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Homework Help: Very confused by Lie Algrbera

  1. Feb 6, 2010 #1
    What is the definition of the lie algebra [itex]\mathfrak{so}(2,1)[/itex]?

    i want to say that it is the set

    [itex]\{ X \in GL(3,\mathbb{R}) | X^t=-X \text{ and } \text{trace}(X)=0 \}[/itex]

    is that true?
     
  2. jcsd
  3. Feb 6, 2010 #2

    Dick

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    You derive the properties of so(2,1) from the definition of SO(2,1), as in your other post. a is in SO(2,1) if a^T(eta)a=eta, right? Set a(t)=exp(t*X) and differentiate at t=0 to get the properties of X. X^T=(-X) isn't quite right.
     
    Last edited: Feb 6, 2010
  4. Feb 6, 2010 #3
    i get

    [itex]\frac{d}{dt} \left( e^{t X^T} \eta e^{tX} \right)_{t=0}= \left(X^T e^{tX^T} \eta e^{tX} + e^{tX^T} \eta X e^{tX} \right)_{t=0} = X^T \eta + \eta X[/itex]

    and since [itex]\frac{d}{dt} \left( \eta \right)_{t=0}[/itex]

    we get

    [itex] X^T \eta + \eta X = 0 \Rightarrow X = \eta^{-1} X^T \eta[/itex]

    which would mean

    [itex] \mathfrak{so}(2,1)= \{ X \in \text{ Mat}(2,\mathbb{R}) | \text{ trace}(X)=0 \quad \text{and } X=-\eta^{-1} X^T \eta[/itex]

    is that right?

    how would one go about finding a basis for [itex]\mathfrak{so}(2,1)[/itex] that satisifes [itex] [T_i , T_j]= \epsilon_{ijk} T_k[/itex]?
     
  5. Feb 6, 2010 #4

    Dick

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    That looks right. I haven't looked at Lie algebra stuff for a while. But since so(2,1) is only 3 dimensional, I'd look for an explicit basis and see what the commutators look like, to start.
     
  6. Feb 6, 2010 #5
    so the condition [itex]X=-\eta^{-1} X^T \eta[/itex] tells us [itex]X_{11}=X_{22}=X_{33}=0[/itex] as well as [itex]X_{12}=-X_{21},X_{13}=X_{31},X_{23}=X_{32}[/itex]

    so we could have something like

    [itex]T_1= \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right), T_2=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{array} \right), T_3=\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right)[/itex]

    these work except some of the commutator relationships have factors of -1 in front of them that i can't get rid of.

    e.g. [itex][T_2,T_3]=-T_1[/itex]. swapping the entry in the matrix that is negative doesn't make a difference (as it shouldn't since the basis isn't affected by scalar multiplication). any ideas?
    thanks.
     
  7. Feb 6, 2010 #6

    Dick

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    Not really. Like I said, this isn't something I do everyday. I think you've got the Lie algebra right. You've got one rotation generator T1 and two boost generators T2 and T3. Do you have to go the complexification of the Lie algebra to get the epsilon in the commutation relation? Don't know offhand.
     
  8. Feb 6, 2010 #7
    i assume my T1,T2,T3 are ok as they satisfy the properties required, right?

    sorry but i don't understand what you mean by the complexification of the lie algebra to get an epsilon?
     
  9. Feb 6, 2010 #8

    Dick

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    They look ok to me. By complexification I mean multiplying some of your generators by 'i'.
     
  10. Feb 6, 2010 #9
    yes but regardless of whether i complexify 1,2 or all 3 of them, that means the commutation relation will have to change to something like

    [itex][T_l,T_m]=i \epsilon_{lmn}T_n[/itex] and i don't want that factor of i at the front.
     
  11. Feb 6, 2010 #10
    actually, scratch that i multiplied T_1 and T_2 by i and left T_3 alone and it's working fine. thanks for the tip!
     
  12. Feb 6, 2010 #11

    Dick

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    Hmm. Really? I had to put i's on T2 and T3 and flip their order. But you've got the idea.
     
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