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There are two Lie algebra (up to isomorphism) of dimension two. One is abelian and other is as follows: $$ L=\text{span}\{x,y\}, [x,y]\neq 0, \ \text{say, } z .$$ Now $[x,y]=\alpha x+\beta y$ where $\alpha$ and $\beta $ are not simultaneously zero. Let $\beta\neq 0.$ Then \begin{align*} [x,z] &= [x,\alpha x+\beta y]=\frac{1}{\beta}[x,y]=\frac{1}{\beta}z. \end{align*} So this Lie algebra is isomorphic to the algebra with basis $\{h,e\}$ and whose bracket is characterized by $[h,e]=e$. I was reading an article Lie algebra of dimension 1,2 and 3, in which I found that the isomorphism is given as $$ x\mapsto \beta h,\ \text{and}\ y\mapsto -\alpha h+\frac{1}{\beta}e. $$ Now I am unable to get that how this isomorphism is coming, is it just by inspection or is there any way to get it.

Thanks.

Travis Willse
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Sachchidanand Prasad
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  • Perhaps it's obvious, but it's maybe worth mentioning why we can assume $\beta \neq 0$: In the case $\beta = 0$, we can interchange the roles of $x$ and $y$ to get $[x, y] = -\alpha y$. Now, the coefficient of $y$ cannot be zero (because in that case $[x, y] = 0$, that is, the Lie algebra is abelian). – Travis Willse Feb 08 '17 at 15:54

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Hint One can proceed naively by viewing both Lie algebras as defined on the same underlying vector space, $\Bbb F^2$. Then, we look for a linear transformation $\phi : \Bbb F^2 \to \Bbb F^2$ that is a Lie algebra isomorphism. Fixing bases---the natural ones are $(x, y)$ and $(h, e)$ for the source and target, respectively---determines a matrix representation $$[\phi] = \pmatrix{A&B\\C&D}$$ of $\phi$.

Now, since $\phi$ should be a Lie algebra isomorphism, we must have $$\phi([x, y]) = [\phi(x), \phi(y)]$$ (and on the other hand, this is the only additional requirement that a linear isomorphism $\phi$ must satisfy to be one).

On one hand, we have $$\begin{align*}\phi([x, y]) &= \phi(\alpha x + \beta y) \\ &= \alpha\phi(x) + \beta\phi(y) \\ & = \alpha (Ah + Ce) + \beta(Bh + De) \\& = (\alpha A + \beta B) h + (\alpha C + \beta D) e . \end{align*}$$ On the other, we have $$[\phi(x), \phi(y)] = [Ah + Ce, Bh + De] = (AD - BC) e .$$ Comparing coefficients then gives the quadratic system $$\left\{\begin{array}{rcl} \alpha A + \beta B &=& 0 \\ \alpha C + \beta D &=& AD - BC \end{array}\right. .$$ One solution is $$ A = \beta, \quad B = -\alpha, \quad C = 0, \quad D = \beta^{-1} , $$ and substituting yields the given isomorphism $\phi$. This choice is distinguished by the properties that it both (1) maps $x$ to some multiple of $e$ (equivalently, it is upper triangular w.r.t. the bases $(x, y)$ and $(e, h)$) and (2) preserves volume in the sense that $(\det \phi)(x \wedge y) = e \wedge h$ (equivalently, that $\det [\phi] = 1$).

Travis Willse
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