= Solution
Represent quasiparticle worldlines by conserved currents $j_I^\alpha$ and add the minimal-coupling term
$$
\mathcal L_{\rm src}=-a_{I\alpha}j_I^\alpha.
$$
Up to the corresponding sign convention for charge, variation of the <Abelian Chern--Simons theory> gives
$$
\boxed{
\frac1{2\pi\hbar}K_{IJ}\epsilon^{\alpha\beta\gamma}
\partial_\beta a_{J\gamma}=j_I^\alpha.}
$$
For a static quasiparticle of integer gauge-charge vector $q$, integration over a small surrounding disk yields the <Chern--Simons flux attachment> relation
$$
\boxed{\Phi_I:=\int d^2x\,\epsilon^{0ij}\partial_i a_{Ij}
=2\pi\hbar(K^{-1}q)_I.}
$$
A quasiparticle of charge $q'$ carried once around this flux acquires the <Aharonov-Bohm effect>[Aharonov--Bohm phase]
$$
\frac1\hbar q'^T\Phi
=2\pi q'^TK^{-1}q=2\theta_{q'q},
$$
so
$$
\boxed{\theta_{q'q}=\pi q'^TK^{-1}q,
\qquad
e^{2i\theta_{q'q}}=e^{2\pi i q'^TK^{-1}q}.}
$$
The constituent-particle current has charge vector $t=(1,\ldots,1)^T$. Its integral over the same disk is
$$
\boxed{N=\frac1{2\pi\hbar}t^T\Phi=t^TK^{-1}q.}
$$
For $K=I_n$ and $q=e_J$, this gives $N=1$. The self-exchange angle is $\theta_{qq}=\pi e_J^Te_J=\pi$, so the excitation is a <fermion> with one unit of microscopic particle number and may be identified with an electron.
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