Seeking for abelian subalgebra of fixed dimension in finite Lie algebra The problem is: I want to know if there is abelian subalgebra of dimension $k$ in Lie algebra of dimension $n$. My Lie algebra is given by its structure constant table. There are some algorithms around, like version of method of undefined coefficients. It reduces the problem to system of polynomial equations and then to computation of Groebner Bases. However, it seems to be not very efficient, because complexity of algorithm for computing Groebner Bases is very high...
How do you think: is there any more thin method? Probably, some more advanced things from theory of Lie algebras?
 A: If $L$ is a solvable Lie algebra over an algebraically closed field of characteristic zero, then the two invariants
$$
\alpha(L)  = \max \{\dim (\mathfrak{a}) \mid \mathfrak{a} \text{ is an abelian subalgebra of }L\},\\
\beta(L)  = \max \{\dim (\mathfrak{b}) \mid \mathfrak{b} \text{ is an abelian ideal of }L\}.
$$
coincide, see Proposition $2.6$ here. For an ideal, the question is easier to decide than for a subalgebra. However in general, 
to decide whether or not there is an abelian subalgebra of given dimension in $L$
the algorithms mentioned use Gröbner bases in one or the other way. 
On the other hand, for semisimple Lie algebras the maximal dimension of an abelian subalgebra is known, see the article "Abelian ideals in a Borel subalgebra of a complex simple Lie algebra" by R. Suter (2004) and the paper cited above. There it is also proved that the maximal dimension of an abelian ideal in the standard Borel subalgebra $B$ of a simple Lie algebra $L$  coincides with the
maximal dimension of a commutative subalgebra of $L$.
