Isomorphism classes of nilpotent Lie algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T13:10:06Z http://mathoverflow.net/feeds/question/21279 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21279/isomorphism-classes-of-nilpotent-lie-algebras Isomorphism classes of nilpotent Lie algebras senti_today 2010-04-14T00:40:11Z 2010-04-14T14:12:18Z <p>I will begin by stating my question, and then write down some related thoughts.</p> <p>Let \$\mathfrak{g}\$ be a finite dimensional nilpotent Lie algebra over \$\mathbb{C}\$. Choose an ideal \$\mathfrak{h}\$ in \$\mathfrak{g}\$ of codimension 2. The quotient \$\mathfrak{g}/\mathfrak{h}\$ is then abelian. If \$L\$ is any 1-dimensional subspace of \$\mathfrak{g}/\mathfrak{h}\$, we can form its preimage in \$\mathfrak{g}\$, which is a Lie subalgebra.</p> <p>True/False? The set of isomorphism classes of Lie algebras obtained in this way is finite. (Here I am disregarding the embedding into \$\mathfrak{g}\$ and asking about isomorphisms as abstract Lie algebras.)</p> <p>EDIT. To clarify the statement: both \$\mathfrak{g}\$ and \$\mathfrak{h}\$ are fixed. Only the 1-dimensional subspace \$L\$ of the quotient \$\mathfrak{g}/\mathfrak{h}\$ varies.</p> <p>Remark 1. A counterexample, if there is one, could only exist in dimension \$\geq 8\$. This is why it's pretty difficult to "get my hands on" this problem. I don't see an easy way to prove the statement, nor do I see any obvious counterexamples.</p> <p>Remark 1'. I tried to assume that the answer is positive and get some kind of contradiction with the statement that the set of isomorphism classes of nilpotent Lie algebras of dimension \$n\$ (where \$n\geq 7\$) is infinite, but I couldn't find one.</p> <p>Remark 2. This is not very helpful, but at least the answer to my question is positive when \$\mathfrak{h}\$ is abelian. (This does severely limit the possibilities for \$\mathfrak{g}\$, but at least it does not limit the nilpotence class of \$\mathfrak{g}\$.)</p> <p>Any information about this would be much appreciated!</p>