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The statement you cited does not imply that a complex representation of a gauge group implies a chiral gauge theory in general. This only holds true if the gauge group corresponds to a chiral symmetry in the first place. A chirally symmetric theory contains massless fermions. Regarding your counterexample: it is true that QCD contains fermions in the ...

3

The eigenvalue of the generator $t_a$ are integer multiples of $g_{min}$ because $t_a$ is a generator of a (cyclic) $U(1)$ group and $$\exp(2\pi i t_a/ g_{min}) = 1$$ holds as an operator equation. This equation says that the exponentiation of the generator with some imaginary coefficient that I parameterized as $2\pi i / g_{min}$ is equal to the identity. ...

0

From the question I see that you are confused by the meaning of "Normally states are vectors in infinite dimensional spaces", not by spinors. Function is a good representation of a vector in infinite space. Let us consider function $\psi(\bf{r})$. This is a vector from infinite dimensional space. What will be with this function when we rotate the space? Not ...

2

You should also specify the Representation. The Representation requires SU(N) Lie group with N×N matrix is called Fundamental Representation. Which is used in Standard model U(1) x SU(2) x SU(3). You can surely have SU(N) Lie groups with other Representation, such as Adjoint Representation, then in this case SU(N) are represented by a matrix with a rank of ...

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The way I understand it, there is actually no shortcut method to write the explicit tensor components from the Young tableaux. We just need to symmetrize and antisymmetrize in all upper and lower indices, and whenever we come across mixed indices, remove traces as well. Traceless symmetric tensors, antisymmetric tensors and traces transform irreducibly. To ...

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In general, the tensor product of tensors with both being in the same representation (e.g. fundamental) results in a decomposition to symmetrical and antisymmetrical parts. This can be seen that for example from the fact that the expression you wrote down is equal to (after rewriting the epsilon-tensors in terms of Kronecker-deltas) ...

3

Not all irreducible representations (irrep's for short) of the Poincaré group lead to a Lagrangian. One example (see my comment to Julio Parra's answer) are the zero-mass, "continuous-helicity" (sometimes called "infinite-helicity") representations. There is, however, a way to begin from a positive energy irrep of the Poincaré group (i.e. a 1-particle ...

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I'm not sure such a thing exists. Usually reps only helps you classify the kind of particles you have (i.e the quantum numbers that identify them) and how they transform under the corresponding group. I believe how to represent this particles mathematically and what is their dynamics is a different matter. The only thing similar I know about is that some ...

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No, you do not want representations of the diffeomorphism group for the same reason that you do not want representations of the gauged Lie group in Yang-Mills. The diffeomorphisms are a gauge symmetry, not a real symmetry of the theory. Gauge transformations act trivially on physical states, they map one redundant description of a state onto another. They ...

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The difference is that Poincare invariance is a global symmetry, so it acts nontrivially on the physical states. This has real physical consequences; for instance, if you act with a translation operator on the state of a particle localized at the origin, you get the state of a particle localized at some position other than the origin. Poincare invariance ...

2

Good Lord! Is Schwinger still worth reading? A top physicist of course, but unfortunately right over my head! You're talking about the big A Adjoint representation as in Hunter's Answer, and there are a great many more modern, rigorous and way clearer texts on this topic. The Wikipedia page is a great start. Also see Rossmann, "Lie Groups, An Introduction ...

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As I mentioned in my comment, I believe you are talking about the adjoint representation of a Lie Group $G$ with a Lie algebra $\mathfrak{g}$: $$\forall x \in \mathfrak{g}, \;\; \mathrm{Ad} \: D(g) : x \mapsto D(g) x D^{-1}(g) \in \mathfrak{g}$$ where $D(g)$ denotes a represention of $g \in G$. One method you can see why $D(g) x ... 1 As has been mentioned in the comments, it is an assumption that the QFT has a vacuum state which is annihilated by$P^\mu$. This is actually a very important point, since it is one of the crucial differences between flat space QFT and QFT in curved spacetime. This is explained in Wald's book QFT in Curved spacetime. Essentially in quantum theories, the ... 5 Yes, this is a result rigorously stated as: There's a proper subgroup of$O(1,3)$isomorphic to$SO(3)$. It's made up of the set of Lorentz transformations of the form: $$\left(\begin{array}{cc} 1 & 0\\ 0 & R(3) \end{array}\right)$$ where$R(3)\in SO(3)$, together with the internal operation of matrix multiplication. 7 One can embed the$3\times3$rotation matrices $$R~\in~ SO(3)~:=~\{R\in{\rm Mat}_{3\times 3}(\mathbb{R}) \mid R^tR~=~{\bf 1}_{3\times 3}~\wedge~ \det(R)=1 \}$$ into the$4\times4$Lorentz matrices $$\Lambda~\in~ O(1,3)~:=~\{\Lambda\in{\rm Mat}_{4\times 4}(\mathbb{R}) \mid\Lambda^t\eta \Lambda~=~\eta \}$$ as $$SO(3)~\ni~R~\stackrel{\Phi}{\mapsto}~ ... 8 I) Firstly, we are talking about the direct or Cartesian product SU(2)\times SU(2) of groups, not the tensor product^1 SU(2)\otimes SU(2) of groups. II) Secondly, SU(2)\times SU(2) is not isomorphic to the Lorentz group SO(3,1) but rather to a compact cousin$$[SU(2)\times SU(2)]/\mathbb{Z}_2~\cong~ SO(4).$$In particular, a ... 4 The problem here is with the identification of the$(A,B)$values of a representation with spin.$A$and$B$do not correspond to spin (they are not even Hermitian!), they just happen to obey$SU(2)$Lie algebras, and as such they add up in the same way that spins do. When we say that$A_\mu,J_\mu,p_\mu,...$are all in the$(\frac{1}{2},\frac{1}{2}) $... 0 The presence of the external electric field breaks$SO(3)$to$SO(2)$. Suppose that$\mathbf{E}$is oriented along the$z$-axis, then rotations about the$z$-axis (of course, chosen to pass through the center of the sphere) is a symmmetry in the problem. This$SO(2)$invariance only implies that the potential is independent of$\varphi$, the polar angle. ... 4 Every Lie group has an adjoint representation. I'm not sure what definition you come at the adjoint representation from, but here's the fundamental one which I'm sure you'll see is always meaningful. Think of a$C^1$path$\sigma:[-1,1]\to\mathfrak{G}$through the identity in a Lie group$\mathfrak{G}$with$\sigma(0) =\mathrm{id}$and with tangent$X\$ ...

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