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## New answers tagged group-theory

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In the absence of Yukawa couplings (only kinetic terms), the SM has the global flavor symmetry: $$G_{y=0} = U(N_f)^5=U(3)^5$$ Because there are 5 distinct representations in the SM (3 for quarks: $u_R$, $d_R$, $Q_L$; and 2 for leptons: $e_R$, $L_L$). However, $U(N) \sim SU(N)\times U(1)$, so the group can also be written as: $$G_{y=0} = SU(3)^5 \times ... 0 I often see \mathrm{SU}{(3)}_\text{flavor}. However, I have seen$$\mathrm{U}(3)_\mathrm L \times \mathrm{U}(3)_\mathrm R = \mathrm{SU}(3)_\mathrm L \times \mathrm{SU}(3)_\mathrm R \times \mathrm{U}(1)_\text{vector} \times \mathrm{U}(1)_\text{axial}$$where the last one is broken by the quantum anomaly. See slide 14 in this lecture summary of Theoretical ... 1 Your SU(3)\otimes SU(3)={\bf 1}\oplus {\bf 8} above is a chimaeric typo from hell. OK, I'll just give you the self-evident answers, but they would be meaningless junk numbers if you failed to reproduce them directly on the basis of your SU(3) text or the WP article which explains the rules and the Dynkin representation notation, D(p,q), which connects to ... 1 Well, you might have spared yourself confusion and grief by checking your peculiar language in SO(3), which any undergraduate is familiar with. Let me illustrate this for SO(3), before moving on to the much messier SU(3). For SO(3), Kronecker-composing two vectors (spin 1, so 3 s) yields a spin 2 quintet (call it φ, so 5), a triplet (π) and a singlet (s), ... 3 The way to do this is using the Wigner-Eckart theorem. The way it is applied to your problem is as follows:$$ \left\langle nlm |\vec{r}| n'l'm'\right\rangle = \left\langle nl ||\vec{r}|| n'l'\right\rangle \left\langle l' m' 1 q | l m\right\rangle  where the second factor is a Clebsch-Gordan coefficient and $q=-1,0,1$ indicates the type of transition. For ...

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Subalgebras whose root system is not a subset of the root system of the original algebra are called special subalgebras. Therefore, the generators are not a subset of the original's group generators. That is not quite right. A special subalgebra is one such that their step operators do not form a subset of the algebra step operators.That is what is ...

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It results from the combination of two facts: $i$) the embedding of $U(1)_Y$ into $SU(2)_R \times U(1)_{B-L}$, and $ii$) the normalization of the $U(1)$ charges. Let me take the simpler chain: $SU(2)_L \times SU(2)_R \times U(1)_{B-L} \to SU(2)_L \times U(1)_Y$. $i$) At the scale of the left-right symmetry, the hypercharge gets merged into $SU(2)_R \times ... 3 Apologies for not producing a most general answer for arbitrary Lie groups, (which you might tease with great effort out of WP ), but only a trail-map for your particular (charmed!) problem. I call it charmed because it should remind you of the Lorentz group, with a,b,c parameterizing Kx,Ky,Kz boosts and d,e,f the three J rotation angles. Decent treatments ... 2 This is the prototypical example of a superselection rule. The operator$U(2\pi)$commutes with all observables (because it represents a full rotation, and is hence physically a "do nothing" operator), and yet is not a multiple of the identity (because it is -1 on the fermionic and 1 on the bosonic parts of the Hilbert space). Therefore, the representation ... 2 The spin$s$of a particle characterizes how the rotation generators act on it. In$D$dimensions, you represent the little group$SO(D-1)$for massive particles and$SO(D-2)$for massless ones. In fact, you really need to consider its universal cover$\textrm{Spin}(n)$which happen to be just its double cover. Now, you can define the spin to be the largest ... 1 To a given Lie algebra$\mathfrak g$there is a unique group$\tilde G$, called the universal covering group, with the property of being simply connected. For example, the covering group of the algebra$\mathfrak{su}(2)$is$SU(2)$. The other groups,$\{G\}$, associated to the same algebra can be obtained from the covering group in the following way ... 4 The vector$(0,0,0,v)is left invariant by the set of matrices of the form \begin{align*} M=\begin{bmatrix} R & \vec 0 \\ \vec 0^T & 1\end{bmatrix} \end{align*} where\det(M)=\det(R)=1$and$M^{-1}=M^T$implies$R^{-1}=R^T$. By definition,$SO(3)$is the group of 3 by 3 orthogonal matrices with determinant 1. In general, you need to know the Lie ... 0 Yes, any group can be a gauge group. To each oriented edge of the lattice you assign a group element, with the opposite orientations of an edge associated to inverse group elements. The observables are conjugacy classes of products of group elements around a loop of oriented edges. The most basic such observable is the product of group elements around an ... 0 (I) Assuming there are$N$distinct fermions in your Lagrangian (e.g. quarks in the standard model), the kinetic term will have the$U(N)\times U(N)$symmetry. This symmetry is broken by the mass term, however, which couples fermions with different handedness. (II) If$V$is the fundamental representation of a Lie group (with dual/conjugate representation ... 1 Yes, the first part of your question is appreciated and answered soundly. The fermion kinetic term splits into two independent parts involving left and right Weyl spinors respectively, so each is independent under a separate U(N) as your wrote down. The second question is a matter of language. A generator is a matrix with one adjoint index, a here, ranging ... 0 I will talk about$SU(3) \times SU(2)$. First, a matrix$T_3 \in SU(3)$acts in the fundamental representation on$\mathbb C^3$in the following way: A vector$\vec v \in \mathbb C^3$with components$v_i$is mapped to$v'_i = (T_3)_{ij} v_j$. Similarly, a$T_2 \in SU(2)$acts on$\vec w \in \mathbb C^2$as$w'k = (T_2)_{kl} w_l$. The bifundamental ... 2 Anthony Zee just came out with Group Theory in a Nutshell for Physicists - covers most of what a undergrad physics student needs including finite groups and representations, except Young diagrams. 2 Let$\phi : \mathbb{R}^4\to V$be a field with (complex) target vector space$V$, transforming in a finite-dimensional projective representation$\rho_\text{fin} : \mathrm{SO}(1,3)\to\mathrm{U}(V)$. As it is a field, the representation of the translations$\mathbb{R}^4$on$V$is the trivial one, since the field transforms as$\phi(x)\overset{x\mapsto ...

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I cannot quite vouch for exhaustive panoramas, but the crucial point is that GL(N), SU(N) matrices are representable in a nonhermitean basis discovered by Sylvester in 1882, the clock and shift matrices which he called nonions for N=3 (long before the Gell-Mann basis!), sedenions, etc. Their braiding relations, and maximal grading, and hence commutators, ...

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