Lie Algebra Structures and the Classification of 2-Dimensional Lie Algebras via wedge product and group action

Let $V$$V$ be a vector space over $k$$k$ with $\dim V=2$$\dim V=2$. We claim that up to isomorphism, there are only two types of Lie algebra structure on $V$$V$.

Let us first consider an algebraic variety. Let $V$$V$ be a finite-dimensional vector space, then the Lie bracket is totally determined by

$$[e_i,e_j]=\sum _{l=1}^n{a}_{i,j}^l{e}_l,a_{i,j}^l\in k,1\le i,j\le n$$

The $a_{i,j}^l\in k,1\le i,j,l\le n$$a_{i,j}^l\in k,\ 1\le i,j,l\le n$, are called the structure constants of $\mathfrak{g}$$\mathfrak g$ relative to the given basis.

The Lie bracket is bilinear and antisymmetric, so it belongs to

$${\mathrm{Hom}}_k(\stackrel{2}{\bigwedge }(V),V)\cong k^{n\cdot (\genfrac{}{}{0ex}{}{n}{2})}$$

In addition, the structure constants have to satisfy the Jacobi identity, which is a series of polynomial equations.

Hence all the structure constants of $\mathfrak{g}$$\mathfrak g$ relative to the given basis form an algebraic variety, which we denote as $\mathrm{Lie}(V)$$\mathrm{Lie}(V)$.

Structure constants depend on the choice of basis. After changing the basis, the same Lie algebra will in general have different structure constants. Therefore, to truly classify Lie algebras, we cannot simply look at the parameters themselves; we must consider whether they are equivalent under change of basis.

Let $\mu \in \mathrm{Lie}(V)\subseteq {\mathrm{Hom}}_k(\stackrel{2}{\bigwedge }(V),V)$$\mu\in \mathrm{Lie}(V)\subseteq \mathrm{Hom}_{k}\left(\bigwedge^2(V),V\right)$, the $GL(V)$$GL(V)$ action is defined by $g\mu \stackrel{2}{\bigwedge }(g^{-1})$$g\mu \bigwedge ^2(g^{-1})$.

i.e.

$$g\cdot \mu (x\wedge y)=g\cdot \mu (g^{-1}x\wedge g^{-1}y)$$

Then

$$g:(V,\mu )\to (V,g\cdot \mu )$$

is a Lie algebra isomorphism since

$$g\cdot [x,y{]}_{\mu }=[g(x),g(y){]}_{g\cdot \mu }$$

So the Lie algebra structure over $V$$V$, up to isomorphism, is given by $\mathrm{Lie}(V)/GL(V)$$\mathrm{Lie}(V)/{GL(V)}$.

Now let us consider $\dim V=2$$\dim V=2$.

Then $\mathrm{Lie}(V)={\mathrm{Hom}}_k(\stackrel{2}{\bigwedge }(V),V)$$\mathrm{Lie}(V)=\mathrm{Hom}_{k}\left(\bigwedge^2(V),V\right)$ since the Jacobi identity is a map from $\stackrel{3}{\bigwedge }(V)$$\bigwedge^3(V)$ to $V$$V$, but $\stackrel{3}{\bigwedge }(V)=0$$\bigwedge^3(V)=0$.

Now since $\stackrel{2}{\bigwedge }(g^{-1})=(\det g{)}^{-1}$$\bigwedge^2(g^{-1})=(\det g)^{-1}$, hence $g\cdot \mu$$g\cdot\mu$ is just $\det {\left(g\right)}^{-1}g\mu$$\det(g)^{-1} g\mu$.

$$\stackrel{2}{\bigwedge }(V)\cong k$$

hence

$$\mathrm{Lie}(V)\cong V$$

The action $\mu \to \det {\left(g\right)}^{-1}g\mu$$\mu\to \det(g)^{-1} g\mu$ has only two orbits: $\{0\}$$\{0\}$ and $V-\{0\}$$V-\{0\}$.

That is the classification of $2$$2$-dimensional Lie algebras.