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A vector space $\mathfrak{g}$ over some field $F$ and kitted with a bilinear, antisymmetric and Jacobi-identity-fulfilling product ("Lie Bracket" or "commutator"). In physics, most often arises as the Lie algebra (tangent space to the identity) of a Lie group; in gauge theories, basis vectors of the gauge group's Lie algebra correspond to Noether currents and conserved quantities.

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Angular velocity is $\dot{g}$ carried to the identity element of the group

Given a group element $g(x)$ the quantity $$ \omega= g^{-1} \frac{\partial g}{\partial x^\mu}dx^\mu $$ is the associated Maurer-Cartan form. Whenever you have a parameter-dependent group element …
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What do the different "levels" in a Kac-Moody algebra tell us physically?

The scale of the algebra is always chosen so that the length of the longest root in the underlying finite Lie algebra is 2. This choice simplifies a number of formulae, and in particular ensures that …
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Jacobi Identity of structure constants of ${\rm SU}(N)$ group

I've not seen that identity before. The only think that comes to mind is to try taking a trace with an extra generator of the identity $$ [T_1,\{T_2, T_3\}]+ \{T_2,[T_3, T_1]\}- \{T_3,[T_1, T_2]\}=0 $ …
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How can the exponential generator apply to all Lie groups (not just rotation)?

It's not even true. Many Lie groups are not of "exponential type," an example being the symplectic group ${\rm Sp}(2n,{\mathbb R})$ that describes Gaussian optics. Even in the part of the group tha …
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2 votes

Product of traceless hermitian $N \times N$ matrices

Set $i=l$ and $j=k$ in 80.17 and sum over the repeated indices. The LHS becomes $\sum_a {\rm tr}\{T^a T^a\}$ the RHS becomes $N^2-1$. There are $N^2-1$ traceless hermitian matrices, so if ${\rm tr}\{T …
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What is the Eigenvalue of $T^2$ ($SU(3)$ Casimir)?

The finite dimensional unitary representations of $\mathfrak{su}(3)$ have highest weights (eigenvalues of $\hat \lambda_3$ and $\hat \lambda_8$) that are non-negative integer linear combinations ${ …
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Rotation of $su(2)$ generators

I think that you are groping for the equation $$ U(R) \sigma_i U^{-1}(R)= \sigma_j {R^j}_i, $$ where ${R^j}_i$ is an ${\rm SO}(3)$ rotation matrix, the $\sigma_i$ are the matrix generators of $\mathf …
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Why doesn't the second Chern form $C_2$ vanish in 4D Euclidean space?

The Chern form does not include a commutator. It has a trace instead: $$ C_2 = -\frac 1 {8\pi^2}{\rm tr}(t_at_b)F^a\wedge F^b. $$ Note that ${\rm tr}(t_a t_b)={\rm tr}(t_bt_a)$, so it has no reason to …
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Lie algebra adjoint representation

It's easier if you do this index free. Define the "ad" action of $A$ on $B$ by ${\rm ad}(A) B= [A,B]$. Then rewrite the usual form of Jacobi $$ [A,[B,C]]+ [B,[C,A]]+[C,[A,B]]=0 $$ as $$ [A,[B,C]]- [ …
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Lorentz Group Representations

The six generators are $S^{ab}\equiv i[\gamma^a,\gamma^b]/4$, where $(a,b)$ are the three boosts $(0,1), (0,2) (0,3),$ and the three spatial rotations $(1,2), (2,3), (3,1)$. The Dirac gamma matrice …
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Parameter space of $SO(3)$ and $SU(2)$

The group manifold of ${\rm SU}(2)$ is the three sphere $S^3$. The group manifold of ${\rm SO}(3)$ is the three-sphere with antipodal points identified. The two spaces have different connectectness (m …
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Eigenvalues of quadratic Casimirs of simple Lie groups

Yes. For any simple Lie algebra, if $\lambda$ is the highest weight in a representation and $\rho$ is the Weyl vector (the sum of fundamental weights or 1/2 the sum of the positive roots) then the Cas …
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1 vote

Understanding the inverse in the definition $(\tilde{\Pi}_gf)(v)\equiv f(\Pi^{-1}_gv)$

The inverse is needed to get a homomorphism, let $$ \tilde f(x)= f((\pi(g_1)^{-1} x). $$ Then $$ \pi(g_2) (\pi(g_1) f(x))\\ = \pi(g_2) f(\pi(g_1)^{-1} x)\\ = \pi(g_2) \tilde f(x)\\ = \tilde f(\pi(g …
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Evaluating the $A \land A \land A$ in the Chern-Simons action

Does this help: Consider a lie-lagebra valued form $A=A_\mu^a \lambda_a dx^\mu$ then $$ {\rm tr}(A^3)= {\rm tr} \{\lambda_a\lambda_b \lambda_c\} A^a_\alpha A^b_\beta A^c_\gamma dx^\alpha \wedge dx^\b …
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How can we go from a 4-dimensional representation of $SO(4)$ to the 3-dimensional one of its...

It's not quite what you say. Yes ${\rm SO}(4)$ has an obvious collection of ${\rm SO}(3)$ subgroups where one rotates in 3-dimensional subspaces of of the 4-dimensional space--- but Greiner's ${\rm S …
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