Past exam of the anyons and topological quantum field theories course of the University of Oxford / 2025 / 2

Cut one nonseparating circle on each handle of the closed orientable surface ${\Sigma }_g$ of genus $g$. The cut surface is a sphere with $2g$ boundary circles. Gluing a pair back together requires an anyon label $a_i$ on one boundary and the antianyon label ${\bar{a}}_i$ on the oppositely oriented boundary. Sum over these labels independently for the $g$ handles.

For a fixed labeling, the dimension of the resulting anyon fusion space on the sphere, with total vacuum anyon charge, isThe gluing rule of the topological quantum field theory therefore gives the handle-cut formula for anyon ground-state degeneracy:Here $1$ denotes the vacuum anyon, and the anyon fusion matrices have entries $(N_b{)}_{ac}=N_{ab}^c$, as in the source. At $g=0$, the empty product is the identity matrix, giving one ground state. At $g=1$, $(N_a{N}_{\bar{a}}{)}_{11}=1$ for each simple charge, so the ground-state degeneracy of the torus is the number of anyon types.

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 $2g-2$.

Let $B_c^{ab}$ denote the split-fuse diagram with a charge-$c$ edge joining the lower fusion vertex to the upper splitting vertex, and outer strands $a,b$. Completeness in the isotopy normalization of anyon diagrams readsResolve the two strands inside the encircling normalized Kirby color loop in this way. Move that loop onto the intermediate charge-$c$ edge. The Kirby killing property multiplies it by $\mathcal{D}{\delta }_{c1}$, leavingThe vacuum anyon channel is allowed only for $b=\bar{a}$, and then $d_b=d_a$ and $d_1=1$. Erasing the vacuum edge turns $B_1^{a\bar{a}}$ into the cup-cap diagram $E_a$, which connects the incoming pair by a cap and the outgoing pair by a cup. Thus the requested two-strand Kirby projector isHere $P_1^{a\bar{a}}=E_a/d_a$ is the orthogonal projection onto total vacuum anyon charge; for $b\ne \bar{a}$ it is zero. Both strands in the source point upwards, so the condition is $b=\bar{a}$, not $b=a$ unless $a$ is self-dual.

The coefficient can be checked without a convention guess: stacking two cup-cap diagrams creates one charge-$a$ loop, so $E_a^2=d_a{E}_a$. Consequently $(E_a/d_a{)}^2=E_a/d_a$, as required of an orthogonal projection. Omitting $1/d_a$ would fail this check.

Use the Witten-Reshetikhin-Turaev invariant with the source's normalized Kirby color on all $2g+1$ surgery components. Since the signature of the linking matrix is zero, the framing-anomaly factor is one, and the TQFT partition function isAll evaluations in this part use the isotopy normalization of anyon diagrams and preserve the indicated blackboard framing.

Expand the long component as ${\mathcal{D}}^{-1}{\sum }_a{d}_{a}a$ and fix its charge $a$. Each pair of auxiliary components is in the Borromean rings arrangement with this component. Eliminate such a pair using the two-strand Kirby projector from c: bend the long component to expose its two oppositely oriented portions, viewed as equally oriented strands labeled $a,\bar{a}$. The first encircling normalized Kirby color loop replaces these by a vacuum cup-cap with coefficient $\mathcal{D}/d_a$. This vacuum connection opens the second encircling loop so that the same two-strand Kirby projector applies once more. Removing the resulting vacuum edges and straightening the bends restores the original charge-$a$ strand. Thus the Borromean handle-pair evaluation multiplies the bare strand byThere is no residual twist. This is a relative strand evaluation: the final closure of the long component is counted separately. As a normalization check, when $a=1$ the two auxiliary components become two unlinked normalized Kirby color loops, each of value $\mathcal{D}$, giving $H_1={\mathcal{D}}^2$.

The $g$ pairs occupy disjoint portions of the long component, so their factors multiply. Closing that component gives its quantum dimension $d_a$. Combining this with its normalized Kirby color coefficient $d_a/\mathcal{D}$ givesApplying the overall surgery prefactor yieldsFinally, the topological quantum field theory assigns the identity map to a cylinder over ${\Sigma }_g$. Closing its time direction takes the trace of that identity, soThis independently reproduces b. For $g=0$ the surgery link is one zero-framed normalized Kirby color unknot: its value $\mathcal{D}$ cancels the surgery prefactor and gives $Z(S^2\times S^1)=1$. For $g=1$ the result is the number of simple anyons, as required for the torus.