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9

We do if there is a color trace. The term $D_{\mu}D_{\nu}F^{\mu\nu}=\frac{1}{2}[D_{\mu},D_{\nu}]F^{\mu\nu}$ is proportional to $F_{\mu\nu} F^{\mu\nu}$.


5

Multiplying by $e^{i\theta}$ is a rotation of $\theta$ in the complex plane. Physically it changes the phase of a plane wave by an angle $\theta$. This is a global symmetry because we arbitrarily choose a reference point for measuring the phase of plane waves. If we change the phase of all plane waves by an equal amount then this is equivalent to just moving ...


3

There are obviously differeing genus types according to which partition function in d-dimensional QFT. At the very outset of $0$-d QFT the index is the push forward in ordinary de Rahm Cohomology; in other words, the integration of differential forms. The genus as you put it, is non-existant here. In $1$-d QFT the index of the Dirac Operator is ...


3

Commutators are Lie brackets, in this case on the algebra of differential operators. Asking whether $[\dot{},\dot{}]$ is a Lie bracket doesn't really make sense (since, for matrix groups, the Lie bracket is the matrix commutator, anyway). The field strength is not "just a definition", it is the natural curvature associated to the principal connection that ...


3

Every Hermitian traceless matrix $H$ is in $\mathfrak{su}(N)$ since $\mathrm{Tr}(H) = 0$ and so $$ \exp(\mathrm{tr}(\mathrm{i}H)) = \det(\exp(\mathrm{i}H)) = 1$$ so $\exp(\mathrm{i}H)$ is unitary with determinant $1$, hence in $\mathrm{SU}(N)$. The gauge field is always in the Lie algebra of the gauge group since it is introduced to cancel terms that are ...


2

The gauge potential is an object that, when introduced in the covariant derivative, is intended to cancel the terms that spoil the linear transformation of the field under the gauge group. Every gauge transformation $g:\Sigma\to G$ (on a spacetime $\Sigma$) connected to the identity may be written as $\mathrm{e}^{\mathrm{i}\chi(x)}$ for some Lie algebra ...


1

No. Gauge invariance is not a real physical symmetry but a mathematical property of the formalism while renormalization is more deep property related to the scaling of the coupling constant. One can think interaction that breaks gauge but is still renormalizable. It is fact that QED and QCD are gauge theories but gravity is a gauge theory and not ...


1

The bracket you have written is of the form $$[\delta,\delta] A_\mu = v^\nu \partial_\nu A_\mu + \partial_\mu (v^\nu A_\nu).$$ As you pointed out, the first term corresponds to a translation by $v^\mu$. The second term corresponds to a gauge transformation $\delta A_\mu = \partial_\mu \lambda$ with $\lambda = v^\nu A_\nu$. So the algebra closes up to a ...


1

I think that the problematic part here is the notion of what demands what. For example, you state This is the standard argument by which Lorentz invariance is found to demand gauge invariance for massless particles. but I am not completely sure if I agree with this, or at least to the interpretation you are carrying with it. Taking a look at ...


1

As the more upvoted answer said, if there are color indices then the covariant derivative doesn't commute with itself and the expression you wrote makes sense. If not, symmetry arguments about the symmetric nature of the derivative and the connection and the anti-symmetric nature of the curvature tensor are enough to reason like I did below that the ...


1

In fact, global gauge transformations are a subset of local gauge transformation: changing the same amount everywhere is a special case (ie, more restricting) of changing the phase of each point independently. In the Dirac Lagrangian $$\mathcal{L} = \bar{\psi}(i\gamma^\mu\partial_\mu - m)\psi$$ you have to derive $\psi$. If you make a global transformation ...


1

Here is an outline of the reduction from the Nambu-Goto (NG) action to the light-cone (LC) formulation from a Hamiltonian perspective: The starting point is the Hamiltonian formulation of the NG string, cf. e.g. this Phys.SE post. The Hamiltonian density is of the form "Lagrange multipliers times constraints"$^1$ $$ {\cal H}~=~\lambda^{\alpha} ...


1

For clarity let's work with a Lorentzian signature. Our $g$ is a metric for a 2 dimensional Lorentzian manifold $M$. It is well-known that any two dimensional pseudo-Riemannian manifold is conformally flat, that is $$g = e^{2\omega}\eta$$ Where $\eta$ is the flat 2D Minkovski metric. Your Lightcone gauge example You didn't define your $x^\pm$s but I ...



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