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$\mathcal{O}(-n)=\mathcal{O}(-1)^{\otimes n}$ is a tensor product of line bundles, so the connection is simply $n$ times the connection on $\mathcal{O}(-1)$. If that wasn't obvious, I think it will be useful if I review the basic ingredients involved here, which will hopefully clear up your confusion. Consider a principal $G$-bundle $P$ over a manifold $X$. ...

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Right now, I have no idea about what you actually mean by gauge fields, or what anyone ever means by gauge fields. According to Danu, the gauge field is the connection itself, while if I remember well, Bleecker defines gauge fields as sections of those vector bundles that are associated to principal bundles. However, I can, I think, give a definitive answer ...

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Well, the main point is that Yang-Mills theory is just one out of many gauge theories, cf. e.g. this Phys.SE post. E.g. SUGRA is a gauge theory. In fact, theoretically there are relativistic gauge theories with gauge fields transforming in virtually any possible representation of the Lorentz group. Whether they are realized in Nature is another story.

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If the graviton exists it's not a 4 vector, but a tensor.

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The Lagrangians are not identical, but they only differ by a total derivative. In other words, you get from the one to the other using partial integration. For example, for the first term:  -\frac 1 2 (\partial_\mu A_\nu) (\partial^\mu A^\nu) = -\frac 1 2 \partial_\mu \left( A_\nu \partial^\mu A^\nu \right) + \frac 1 2 A_\nu\, \partial_\mu \partial^\mu A^\...

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Answer posted by Lubos Motl in the comments; I reproduce most of it here. This answer was posted in order to remove this question from the "unanswered" list. Some (sketches of) answers to your questions, one by one: Physical states have to be invariant under gauge symmetries, so all of them are singlets and there are no nontrivial representations, (and 3....

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You need to look up the Helmholtz Theorem and similar results that will basically give you ACuriousMind's Answer. But a way I like to visualize this is through the Fourier transform; in Fourier space the curl $X\mapsto\nabla\times X$ and divergence $X\mapsto \nabla\cdot X$ become simply the cross $\tilde{X}\mapsto k\times\tilde{X}$ and scalar$\tilde{X}\... 1 Restrictions can be imposed on the anomalous terms from general considerations even without fully solving the Feynman diagrams. In fact, the restrictions on the Schwinger term in the commutator given in the question are explicitly described in detail by Roman Jackiw in his review article: Field theoretic investigations in current algebra (section 2.2.). His ... 2 It seems OP's main question is how to understand the representation of the matter fields of YM theory. The matter fields can in principle transform in any representation$\rho:G\to {\rm End}(V)$of the local gauge group$G=SU(N)$, e.g. the fundamental, or adjoint representation. Here${\rm End}(V)$denotes the algebra of endomorphisms on the vector space$...

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@Michael Brown is right. The SM has 12 exactly conserved charges. All local invariances, a fortiori also imply global invariances, if you ignore (for the sake of argument) the spacetime variability of transformation parameters/angles. So SU(3) has 8, not 3 conserved charges, RG, BG, .... The group has 8 generators. Likewise, SU(2) has 3, not 2 conserved ...

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$A_\mu$ is introduced simply as a tool to assert gauge invariance of the fermion's (in this case) kinetic term. Once this field is added to our lagrangian, we recognize that we must add a kinetic term for the gauge field itself. This is the point where we will make contact with the vector potential of electromagnetism. We introduce a field strength tensor,...

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The wave equation needs to stay invariant under local changes of phase. The gauge field $A_{\mu}$ that is introduced to enforce local gauge invariance is NOT an arbitrary function, it needs to represent something and it represents the possibility that the particle either emits or absorbs a photon, a quantum of the EM field. The probability that it does so,...

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Note that, for example, \begin{align} [A_\mu,\partial_\nu]f&=A_\mu\partial_\nu f-\partial_\nu(A_\mu f)\\ &=A_\mu\partial_\nu f-\partial_\nu(A_\mu)f-A_\mu\partial_\nu f\\ &=-f\partial_\nu A_\mu\,. \end{align} So you don't get terms like $A_\mu\partial_\nu$.

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