Because $t^TM=n t^T$,
$$
t^TK^{-1}=t^T\left(I_n-\frac p{1+np}M\right)
=\frac1{1+np}t^T.
$$
Thus a fundamental quasiparticle has fractional constituent number
$$
\boxed{N(e_J)=t^TK^{-1}e_J=\frac1{1+np}.}
$$

Let $q$ be the gauge-charge vector of a local electron. Trivial full braiding with every integer quasiparticle $q'$ requires
$$
q'^TK^{-1}q\in\mathbb Z
\quad\text{for all }q'\in\mathbb Z^n.
$$
Equivalently $K^{-1}q=\ell$ for some $\ell\in\mathbb Z^n$, so the allowed local charges lie in the local-particle lattice
$$
q=K\ell.
$$
Unit constituent number requires
$$
1=t^TK^{-1}q=t^T\ell,
\qquad\text{that is,}\qquad \sum_I\ell_I=1.
$$
Finally its self-exchange angle is
$$
\theta_{qq}=\pi q^TK^{-1}q
=\pi\ell^TK\ell
=\pi\left(\sum_I\ell_I^2+p\left(\sum_I\ell_I\right)^2\right).
$$
For integer $\ell_I$, one has $\ell_I^2\equiv\ell_I\pmod2$, and hence
$$
\frac{\theta_{qq}}\pi\equiv1+p\pmod2.
$$
The exchange phase is fermionic precisely when this integer is odd. Therefore
$$
\boxed{q=K\ell,\quad\ell\in\mathbb Z^n,\quad\sum_I\ell_I=1,
\qquad p\text{ must be even}.}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 342 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-342/1/c/solution