Abelian finite BF counts on S³ and T³

Give $S^3$ its CW structure with one 0-cell and one 3-cell. Its 1-skeleton has no non-tree edges and it has no 2-cell relators, so [F2] gives ${\pi }_1(S^3)=1$. There is exactly one homomorphism from the trivial group to $G$, and [F1] gives $Z_G(S^3)=1\mathrm{/}\mathrm{\mid }G\mathrm{\mid }$.

Give $T^3$ its CW structure with one 0-cell, three 1-cells $a,b,c$, three 2-cells attached along the commutators $ab{a}^{-1}{b}^{-1}$, $bc{b}^{-1}{c}^{-1}$, and $ca{c}^{-1}{a}^{-1}$, and one 3-cell. By [F2], ${\pi }_1(T^3)=⟨a,b,c\mid [a,b]=[b,c]=[c,a]=1⟩\cong {\mathbb{Z}}^3$. The isomorphism sends $(1,0,0),(0,1,0),(0,0,1)$ to $a,b,c$: it is surjective because these generators generate the presented group, and its inverse sends $a,b,c$ to the standard basis because the relators are satisfied.

A homomorphism ${\mathbb{Z}}^3\to G$ is determined by the images of its three standard generators. Every triple in $G^3$ satisfies the commutation relations because $G$ is abelian, so $\mathrm{\mid }\mathrm{Hom}({\mathbb{Z}}^3,G)\mathrm{\mid }=\mathrm{\mid }G{\mathrm{\mid }}^3$. Equation [F1] now gives $Z_G(T^3)=\mathrm{\mid }G{\mathrm{\mid }}^3\mathrm{/}\mathrm{\mid }G\mathrm{\mid }=\mathrm{\mid }G{\mathrm{\mid }}^2$. For $G=\mathbb{Z}\mathrm{/}2\mathbb{Z}$, the two amplitudes are $1\mathrm{/}2$ and $4$, respectively, so neither normalization forces every scalar amplitude to be one.