Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 342 / 1

Varying the Abelian Chern--Simons theory with respect to $a_{\alpha }^I$ givesFor a stationary quasiparticle of charge vector $q$, the time component implies that its enclosed gauge flux isup to the orientation convention. Coupling a particle of charge $p$ to this flux produces the full-braid phaseAn exchange is half of the corresponding full braid. Reversing the orientation conjugates the phase.

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Under $a_{\alpha }^I\mapsto a_{\alpha }^I+{\partial }_{\alpha }{\Lambda }^I$, the Chern--Simons term changes bybecause the contraction with two commuting derivatives vanishes. The source term changes byCurrent conservation removes the last term, so

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For a worldline winding once around a noncontractible torus cycle, a large gauge transformation can satisfyThe source action changes byGauge invariance of the path-integral phase for every integer vector $n$ requiresTherefore

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If $W\in SL(n,\mathbb{Z})$, then both $W$ andhave integer entries. Thus $q^{\prime }=Wq$ maps ${\mathbb{Z}}^n$ into itself, and the inverse map shows that it maps onto all of ${\mathbb{Z}}^n$.

Writing the source contraction as $j^{T}a$, with $j^{\prime }=Wj$, invariance requiresHencefor every current, so

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Substituting $a=W^T{a}^{\prime }$ into the Chern--Simons term givesThis matrix is integral and symmetric. It is invertible becauseFor $p^{\prime }=Wp$ and $q^{\prime }=Wq$,Thus every anyon braiding phase is unchanged, and bijectivity of $W$ on the charge lattice shows that the full sets coincide. The torus ground-state degeneracy of an Abelian Chern--Simons theory is also invariant:

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Forthe two basic particles have bosonic self-statistics and full mutual-braid phase $-1$. Moreover $|\det K_1|=4$. These are the electric and magnetic particles of the surface code.

Forthe first basic particle is a fermion, the second is a boson, and they are mutual semions; again $|\det K_2|=4$. They can be identified with $f=e\times m$ and $m$, so this is merely another integral basis for the same surface-code phase.

Finally,which already rules it out. Therefore

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