Free fermion dual to monopole operator in scalar $QED_3+$ Chern-Simons term equivalence proof? In most papers discussing 3D Abelian bosonization duality, they say that monopole operator in scalar $QED_3+CS$ is dual to free fermions.
How do they know it, because I have never seen an actual proof in single flavor systems?
https://arxiv.org/abs/1606.01989  see 2.2 for example, it just comes from air no proof whatsoever.
edit:
they have some justification but it is not at all a proof, and pretty hand waving.
 A: I don't think that there exists such a thing as proof in this business but some examples may substantiate some claims.
To explain why it may be true we should invoke two well-known facts:

*

*Particle-vortex duality (the Abelian Higgs model dual to XY-model), which is an IR duality widely believed to be true.

*Flux attachment. The mechanism changes the statistics of particles by adding a Chern-Simons term to an action. When the level k equals one fermions get switched to bosons and vice versa.

The duality asserts that $M(x) \leftrightarrow \phi(x)$. Then we include the Chern-Simons term and the Gauss law constraint is modified to
\begin{equation}
\rho + \frac{1}{2\pi} F_{12} = 0
\end{equation}
and a monopole operator is not gauge invariant anymore.
So, informally people say that a monopole acquires electric charge k(=1 in this case).
Now, a monopole is charged under the gauge $U(1)$ and, by the second fact, it would get a spin $\sim k/2 = 1/2$, which is schematically written as a dressed monopole $\phi M$.
Altogether, the new IR duality would take a different form
\begin{equation}
 \phi M \leftrightarrow \psi,
\end{equation}
which is more or less the result.
These steps could be made more rigorous, but, of course, not to the level of mathematical rigor. Even the first fact lacks a firm anylitical proof.
