Homework Help: All possible inequivalent Lie algebras

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1. Dec 22, 2017

Azure Ace

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. Dec 22, 2017

Staff: Mentor

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.

3. Dec 22, 2017

Azure Ace

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 $\frac{1}{\alpha}[X,Y] = X + \frac{\beta}{\alpha}Y$?

4. Dec 22, 2017

Staff: Mentor

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 $1$?

5. Dec 22, 2017

Azure Ace

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
6. Dec 22, 2017

Staff: Mentor

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.

7. Dec 22, 2017

Azure Ace

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!! ^_^