On page 10, starting from $QED_{3}$ with $N_{f}$ fermions of charge 1, the monopole operator is defined in the following way. Let $\left\{a_{i},a_{j}^{\dagger}\right\}=\delta_{ij}$ be the annihilation and creation operators for the zero-modes of the Dirac fermion in the monopole background. Let $\left|0\right>$ be the bare monopole state. Then the monopole operator is defined to be associated with the state. $$\left|\mathfrak{M}_{i_{1} i_{2}\cdots i_{l}}\right>=a_{i_{1}}a_{i_{2}}\cdots a_{i_{l}}\left|0\right>$$ This state transforms in totally anti-symmetric representation of $SU(N_{f})$, with $l$ indices, and is bosonic.
Could anyone please help me understand this definition? Why is monopole operator defined in such a way? How do I see that it transforms in representation of $SU(N_{f})$? Why is it bosonic?
For starters the "bare" monopole state is not gauge invariant in the monopole background since it carries a charge for the Gauss Law constraint $$G |0\rangle\neq0$$. To fix this we really should dress the state with the appropriate number of zero modes to cancel this charge. That is why the monopole operators are defined like that. To see that it transforms as an anti-symmetric tensor representation of $$SU(N_f)$$, you have to notice that the matter fields transform in the (anti-)fundamental of the flavor symmetry group, and so do the raising/lowering operators. Because the creation operators for a fermion field anti-commute the monopole operator transforms as an anti-symmetric tensor of $$SU(N_f)$$. The monopole operators in this case are bosonic because the creation/annihilation operators have no spin!