A p-section with no local block inducing to the chosen global block

Put $a=(123)$, $C=a+a^2$, and let $T$ be the sum of the three transpositions. The center of $kG$ has basis $1,T,C$, and in characteristic $2$ one directly obtains $T^2=1+C$, $C^2=C$, and $TC=0$. Thus its only nonzero primitive central idempotents are $e=1+C$ and $f=C$. The first acts as $1$ on the trivial module, so it is the principal block idempotent for $B_0$; F1 then identifies $f$ with the remaining block $B_1$. In $\mathcal{O}G$, the idempotents $\widehat{e}=(1+a+a^2)\mathrm{/}3$ and $\widehat{f}=1-\widehat{e}$ reduce respectively to $e$ and $f$. By uniqueness in F2 they are the integral block lifts. Let $W=\{(x_1,x_2,x_3)\in K^3:x_1+x_2+x_3=0\}$ with the coordinate-permutation action. The operator $\widehat{e}$ averages over $⟨a⟩$ and projects onto $W^{⟨a⟩}$. An $a$-fixed vector has equal coordinates, and its coordinate sum is $3x_1$, so $W^{⟨a⟩}=0$. If a line in $W$ were $G$-stable, the scalar $\lambda$ by which $a$ acted would satisfy ${\lambda }^3=1$ and, from $tat=a^{-1}$, $\lambda ={\lambda }^{-1}$; hence $\lambda =1$, contrary to $W^{⟨a⟩}=0$. Thus $W$ is irreducible and affords ${\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}}$. Now $\widehat{e}W=0$ and $\widehat{f}W=W$, making ${\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}}$ a row of $B_1$ by F2.

By F1, the only local block over $u=t$ induces to $B_0$, whereas step 1.1 puts the row ${\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}}$ in $B_1$. F3 therefore gives $d_{{\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}},\phi }^t=0$ for every $\phi \in \mathrm{IBr}(C_G(t),c)$.

More explicitly, $k{C}_G(t)\cong k[X]\mathrm{/}((X-1{)}^2)$ has the single simple quotient $k$, so there is exactly one local irreducible Brauer character $\phi$, with $\phi (1)=1$. The only $2$-regular element of $C_G(t)\cong C_2$ is $v=1$. The full generalized-decomposition identity at $tv=t$ therefore is ${\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}}(t)=d_{{\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}},\phi }^t\phi (1)=0.$ Equivalently, F3's restricted sum for the block $B_1$ is empty.

There is also a direct characteristic-zero check. On the basis $v_1=(1,-1,0)$, $v_2=(0,1,-1)$ of $W$, the transposition $t=(12)$ satisfies $t{v}_1=-v_1$ and $t{v}_2=v_1+v_2$, so its matrix has trace $0$. This agrees with step 3.1. The example is a genuine empty-local-support boundary case; it does not purport to compute a larger generalized-decomposition table. Algebraic closedness and AC are used only through F3 and the preceding AC-stated example.