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In the linear sigma model, the chiral action on the pion fields can be implemented on the following matrix combination of the fields: $$U(2) \ni \Sigma = \sigma + i \tau^a \pi_a$$ An element $(U_L = exp(\frac{i}{2}\theta^{(L)}_a \tau^a), U_R = exp(\frac{i}{2}\theta^{(R)}_a \tau^a)) \in SU(2)_L \otimes SU(2)_R$ acts on \Sigma as follows: $$\Sigma ... 2 You should also specify the Representation. The Representation requires SU(N) Lie group with N×N matrix is called Fundamental Representation. Which is used in Standard model U(1) x SU(2) x SU(3). You can surely have SU(N) Lie groups with other Representation, such as Adjoint Representation, then in this case SU(N) are represented by a matrix with a rank of ... 1 The way I understand it, there is actually no shortcut method to write the explicit tensor components from the Young tableaux. We just need to symmetrize and antisymmetrize in all upper and lower indices, and whenever we come across mixed indices, remove traces as well. Traceless symmetric tensors, antisymmetric tensors and traces transform irreducibly. To ... 4 In general, the tensor product of tensors with both being in the same representation (e.g. fundamental) results in a decomposition to symmetrical and antisymmetrical parts. This can be seen that for example from the fact that the expression you wrote down is equal to (after rewriting the epsilon-tensors in terms of Kronecker-deltas) ... 1 Below follows the proof which Howard Georgi seems to have in mind. Let us call the root vector(s) in the Dynkin diagram (a) corresponding to the single 3-vertex for \vec{\gamma}, the three 2-vertices for \vec{\beta}_1, \vec{\beta}_2, \vec{\beta}_3, and the three 1-vertices for \vec{\alpha}_1, \vec{\alpha}_2, \vec{\alpha}_3. Since ... 2 Good Lord! Is Schwinger still worth reading? A top physicist of course, but unfortunately right over my head! You're talking about the big A Adjoint representation as in Hunter's Answer, and there are a great many more modern, rigorous and way clearer texts on this topic. The Wikipedia page is a great start. Also see Rossmann, "Lie Groups, An Introduction ... 2 As I mentioned in my comment, I believe you are talking about the adjoint representation of a Lie Group G with a Lie algebra \mathfrak{g}: $$\forall x \in \mathfrak{g}, \;\; \mathrm{Ad} \: D(g) : x \mapsto D(g) x D^{-1}(g) \in \mathfrak{g}$$ where D(g) denotes a represention of g \in G. One method you can see why D(g) x ... 2 Joshphysics beat the answer I was writing and which I left off to eat my tea. Just a couple of further points: The vector y is an arbitrary vector in \mathbb{R}^4. Hall's rendering of a general Lie algebra member:$$\left(\begin{array}{cc}Y&y\\0&0\end{array}\right)$$exponentiates to: ... 4 Preliminary remarks. As Danu writes in his comment, the physics of the other four generators has to do with spacetime translations, one for each spatial direction, and one for time. But how do we see this explicitly in the math behind the somewhat odd-looking presentation of the Poincare group and its Lie algebra that Hall discusses. First, recall that ... 2 You've got two very good answers from Hunter and NowIGetToLearnWhatAHeadIs. However, it's probably useful to know that this beast O(1,3) is isomorphic or locally isomorphic (i.e. has the same Lie algebra) to a surprising number of other interesting groups, which each give you a slightly different way to think about it. First note that its identity ... 3 It's the same way you know there are three parameters in SO(3). The equation \Lambda^T \eta \, \Lambda = \eta has (n^2+n)/2 independent scalar equations. To see this, write the equation in component form: \Lambda^{\mu\nu} \Lambda_\mu{}^\rho = \eta^{\nu\rho}. Now we see there are n^2 scalar equations equations, but because \eta is symmetric and ... 7 From special relativity we know that a Lorentz transformation: $$x'^\mu = \Lambda^\mu {}_\nu x^\nu$$ preserves the distance: $$g^{\mu \nu} \Delta x_\mu \Delta x_\nu = g^{\mu \nu} \Delta x_\mu' \Delta x_\nu'$$ The above two equations imply: g^{\mu \nu} = g^{\rho \sigma}\Lambda_\rho ... 5 Yes, this is a result rigorously stated as: There's a proper subgroup of O(1,3) isomorphic to SO(3). It's made up of the set of Lorentz transformations of the form:$$\left(\begin{array}{cc} 1 & 0\\ 0 & R(3) \end{array}\right)$$where R(3)\in SO(3), together with the internal operation of matrix multiplication. 7 One can embed the 3\times3 rotation matrices$$R~\in~ SO(3)~:=~\{R\in{\rm Mat}_{3\times 3}(\mathbb{R}) \mid R^tR~=~{\bf 1}_{3\times 3}~\wedge~ \det(R)=1 \}$$into the 4\times4 Lorentz matrices$$\Lambda~\in~ O(1,3)~:=~\{\Lambda\in{\rm Mat}_{4\times 4}(\mathbb{R}) \mid\Lambda^t\eta \Lambda~=~\eta \}$$as$$SO(3)~\ni~R~\stackrel{\Phi}{\mapsto}~ ...

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General Remarks. In general, you cannot "derive" a representation of a given group $G$ on the objects you're considering, but there are some really standard definitions of certain group representations which are given special names like "scalar," "vector," and so on. However, given the representation of a Lie group $G$, this induces a representation of its ...

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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 ...

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