Making sense of this seemingly circular argument about $2$-dimensional Lie algebras

Question

From Introduction to Lie Algebras by Humphreys:

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This argument is somewhat confusing as the author is replacing basis elements but keeping the same notation for basis elements throughout the argument.

When $x$ is replaced in the basis, is the bracket being redefined with a "new" $x$?

He mentions "original $[xy]$". Does this imply the latter $[xy]$ are not original and are redefined?

How is arriving at $[xy]=ax$?

How is he arriving at $[xy]=x$? Is the $x$ inside the bracket different than the $x$ outside the bracket? Is the $x$ inside the bracket the new $x$ and the $x$ outside the bracket the old $x$?

Can someone clarify this argument?

Answer 1

In dimension $2$, start with a basis $(x,y)$ of $L$, and suppose $[x,y] \ne \mathbf{0}$.

For any two vectors $v, w \in L$, we can write $v=ax+by$ and $w=cx+dy$. So, by the bi-linearity and alternation in the definition of a Lie algebra bracket: $$[v,w]=[ax+by, cx+dy]=[ax, dy]+[by,cx]=ad[x,y]-bc[x,y]=(ad-bc)[x,y].$$

So, all products in $L$ yield scalar multiples of $[x,y]$.

If $[x,y]$ is not the zero vector of $L$, we can choose a new basis, for example: $([x,y], y')$ for some $y' \notin span\{[x,y]\}$.

Note that we are not redefining the bracket operation. It is already defined independent of any bases. However, computing the bracket is going to be easiest when we know what it does to basis vectors (since the bracket is bi-linear and alternating). So, let's look at what the bracket does on these two new basis vectors.

Since for any two vectors $v,w \in L$, we know $[v,w]=k[x,y]$ for some scalar $k$, then on the new basis: $$[[x,y], y']=a[x,y], \quad \text{for some scalar } a \ne 0.$$

Now, again, choose a new basis, $([x,y], (1/a)y')$, so that finally we get: $$[[x,y], (1/a)y']=(1/a)[[x,y], y']=[x,y].$$

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Basically, what this says is that if you have a two dimensional non-abelian Lie algebra, you can find a basis (for instance, by following the procedure) so that the bracket operation on that basis is nice and simple, i.e., you can find a basis $(x,y)$ so that $[x,y]=x$.

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Since this procedure can be done with any two dimensional non-abelian Lie algebra, then any two non-abelian two dimensional Lie algebras will be isomorphic:

If $L$ and $L'$ are non-abelian two dimensional Lie algebras, by the above, we can find a basis $(x,y)$ of $L$ with bracket $[xy]=x$, and we can find a basis $(x',y')$ of $L'$ with bracket $[x'y']=x'$. Then the linear map defined on basis elements $x \mapsto x'$, $y \mapsto y'$ will be a Lie algebra isomorphism.

Answer 2

I prefer to write $[x,y]$ instead of $[xy]$. More precisely, in the basis $([x,y],y)$, one can write $[xy]=ax+by$ and $[[x,y],y]=[ax+by,y]=a[x,y]$.

This implies that in the basis $[x,y],a^{-1}y$,

$[[x,y],a^{-1}y]=[ax+by,a^{-1}y]=[ax,a^{-1}y]=[x,y]$.

We have defined a linear map $f(x)=[x,y], f(y)=a^{-1}(y)$ with defines an isomorphism between the original Lie algebra and an algebra generated by $u,v$ with $[u,v]=u$.

Answer 3

Since the Lie algebra is $2$-dimensional, $[x,y]$ must be a linear combination of the basis $(x,y)$, i.e., we can write $$ [x,y]=ax+by. $$ If $(a, b)\neq (0,0)$, then we may apply a Lie algebra isomorphism so that $[[x,y]]=x$ for the new, but isomorphic Lie bracket $([[, ]]$ on the vector space.