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Since you start with a group $G$ which is a symmetry of the theory (ie of Lagrangian) there will be some generators for $G$. Call them $T_1,\ldots,T_N$. In the original Lagrangian, (before the Higgs gets a vev) you can do a transformation with any of the $T_i$'s and the Lagrangian will not change. After the field gets a vev, you can try the same and you will ...


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There is a real Lie group $\tilde{Diff}(S^1)$ which is a $U(1)$ central extension of the real Lie group $Diff(S^1)$, and the Virasoro algebra is the Lie algebra of this Lie group. The central extension $\tilde{Diff}(S^1)$ can be realized geometrically in two ways. The first is via a Hilbert space embedding (as in the book of Pressley-Segal), and the second ...


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Because you are looking only at the so-called global part, i.e. the part of the gauge transformation which resembles a group action. Recall that the vector bosons transform as $$ A_\mu \to g A_\mu g^{-1} - (\partial_\mu g) g^{-1}$$ where the first part is the global part of the gauge transformation, which tells you that $A_\mu$ transform in the Adjoint ...


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Usually the first step in deriving the reps of Poincaire is to go to the rest frame of the particle. This amounts to choosing a basis where $P^0$ acting on the state is nonzero, and where the eigenvalue a of $P^i$ are zero. We can do this if the momentum is timelike, that is if the eigenvalue of $P_\mu P^\mu$ is negative (in -+++ signature). Furthermore the ...


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There is a new book called Physics From Symmetry which is written specifically for physicists and includes a long, very illustrative introduction to group theory. I especially liked that here concepts like representation or Lie algebra aren't only defined, but motivated and explained in terms that physicists understand. Plus no concepts are introduced which ...


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Physics from Symmetry is a book that explains all group theoretical concepts needed to understand the foundations of QFT in great detail and is written specifically for physicists. It's not very technical, but it's great if you want to understand quickly what concepts are really important for modern physics and why. For example, it explains why things ...


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\begin{equation} \boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}= \boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{10}\boldsymbol{\oplus} \boldsymbol{8}^{\boldsymbol{\prime}}\boldsymbol{\oplus}\boldsymbol{8} \end{equation} We talk about this because it explains the structure of a number of baryons in Particle Physics made ...


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If I understand your question correctly, you have two systems $A$ and $B$, on both of which a symmetry group $G$ acts, so we have mappings $$G\times A\to A,\ \ \ (g,a)\mapsto ga$$ and $$G\times B\to B,\ \ \ (g,b)\mapsto gb$$ If the composed system is $A\times B$, then it symmetry group obviously includes $G\times G$ by $(g,h)(a,b) = (ga, hb)$. We have an ...


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Let me attempt to answer your question, since your question is about SO(10) GUT model, so I will assume that you have the knowledge of simpler version of GUT namely SU(5) GUT model and also little of group theory. You have 4 different questions >>> 01. Isn't this term ($\psi^{T} C \psi$) already invariant under SO(10)? 02. Doesn't this term ($\psi^{T} C ...


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A single transformation is not strictly speaking discrete, but a group of transformations can be. Every transformation of finite order (i.e. $\Lambda^n = I$ for some $n$) generates a discrete group of transformations, so elements of finite order are sometimes called discrete transformations. As Qmechanic explains, $T$ and $P$ generate a subgroup of coset ...


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Comments to the question (v2): Recall that the Lorentz group $G=O(3,1)$ has 4 connected components $$ G~=~G_0 ~\cup~ P\cdot G_0 ~\cup~ T\cdot G_0 ~\cup~ PT\cdot G_0. $$ Here the connected components $G_0$ that contains the identity is the restricted Lorentz group $G_0=SO^+(3,1)$. It is straightforward to see that $G_0$ is a normal subgroup of $G$. ...



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