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23

There are a number of imprecisions in your question, mostly having to do with confusing the Lie group and its Lie algebra. I suppose this will make it hard to read the mathematical literature. Having said that, the first volume of Kobayashi and Nomizu is probably the canonical reference. Let me try to summarise. Let me assume that $H$ is connected. The ...


21

Color charge is the representation of the SU(3) gauge group. The representation theory of SU(3) is described below: The basic representation is called the "3" or the fundamental, or defining, representation. It is a triplet of complex numbers $V^i$, which transform under a 3 by 3 SU(3) matrix by getting multiplied by the matrix. The value of "i" is ...


19

Great, important question. Here's the basic logic: We start with Wigner's Theorem which tells us that a symmetry transformation on a quantum system can be written, up to phase, as either a unitary or anti-unitary operator on the Hilbert space $\mathcal H$ of the system. It follows that if we want to represent a Lie group $G$ of symmetries of a system via ...


15

UPDATE - Answer edited to be consistent with the latest version of the question. The different definitions you mentioned are NOT definitions. In fact, what you are describing are different representations of the Lorentz Algebra. Representation theory plays a very important role in physics. As far as the Lie algebra are concerned, the generators ...


14

I don't know if these rise to the level of "useful," but: Yang-Mills theory with gauge group $G_2$ is interesting because $G_2$ has trivial center. So people simulate it on a lattice, try to understand in what sense it might be confining, how string tensions scale, if it has a deconfinement phase transition, and so on. The idea is that looking at a group ...


14

$F_4$ is the centralizer of $G_2$ inside an $E_8$. In other words, $E_8$ contains an $F_4\times G_2$ maximal subgroup. That's why by embedding the spin connection into the $E_8\times E_8$ heterotic gauge connection on $G_2$ holonomy manifolds, one obtains an $F_4$ gauge symmetry. See, for example, http://arxiv.org/abs/hep-th/0108219 Gauge theories and ...


14

After the answers by joshphysics and user37496, it seems to me that a last remark remains. The quantum relevance of the universal covering Lie group in my opinion is (also) due to a fundamental theorem by Nelson. That theorem relates Lie algebras of symmetric operators with unitary representations of a certain Lie group generated by those operators. The ...


12

The elements of the gauge transformations belong to a gauge group. In physics, it's most typically $SU(N)$ (both the electroweak theory, with its $SU(2)$, and the QCD for quarks, $SU(3)$, use these $SU(N)$ groups; $U(1)$ we first learn in electromagnetism – but we must reinterpret the charge as the "hypercharge" when we study the electroweak theory – is the ...


12

Yes. G2 shows up often, starting with atomic physics (perhaps Racah is the first; see R. E. Behrends, J. Dreitlein, C. Fronsdal, and B. W. Lee, “Simple groups and strong interaction symmetries,” Rev. Mod. Phys. 34, 1 (1962).). You will find some refences in my 1976 Phys rev paper on cns.physics.gatech.edu/GroupTheory/refs . I have whole folder of physics ...


11

In this relatively simple example, one can observe that the subalgebras $\{\sigma_a \otimes \frac{1\mp\eta_1}{2}\}$ are the two commuting copies of $su(2)$. For more complicated situations, one actually has an algorithm to veify the simplicity of a Lie algebra. This is because (the root systems of) simple Lie algebras are classified by Cartan, thus one just ...


11

I) The Casimir invariants of a Lie algebra $L$ over a field $\mathbb{F}$ are the central elements of the universal enveloping algebra $U(L)$. Example: The angular momentum square $\vec{J}^2$ is a quadratic Casimir invariant of the Lie algebra $L=sl(2,\mathbb{C})$. II) Given a bilinear associative/invariant form $B:L\times L\to \mathbb{F}$, one can create ...


11

I'll give you enough hints to complete the proof yourself. If you're desperate, I'm following the notes by Zuber, which are available online, IIRC. Let's start with some notation: pick some basis $\{t_a\}$ of your Lie algebra, then $$ [t_a,t_b] = C_{ab}{}^c t_c$$ defines the structure constants. If you define $$ g_{ab} = C_{ad}{}^e C_{be}{}^d,$$ then this ...


11

The general idea. Let's restrict the discussion to matrix Lie Groups for simplicity. Determining the generators of a given Lie group $G$ simply means (by definition) determining a basis for its Lie algebra $\mathfrak g$. Here's a standard method for finding such a basis: Recall that the Lie algebra $\mathfrak g$ of a matrix Lie group $G$ is defined as ...


10

Nice question! The short answer is that the group is not $SU(2)\times U(1)$, it is $SU(2)_L \times U(1)_{em}$. In other words the two groups act on different standard model particles differently. For example the left handed neutrino does interact weakly and so transforms under the $SU(2)_L$, but is electrically neutral so it doesn't transform under the ...


10

QuantumMechanic's links turn up a dizzying array of meanings for $SU(2)$ in physics, so your question probably turns out to be too broad for a simple answer. Nonetheless, I do like it, and similar questions that grope for pithy overviews of things, so I'll try to answer it with my non particle physicist's understanding. Probably the "main" meaning of ...


