Generalized decomposition numbers exist and are unique

By F2, restriction to $H$ gives the finite decomposition ${\mathrm{Res}}_H^G\chi ={\sum }_{\zeta }{n}_{\chi ,\zeta }\zeta$. Since $u\in Z(H)$, its representing operator commutes with $H$; F3 and the splitting condition give ${\rho }_{\zeta }(u)={\lambda }_{u,\zeta }\mathrm{id}$. Moreover ${\rho }_{\zeta }(u{)}^{\mathrm{\mid }u\mathrm{\mid }}=1$, so ${\lambda }_{u,\zeta }^{\mathrm{\mid }u\mathrm{\mid }}=1$ without any algebraic-closedness assumption.

Let $v\in H$ be $p$-regular. Because $u$ and $v$ commute, trace gives $\zeta (uv)=\mathrm{tr}({\rho }_{\zeta }(u){\rho }_{\zeta }(v))={\lambda }_{u,\zeta }\zeta (v)\mathrm{.}$ By F4, $\zeta (v)={\sum }_{\phi }{d}_{\zeta ,\phi }\phi (v)$. Substitution into the restriction formula and reordering its finite sums yields $\chi (uv)={\sum }_{\phi }\left({\sum }_{\zeta }{n}_{\chi ,\zeta }{\lambda }_{u,\zeta }{d}_{\zeta ,\phi }\right)\phi (v)\mathrm{.}$ The coefficient in parentheses is exactly $d_{\chi ,\phi }^u$ by F1.

Let $v_1,\dots ,v_r$ represent the $p$-regular conjugacy classes of $H$, and list the irreducible Brauer characters as ${\phi }_1,\dots ,{\phi }_r$; equality of the two cardinalities follows from [F6]. By [F5], the evaluation matrix $M=({\phi }_j(v_i){)}_{i,j}\in M_r(K)$ has entries in the cyclotomic subfield generated by the Teichmüller lifts that occur. Applying the standard complex realization entrywise gives the Brauer-character table from [F6], whose determinant is nonzero by complex linear independence. Hence $\det M\mathrm{\ne }0$ already in that cyclotomic subfield, and therefore $M$ is invertible over $K$. If another family in $K^r$ gave the same values, subtraction would yield $Ma=0$; invertibility forces $a=0$. This proves uniqueness without applying a complex-linear independence assertion directly to $K$-coefficients.

If $u=1$, then $H=G$, the restriction has only the constituent $\chi$ with multiplicity one, and ${\lambda }_{1,\chi }=1$. The formula in F1 becomes $d_{\chi ,\phi }^1=d_{\chi ,\phi }$. The case $v=1$ in step 2.1 is valid and gives the corresponding value at $u$. All decompositions and sums are finite, so no choice principle is used.