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1

A particle is defined as an irreducible representation of the Poincar\'e group. These were classified by Wigner in 1939. This was done via the little group construction. The important representations are (metric signature $(-,+,+,+)$ $p^2 = 0$, $p^0 < 0$ - The little group is ISO(2). All finite dimensional representations of this group are ...

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Your understanding of reducible and irreducible representations is a little bit muddled. Let me try to clarify this a bit: A reducible representation $D:G\to \text{GL}(V)$ is one that has a nontrivial invariant subspace $W$. That is, there exists a nonzero $W<V$ such that for all $g\in G$ and all $w\in W$, the action $D(g)w\in W$ remains in the ...

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Your idea 1) is the right idea: It's just the law of transformation of matrices generalized from the transformation of matrices: If we apply a general linear transformation $U : V \to V$ on a vector space, the matrices/operators on it transform as $M \mapsto U^\dagger MU$ For unitary operators $U^\dagger = U^{-1}$, so the transformation law becomes $M ... 1 Recall that when classifying representations of the Lorentz group, we consider $$\textbf{N}_{\pm} = \frac{\textbf{J} \pm i\textbf{K}}{2}, \tag{1}$$ where$\textbf{J}$is the angular momentum (rotation generator) and$\textbf{K}$is the boost generator. The generators$\textbf{J}$and$\textbf{K}$satisfy ... 0 As @TwoBs and @Trimok mentioned, in the case of the breaking$U(1)^n\to U(1)^{n-k}$, the charges don't change. This is however true only in a basis the broken fields are diagonals (only charge under one U(1). As an example, consider$U(1)^3and the following three fields with their charges: \begin{aligned} \Phi_1:& (1,1,0)\\ \Phi_2:& (1,-1,0)\\ ... 2 Here is a partial answer: DefineSp(2N,\mathbb{R})$as the group of matrices$S$such that$S \cdot\Omega\cdot S^T=\Omega$where$\Omega_{ij}$is a non-degenerate anti-symmetric matrix. Then$\Omega_{ij}$is an invariant tensor similar to the Kronecker delta for orthogonal transformations. I don't think there are any more (not 100% sure). For$E_7$:$E_7$... 5 There is a definition that$\left( \frac{m}{2}, \frac{n}{2}\right)$representation is equal to spinor tensor $$\psi_{a_{1}...a_{m}\dot{b}_{1}...\dot{b}_{n}},$$ where$\psi_{\dot{b}}$transforms as complex conjugation of$\psi_{b}$. Why do we assume that$\left( \frac{1}{2}, 0\right)$and$\left( 0, \frac{1}{2}\right)$represent spinors? You can think about ... 1 The stationary action principle and the Euler-Lagrange (EL) equations are very broad and general constructions. The field variables in the variational principle could in principle map into some generic manifold$M$. On the other hand, Euler-Poincare (EP) equations appear in the special situation where the manifold is a Lie group$M=G\$, and the action is ...

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