Show this two Lie algebras are isomorphic

Question

Let $\mathfrak{g}_{\alpha}$ be a three dimensional Lie algebra, with basis $\{X,Y,Z\}$, and the Lie bracket is described as $$ \begin{aligned} {[X,Y]}&=Y\\ [X,Z]&=\alpha Z\\ [Y,Z]&=0 \end{aligned} $$

I need to show $\mathfrak{g}_{\alpha}\cong\mathfrak{g}_{\alpha'}$ if and only if $\alpha=\alpha'$ or $\alpha=\frac{1}{\alpha'}$.

Clearly if $\alpha=\alpha'$ or $\alpha=\frac{1}{\alpha'}$ then $\mathfrak{g}_{\alpha}\cong\mathfrak{g}_{\alpha'}$. But for conversely, I choose a Lie algebra isomorphism $\varphi$ and write $$ \varphi(X)=a_{11}X+a_{12}Y+a_{13}Z $$ and so on. I want to use the property $\varphi([X,Y])=[\varphi(X),\varphi(Y)]$ to get the relation about $\alpha$ and $\alpha'$.

But I only get lots of meaningless equations, is there anything wrong with my idea? and is there any easy way to show this?

Thanks in advance!

Answer 1

Write $\varphi([X,Y])=[\varphi(X),\varphi(Y)]$ in matrix form $$ \phi\circ {\rm ad}(X)={\rm Ad}(\phi(x))\circ \phi. $$ In particular we have $$ \phi\circ {\rm ad}(X)\circ \phi^{-1}={\rm Ad}(\phi(x)), $$ so that the operators ${\rm ad}(X)$ and ${\rm Ad}(\phi(x))$ are similar, so they have the same characteristic polynomial, the same trace, the same determinant. These conditions give equations in the parameters of $\phi$ which immediately give $\alpha-\alpha'=0$ or $\alpha\alpha'=1$.

Answer 2

Let $g = aX+bY+cZ$ be a general element. The inner derivation $\text{ad}_g : y\mapsto [g,y]$ is represented, in the basis $(X,Y,Z)$ by : $$\text{ad}_g = \begin{pmatrix} 0&0&0 \\ -b & a & 0 \\ -c\alpha&0&a\alpha \end{pmatrix}$$

Therefore,

• if $\alpha = 0$, the inner derivations of $\mathfrak g_\alpha$ have at most $1$ non-zero eigenvalue.
• if $\alpha\neq 0$, the inner derivations of $\mathfrak g_\alpha$ have zero or two non-zero eigenvalues, whose ratio is $\alpha$ or $1/\alpha$.

This is enough to show that if $\alpha \neq \alpha'$ and $\alpha \neq 1/\alpha'$,we have $\mathfrak g_\alpha \not\simeq \mathfrak g_{\alpha'}$.

Answer 3

The derived algebra of $\mathfrak{g}_{a}$ is two dimensional and commutative. Therefore, the abelianization $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ is $1$-dimensional and acts on $[\mathfrak{g},\mathfrak{g}]$ via operators with eigenvalues with ratio $\alpha$. Hence, $\{\alpha, 1/\alpha\}$ is an invariant.