I am interesting to know that if a finite dimensional Lie algebra $L$ has two bases $\beta_1$ and $\beta_2$, how can we compare the cardinal of two sets $\{(x,y)\in \beta_1\times \beta_1~|~[x,y]=0\}$ and $\{(x,y)\in \beta_2\times \beta_2~|~[x,y]=0\}$?
Thanks for any comments and suggestions.
Takjk
These cardinalities are not independent of the base, i.e., they can be quite different. For example, consider the Heisenberg Lie algebra in the basis $(e_1,e_2,e_3)$ with $[e_1,e_2]=e_3$. Then your cardinality is $7$ in this base, since $$ [e_1,e_1]=[e_2,e_2]=[e_3,e_3]=[e_1,e_3]=[e_2,e_3]=[e_3,e_1]=[e_3,e_2]=0. $$ Only $2$ brackets in this base are nonzero, namely $[e_1,e_2]$ and $[e_2,e_1]$. Now consider the new basis $f_1=e_1$, $f_2=e_2$ and $f_3=e_2+e_3$. In this base the cardinality is only $5$, namely $$ [f_1,f_1]=[f_2,f_2]=[f_3,f_3]=[f_2,f_3]=[f_3,f_2]=0. $$ Note that $[f_1,f_3]=f_3-f_2$. For this reason one usually considers invariants in Lie theory, which are base-independent, like the dimension of the center, the annihilator, or the derived algebra.
