Cooking up examples of lie algebras - quickly check using strucutre constants

Question

A Lie algebra is is a vector space $L$ equipped with a Lie bracket $[\cdot,\cdot]$.

Given linearly independent vectors $e_1,...,e_n$ and structure constants $[e_i,e_j]=\sum_{k=1}^n a^k_{ij}e_k$ is there a quick way to check if this is a lie algebra

I was trying to show that the sum of two subalgebras need not be a subalgebra and couldn't think of the obvious example of taking an sl-2 tripple. So I tried to define $L=\text{span}(s_1,s_2,t_1,t_2,e_1,e_2)$ by $[s_1,s_2]=s_1$, $[t_1,t_2]=t_1$,$[s_i,t_i]=e_i$. This would give that $S+T$ was not a subalgebra. One can go through the motions of checking all the relations by hand and seeing what we should define the bracket on $e_1,e_2$ but is there a nicer way to ensure that something will be a lie algebra?

Answer 1

There are two questions, it seems. The first one is about checking skew-symmetry and Jacobi identity for given structures constants. For a fixed basis $(x_1,\ldots,x_n)$ of a vector space $V$ a Lie bracket $\mu$ is determined by the point $(c_{ij}^r)\in k^{n^3}$ of structure constants with \begin{equation*} \mu(x_i,x_j) = \sum_{r=1}^n c_{ij}^r x_r \end{equation*} satisfying the polynomial conditions \begin{align*} 0 & =c_{ij}^r + c_{ji}^r, \\ 0 & =\sum_{r=1}^n (c_{ij}^r c_{lr}^s+c_{jk}^r c_{ir}^s+c_{ki}^r c_{jr}^s). \end{align*} for $1\le i<j<k\le n,\; 1\le s\le n$, given by skew-symmetry and Jacobi's identity.

There is no shortcut to checking or solving these equations for the structure constants in general. In special cases there is nothing to check. For example, they are automatically satisfied, if $[e_i,e_j]=e_i\cdot e_j-e_j\cdot e_i$, with a Lie-admissible $K$-bilinear product $V \times V\rightarrow V$.

Reference: What does the Jacobi identity impose on structure constants?

The second question is, whether or not $S+T$, the sum of two subalgebras, is again a subalgebra. This need not be the case. Let $L=\mathfrak{sl}_2(K)$ with standard basis $(e_1,e_2,e_3)$ and $[e_1,e_2]=e_3$, $[e_1,e_3]=-2e_1$, $[e_2,e_3]=2e_2$. Then take $S=\langle e_1\rangle$ and $T=\langle e_2\rangle$ as the abelian $1$-dimensional subalgebras generated by $e_1$ respectively $e_2$. Then $S+T=\langle e_1,e_2\rangle$ is not a Lie subalgebra, since $[e_1,e_2]$ is not in the subspace.