I am reading Humphrey's book on Lie algebras and now the construction of Lie algebras by generators and relations was just presented.
We simply get a set $X= \{\hat x_i, \hat y_i, \hat h_i: 1\leq i \leq l\}$ and a subset $R$ of the free lie algebra $\hat L$ generated by $X$, given by
$R = \{[\hat h_i, \hat h_j], [\hat x_i, \hat y_j] - \delta_{ij}\hat h_i, [\hat h_i, \hat x_j]-c_{ji}\hat x_j,[\hat h_i,\hat y_j] + c_{ji}\hat y_j$}
and then consider $\hat K$ the ideal generated by $R$.
The lie algebra generated by $X$ with relations $R$ is then the quocient $L_0 = \hat L/\hat K.$ In the book Humphreys states that $L_0$ is usually infinite dimensional.
I wonder why is that. Why usually dim $L_0 = \infty$ ?I tried to see it by finding explicitly its root system, but with no sucess.
As said, "usually" is too vague. It really depends on the relations. For example, the Lie algebra generated by $e_1$ and $e_2$, and Lie brackets $$ [e_1,e_i]=e_{i+1}, \forall\; i\ge 1,\; [e_2,e_3]=e_5,\; [e_2,e_5]=re_7 $$ is infinite-dimensional for $r=1$ and $r=9/10$, but $11$-dimensional for $r\neq 1,9/10$, i.e., with $e_i=0$ for $i\ge 12$ ( it is a nilpotent Lie algebra). So it is "usually" a finite-dimensional Lie algebra, and not an infinite-dimensional one.