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You may decompose (irreductible) representation of groups as a sum of (irreductible)representations of subgroups. Starting from a traceless symmetric irreductible representation of $SO(D-1)$: $$R_{ij} = \frac{1}{2} (v^iv^j+v^jv^i) - \frac{1}{D-1} \delta_{ij} ( \sum\limits_{i=1}^{D-1} v_k v_k),\, with \, (i,j) \, in \,[1, D-1] \tag{1}$$ You consider ...


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First of all, the problem is technically difficult due to the fact that generally unbounded self-adjoint operators like those used in general QM have domain smaller that the whole Hilbert space. For this reason I will consider here only bounded self-adjoint operators whose domain, as is well known, is the full Hilbert space. Proposition. The elements of ...


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If $\mathcal H$ is the Hilbert space of the QFT, then \begin{align} U:\mathrm{SO}(3,1)\to \mathscr U(\mathcal H) \end{align} where $\mathscr U(\mathcal H)$ is the set of unitary operators on $\mathcal H$. In other words, $U$ is a unitary representation on the Hilbert space of the theory. If $V$ is the target space of the fields begin considered, then ...


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It is not surprising that you found one of the Pauli matrices as the generator of your rotation. Let us see how it can be seen algebraically that it is to be expected: Observe that 2D rotations embed naturally into 2D unitary matrices, as $\mathrm{SO}(2) \subset \mathrm{SU}(2)$, corresponding to the subgroup of real matrices. Now, as we know, the ...


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Good luck. To check the cancellation for particular groups like $E_8\times E_8$ and $SO(32)$, you will indeed have to get through similar group-theoretical tasks. Similar trace formula for the traces of $E_8$ transformations are especially yummy, including the factor of $1/30$. The orthogonal case is easier even if one is not an intimate friend of all ...


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Every Lie group has a set of generators, and typically a group element is found by exponentiate (linear combinations) of these generators. Since the fundamental definition of say $SU(N)$ [similarly $SO(N)$] is something like The group of unitary (orthogonal), $n$ by $n$ matrices with unit determinant then, the fundamental representation is given by ...


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Congratulations, you made me look into this for the last hour! And, unfortunately, I believe the answer is: Nope We are looking for a Ricci-flat Riemannian symmetric space, since your isometry group is a Lie group. I spent some time trying to construct the Ricci-flat manifold from the irreducible symmetric spaces given there, but couldn't figure out a good ...


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There are 5 standard model (SM) multiplets per generation of fermions. The SM gauge group is $\mathcal{G}_\text{SM} = SU(3)_C \times SU(2)_L \times U(1)_Y$. Various multiplets can then be written as $\mathcal{G}_\text{SM} \ni x = (C,T)_{(Y)}$, where $C$ denotes colour multiplet, $T$ weak isospin multiplet and $Y$ hypercharge value. Multiplets (1st ...


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First of all, we are dealing with unitary representations, so that the $T^a$s are always self-adjoint and the representations have the form $$U(v) = e^{i \sum_{a=1}^Nv^a T_a}$$ with $v \in \mathbb R^N$. When you say that $U$ is real you just mean that the representation is made of the very real, unitary, $n\times n$ matrices $U$. This way, the condition ...


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The names of these creatures are a true mess and there are mainly two independent notation schemes: the mathematical and the physical one. Let $P \to M$ be a $G$-principal bundle. Then $G$ is called the structure group by mathematicians and the gauge group by physicians The (infinite-dimensional) group of automorphism of $P$, or equivalently the group of ...


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A group $G$ by itself is not a group of linear transformations, it is an abstract algebraic object. Only its representations map its elements (injectively if the representation is faithful) to elements $\mathrm{Aut}(V)$ of some vector space $V$. Now, physics seems to have no need of such abstract language at first. Our "vector space" is pretty much our ...


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Here is what I thought: In particular physics system a group of symmetric operators (let's say acting on the Hilbert space V) is a subgroup of the group L(V). Therefore with representation theory instead of dealing with L(V) we can deal with an irreducible representation L(A) which is substantially simpler than L(V). It's quite f* awesome to figure out ...



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