3-dim simple complex Lie algebra

If we know that a 3-dimensional Lie algebra $L$ with $[L,L]=L$ is simple.

How to prove that the only (up to isomorphism) 3-dimensional complex Lie algebra $L$ with $L=[L,L]$ is $sl_2(\mathbb C)$?

Solution 1:

Let $(e_1,e_2,e_3)$ be a basis of $L$. Since $[L,L]=L$, $$ (f_1,f_2,f_3):=([e_2,e_3],[e_3,e_1],[e_1,e_2]) $$ is again a basis of $L$. Hence we may write the $f_i$ as linear combinations of the $e_i$, i.e., $f_i=\sum a_{ij}e_j$, with invertible coefficient matrix $A=(a_{ij})$. One checks that $A$ must be symmetric, and that every invertible symmetric matrix $A$ determines a Lie algebra $L_A$. Then two such Lie algebras $L_A$ and $L_B$ are isomorphic if and only if there is a matrix $M\in GL_3(\mathbb{C})$ such that $$ B=\det(M)(M^{-1})^tAM^{-1}. $$ In other words, $L_a$ and $L_B$ are isomorphic if and only if $A$ and $B$ lie in the same $G$-orbit under the action of $G:=\mathbb{C}^*\times GL_3(\mathbb{C})$ given by $(t,C)A=tCAC^T$ on the space of symmetric $3\times 3$ matrices. By linear algebra (Principal axis theorem), the system of representatives is just the identity. Hence there is only one simple complex Lie algebra of dimension $3$ (we only need that $L$ is perfect, because in dimension $3$ this means simplicity).

Solution 2:

Consider the adjoint representation. Since the algebra is perfect, we have $tr\ ad(x)=0$ for all $x \in L$. Since $ad(x)$ has determinant zero, because $[x,x]=0$, we see that the characteristic polynomial of $ad(x)$ is of the form $f(t) = t^3 + c(x) t$ for some scalar $c(x)$ depending on $x$.

If we had $c(x) = 0$ for all $x$, then $ad(x)$ would be nilpotent for all $x$, By Engel's theorem, then, L is a nilpotent Lie algebra and, in particular, not perfect. This is absurd.

It follows that there exists a nonzero h in L such that $c(x)$ is not zero. This implies that $ad(x)$ is diagonalizable with three distinct eigenvalues. One of them is zero, and $h$ is an eigenvector for zero. Let x and y be eigenvector for the other two eigenvalues. Since $tr \ ad(x)$ is zero, these two eigenvalues are $a$ and $-a$ for some complex number $a$.

Compute now that $[x,y]$ is an eigenvector for $ad(h)$ of eigenvalue zero, so it is a scalar multiple if $h$, say $b, h$. If $b$ is zero, then clearly $[L,L]$ is spanned by $x$ and $y$, which is against our hypothesis. Therefore $b$ is not zero.

The Lie algebra is therefore completely determined by two nonzero scalars with brackets given by $[h,x]=ax, [h,y]=-at, [x,y]=bh$

Now play a bit with rescaling the three elements $x$, $y$ and $h$ to get an isomorphism with $sl_2$: consider the basis $\{\frac{2h}{a}, \frac{2x}{ab}, y\}$.