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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?

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    $\begingroup$ Have you looked into the book (or related papers) by Willem de Graaf? He has studied computational problems for finite dimensional Lie algebras extensively, though my own experience here is very limited. Anyway, you need to specify more clearly what kind of field you are working over, since prime characteristic often gets more complicated. [Does"finite Lie algebra" in the header mean "finite dimensional Lie algebra"?] $\endgroup$
    – Jim Humphreys
    Commented Nov 10, 2015 at 14:33
  • $\begingroup$ Right, I mean finite dimensional Lie algebra over real or complex numbers. Thank you for reference. I will have a look. $\endgroup$
    – user47116
    Commented Nov 10, 2015 at 15:02
  • $\begingroup$ P.S. This is the book I mentioned: ams.org/mathscinet-getitem?mr=1743970 (he has also written some relevant papers, though I recall that he relies a lot on Groebner bases). $\endgroup$
    – Jim Humphreys
    Commented Nov 10, 2015 at 15:56

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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 $L$.

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    $\begingroup$ Thank you. It's rather interesting that for solvable Lie algebra the question is easier. How do you think: could Levi decomposition be useful in general case? Since, it decomposes Lie algebra into semi-direct product of radical ideal and semi-simple algebra... $\endgroup$
    – user47116
    Commented Nov 11, 2015 at 12:49
  • $\begingroup$ Any abelian ideal of $L=\mathfrak{s}\ltimes {\rm rad}(L)$ lies of course in the solvable radical ${\rm rad}(L)$. This may already be helpful. However, also the algorithm to compute the Levi decomposition has to be taken into account. $\endgroup$
    – Dietrich Burde
    Commented Nov 12, 2015 at 9:34

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