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The short answer is that only if the Berry curvature is defined by: (in matrix notation): $$F_{\mu \nu} = \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu} + [A_{\mu}, A_{\nu}]$$ it becomes gauge covariant, i.e., for a gauge transformation: $$A_{\mu} \rightarrow g^{-1}A_{\mu} g+g^{-1}\partial_{\mu}g$$ $g \in U(N)$ ($N$ is the degeneracy of the level), the ...


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They're not different at all! Seiberg has used the term electric-magnetic duality in the title of his famous paper, too. His duality is an electric-magnetic duality because the duality relates two descriptions and objects that are electrically charged under the gauge group on one side (e.g. quarks and gluons) are mapped to solitons (forms of magnetic ...


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We do not start from the gauge fixed path integral in the BRST construction. What you describe (once one adds the missing Faddeev-Popov determinant) is the original Faddeev-Popov trick to get the ghosts, not the systematic BRST construction. The (Hamiltonian) BRST construction crucially first introduces the ghosts as parts of the extended phase space, and ...


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The BRST symmetry cannot be seen without introducing auxiliary variables. The fastest way to realize the BRST symmetry is to "exponentiate" the delta function $$\delta(G)~=~\int \!{\cal D} B ~\exp\left[iB_{\alpha}G^{\alpha}\right]$$ and the Faddeev-Popov (FP) determinant $$\det\Delta ~=~\int \!{\cal D} c ~{\cal D} \bar{c} ~\exp\left[\bar{c}_{\alpha} ...


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It depends on what you mean by vacuum. If you mean a field configuration that has $F=0$, then the gauge potential is locally pure gauge (cf. this answer by Qmechanic), so there can only be global obstructions to the gauge equivalence class of the $A$ corresponding to $F=0$ being $A=0$ everywhere. On $\mathbb{R}^{1,3}$, there is no such obstruction. If ...


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Because the static solution considered is the 't Hooft and Polyakov monopole. There is no (non-Abelian) electric field in this case. A more general solution would be a dyon and then the electric field contributes to the mass of the particle.


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$$ F^2 \to F^2 + \alpha_i f^{ijk} F_{j\mu\nu} F_k^{\mu\nu} $$ $f^{ijk}$ is completely antisymmetric in all its indices and is being contracted with something that is symmetric in $jk$. Thus, the action is invariant. We can see that $f^{ijk}$ is totally antisymmetric as follows $$ [T^i , T^j] = f^{ij}{}_k T^k \implies \text{tr} \left( [ T^i , T^j ] T^k ...


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The field stregth tensor of a Yang-Mills theory is defined as $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu+ie\left[A_\mu,A_\nu\right].$$ In general, the gauge field is in the adjoint representation of the gauge group (we normally say it takes value in the algebra) so it is written as $$A_\mu=A_\mu^aT_a,\quad a=1,2,\ldots dim(G),$$ where $dim(G)$ ...



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