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Homework Help: Lie Bracket

  1. Mar 8, 2006 #1
    Could someone check if I have done this right.

    [tex]R_1 = x^2\partial_3 - x^3\partial_2[/tex]
    [tex]R_2 = x^3\partial_1 - x^1\partial_3[/tex]
    [tex]R_3 = x^1\partial_2 - x^2\partial_1[/tex]

    Where [itex]x^i[/itex] are coordinates.

    I need to calculate the commutator [itex][R_1,R_2][/itex].

    [tex][R_1,R_2] = x^2\partial_3x^3\partial_1 - x^3\partial_1x^2\partial_3 + x^3\partial_2x^1\partial_3 - x^1\partial_3x^3\partial_2
    - (x^3\partial_2x^3\partial_1-x^3\partial_1x^3\partial_2) - (x^2\partial_3x^1\partial_3-x^1\partial_3x^2\partial_3)
    .\quad\quad\quad = x^2x^3\partial_3\partial_1 - x^2x^3\partial_1\partial_3 + x^1x^3\partial_2\partial_3-x^1x^3\partial_2\partial_3
    - x^3x^3\partial_2\partial_1 + x^3x^3\partial_1\partial_2-x^1x^2\partial_3\partial_3+x^1x^2\partial_3\partial_3
    [tex].\quad\quad\quad = x^2\partial_1-x^2\partial_1+x^1\partial_2-x^1\partial_2 -x^3x^3\partial_2\partial_1 + x^3x^3\partial_1\partial_2 - x^1x^2\partial_3\partial_3 + x^1x^2\partial_3\partial_3
    .\quad\quad\quad = 0

    And as a result,

    [tex][R_1,R_2] = [R_2,R_3] = [R_3,R_1] = 0[/tex]

    by cyclically permuting the indices.
    Last edited: Mar 8, 2006
  2. jcsd
  3. Mar 9, 2006 #2


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    Homework Helper

    I think you're wrong...Check the angular momentum so(3) algebra...

  4. Mar 9, 2006 #3
    You're right. But I dont know where I went wrong. I would have expected

    [tex][R_1,R_2] = R_3[/itex]
    Last edited: Mar 9, 2006
  5. Mar 9, 2006 #4
    I think my problem is evaluating the differentials. Here is an example of what Im doing in a simpler problem.

    let [itex]A=\partial_x[/itex], [itex]B=\partial_y[/itex], and [itex]C=x\partial_y-y\partial_x[/itex].

    Im going to compute [itex][A,C][/itex] and let me know if Ive done something wrong...

    [tex][A,C] = [\partial_x,x\partial_y-y\partial_x][/tex]
    [tex]= \partial_x(x\partial_y - y\partial_x)-(x\partial_y-y\partial_x)\partial_x[/tex]
    [tex]= \partial_xx\partial_y-\partial_xy\partial_x - x\partial_y\partial_x+y\partial_x\partial_x[/tex]

    Now Im pretty sure [itex]\partial_xx = 1[/itex] so the first term is [itex]\partial_y[/itex].

    The second term [itex]\partial_xy\partial_x = \partial_x\partial_xy[/itex] since the partial derivatives commute. The second term equals the fourth term so they cancel.

    This leaves the third term. [itex]x\partial_y\partial_x=\partial_y(\partial_xx) = \partial_y(1) = 0[/itex]

    So I would say the answer is

    [tex][A,C] = \partial_y = B[/itex]

    how does this look?
    Last edited: Mar 9, 2006
  6. Mar 9, 2006 #5
    To be honest I dont think Ive done this right at all.

    My basic question is what does


    equal? And is it the same thing as


    EDIT: Could

    [tex]\partial_yy\partial_x = \partial_y(y)\partial_x + y\partial_x\partial_y[/tex]?

    If this calculation is right could someone tell me how.
    Last edited: Mar 9, 2006
  7. Mar 9, 2006 #6
    What [tex]\partial_yy\partial_x[/tex] means depends on how you are applying the [tex]\partial_y[/tex]. From a commutator, for example:
    [tex] [\partial_y, y\partial_x] = \partial_y(y\partial_x)-y\partial_x(\partial_y)[/tex]
    In this case the y operator acts on the whole expression so you use the product rule.

  8. Mar 14, 2006 #7
    Ah, of course! So in fact

    [tex][R_1,R_2] = -R_3[/tex]


    [tex][R_2,R_3] = -R_1[/tex]
    [tex][R_3,R_1] = -R_2[/tex]
    Last edited: Mar 14, 2006
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