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The question of classification of these two types of Lie algebras has been treated here: Classsifying 1- and 2- dimensional Algebras, up to Isomorphism. What I am trying to understand is how they found what Lie algebras they are isomorphic to.

For the one-dimensional ones, since we easily find they are Abelian, it sounds fair to me to think of them as $\mathbb{R}$, one-dimensional, Abelian, everything corresponds. Why then express it in terms of $2\times 2$ matrices?

For Abelian two-dimensional ones, again, $\mathbb{R}^2$ sounds like an intuitive choice. However, I have absolutely no idea how one would think to find the last category. Also, has the equality $[X,Y]=Y$ a particular meaning?

Spida
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1 Answers1

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Abstract classification

A Lie algebra $\mathfrak{g}$ is a vector space (over some field $k$) with a bilinear pairing $[\cdot, \cdot]\colon \mathfrak{g}\times\mathfrak{g}\to \mathfrak{g}$ called a bracket, which satisfies

  • $[x,x]=0$ for $x\in \mathfrak{g}$
  • $[x,[y,z]]+[z,[x,y]]+[y,[z,x]]=0$ for $x,y,z\in \mathfrak{g}$

If the dimension of $\mathfrak{g}$ is $n$, then as a vector space this is just $k^n$. So the question is really what brackets can we define on a $1$ and $2$ dimensional vector space?

n=1. For $x,y\in\mathfrak{g}$ we have $cx=y$ for some $c\in k$ and hence $[x,y]=[x,cx]=c[x,x]=0$. In particular, the only bracket one can put on a 1 dimensional vector space is the trivial one: $[x,y]=0$ for all $x,y\in \mathfrak{g}$

n=2. Let $e_1,e_2$ be a basis for this vector space. Carrying out the same computation one finds $$[ae_1+be_2,ce_1+de_2]=(ad-bc)[e_1,e_2].$$ If $[e_1,e_2]=0$ we get the trivial bracket, and if it is non-zero we may, after a change of basis we may assume that $[e_1,e_2]=e_2$. The bracket is fully determined by this and it is easy to check that it satisfies condition 2 above and hence form a valid Lie algebra.

Also, has the equality [,]= a particular meaning?

When we view these as abstract Lie algebras, no, not that I know of. It is algebraically one possible choice of bracket. If you want to interpret this as vector fields on a Lie group, then you might find a geometric interpretation somewhere.

Representations: If you are happy with the fact that there is a single 1-dimensional and two 2-dimensional Lie algebras up to isomorphism then the above is a fine classification. As I understand your question, you want to find a realization as a Lie subalgebra of $\mathfrak{gl}_2$ of $2\times 2$ matrices with the usual commutator bracket.

Warning: It is not true in general that a $n$-dimensional Lie algebra can be realized as a Lie subalgebra of $\mathfrak{gl}_n$.

For the abelian ones, any 1 and 2-dimensional sub vector space of commuting $2\times 2$ matrices will be isomorphic as a Lie algebra to the abelian ones we found (the bracket is zero on both of them so any linear isomorphism is a lie algebra isomorphism).

  • For the $1$-dimensional one, there are many choices, any one-dimensional subspace of the diagonal matrices will do.
  • For the $2$-dimensional one, note that diagonal matrices commute, so the two-dimensional subspace of diagonal matrices will suffice.

For the non-abelian one, let us denote it by $\mathfrak{g}$, we only need to find non-zero matrices $A_1,A_2$ such that $[A_1,A_2]=A_1A_2-A_2A_1=A_2$. Then the subalgebra $\mathfrak{h}$ of $\mathfrak{gl}_2$ spanned by these is isomorphic to $\mathfrak{g}$ via the map $\mathfrak{g}\to\mathfrak{h}$ where $e_i\mapsto A_i$. One choice here is

$$A_1=\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}, ~~ A_2=\begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}.$$

You can find this by trial and error, or try to find a solution to the 4 equations you get from $[A_1,A_2]=A_2$.

Smn
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