I will begin by stating my question, and then write down some related thoughts.

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.

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.)

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.

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.

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.

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 $\mathfrak{g}$.)

Any information about this would be much appreciated!

  • 1
    $\begingroup$ By "obtained in this way" I assume you mean that you want: For fixed nilpotent $\mathfrak g$, the number of isomorphism classes of subalgebras that are the preimages of one-dimensional subalgebras of a two-dimensional quotient algebra. But it's not clear to me what is fixed and what is varying. $\endgroup$ – Theo Johnson-Freyd Apr 14 '10 at 5:50
  • $\begingroup$ Thanks for pointing this out; I added a comment right after the question to clarify the issue. Both $\mathfrak{g}$ and $\mathfrak{h}$ are fixed. $\endgroup$ – senti_today Apr 14 '10 at 14:13
  • $\begingroup$ If I am not mistaken, if we did not fix $\mathfrak{h}$ and the answer to my question were affirmative, then one could show that the set of isomorphism classes of all Lie subalgebras of $\mathfrak{g}$ is finite. In turn, this would imply that the collection of isomorphism classes of all finite dimensional nilpotent Lie algebras over $\mathbb{C}$ is at most countable. However, this is known to not be the case. $\endgroup$ – senti_today Apr 14 '10 at 14:35

I have found a counterexample while studying Lie algebra degenerations. Consider the following filiform nilpotent Lie algebra $\mathfrak{g}$ of dimension $13$, given by the brackets with respect to a basis $(e_1,\ldots ,e_{13})$: \begin{align*} [e_1,e_i] & = e_{i+1},\quad 2\le i\le 12 \\[0.2cm] [e_2,e_3] & = e_5 \\ [e_2,e_4] & = e_6 \\ [e_2,e_5] & = \frac{9}{10}e_7-e_9 \\ [e_2,e_6] & = \frac{4}{5}e_8-2e_{10} \\ [e_2,e_7] & = \frac{5}{7}e_9-\frac{335}{126}e_{11}+ \frac{22105}{15246}e_{13}\\ [e_2,e_8] & = \frac{9}{14}e_{10}-\frac{125}{42}e_{12}\\ [e_2,e_9] & = \frac{9}{12}e_{11}-\frac{4421}{1452}e_{13}\\ [e_2,e_{10}] & = \frac{8}{15}e_{12}\\ [e_2,e_{11}] & = \frac{27}{55}e_{13}\\[0.5cm] [e_3,e_4] & = \frac{1}{10}e_{7}+e_{9}\\ [e_3,e_5] & = \frac{1}{10}e_{8}+e_{10}\\ [e_3,e_6] & = \frac{3}{35}e_9+\frac{83}{126}e_{11}-\frac{22105}{15246}e_{13}\\ [e_3,e_7] & = \frac{1}{14}e_{10}+\frac{20}{63}e_{12}\\ [e_3,e_8] & = \frac{5}{84}e_{11}+\frac{697}{10164}e_{13}\\ [e_3,e_{9}] & = \frac{1}{20}e_{12}\\ [e_3,e_{10}] & = \frac{7}{165}e_{13}\\ [e_4,e_5] & = \frac{1}{70}e_9+\frac{43}{126}e_{11}+\frac{22105}{15246}e_{13}\\ [e_4,e_6] & = \frac{1}{70}e_{10}+\frac{43}{126}e_{12}\\ [e_4,e_7] & = \frac{1}{84}e_{11}+\frac{7589}{30492}e_{13}\\ [e_4,e_8] & = \frac{1}{105}e_{12}\\ [e_4,e_9] & = \frac{1}{132}e_{13}\\[0.5cm] [e_5,e_6] & = \frac{1}{420}e_{11}+\frac{313}{3388}e_{13}\\ [e_5,e_7] & = \frac{1}{420}e_{12}\\ [e_5,e_8] & = \frac{3}{1540}e_{13}\\[0.5cm] [e_6,e_7] & = \frac{1}{2310}e_{13} \end{align*} Let $\mathfrak{h}$ be the ideal generated by $e_3,\ldots , e_{13}$ of codimension $2$. Then $\mathfrak{g}/\mathfrak{h}=\{ [e_1],[e_2]\}$, and we may write $L=\{ \alpha[e_1]+\beta[e_2]\}$, where $\alpha,\beta$ vary over all complex numbers. Hence the preimage is $L(\alpha,\beta)=\{\alpha e_1+\beta e_2,e_3,\ldots ,e_{13} \}$, which is a $1$-codimensional ideal in $\mathfrak{g}$. Now it is easy to see that $$ L(1,\alpha)\simeq L(1,\alpha') \text{ if and only if } \alpha=\pm \alpha'. $$ So we obtain infinitely many non-isomorphic Lie algebras.


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