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Homework Help: All possible inequivalent Lie algebras

  1. Dec 22, 2017 #1
    1. The problem statement, all variables and given/known data
    How can you find all inequivalent (non-isomorphic) 2D Lie algebras just by an analysis of the commutator?

    2. Relevant equations
    $$[X,Y] = \alpha X + \beta Y$$

    3. The attempt at a solution
    I considered three cases: ##\alpha = \beta \neq 0, \alpha = 0## or ##\beta = 0, \alpha = \beta = 0##. I found that there can only be one 2D Lie algebra with commutator ##[X,Y] = X##. I'm not sure how to arrive at this answer. Any help is appreciated T_T. Thanks!!!!
     
  2. jcsd
  3. Dec 22, 2017 #2

    fresh_42

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    Where did the Abelian case get lost?
    Now, say w.l.o.g. that ##\alpha \neq 0##. Then we have ##[X,\frac{1}{\alpha}Y]=X+\frac{\beta}{\alpha}Y## or ##[X,Y]=X+\gamma Y##. Can you find an isomorphism to ##[U,V]=V\,?## I switched the names of the basis vectors here, to avoid confusion between the two copies of the Lie algebra.
     
  4. Dec 22, 2017 #3
    Sorry, I still fail to see how the the Abelian case was gone and I'm a little bit new when it comes to this so I'm a little bit slow. Could you explain how you arrived at that equation? Is it something like ##\frac{1}{\alpha}[X,Y] = X + \frac{\beta}{\alpha}Y##?
     
  5. Dec 22, 2017 #4

    fresh_42

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    We have ##[X,Y]=\alpha X + \beta Y## and assumed ##\alpha \neq 0##. Thus we can divide the equation by ##\alpha## and get ##\frac{1}{\alpha }[X,Y]= [X,\frac{1}{\alpha }Y]=1\cdot X + \frac{\beta}{\alpha }Y##. With ##Y':=\frac{1}{\alpha}Y## this means ##[X,Y']=X+\beta Y'##. I didn't do this calculation, so I didn't bother the coefficient and wrote ##\gamma ## instead of ##\beta ## and ##Y## again instead of ##Y'##.

    Now which multiplication do you get for ##U=X## and ##V=X+\beta Y'## and how can we turn the factor ##\beta ## into a factor ##1##?
     
  6. Dec 22, 2017 #5
    One last set of questions (I think). Can we really just ignore ##\alpha = \beta = 0##? To me, it seems that ##\alpha \neq 0## is more of an initial assumption. Also, can you clarify what do you mean by "which multiplication"? Do you mean what constant should I multiply to make the factor ##\beta## to 1? Or is this something like if ##V=X+\beta Y'##
    $$[V - \beta Y', Y'] = [V,Y] - \beta[Y', Y'] = V$$ $$[V,Y'] = V$$

    Thank you very much for your help and patience ^_^
     
    Last edited by a moderator: Dec 22, 2017
  7. Dec 22, 2017 #6

    fresh_42

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    No, we can't. This leads to the Abelian solution. There are two two dimensional Lie algebras.
    There are all four cases you have listed. ##\alpha = \beta = 0## (Abelian Lie algebra) and all other three cases ##\alpha \neq 0 = \beta\; , \;\alpha =0 \neq \beta\; , \;\alpha \neq 0 \neq \beta## should lead to isomorphic copies of ##\mathfrak{g}=\langle X,Y\,\vert \,[X,Y]=Y\rangle##, the two dimensional non-Abelian Lie algebra.
    ##\{X,Y\}## is a basis. Now basis transformations don't change the Lie algebra, only the way we write it down. We arrived at ##[X,Y']=X+\beta Y'## by setting ##Y'= \frac{1}{\alpha}Y##, which is simply another basis. Now let ##Y''=V=X+\beta Y'##. Then ##[X,Y'']=\ldots =\beta Y''##. Here we have again two possibilities. The Abelian solution for ##\beta =0## and with which substitution ##X'=\ldots## do we get ##[X',Y'']=Y''\,?## By multiplication I meant these Lie multiplications.

    Until now we have made only basis transformations within a given Lie algebra. What is left, is the case ##[X,Y]=\alpha X \neq 0## with ##\beta = 0##. Which transformations lead from there to ##[U,V]=V## or if you like it better ##[X',Y']=Y'\,?## The last multiplication is the one we want to arrive at.
     
  8. Dec 22, 2017 #7
    OOOHHH!!!! Now I see!! Thank you very much!! I understand it clearly now. Actually, at first I thought there could only be one possible 2D Lie algebra (from what I saw from Wikipedia). Either it was wrong and misleading or I misunderstood, but I am more convinced that there should be an Abelian case. Your answer was more than enough. Thank you very much!! ^_^
     
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