantzen pieces holds if and only if the corresponding equalities of weight Gram valuations hold. No Schur functor, residue-sequence character, rationalization or semisimple Hecke Grothendieck group is used anywhere.