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Dirac gammology - dimension of the algebra

  1. Oct 13, 2014 #1
    Dirac matrices satisfy the relations:

    [itex]\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=2g^{\mu\nu}[/itex]

    I would like to understand why the dimension of this algebra in 3+1 dimensions is 4.

    If we're looking for all possible sets {[itex]\gamma^0,\gamma^1,\gamma^2,\gamma^3[/itex]} of 4x4 matrices that satisfy this, how do I show that when I find just one set, it already forms a complete basis?
     
  2. jcsd
  3. Oct 13, 2014 #2

    dextercioby

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    You know that there are 4 matrices to begin with (the nr of matrices is equal to the dimension of space-time, i.e 4).. You need to show they are 4*4 and not 3*3, 5*5, 6*6, etc. You can show that starting with these 4, there are 12 more linear independent matrices. Then the dimension of representation follows from the theorem of Schur and Frobenius.
     
    Last edited: Oct 13, 2014
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