Lie algebras of same dimension are isomorphic

Question

I just began studying Lie algebras and I'm trying to prove that two Lie algebras $\frak g$ and $\frak h$ are isomorphic if and only if they have the same dimension.

If they're isomorphic, then proving that they have the same dimension goes just like for vector spaces, I guess.

Now, if $\{g_i\}_{i \in I}$ and $\{h_i\}_{i \in I}$ are bases for $\frak g$ and $\frak h$, we can define a linear map $T\colon {\frak g}\to {\frak h}$ by $Tg_i = h_i$, for all $i \in I$, and extending. But then we should check that $T$ also preserves commutators, so $T$ will be a morphism of Lie algebras.

If $[g_i,g_j] = \sum_k c_{ij}^kg_k$ and $[h_i,h_j]= \sum_k d_{ij}^kh_k$, writing $x= \sum_i a_i g_i$ and $y = \sum_j b_jg_j$, I got: $$ T[x,y] = \sum_{i,j,k}a_ib_jc_{ij}^k h_k, \qquad [Tx,Ty] = \sum_{i,j,k} a_ib_j d_{ij}^k h_k. $$How can I prove that $c_{ij}^k = d_{ij}^k$? I guess I should use Jacobi's identity, but I'm not sure of how exactly doing this. Help?

Answer 1

It is not true. Take the set of all $6\times6$ matrices with usual bracket operation $[A,B]=AB-BA$. This is 36-dimensional Lie algebra.

Now look at upper triangular matrices of size $9\times 9$, under the 'same' operation. This is of dimension $9\times8/2=36$. They are not isomorphic. (Second one is solvable, whereas the first is not). Of course it is much easier to give example with one of them with trivial Lie algebra structure.

Answer 2

The minimal counterexample is of dimension $2$. Every $2$-dimensional Lie algebra over a field $K$ is either isomorphic to the abelian Lie algebra $K^2$, or to the solvable, non-abelian Lie algebra $\mathbb{r}_2(K)$, with basis $(x,y)$ and Lie bracket $[x,y]=x$. These two Lie algebras cannot be isomorphic for various reasons. The most obvious property is that $K^2$ has zero brackets, and $\mathbb{r}_2(K)$ has not. Another invariant is that $\dim Der(K^2))=4$ whereas $\dim Der(\mathbb{r}_2(K))=2$ for the derivation algebras. A third argument is that $L=\mathbb{r}_2(K)$ is not nilpotent. Indeed $[L,[L,L]]=[L,L]$ stabilizes. Of course, $K^2$ is nilpotent.