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So, I have been reading Lecture notes on "Supersymmetry and Extra Dimensions" (PDF), taken by Flip Tanedo (notes of the course of SUSY and Extra Dimension taken by Professor Quevedo, Cambridge Uni). In the notes, graded Lie algebra was introduced, it was mentioned that Poincare generators are Bosonic generators. I do not understand how exactly they are commenting that and also I do not understand the $(A,B)$ representation given for the generators. I thought it was only $(1/2, 0)$ direct sum $(0, 1/2)$

I have attached the page for reference.

P.S. I have been following the Lectures by Professor Quevedo, but these things were not mentioned, and since I am a beginner to the subject, a lot of things are quite confusing from time to time.

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Qmechanic
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Alex
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    The Poincare generators have commutation relations of the above form if and only if $(-1)^{\eta_a \eta_b} = 1$. And $(1/2, 0) \oplus (0, 1/2)$ is the representation for a Dirac spinor, not for a vector and anti-symmetric tensor. – Connor Behan Jun 06 '23 at 13:17
  • I understand that (1/2,0)⊕(0,1/2) is the representation for a Dirac spinor, but how is the Poincare generators are having (1/2, 1/2) , (1,0)⊕(0,1) representations? – Alex Jun 07 '23 at 05:12
  • Ah, I got the part about why it's (1/2,1/2) rep, it's basically the Lorentz four-momentum representation. I am still trying to find about the (1,0)⊕(0,1) representation – Alex Jun 07 '23 at 06:11

1 Answers1

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The Poincare generators are explained as follows:

  1. The 4-dimensional vector representation $\mathbb{R}^{1,3}$ (of the double cover) of the restricted Lorentz group in 3+1D is isomorphic to the $(\frac{1}{2},\frac{1}{2})$ representation, cf. e.g. this & this Phys.SE posts. This explains the inhomogeneous Poincare generators, i.e. momenta/translations.

  2. A rank-2 tensor can be decomposed as $$\begin{array}{rcccccl} T_{\mu\nu}&=&A_{\mu\nu} &+&S_{\mu\nu} &+& S \eta_{\mu\nu},\cr (\mathbb{R}^{1,3})^{\otimes 2}&\cong&\cr (\frac{1}{2},\frac{1}{2})^{\otimes 2}&=&\cr (\frac{1}{2}^{\otimes 2},\frac{1}{2}^{\otimes 2}) &=&\cr (1\oplus 0,1\oplus 0) &=&\underbrace{(1,0)\oplus (0,1)}_{\text{antisymmetric}}&\oplus&\underbrace{(1,1)}_{\text{traceless sym.}}&\oplus& \underbrace{(0,0)}_{\text{trace part}}. \end{array}\tag{1}$$ The antisymmetric part $$ (\mathbb{R}^{1,3})^{\wedge 2}~\equiv~ \bigwedge\!{}^2\mathbb{R}^{1,3}~\cong~ so(1,3;\mathbb{R})\tag{2}$$ of eq. (1) is isomorphic to the Lorentz Lie algebra, which has angular momentum as generators.

Qmechanic
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