# Tag Info

## Hot answers tagged lie-algebra

22

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 ...

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 ...

13

$F_4$ 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 ...

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 ...

10

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 $L_{\mu\nu}$ are simply some operators with some defined commutation properties. The choices ...

10

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 ...

9

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 ...

8

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 ...

8

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 ...

8

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' = ... 7 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 ... 7 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 ... 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 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 ... 6 Well, if something is invariant under the action of some group it means that every element leaves it invariant. In particular, every element that belongs to some subgroup. Also, you can get an obvious representation of the subgroup by restriction. If \rho: G \to {\rm Aut}(V) is a representation then so is \rho_{H}: H \to {\rm Aut}(V) for H < G. ... 6 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 Cartan ... 6 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 ... 6 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 ... 5 One answer is that most calculations in modern physics do not actually depend on the explicit realization of the Pauli matrices \sigma_a, a=1,2,3, but rather on the relations$$ \sigma_{a}^{\dagger}~=~ \sigma_{a},\qquad\qquad {\rm tr}(\sigma_{a})~=~0,\qquad\qquad a=1,2,3,  \sigma_a \sigma_b ~=~ \delta_{ab} {\bf 1}_{2\times 2} + ...

5

Dear Leandro, it's because the Pauli matrices, together with the $2\times 2$ identity matrix, form a full real basis of all the Hermitian $2\times 2$ matrices (note that $2\times 2$ Hermitian matrices depend on four real parameters), and the identity matrix is irrelevant in a Hamiltonian because it's just a conventional energy shift that acts on all vectors ...

5

Wow. I just had an amazing experience of discovering the following fact: 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 ...

5

I believe you can, if you try to follow the path of finding representations of the $SO(n)$ group over a given Hilbert space. I really haven't done the calculation, but if it is the same, you would have something like this: $H=L_2(\mathbb R^n,\mathbb C)$ would be the Hilbert Space that would correspond to spin 0 particles, and the representation of the ...

5

Ah, this is a very subtle thing, and it's true that it first occurs for four electrons. First, here's an easy way to tell how many states you should expect: Just use the "individual-electron's spin-basis". With four electrons, each of them could be up or down spin, so we expect a total of $2^4 = 16$ states. So where are you missing the states? Well, ...

5

Visualizing this is, in my opinion, pretty tricky because boosts generally act in four dimensions (time included!) and it's hard to draw 4D on a piece of paper. Fortunately, for this question, you can get around this artistic limitation. Here's what I did right after reading your question to attempt to visualize what's going on. I encourage you to try ...

5

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 ...

4

An infinitesimal generator $X$ is a vector field, which satisfies Leibniz rule $$X[fg]~=~ fX[g]+g X[f].$$ In the $1$-dimensional case, the generator is of the form $X=p(x)\frac{\partial}{\partial x}$, where $p=p(x)$ is some function. Assume furthermore that there exists a bijective smooth function $h=h(x)$, such that $$p(x) h'(x)~=~1.$$ In other ...

4

For practical reasons, physicists like to label the states of the system by a set of "quantum numbers". Technically this means that you are looking for a set of mutually commuting Hermitian operators such that: (i) Every vector from the basis of common eigenvectors of these operators is uniquely characterized by the set of eigenvalues (i.e. the above ...

4

It means that the charge operator $Q$ is a Lie algebra generator for some Lie group $G$. the field $\phi\in V$ takes values/transform in a representation $V$ of the Lie group. (Note that any Lie group representation $V$ is also Lie algebra representation of the corresponding Lie algebra.) the charge operator $\rho(Q)$ in the representation \$\rho: G \to ...

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