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Cutting one nonseparating circle on each handle of a closed orientable surface gives a sphere with $g$ pairs of oppositely oriented boundary circles. The topological quantum field theory gluing rule sums over the paired anyon labels $a,\bar{a}$. The resulting anyon fusion matrix formula becomes $\dim V({\Sigma }_g)={\mathcal{D}}^{2g-2}{\sum }_a{d}_a^{2-2g}$ by the Verlinde formula.

The matrix of Hopf link evaluations divided by the total quantum dimension $\mathcal{D}$. In a unitary modular tensor category it is a symmetric unitary matrix, with $S_{1a}=d_a/\mathcal{D}$ and $S_{\bar{a}b}=S_{ab}^{\ast }$. Its columns diagonalize the anyon fusion matrices through the Verlinde formula.

Put $K={\sum }_a{N}_a{N}_{\bar{a}}$. The Verlinde formula diagonalizes every anyon fusion matrix using the same modular S-matrix:Because $S_{\bar{a}x}=S_{ax}^{\ast }$, the eigenvalue of $K$ in column $x$ isThe last equality uses unitary matrix orthogonality. ThereforeFinally, insert $S_{1x}=d_x/\mathcal{D}$, where $d_x$ is the quantum dimension and ${\mathcal{D}}^2={\sum }_x{d}_x^2$ is the squared total quantum dimension:Although only quantum dimensions remain, the result is an integer because a counts an anyon fusion space.

The sphere value is ${\mathcal{D}}^{-2}{\sum }_x{d}_x^2=1$, and the torus value is ${\sum }_{x}1$. If all $n$ simple anyons have quantum dimension one, the answer is $n^g$, agreeing with the Abelian handle count. These checks also fix the exponent $2g-2$.