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Susy algebra trivial question

  1. Feb 18, 2016 #1
    Dear All
    Am trying to study supersymmetry algebra. We start by constructing the graded poincare algebra by considering the direct sum of the ordinary Poincare algebra L0 with a subspace L1 spanned by the spinor generators Qa where a runs from 1 to 4 ( L1 is not a lie algebra ). So now we have only 4 spinor generators. Then, i want to add the internal symmetry group to get the maximal symmetry of the S matrix. Here is my question. The internal symmetry group has dimension N of N generators Bl. Now we add N spinor generators . Then Q will have 2 indices one for the spinor part a and the other for the internal one. but initially L1 is 4 dimension so these spinors to which space belong?
    Or should i from the begining suppose that the L1 is spanned by 4N generators.
    Thank you
     
  2. jcsd
  3. Feb 19, 2016 #2

    haushofer

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    Hi Shereen,

    it helps if you give a reference of the text you use to study it. I´m not sure if I can follow your analysis, but this is how I understand it:

    1) You start with the Poincaré algebra {M,P} in D dimensions
    2) You want to add fermionic generators to avoid the Coleman-Mandula theorem
    3) These fermionic generators sit in the spin-1/2 representation of the Lorentz-algebra. The corresponding Dirac spinor has 2^[D/2] components, where [...] is the integer part, and it depends on D whether you can choose a Weyl- or Majorana representation to make these spinors irreducible (see e.g. Van Proeyen's Tools for SUSY). E.g. for D=4 you can choose Weyl or Majorana spinors, whatever you prefer.
    4) The commutator [M,Q] ~ Q is fixed by the fact that Q is a spinor under M. The Jacobi's now fix [Q,Q]~P
    5) Now, if you want, you can introduce more than one set of Q's, which gives them an extra index running from 1,...N. As such you make N copies of the algebra, so to speak. This extension also allows you to introduce other extensions to the algebra, e.g. central extensions, again by studying the Jacobi's. Physically, this means you enlarge the multiplets (see e.g. Bilal's notes)
     
  4. Feb 23, 2016 #3
    Thank you haushofer.
    In fact i was wondering if this number N is related to the internal group G ( We have said that the maximum symmetry of the S matrix is P*G since they commute) which i can later introduce. I thought that this N must be equal to the dimension of G.
    Thank you haushofer alot.
     
  5. Feb 23, 2016 #4
    I just want to know the exact meaning of the N, is there is an internal group for the Q's , or should it be related to G.
    Thank you alot
     
  6. Feb 24, 2016 #5
    Sorry for replying several time but i think i just find the answer in David Bailin and Alexander Love book (1 st edition) page 27. It is noted that if we have Q _alpha ^A; alpha spinor index; while the A labeled some internal symmetry if the Q; A=1....N;. So A labeles the representation of the internal symmetry group to which Q belong. I just wanted to ask if this N is related to G the previous internal symmetry group.
    Thank you
     
  7. Feb 24, 2016 #6

    haushofer

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    You mean the R-symmetry group? Yes, this group is related to N, because it is an automorphism (i.e. an isomorphism from the group to itself) of the group. It 'rotates' the supercharges into themselves, so to speak. From the top of my head, I believe that e.g. for D=4 this group is SU(N), where N is the amount of SUSY. So for N=2 SUSY you get the automorphism group SU(2). which means you can rotate the two supercharges into each-other without changing the algebra.
     
  8. Feb 24, 2016 #7

    haushofer

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    Or do you mean gauge symmetries? In that case I'm not sure if they are related.
     
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