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SO(2,1) invariance algebra

  1. Nov 5, 2014 #1
    Hello,
    Please excuse me about my ignorance.
    I would like to know how SO(2,1) Lie algebra, is derived from operators and commutators.
    I have some notes, that the Lie algebra of SO(2,1) is derived from:
    [D,H]=-iH
    [K,D]=-iK
    [H,K]=2iD
    where D, H, and K are the "generators".
    I have no clue what does the word "generators" mean, nor how commutators derive an algebra.
    Please let me know.
    Thank you,
    Askalot.
     
  2. jcsd
  3. Nov 5, 2014 #2

    ShayanJ

    User Avatar
    Gold Member

    Imagine a member of SO(2,1) (M) which is very close to identity. So you can taylor expand it:
    [itex]
    M \approx I+\varepsilon^\mu L_\mu
    [/itex]
    Where I have used the Einstein summation convention and [itex] \mu [/itex] runs through the dimensions of the lie group. The operators(marices) [itex] L_\mu [/itex] are called the generators.
    So just find out the dimensions of your lie group, then consider all its parameters to be infinitesimals and then expand w.r.t. those parameters to first order. The operator multiplying the infinitesimal corresponding to a dimension, is the generator in that dimension.
    Then you should only determine the commutators of these operators.
     
  4. Nov 5, 2014 #3
    I found in some notes, that H=Hamiltonian= \frac{1}{2} \nabla^2 + \upsilon\delta(r) and it does not contain dimensional parameters,
    D = dilation operator = tH - (rp + pr) and $\frac{dD}{dt} = 0$, and
    $K = -t^2H + 2tD + \frac{1}{2}r^2$ and $\frac{dK}{dt} = 0$.
    It states that the invariance algebra, following from the above is SO(2,1).

    Still I cannot understand why and how, were these operators D, K defined and also, how you can derive SO(2,1) out of them!!
    Thank you for your time,
    Askalot.
     
  5. Nov 5, 2014 #4

    samalkhaiat

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    Science Advisor

    See this post
    https://www.physicsforums.com/threads/why-is-lorentz-group-in-3d-sl-2-r.764072/#post-4815039
     
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