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Point symmetry generators

  1. Mar 2, 2008 #1
    I am taking a first course in Lie algebras and currently working with this problem (see attached file). I understand that the product of the two operators should be regarded as composition. How to explain the final expression?
    Regards
    Staffan
     

    Attached Files:

  2. jcsd
  3. Mar 2, 2008 #2

    HallsofIvy

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    Haven't you tried just doing the computation? That's how you learn mathematics- not just by looking at formula and expecting to understand them, but by actually doing the calculations!

    You are given that
    [tex]Z_j= \xi_j(x,u)\frac{\partial}{\partial x}+ \chi_j(x,u)\frac{\partial}{\partial u}[/tex]
    for j= 1 and 2. In other words, the difference between Z1 and Z2 is the functions multiplying the derivatives.

    If f is any function of x and u (any reason for using x and u instead of x and y?) then
    [tex]Z_1 Z_2(f)= \xi_1(x,u)\frac{\partial}{\partial x}+ \chi_1(x,u)\frac{\partial}{\partial u}[ \xi_2(x,u)\frac{\partial f}{\partial x}+ \chi_2(x,u)\frac{\partial f}{\partial u}][/tex]
    [tex]= \xi_1(x,u)\frac{\partial}{\partial x}[\xi_2(x,u)\frac{\partial f}{\partial x}+ \chi_2(x,u)\frac{\partial f}{\partial u}]+ \chi_j(x,u)\frac{\partial}{\partial u}[\xi_2(x,u)\frac{\partial f}{\partial x}+ \chi_2(x,u)\frac{\partial f}{\partial u}][/tex]
    [tex]= \xi_1\xi_2\frac{\partial^2 f}+ \xi_1 \frac{\partial \xi_2}{\partial x}\frac{\partial f}{\partial x}+ \cdot\cdot\cdot[/tex]

    Finish that, then do the same for [itex]Z_2Z_1[/itex] and subtract. All the second derivative terms (those not involving derivatives of [itex]\xi_1[/itex], [itex]\xi_2[/itex], [itex]\chi_1[/itex], or [itex]\chi_2[/itex]) will cancel leaving only first derivative terms.
     
  4. Mar 2, 2008 #3
    Yes, I *have* calculated but I didn't manage to get the 1st derivatives right. Thanks a lot!
     
  5. Mar 2, 2008 #4

    HallsofIvy

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    Yeah, I hate tedious calculations like that!
     
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