10

By definition, the metric tensor $\eta_{ij}$ transforms trivially under the defining rep of $SO(n,m)$. $$ \eta_{ij}=[D(g^{-T})]_{i}^{\ k}[D(g^{-T})]_{j}^{\ l}\eta_{kl} =[D(g^{-1})]^{k}_{\ i}[D(g^{-1})]^{l}_{\ j}\eta_{kl} $$ and this holds for all $g\in SO(n,m)$. Consider a one-parameter subgroup of the defining rep with matrices $D(g)=e^{tJ}$ where ...


10

Here we will sketch a possible derivation. Let $\eta\in {\rm Mat}_{n\times n}(\mathbb{R})$ be a real symmetric matrix of signature $(p,q)$, where $n=p+q$. Define the Lie group $$O(p,q)~:=~ \{ \Lambda\in {\rm Mat}_{n\times n}(\mathbb{R}) \mid \Lambda^T\eta \Lambda = \eta \}, $$ where $\Lambda^T$ denotes the transposed $\Lambda$ matrix. Prove for fun that ...


10

First of all, note that the real Abelian Lie group $U(1)$ comes in two (multiplicatively written) versions: Compact $U(1)~\cong~e^{i\mathbb{R}}~\cong~S^1$, and Non-compact $U(1)$ $~\cong~e^{\mathbb{R}}\cong~\mathbb{R}_+\backslash\{0\}$. Also note that in the physics literature, we often identifies charge operators with Lie algebra generators for a ...


10

I apologize, this is my third correction to my answer. This question is very subtle indeed. I hope this answer is the ultimate one! First of all, if you want to take advantage of Lie's theorem you mention (some time called third Lie theorem), the Lie algebra has to be real, as it must be the Lie algebra of a real Lie group. Then, if you are interested in ...


9

User twistor59 has addressed the part regarding the "generator" terminology, but let me give a bit more detail on the second part of the question. I'm going to restrict the discussion to matrix Lie groups for simplicity. Some background. Given a Lie group $G$ with Lie algebra $\mathfrak g$, there exist two mappings $\mathrm{Ad}$ and $\mathrm{ad}$, both ...


9

What we want to achieve is an invariance of the Lagrangian under a certain symmetry. These symmetries are described by Lie groups (if they are continuous). Let's take QCD as a working example: We want our Lagrangian to be invariant under certain redefinitions of color, i.e. $$ \psi = \begin{pmatrix} q_r \\ q_b \\ q_g \end{pmatrix} \mapsto \psi' = ...


8

No, the group $U(1)\times SU(2)\times SU(3)$ isn't a vector space of any kind because it doesn't have any (commuting) addition operation (curved group manifolds can rarely have such a structure). The article "group extension" you linked to makes it very clear that the group extension does not have to be a vector space and the group operation does not have ...


8

Since I have gleaned some more idea of "where you are" in your learning and shown considerable enthusiasm for getting a thorough grasp of fundamentals, I'd like to add some more details (from a non-particle physicist, mind you, so there are many aspects of your question I must steer clear of) to Lubos's excellent answer. Also, in your question you spoke of ...


8

Here we will only discuss the case of finite-dimensional irreducible representations (irreps) of a complex semisimple Lie algebra $L$. Recall that the set $Z$ of Casimir invariants is the center $Z(U(L))$ of the universal enveloping algebra $U(L)$, cf. e.g. this Phys.SE post. OP's question is answered without proof on p. 253 in Ref. 1: Theorem 2. For ...


8

It is known that for an element $U$ of the group, in matrix sence: $$ Ad_Ux=UxU^{-1}.\,\,(1) $$ Now, we note that the target space of the adjoint rep is spanned by $N^2-1$ traceless matrices $t_a$. So, if we add the unity matrix, we get a full basis in $\mathrm{Mat}_N(\mathbb{C})$. We now note that that the adjoint action is trivially extended to this space, ...


8

Here's my two cents worth. Why Lie Algebras? First I'm just going to talk about Lie algebras. These capture almost all information about the underlying group. The only information omitted is the discrete symmetries of the theory. But in quantum mechanics we usually deal with these separately, so that's fine. The Lorentz Lie Algebra It turns out that the ...


7

There is one more option. You can check that $aa$, $\{a,a^+\}$ and $a^+a^+$ form Lie algebra $sp(2)\sim sl(2)$. Then you can add $a^+$ and $a$ treating them as supergenerators. These are words that tell you to take anticommutators of $a$ and $a^+$ as I did in the first line. Then you get a $5$-dimensional superalgebra, which is $osp(1|2)$. There is a ...


7

People have essentially explained the details, but let me make an attempt to formulate it in a language more familiar to a mathematician. I will ignore subtleties that enter for more general Lie superalgebras. Let $\mathfrak g$ be a Lie superalgebra with the $\mathbb Z_2$ grading $\mathfrak g = \mathfrak g_e\oplus\mathfrak g_o$, where the two factors are ...


7

Background (skip this if you know it all)! I too worried about this when I first learned it. Basically I think it's easiest to think of the Eightfold Way quantum mechanically first and worry about QFT later. So that's what I'll do in this answer. In quantum mechanics (at least according to Wigner) a particle is a basis vector in some representation of the ...



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